EEVblog 1427 - An INFURIATING Electronics Exam Question!

[@EEVblog]

There is a reply from the OP: "Thank you, everyone. Especially Dave for even making a whole video regarding the problem. Very interesting to read your different approaches. So, the backstory of the problem. The circuit got handed to us in a high school entry course in electronics. The circuit was supposedly made by our teacher. Our teacher seems to have lacking knowledge regarding LEDs and current in a series circuit. By his means, the LED drops voltage AND current. I had to prove him wrong 8). We got to both calculate and elaborate. When elaborating, I got similar results as Dave in his video. But when I had to calculate the right branch, things got a little confusing, thus creating the thread. The circuit that stood out to me as my first thought was that insufficient information was given. But I made the assumption that Vf = 2.0 at all given currents(which would not be the case in reality). My conclusion was, half of the current goes through each LED which is in pairs. But the circuit seemed odd anyway. I see some of you made some good realistic calculations, but it is not near the level of this course. Ohms law and Kirchoff's are pretty much the only formulas used. If my teacher is watching this thread, maybe he could post the sole purpose of this circuit with calculations included. The circuit in my opinion is just a brain itch and would never be used in reality(hopefully). An LED circuit is never seen as a theoretical circuit for me."

[@ΣπυροςΓιαννας]

Can you make a video about faraday's law where the kvl doesn't hold and emf equals -NdΦ/dt = closed intergal of E•dl. I'm learning electromagnetism by the Walter lewin MIT courses here on KZbin and I know he's right but it's so counter intuitive where i can't get a feel for the explanation since he's the one of two people who have videos on this topic. His video is kvl for the birds. I would like to see your take on it since I'm sure many students are confused about the topic. Thank you for the informative content

[@ramonbenavides7125]

I might be under thinking this but assuming this is a purely resistive circuit wouldn't you get the resistance of each LED and add them up to your combination circuit? if that's the case I ended up with A.) 0.0364A B.)0.2667A and c.)0.1333A

[@westelaudio943]

@@ramonbenavides7125 No. LEDs don't act like resistors.

[@passwordpassword2199]

​@@ramonbenavides7125 I guess you mean 20mA for A?

[@paulkocyla1343]

I actually have seen such a circuit in reality: In cheap outdoor solar light chains. The individual LEDs have been wired in parallel. Due to the cable resistance from LED to LED, the following LED was sligthly dimmer than the last. However, no one would drive LEDs without a current limiter on a low impedance source. Even if you carefully match the supply voltage with the LEDs Vf and use the battery/wire impedance for current limiting, the Vf of the LEDs is dropping with temperature, so they would quickly burn through.

[@BogdanTheGeek]

The only correct answer is A = 20mA, B=C=0mA because that LED no longer has any magic smoke in it.

[@teardowndan5364]

Depends on the diodes' VI curves. The B-diode will operate higher on its VI curve from passing the C-diodes' combined current and that will cause the C-diodes to operate below their 20mA@2V point. Depending on the curves, the B-diode could be passing less than 30mA and last a fairly long time.

[@ScottGrammer]

Nope. A = 10mA, B = 20mA, and C = 20mA. The two top-right LED's will simply glow dimly. I'm doing this in my head, as I have not yet watched the video. Let me see now what our host says. EDIT: XRunner is right, the actual current at A depends on the matching of the diodes.

[@ulwur]

In practice they'll be goners. The temp co of semiconductors like this is negative, so the forward voltage will lessen leading to smoke.

[@steveschulte8696]

A is 20mAmps, B = 2*C. To find the value of C, I would need the simplified V-I characteristics of the LED's. The LED's are then modeled as a constant Voltage drop plus a resistance (the discrete differential of the curve in the area of 20mA, Delta-V/Delta-I). So split the right side, and double the resistance of the lower LED, subtract the extrapolated 0 current forward voltage drop of the two LED's and solve for the current C. The ideal current meters add no voltage drops. This is an idealized case solution.

[@CurtWelch]

@@ulwur I kinda doubt this. I don't think it will generate enough heat to create a runaway effect. I just hooked up three red LEDs to test my answer and I'm not seeing run away after 10 minutes. Will leave it for hours and see what develops. The three together are drawing 30ma (so 15 though each of the two in parallel) which is right on my estimate. It did start at 28 ma and inch up to 30 but seems stable at 30ma. Will see if the smoke rises!

[@WizardTim]

I’ll never forget in the first assignment of my electronics degree I was asked to find a suitable value power resistor to limit the current in an LED in a big, convoluted yet massively oversimplified circuit of a solar panel directly charging a battery (zero protection). Of course the LED was of the magical 3.65 V ideal type so, instead of doing a big massive current analysis for all of the insignificant wire resistances and unrelated circuit branches they gave I assumed ideal wires and just used the voltage of the power source (an ideal constant voltage battery) and a single ohms law calculation to get the value of 12 Ω, a nice E12 value. But of course I was marked wrong for not solving it ‘correctly’ with three pages of working to get exactly 11.908 Ω. From there I realised that the electronics class I was doing was actually just an electronics themed math class detached from any concept of practicality. There were plenty more massively over simplified circuits in that class. I really do agree that those questions are “pathological” and should be avoided as I have had a number of people who should know better ask how I’m running a 3.3 V white/blue diode off a 3.3 or 3.0 V supply.

[@michamaecki8104]

Yep, thats the reason why i didn't went to college, despite easy way in. Everything depends od people, that you'll meet. There are unfortunately lot of people, that stop thinking after getting a degree, which is even worse, if they teach. Btw, nice channel you have. Only thing that i don't like is frequency of uploads :v

[@Mobin92]

Because studying is not about remembering a few practical shortcuts... It's about understanding everything in detail! Even in unrealistically convoluted setups, because if you understand them, you will understand most things in real life.

[@the_drgoblin]

Asssuming this was your first assignment, they probably just wanted to make sure you understood the basics of circuit analysis, which you managed to mess up lmao.

[@AlienRelics]

@@the_drgoblin in the real world, 12 ohms. No one will order a resistor specified to 3 significant digits to light an LED.

[@almostanengineer]

@@Mobin92 yeah, it sucks when you already know a few bits that your learning, but they want to know you understand the complicated methods they have taught you, even on simple circuits.

[@whiskyguzzler982]

Entry-level hobbiest here and having you analyze questions like this is very satisfying. I get to think my thoughts and then learn the truth! Thanks!!!!

[@MrMaxeemum]

I got this question in an interview and ended up getting arrested for assault, under the instructions of Dave I let them have it. I didn't get the job.

[@wouldntyouliketoknow9891]

A is easy. B and C you would need a characteristic curve to solve.

[@jjmcrosbie]

Yes.

[@szabolcsmate5254]

But even then it would be just a nominal value at a given temperature. Real world part variations would throw it all out of the window.

[@jjmcrosbie]

@@szabolcsmate5254 Yes, again.

[@eDoc2020]

​@@szabolcsmate5254 In my opinion we can assume all information given is absolutely true even if it would conflict with reality. Any LED in the circuit, including any variations, at any temperature, with 2V across its terminals, will pass exactly 20mA. We can only introduce reality when it does not conflict with the given information, therefore all we can say about the parts is that an LED across less than 2V will draw less than 20mA and any LED with more than 2V will draw more than 20mA.

[@OctavMandru]

Then you have to take in consideration the interna resistance of the Ampmeter 😭

[@rdwatson]

Answering test questions like this is an entirely different skill than real-world applications, mostly about understanding the answer they're looking for and assumptions they're making.

[@userPrehistoricman]

Can be great for customer relations

[@SidneyCritic]

When I did mech eng I noticed a lot of the questions were similar, but worded differently, and that's when it clicked that they weren't that interested in the answer but instead teaching you how to think, ie make the question convoluted and see if you can think your way around it. It got so cryptic some questions were unanswerable if you weren't doing another particular subject, ie, the solution wasn't taught in this class, but another classes subject.

[@JasperJanssen]

Even with that, though, the question is nonsensical. When you specify the LED as just one datapoint, the question is impossible. I suppose you could assume they mean “up to 20 mA”, but it’s not what they say.

[@glarynth]

Reverse-engineering other people's assumptions is absolutely a real-world skill.

[@hightttech]

@@JasperJanssen: I've see BOM's and schematics that show a basic part description, just just like in test question: "LED RED 2V 20mA". The problem solver that believes he must get 20mA through ALL LEDs is in trouble 😆.

[@AintBigAintClever]

Dave muttering to himself: "Right, that's question 1 done. Question 2..." Examiner: "Okay, pens down please."

[@fivish]

He took soooooo long on this 30 second max question.

[@christianweagle6253]

@@fivish LOL yeah it doesn't take long to say "this cannot be solved without more model details, and iteratively, at that".

[@thehobe9]

I love this problem! This is a classic Analog IC design question. The intuitive understanding of semiconductor junctions give s a quick answer to this problem. Because the nature of the junction is logarithmic, the two parallel diodes (LEDs) will have a decreased drive exactly cancelling the increased drive to the bottom single diode (LED), a ~30mA current will result in the second path due to the 1.5X effective junction area spread over the 3 diodes in the second path. The kT/q ln I1/I2 is typically18mV for a diode at room temperature, so for a small signal analysis for the LEDs, the same relationship should hold true. I am a former Analog IC designer and Analog guys LOVE the nature of logging characteristic of diodes, using them in multiplying, RMS, logging and other current mirror applications in designing ICs.

[@MinazukiShiun]

This actually corrobrated with curt welch's test result above. Might have been what the question was asking for

[@firefly618]

When I saw that problem, I approached it in a somewhat uneducated way: I approximated the IV curve of a LED with an exponential, choosing coefficients so that the curve on a graphing calculator was as similar as possible to a random LED datasheet, while still passing through 20mA @ 2V. Then I solved the system of equations using kirchhoff's law. The result was ~30mA for the second path! I'm impressed! Stephen, could you recommend me an introductory textbook or other resource to learn the basics (very basics!) of analog IC design? Is there a simulation software I could use for basic circuits? I'm ok with maths (I know my way around exponentials, calculus...) but the more practical the resource, the better. Thank you in advance!

[@ambassadorkees]

I choose to read the Vishay curve, not remembering the logarithmic nature of diodes (40 years in distribution electrics now). My result 28mA on B. Derivation: IF you'd be given the Vishay specific curve, you could read I=13mA @V=1,9V, and I=30mA @V=2,1V and approximate the curve rises from these to 20mA @2V with two straight lines: For V2,0V (the single diode) I [ma] = 100*Ub-180 Now we have an equation where Kirchoff tells us: 2*(70Uc-120)=100Ub-180 and U1+U2=4 Solve for Ub and Uc: Ub=2,083333V and Uc=1,91666V => Ib = 28mA, Ic = 14mA The reading of the curve was in 2-digit accuracy only, the final answer can't be any more exact. A better approximation might be connecting the 3 points with an inverted parabola, but there's no sense in that either, since the Vishay curve is obviously built on a limited amount of datapoints: 0,1mA, 5mA, 10mA, 20mA, 40mA, 60mA and 80mA with linear connections.

[@profradon]

I would have thought it is some kind of "estimation-question": assuming ideal LEDs you would get exactly 20 mA in the right path if there where only two LEDs in series. If you put four ideal LEDs in the right path (two each in parallel an series, connected in the middle) you would geht 40 mA. So only three LEDs has to result in some current larger than 20 mA and lesser than 40 mA. Never thought you would get exactly 30 mA. But it makes sense, if you do the following assumption: for every non linear element you can find a linear replacement at a given current or voltage across them. For these LEDs the linear replacement at 20 mA is a 100 Ohms resistor (surprise!). This linear replacement is 100% correct only for a single point of the characteristic curve. But to get a rough equivalent circuit, you could just replace all LEDs in the right path with 100 Ohm Resistors. That would give a total current of 26,7 mA. Due to the inlinearity of the LEDs the equivalent resistance of the upper LEDs will be slightly higher than 100 Ohms while that of the lower LED will be slightly less than 100 Ohms. Since the resistance rises with falling current and it falls with rising current you will get something like a 110 Ohm replacement for the upper LEDs each and 80 Ohms for the lower LED (I totally made up those numbers to fit your 30 mA). I am too stupid (or lazy) for the exact equations, but 30 mA makes somehow sense to me.

[@muppetpaster]

Kirchoff's current law is litterally the first law you learn after Ohm's law....

[@CurtWelch]

If it's not, you need a better education! :)

[@hausaffe100]

@@CurtWelch Kirchhoffs laws please i don'T want to loose any voltage in my mesh, too

[@CurtWelch]

@@hausaffe100 :)

[@kirepudsje3743]

Agree about the current law, but the voltage law is one of the hardest ones to get rid of when teaching microwave engineering. (voltage only being properly defined for electrostatics).

[@4wdguydrivesby]

Then maybe the Thevenin equivalent ciruits?

[@NikiBretschneider]

I went through the Tomorokoshi's calculations and everything I can say about that is that they are perfectly correct and IMO it is the only correct answer to that question. It's a little bit "speedy" to me, there is no explained how the (3.1) equation was obtained from (1.3), but I didn't have to take a paper and pencil to prove it is correct, so on the other hand, it is not so hard. If you look at the (1.3) equation, there are a lot of variables, but according to (1.1) and (1.2), the η and Vt are constants, because that diodes are the same, they are only at different points of its V-A characteristics, so only the V and I varies. At the next step (1.3) was splitted to two equations with Vb Ib and Vc Ic, the rest variables was untouched because they are constants and that equations was substracted one by another which resulted in (3.1) equation. The rest is simple and easy to understand. In my opinion, this is much beter question than something like "write the Ideal Diode Equation", because this proves that student is capable of using that equation. In fact, I have no experiences in teaching, but it looks like a good idea to me. Maybe someone experienced in teaching (like @TheSignalPath :3 ) could tell us whether exam questions like this are good or bad idea.

[@ChumpusRex]

I agree, it was a good answer to the question. I missed this on the forum, and I would have taken a slightly more abstract approach, so rather than the pragmatic assumption of logarithmic behaviour of the diodes, an inequality solution can be achieved with the simpler assumption of a monotonic V/I curve. Ib = 2 * Ic by Kirchoff's law; V(Ib) > V(Ic) by monotonicity assumption. V(Ib) + V(Ic) = 4 V; therefore V(Ib) > 2 V and Ib > 20 mA and V(Ic) < 2 V and Ic < 20 mA. From which it is possible to give the answer as inequalities: 20 < B < 40 and 10 < C < 20 ma.

[@stephens1393]

@@ChumpusRex Agreed. Tomorokoshi's answer is reasonable, with the caveat of "but real diodes would not guarantee the result," in which case your bounds might be useful. I very much disagree with the OP that this is an awful question. It's not a great question for entry-level students limited to KVL and KCL, but it may be representing a model of a circuit that happens under certain conditions-- it may not be a circuit you would ideally build, but rather one you "end up with" as a result of some non-ideal chip construction. More advanced courses address exactly this kind of issue.

[@Pukkeh]

I don't understand why people seem to think the problem is not suitable for hand calculation or not solvable without knowing the LED specs, because a hand calculation works fine for a decent estimate (assuming matched LEDs), and isn't even difficult. Assuming that the LEDs follow the familiar Shockley diode equation, the I-V relation is I = I0*exp((V - V0)/(n*VT)), where I0 = 20 mA and V0 = 2 V. If the voltage of the center node is 2V + Delta, KCL applied at that node gives 2*I0*exp(-Delta/(n*VT)) = I0*exp(Delta/(n*VT)), or exp(Delta/(n*VT)) = sqrt(2). The current through the bottom LED is then sqrt(2)*I0 = 28.3 mA, and the current through the upper two LEDs is half that: 14.1 mA.

[@JasperJanssen]

Only until the bottom LED burns out, though, given that it is speced as max 20 mA. So the actual answer is 0 mA.

[@therealjustincase]

@@JasperJanssen But it is not spec’ed at 20mA _max_, there are no max specs anywhere. The circuit is of course completely impractical (but perfectly calculable nevertheless, as the commentator above shown), but so are many many other test problems. “What is horse-power?” It is power yielded by a spherical horse in vacuum.

[@JasperJanssen]

@@therealjustincase assuming these LEDs follow the Shockley diode equation is just as big an assumption as any other you can make to get a number out of this.

[@joejane9977]

i choose simple thinking. let me know what you hink : ) imo if you have to consider the LEDS as linear devices then you can say that at 2v and .02 amps then all leds are 100 ohm resistors as R=E/I then 2/.02=100 ohms. so the left is 200ohms and the so 20ma. B=150 ohms at 2v is 13.333ma and C is 150ohms at 2v so 26.667/2=13.333 but this is not the answer so if you consider that b drops 20ma and 2v then the branch C is 1v and 10 ma this is a real shit show this is also assuming electron flows from - to +

[@absurdengineering]

@@JasperJanssen Since verifying whether LEDs behave this way is rather simple - just two multimeters and a rudimentary breadboard and component set - I highly recommend for you to just try it. And once you check over say 6 orders of magnitude of current, you can then also think about why it works where it does, and why it doesn’t where it doesn’t. Don’t forget to keep the LED shielded from any external light! What you will also find (and I recommend you try yourself, for real!), is that LEDs are not magical and unpredictable. In fact they are kinda boring diodes if it wasn’t for their knack for emitting light without being on fire.

[@henrikjensen3278]

The only answer that can be given from the stated data is: A: 20mA B: >20mA and equal to 2xC C:

[@smunaut]

If you assume an ideal diode (as in following ideal diode equation), and that they are matched, it might be possible to workout the current split.

[@pbazarnik]

Yes. Absent the LED V(I) model nothing further can be said. I agree with Dave that this question is an excuse to generate a discussion during interview to allow interviewee show his/her knowledge. I'm disappointed that no SPICE simulation was mentioned (one can use free LTSpice which has models for many LED models - the difficulty would be finding one with V(I) curve passing 2V 20mA) LTSpice is very approachable for even novices.

[@ianhill20101]

They want you to calculate the power across the resistor 20mA, 2v volts across a 100ohm resistor then try catch out can you follow the path of a current, in its context of a high school question its fine.

[@GodmanchesterGoblin]

@Henrik Jensen - best and most concise answer.

[@BrianG61UK]

A real LED isn't going to drop exactly 2V at 20mA. So even if they are all perfectly matched you might find they draw much more or much less than 20mA at 2V.

[@briancox2721]

I had an old school EE prof for my first circuits class. Had him in just about the last term before he went emeritus. He was very practical. He would say that if you couldn't understand whether the answer your calculator had given you was reasonable or not, it wasn't worth the electrons used to calculate it. I'm pretty sure this question would send him into an apoplectic fit.

[@MushookieMan]

Electric charge is a conserved quantity, so calculate away!

[@sayamqazi]

@@MushookieMan well then the movement of electrons was wasted

[@chizzt]

I found I learned more from the 'bad circuit' ideas in the early versions of Horowitz & Hill than i did from many other texts

[@bobsage4963]

Same here! Good old H & H :)

[@therzook]

Did not have old versions of arts, but this book is a class, sadly it sits on my bookshelf not used after I went into plcs and industrial networks...

[@MrFreeElectron]

the correct answer is : all leds carry 20 mA of current and have 2v volt across them , but only in a universe where the laws of physics of that universe allow it.

[@quantumbemusement2947]

The question "How would I calculate ..." appears to provide a philosophical opportunity! First, I'd power up this circuit to let the magic smoke out of that right side of the circuit. Once the smoke had cleared, the answer on the left branch of the circuit would be 20mA and the answers on the left would be a peaceful ZERO. I would choose to ignore the temporary and catastrophic vaporization of material on the right side and only consider this calculation after this circuit was at steady state. It's akin to ignoring what occurred in our Universe during its first few moments before baryonic matter condensed and only pondering life after photons were able to be emitted from these Universal LEDs.

[@ThatEEguy2818]

Once passed by the EE lab and saw a friend in there crying. Her lab was to build a DAC with resistors, but it wasn't giving her the right result. Lab design showed the drive circuits providing 0 and 5V for the logic levels, but the lab provided TTL chips. No 5V out on those, so no accurate to predicted DAC output. I often wondered if any of these professors ever designed a real circuit.

[@waltercomunello121]

CMOS (FETs) =/= TTL (BJTs) as far as I know. they're two completely separate universes. no wonder she got confused. I still am.

[@ambassadorkees]

@@waltercomunello121 I grew up with TTL. Can use CMOS, but have trouble understanding discrete FET's. Must admit I never took the effort to dive deeper into them.

[@fredbloggs5902]

At a job interview once, I was given a multiple choice test they had ‘borrowed’ from Microsoft where one of the questions famously had the wrong answer... ...what did I do?... ...I was congratulated for being their first candidate to ever get 100% 🤣 (I’ve felt vaguely unclean ever since).

[@Aaron36983]

Just out of curiosity do you remember the question?

[@KarlHamilton]

What was the question

[@hinzster]

Ah, Microsoft exams. Do they still care about Netware? I ask that because for a long long time they have been the only ones in the world to do that, leading to generations of people having to learn totally useless stuff about Netware just to take those exams. Not even the retrocomputing community cares about Netware.

[@mozismobile]

I hate those. Partly because there are cram schools where people memorise thousands of those questions, and partly because there's a huge reward for being lucky with your first attempt at a solution (or which answers you memorised). And that approach carries over to the certification exams, which especially for programming often amount to "what would the compiler do here" and I DGAF I let the compiler do what it does best, I'm not sitting there counting brackets and trying to work out order of operations. If the code requires that it's bad code and I will refactor it. (FWIW that is always the wrong answer in those exams).

[@ddegn]

I had a biology quiz which asked "What are the odds of rolling doubles with two 6-sided dice?" The material the quiz covered used the odds of rolling double fours. I knew the teacher expected an answer of *1/36* but I still wrote *1/6.* I was the only person to be marked wrong on this question. I was further disappointed at how long it took me to convince the teacher I was correct (I wasn't able to convince the same day we took the quiz). I eventually got him to change my score on the quiz. I decided I didn't want the rest of the class to hate me so I cowardly didn't insist he reduce the score of everyone who answered 1/36.

[@dhanbirthethi9575]

Hi, so I ran this simulation is TinkerCAD, and my results are as follows: 1. The left side is definitely brighter than the right side, but the right side flickers?? 2. Node A was measured to be drawing 16.9mA whereas both B and C were drawing 3.13mA, which makes sense because they are on the same line and according to KVL will have the same current flow. 16.9+3.13 = 20mA Therefore, A = 16.9mA B = C = 3.13mA

[@Brian_Of_Melbourne]

That table at 15:01 is seriously broken. The headings don't match the collumn data and there are TWO collumns called MIN. I suspect the headings should be Symbol, Minimum, Typical, Maximum, and Unit.

[@jamesnewman9547]

Led characteristics also change depending on the amount of light *hitting* the led itself. Infact, you can use a lit led as a light sensor with sensitive current sensing. I imagine the changes from touching the led from the breadboarded circuit included some of this effect.

[@antilogism]

Yes. Even glass diodes do this but LEDs are much stronger.

[@lukasvanginneken1859]

I would use a substitute circuit for each of the LEDs consisting a voltage source and a small resistor. A linear current / voltage relationship can be assumed within a limited range. The voltage sources can be ignored, so the circuit to solve becomes two resistors in parallel with one in series. This gives current B is 26.6667 mA and current C is 13.3333 mA.

[@chaos.corner]

Had what I think was a question like this in my physics final. The problem with letting them have it in the answer is that it's usually some postgrad working from an answer sheet that's marking it and they won't care or be able to do anything about it. Spend the time improving your answers on other quesitons.

[@MushookieMan]

This is why the education system is broken.

[@BulletMagnet83]

My biggest take-home from this was "always bring whiteboard pens to any technical job interview from now on". I feel like a knobber for never having considered that before.

[@maxine_q]

You think they would have a whiteboard there but no pens to write on it?

[@LeifNelandDk]

@@maxine_q just bring a whiteboard pen. The chance of there being none * the bonus points for having one and be able to draw anyway can only be positive. But if you bring a permanent marker, you'll get many negative points instead ;-)

[@srboromir452]

Can always use a window

[@Spookieham]

Good point about the whiteboard. Before Covid I asked all software job candidates to draw some architecture on the whiteboard. That quickly sorts out the also-rans or the bluffers.

[@fredbloggs5902]

At a job interview for a proof-reader/editor position at a CD-ROM publishing company, I was given a written exam with nonsense questions, so I proof-read and corrected their exam paper... ...I didn’t get the job! Some people have no sense of humour ☹️

[@Jacob-gj8hz]

Damn, even did exactly what you were there to do.

[@MrBCRC]

I had a very similar situation. I was going for a job working on Oracle databases. The interviewer sent me the table creation and insert scripts for a practical query writing task that was for a different database so wouldn't even run on oracle. I wrote straight back and asked if cleaning up the scripts was part of the task because these were for (i named the database they were written for) and won't run on oracle and it will take longer to clean them up than I was allocated for the task. I was told to go ahead and forget about the time limit. I got the job :-)

[@Jacob-gj8hz]

@@MrBCRC awesome

[@stanimir4197]

>Some people have no sense of humour ☹ it's more like you were "overqualified", i.e. their tech lead was too scared to be made obsolete.

[@fredbloggs5902]

@@stanimir4197 Yes, that had occurred to me. Unfortunately that doesn’t get the mortgage paid.

[@byronwatkins2565]

Diodes obey Ebers-Moll: I(V)~I0 exp(V/V_T) and V=V_T ln(I/I0) where V_T=kT/q. The left side does balance; since (100)(0.02)=2V, point A is at 2V at 20mA so the LED voltage is 2V at 20mA... this is the given point. But the right side does not balance. Since I_B=2I_C, we can find the difference between the two LED voltages: V_B-V_C=V_T [ln(I_B/I0) - ln(I_C/I0)] =V_T ln(I_B/I_C) = V_T ln(2) Kirchoff's loop also gives V_B+V_C=4V Adding these we see 2V_B=4V+V_T ln(2) and V_B=2V + V_T ln(2)/2 The node voltage is elevated... Now, use V_B-2V=V_T [ln(I_B/I0) - ln(20 mA/I0) = V_T ln(I_B/20 mA) Combine these V_T ln(I_B/20 mA) = V_B-2V = [2V + V_T ln(2)/2] - 2V = V_T ln(2)/2 Cancel V_T to see I_B = 20 mA * sqrt(2) = 28.3 mA Identical diodes will divide this current equally so that I_C=14.1 mA. Without Ebers-Moll there would be no current mirrors, no log amps, no band gap references, etc. So it is NOT a silly question although small fluctuations will indeed be amplified in the real world. I0 and V_T are TINY, so any ammeter voltage will reduce the current significantly. This, in fact, is what causes the fluctuations to be amplified.

[@ambassadorkees]

How can it be I arrived at the same with reading the Vishay curve and making 2 different linear approximations?! IF you'd be given the Vishay specific curve, you could read I=13mA @V=1,9V, and I=30mA @V=2,1V and approximate the curve rises from these to 20mA @2V with two straight lines: For V2,0V (the single diode) I [ma] = 100*Ub-180 Now we have an equation where Kirchoff tells us: 2*(70Uc-120)=100Ub-180 and U1+U2=4 Solve for Ub and Uc: Ub=2,083333V and Uc=1,91666V => Ib = 28,33...mA, Ic = 14,166..mA, have to answere 28 and 14mA because the reading was in 2 digits only. A better approximation might be connecting the 3 points with an inverted parabola, but there's no sense in that either, since the Vishay curve is obviously built on a limited amount of datapoints: 0,1mA, 5mA, 10mA, 20mA, 40mA, 60mA and 80mA with linear connections.

[@byronwatkins2565]

@@ambassadorkees Log graph paper performs the logarithm for you. Lines on log graph paper are exponentials.

[@ambassadorkees]

@@byronwatkins2565 Yep, but there's a difference between single log (X axis linear) and double log paper

[@michaelmoorrees3585]

Back in the old day's LEDs would actually explode when overcurrent was applied. So those that branch with no resistors would be all kinds of fun, when the supply voltage drifted over 4V. A software engineer wrote that question.

[@eDoc2020]

I'd say that we can only introduce reality when it does not conflict with the given information. The total voltage is exactly 4v and each LED across 2V passes exactly 20mA. It's _impossible_ for current B to be 20mA because that would mean the LEDs up top see exactly 2.0V and thus would, by the nature of the given information, each draw 40mA. That obviously isn't the case because of Kirchoff's Current Law. We know in reality that there is a positive slope relating forward voltage and current and this does not conflict with the given information. The answer ends up being somewhere between the two conflicting "ideals", specifically that B is 20-40 and C is 10-20. This is the answer lacek gave on the forum. Tomorokoshi's solution assumes the diodes follow the ideal curve but this may not be the case, we only know the parameters at one specific point. We don't even know if the LEDs are matched if operating at something other than 2V. Tomorokoshi's solution _may_ be valid, but possibly not because it assumes information not given. Naturally it _does_ does fall into my given range of possibility.

[@Aequitas0Sports0]

Actually Tomorokoshi did it right and his calculations would probably match the experiment. But anyway you can actually easily calculate the approximate values, A brunch current is 20 mA because 2v(LED Vf @20 mA) + 100Ohms + 0.02 = 4V. And using simple logic you may estimate the approximate values for the current in branch B like bottom LED voltage drop should be higher than upper parallel LEDs, because their individual currents are less than the bottom LED current (KCL). And if we assume the top led voltage is about 1.8 volt and the bottom is 2.2 volts we would probably be close to the right answer, next we just estimate the current through the bottom led knowing it's exponential nature at 2.2 v it is higher than 20 and is probably around 28. Then the current through the one of the top LEDs is approximately 14 mA. Even with the slight parameters variation in practice due to temperature and manufacturing process the circuit will still be close to these values, given that the LEDs are of the same type from the same manufacturer. I think an experienced engineer who designed LED drivers a lot and holds the picture of the exponential curve for a "universal LED" in mind could give even more precise estimate for the branch B.

[@BjornMoren]

I agree. I would have done it iteratively. First assume 1.8 V upper and 2.2 V lower, then look at the data sheet and see if the Voltage/Current curve matches. If not, adjust the upper/lower voltages and check again, until it matches. But the question doesn't supply a data sheet, so in the end it is just a stupid question that shouldn't be taken seriously.

[@Aequitas0Sports0]

@@BjornMoren Yes you are absolutely right. The question is just dumb. There is no data for the curve and anyway you never use a circuit like this in real life.

[@6F6G]

The resistance of any meter connected at points B and C would greatly effect the current.

[@jan.tichavsky]

Yeah but you can imagine the meter as ideal meter in SPICE circuit simulation with real world components. That's another problem, living in last century and calculating circuit on paper with fixed points instead of using interactive simulations, with ideal or real world models, then explaining why it happens.

[@Vincent_Sullivan]

I am glad that somebody finally brought up the issue of the voltage burden of the ammeters! Yes, the circuit in this question is very poorly designed and detailed theoretical calculations CANNOT be accurately done with the information provided. Note that the 2 LEDs in parallel are not really in parallel. One has an ammeter in series! The ammeter will have a voltage drop (a low voltage, but NOT zero) which will reduce the current through the LED it is in series with. Bottom line, this circuit is a mess and will be very sensitive to parametric variations in the components! By the way, Dave (and others) have stated that you need to have a resistor or regulated current source in series with LEDs or you let the magic smoke out. This is not necessarily true. If you put an real world LED on a curve tracer the curve you get is typical of a diode. The current VS voltage curve, once the threshold voltage is reached, is NOT vertical but will show some degree of slope. This indicates that the diode has some resistance. If you need to drive a bunch of LEDs and you connect them in series these resistances add up and eventually reach the value of resistor that you would calculate to regulate the current to the value you desire. I have done many designs with 10 to 15 LEDs in series driven directly from a low impedance voltage source with no problems at all. Other diodes, such as power rectifiers, also show this parasitic resistance. This gets people designing low voltage linear power supplies into trouble because they allow for 0.7 Volts drop in the diode when in fact, due to high peak charging currents in the filter capacitor, the drop across the diode will be much larger - commonly in the range of 1.2 to 1.5 volts. The end result is a power supply with no headroom to accommodate low line voltage conditions.

[@vihai]

If they give you the forward voltage at a specific current it means that the LED has to be modeled as a real component with a curve that has a unambiguos voltage/current characteristic. The circuit would work. The current will be equally shared between the two parallel leds. Thus it is impossible that all leds work at 20 mA, than the full curve is needed to determine the actual work point.

[@berndrosgen1713]

Yes, I think that is the point of that excercise.

[@atafakheri8659]

I had a professor that told us when I give you a non linear just use some nominal parameters of Is , Betta , Vt , ..... Maybe it us like that So

[@Magneticitist]

You can't really be wrong if you just treat everything as linear while stating you are doing so ideally and the loads are thermally invincible in imagination land

[@atafakheri8659]

@@Magneticitist yeah but if you never learn non linear analysis in easy circuits you will have a hard time doing it later

[@Magneticitist]

@@atafakheri8659 You're right but it's impossible to do so unless you create your own values in this specific question. Arbitrarily doing so and answering it both ways would be a great response.

[@atalhlla]

My first thought was: I assume that the power supply doesn’t have any particular output protection and delivers an over voltage spike that kills the right branch. From there, either the left branch is fine or the right branch fails closed and possibly pulls down the Vcc or causes a cascading failure.

[@ebenwaterman5858]

Question on exam: Draw up a circuit to light 2 LEDs at full brightness and 2 LEDs at half brightness using only one 100 ohm resistor.

[@MrPhilip796]

Could make it a bit more interesting as well and add in "Make it work at X voltage, each diode's Vf at 20mA is Y volts"

[@sebastianwittmeier1274]

The right LEDs are brighter: The top two are more than half the brightness of the left one and the right bottom one more than the 'full' 20mA brightness of the left one

[@twilightsparkle3157]

assuming stiff power, 5% tolerance, and room temperature: A = 20mA (range: 19mA - 21mA) B = 28.28mA (range: 26.1mA - 30.4mA) C = 14.14mA (range: 12.7mA - 15.5mA)

[@glenslick2774]

This is a Voight-Kampff test for determining whether or not an EE is a replicant.

[@rubikmonat6589]

Words cannot convey how much I like this comment and it's subtleties.

[@MartinE63]

The LED that burns twice as bright burns half as long

[@ThereIsOnly1ArcNinja]

@@MartinE63 If any EE comes up with that kind of circuit - Don't fire them! Burn them!

[@metatechnologist]

He fails Voight Kampf and his memories are not real because it's "ELLE EEE DEE".

[@MuzixMaker]

Like tears in rain.

[@sidneykantor]

The idea of analyzing a simple circuit like this is very helpful and makes me wonder whether you could do a series on well known electronic circuits we all should know. You could go through a circuit per video explaining exactly what is happening. For example, so many examples of "good" circuits in the "Art of Electronics". You could pull circuits from there that you know well and that you think we should all know and understand. I would watch them for sure!

[@warup89]

My hardest professors I had for my electronic courses were either from India or Sweden. As soon as I was done with those classes I just mind dumped everything, I dont know why they pushed so hard.

[@lightningmcqueen1577]

I can answer for the Indian professor, here at home we have really tough exams at university's which test your mathematical aptitude/equation manipulation instead of practical skills

[@lightningmcqueen1577]

@@kingflockthewarrior202 dsp oh my degree stopping paper that is

[@mozismobile]

Sometimes it was because to work in that field you have to have an intuitive feel for what will happen. Just as a lot of us looked at the circuit above and went "LEDs in series with no resistor" and heard klaxons in our heads, you really need to be able to look at the way a grid interconnect is wired and just go "HELLL NO" before you sit down and work out why. Antenna design and microwave circuits are even worse, you can't even start to analyse many of them without a good idea of which option might possibly work. So you just have to grind through 100's of example problems until it's obvious which geometric transform to use on a given antenna design.

[@absurdengineering]

@@mozismobile People have klaxons in their heads but it’s possible to get quite accurate theoretical answers to this question, with just a few measurements of the actual LED in isolation, using very rudimentary techniques. Furthermore, circuits with low impedance tied directly across a diode drop or two are useful. You know the CMOS inverter? You can build the same thing from BJTs as long as you keep the voltage small and add a bit of feedback. With low inductance design you can have inverters switching in nanoseconds, using jellybean parts. This works even with 50-year-old transistors like 2N2222 and 2N2907. Not as fast but much faster than most inverters people can build out of two leaded BJTs. And you can get the insight that this actually has a chance of working by playing a bit with mathematical models of transistors. Even the EM model suggest that it would work :)

[@dukeibzusa]

You can use the lineal approximation of the diode, Id=(20ma/(2-vo))×(vd-vo) where vd is diode voltage, and v0 is the voltage where Id=0. This v0 is unknown and we can give some values a little less than 2v. If we call V1,I1 the voltage drop and current in the upper diodes and V2,I2 in the bottom diode, we have: V1+V2=4 and I2=2×I1. Because we have 3 equations and 4 unknown values, if we give v0=1.8v, we can find the voltage in the upper diodes is 1.93 and in the bottom 2.07, the currents are 13.3 and 26.6ma.

[@SamuelCarrier92]

The guy with the long mathematic proof actually had an amazing answer that wasn't super complex

[@zetacrucis681]

Except he got the ideal diode equation wrong. He has positive current for both forward and reverse bias. The equation should be I = I0 * [exp(-V/V0) - 1], not I = I0 * exp(-V/V0). Yes, he goes on to assume Vcc = 4V, in which case the latter equation would be a good enough approximation of the former, but if you are gonna go to that level of theoretical analysis, you might as well solve for arbitrary Vcc, whose value is not given.

[@warasilawombat]

@@zetacrucis681 I don’t think he got the equation wrong persay - that negative one is commonly dropped because it’s influence on the result is considered negligible. It makes derivations much easier. I typically explicitly call it out as negligible in my proofs personally.

[@smunaut]

One thing I'd point out is that some circuit that seem stupid if you imagine them on a PCB can make perfect sense if you think about them as multiple devices on the same wafer in a chip ... very different conditions and the devices / diode / transistor end up very closely matched (because they're right next to each other on the same wafer).

[@supernumex]

Uh, nope. Even on wafers they are very often not matched. which is why there is sometimes a large percentage that are rejected. Also partly why binning exists.

[@Laogeodritt]

@@supernumex I think you're thinking of a larger scale than Sylvain is-my interpretation of Sylvain's comment is that in the case of multiple devices _on a single die_, the short distance means that you have much more limited across-chip parameter variation. I would add that you can use various layout techniques (like interdigitation, dummy fingers, etc.) to reduce that even further, so it is possible to have closely matched devices operating in parallel or in a differential configuration (think the two input transistors of your basic differential amplifier). Of course, that depends on how close _close_ is. Clearly we still have laser-trimmed devices for a reason, despite the availability of matching techniques.

[@tolkienfan1972]

This circuit is pointless. If you add the necessary missing resistances and characteristic curve then you get a useful result. Without these the result is meaningless. You can't put an ideal voltage source across an ideal diode. So these characteristics must be specified in the question. If they are specified it makes no difference whether it's a pcb, a breadboard or a silicon wafer.

[@user-hk3ej4hk7m]

If you have V I graphs in hand you can, assuming all LEDs are identical, multiply by two the current on the graph (for the parallel branch), sum the voltages (for the series branch) and then intersect that with 4V. That will give you the current for that branch, divide by two to get the current for the parallel branch.

[@markholm7050]

I had exactly one college course in basic electronics. Kirchoff’s current law was definitely in the course. As soon as I looked at that right hand branch I looked for any current limitation at the input, and seeing none, realized that this circuit would be, at best, highly dependent on small variations and, more realistically, likely to blow out the bottom right LED.

[@edgaraskorsakas5703]

My answer would have been A = 20mA, B = C = 0mA. Without even thinking.After a second of thinking, right side blown.

[@woopsserg]

If you would actually think at least one time, the right side won't be blown, at least in short term. Try to use your brain and actual numbers before thinking that LED with no resistor = blown LED. Even with real parts, they would not be blown. If Vf is 2V, And you put LEDs in series, with 4V voltage source current would not be so high that it immediately blows the LEDs. With real parts with such spec total current most likely would be somewhere around 30mA. And there will be some imbalance between parallel LEDs.

[@npiper]

@@woopsserg Cool the hostility, we are all friends here. Second, you should watch the video before you attack people in the comments. Third, We all have things to learn and we all have things to teach and being rude is a fast track to being completely ignored by others for both of those things.

[@Landrew0]

When I was an instructor, I could always jig the curve by making about 65% of the questions easy, about 30% challenging, and 5% ridiculously hard.

[@vencibushy]

Theoretical electronics is all about the theory. At the uni we were forced to do calculation on absurdly complex schematics using variety of methods. Basically it took us 30-40 minutes to calculate something a computer does in less than a millisecond.

[@ohmslaw6856]

Sounds like a waste of time

[@scunnerdarkly4929]

Computers are great at giving us answers without us needing to know how the answers are derived. That’s pretty much the point here. Sure, you may never need to manually do those calculations ever again, but then again you might, say at a job interview with nothing but a whiteboard and a marker for company.

[@Cynthia_Cantrell]

Computers are great at giving you answers. It has no idea whether those answers are any good or not. The hard part about being a good engineer these days is knowing when the computer is lying to you. For that, you need to understand how the math works behind the models... and yeah, that gets complicated sometimes.

[@tbriceno]

This is actually an amazing question and one I find very typical. As an instructor, I may actually use this question in my Analog Electronics course where we cover semiconductor based components. To "properly" analyze this, we would have to do iterative analysis using ideal assumptions to get a ballpark answer, similar to as Tomo* in the video did. Assume 2V @ 20mA for the single LED. KCL says 10mA through the parallel LEDs. Then using the 10mA, calculate the voltage drop across an ideal diode. Then use the voltage drop to calculate a new voltage drop for the single LED. Then calculate the current through that LED with that voltage drop. Then divide that current and calculate the voltage drop across the parallel LEDs. Repeat this process a couple more time to allow the variations to stabilize.

[@timmy7201]

This gives me flashbacks to high-school. We had to solve theoretical questions like these for two years, without once holding a real component. When we were allowed to build our first breadboard setup after those two years, the question included a similar illogical theoretical schematic... Not that it mattered anyways... Besides being able to name the symbols on an electrical schematic, most people in our classroom didn't know how the real component looked like anyways... Then our government and industry wonders why there is a lack of technical skilled people...

[@volodymyrzakolodyazhny]

A very common problem all over the world I think.

[@Neyna4ever]

@@volodymyrzakolodyazhny not true. In my country i took electronics option in high school and we were toying with components after a few weeks (in the first year of high school)

[@1kreature]

I don't see the problem. The question uses the LED in 2 ways, the stable way that gives the expected result and the unstable way. The stable way allows you to say that given the single datapoint for the LED's characteristic of 2v, 20mA you will get the expected behavior of 20mA since the LED and the resistor both get 2v and thus resistor will set the current correctly for the 4v supply. That takes care of A. The right hand branch on the other hand, assuming the LED's are matched, will have the two upper LED's deliver a lower drop for the current passing through the lower LED. Thus the operating point for the two upper LED's will be pushed lower, and the lower LED will have it's operating point pushed up. Exact values would have to be based on the not supplied characteristic curve but one can safely assume the current in B will be higher than 20mA and the current in C will be less than 20mA, and half of the current in B. The only wrong answer is 10mA+10mA = 20mA for the LED's in right hand branch.

[@derkeksinator17]

Worse, I've recently seen a design which drives a couple of LEDs in series with an npn to ground(open collector, driven by a pwm signal). No series resistance whatsoever. Guess what, the LED Type changed due to availability and now the current draw as well as the light output dropped by a factor of ~11.

[@Scrogan]

With a base resistance? A well tuned base resistance to keep the transistor in the linear region could theoretically give a constant current to the LEDs regardless of forward voltage if the hFE remains relatively still. I’d have thought such a circuit would be more dependant on the transistor than the LED.

[@derkeksinator17]

@@Scrogan nope, just the obligatory base resistance for protection. What you mentioned is rather touchy, so it doesn't scale too well for mass production.

[@petersage5157]

My answer would be something like "Because LEDs are not constant voltage devices, the two LEDs in parallel would have something less than 2V across them; therefore, the fourth LED would have something greater than 2V across it. B would be >20ma and C must be B/2."

[@rayceeya8659]

Would it be fair to say the result is 3 out of four burned out LEDs?

[@JoQeZzZ]

Not really because of the strict 4V VCC. There is some negative feedback. Lets start with inital value of V=2,I=20mA on the bottom right. Current not limited to 20A so current increases. Current increases so forward voltage increases. Forward voltage cant increase so current decreses. You could calculate the equilibrium using the ideal diode law, but that probably won't give you a physically correct answer due to the 3 right diodes interacting.

[@jg374]

​@@JoQeZzZ I agree with the negative feedback part. If we wanted to make the question more complex though, the forward voltage would drop as the temperature of the leds increases, leading to more current and possible thermal runaway in certain situations depending on the LEDs and their surrounding environment, so I would say Ray is correct as well. I think the only conclusion is that in real life there are many things I would try to use before this :)

[@hanelyp1]

ASSUMING LEDs exactly on spec, dropping 2V at 20mA, the current through B is trivially between 20mA and 40mA. Exact value depending on the characteristic curve. Assuming resistors in place of the LEDs for the same voltage/current operating point given, ~26.7mA, giving an upper bound estimate for current on assumptions given. Real world components will be less predictable.

[@robertbox5399]

The parallel LEDs must be half the bottom. Hence the bottom will have more than 20mA (2.1V?). The top two will be less (1.9V?). This would give currents either side of the datasheet figures, giving a very unpredictable current down the right leg.

[@chrisheath2637]

My response is close to that ! We know that the current in the whole right leg is approx 20mA, since we also know that the voltage change will be small around 20mA (ie approx 2volts across all 3 leds) . But the current in the upper right hand branch is exactly 1/2 the entire right branch (assuming the 2 parallel leds are exactly matched.) So the voltage across the top 1/2, will be slightly less than 2volts, while the lower led will have a voltage slightly greater then 2 volts, ( to make 4volts) which implies that the current in the whole branch will be slightly greater than 20mA (rather than very unpredictable).

[@ScottGrammer]

Yes. That's why one usually adds a series resistor to LED's.

[@c_b5060]

Using Ohm's Law, each LED has an apparent resistance of 100 Ohm. Note that the left leg of the circuit and the right leg of the circuit could be separated into two separate circuits and the results would be the same for meters A, B, C. Solving for the series and parallel resistance circuit values yields: A = 20.0 mA, B = 26.6 mA, C = 13.3 mA

[@ambassadorkees]

Ohm's law is linear. The LED not. See the Vishay datasheet. Resembles an upside down parabola but the physics of LED are logarithmic. You could use linear approximation on a short stretch, but the datasheet is clearly showing logarithmic currents!

[@mdfyui8000]

Thought: '... ah, you're leading me down the garden path... or are you?'. One of those questions where I'd end up possibly using it as leverage to suss out how competent the interviewers are (if I knew their claimed backgrounds were relevant, not HR or they were hiring because they had no idea about the field). Depending on the circumstances, it might be the case that you find out the real reason why they have a position open and maybe that position isn't for you even if you are offered it.

[@MichaelOfRohan]

And gives you the opertunity to abuse their lack of ___insert_strength_here____.. Use your powers for good

[@Scrogan]

Actually it’s fine to put a voltage source directly across some nonlinear elements. LEDs are usually a no-go (IIRC some deliberately have a high ESR or other on-die current limiting) but gas discharge tubes are considerably nonlinear but can be stable with a constant voltage. What really matters is if a small change in voltage means a large change in current, or the other way around. And if they’re positive or negative tempco.

[@laharl2k]

an ideal diode/LED has a fixed Vf so even at exactly 2V you have an infinite current, "infinite" only limited by the impedance of the source, but again ideal sources have 0 ESR so.....in any case the question is flawed.

[@petehiggins33]

It's notable that these comments include a wide spread of confident answers ranging from total ignorance about diodes to neat solutions using Shockley's diode equation. But its rather sad to see how many of the former there are on the EEV channel where I would have expected a better educated audience especially considering Dave's previous videos on fundamental principles. I don't expect the average electronics enthusiast to know the diode equation but they should at least understand the general shape of a diode V-I curve and be able to see that the current C will be less that 20mA and current B will be greater than 20ma.

[@ovalteen4404]

The column headers on that datasheet are pretty much all screwed up. The first "MIN" is the shorthand name of the parameter. The next "TYP" should probably be "MIN". Then TYP, then MAX, and finally UNIT.

[@alexartiushkin2865]

This is a theoretical question. Schema parameters are chosen so that the result does not depend on the nonlinearity of the diodes. Ia = 20mA, Ib = 40mA, Ic = 20mA

[@jackevans2386]

Wrong wrong wrong ! Ib = 20mA (2 x 10ma ). Were you not listening to Dave ?

[@therealjammit]

I think this schematic was drawn from a previous test where the LED's were originally incandescent bulbs. I think some guy decided making it with LED's would make it more "modern". Edit: I think the original schematic also assumes the bulbs are a constant resistance.

[@JasperJanssen]

That would make sense, yeah.

[@keeperofthegood]

Although a diode is an active device, operated on DC it can be treated as a resistor in idealized case. 2v/0.02a = 100 ohms. Just replace the LED's with 100 ohm resistors and give the idealized answer.

[@ferrumignis]

Fair point, I wonder if that's what they were expecting? It would give students a very wrong idea of diode operation thought.

[@keeperofthegood]

@@ferrumignis Probably. It does have limitations though. 2 x 100 ohms in parallel would be 50 ohms, 150 ohms across 4v would be 0.027ma, which Dave did measure, but the voltage division isn't 1.35v and 2.7v and that was where my college program then went into the discussion on what makes an active component an active component. Man that takes me back more than 30 years ago now >.< and I am 100% sure my memory is fluid and flawed but close enough to comment on an Australian video :)

[@stephanweinberger]

Unfortunately the real world is not idealized. :-) With half the current the voltage drop across the two upper LEDs will only be ~1.8V, thus the voltage acrosse the lower one will be ~2.2V. This will in turn drive up the current to ~30mA - which will in turn change the voltage drop across the other two... So depending on the impedance of the wires you could even end up with a nasty oscillator.

[@CurtWelch]

It's not a bad starting point and acts as a good sanity check, but won't give a very good answer because the slope of the VI curve for a resistor is quite a bit different than for the LED. The LED has a steeper curve so the voltage across the diodes will be closer to 2V than with the resistors. The resistor has a straight-line curve that intersects with the origin. The LED curve is not so straight and intersects at the forward voltage drop point that is a good bit above 0. With the resistors, the current through the right side diodes is 26.7 ma with half of that through the parallel set. So the voltage across the bottom single diode would be 2.67V with 1.33V across the top pair. With the real 5mm RED LEDs I tested, the current was 30 ma total and the voltages were 2.05 across the bottom and 1.95 across the top. So a 50% drop in current only resulted in a 0.1 voltage drop across the LEDs. We can use that change in current and voltage to calculate the effective impedance of the LED ad we get .1V/15ma or 6.7 ohms. Not very close to 100 ohms at all. Model the LED as a 1.9V attery in series with a 6-ohm resistor and you will get a better result than trying to model it as a 100-ohm resistor.

[@Magneticitist]

I believe you are right because there isn't any other sensible way to look at it honestly. It's based on a basic formula of Vs-Vf/If used in led series resistor calculation. To make it work in reality there is the automatic leeway given in extra power rating where it is ideally assumed the components can handle it.

[@gandalf87264]

No, no, no, that is all wrong: You can see that VCC and GND are offset to the left on the schematic. Due to the nature of the toner used in the printer (Assuming it was printed with a lazer printer), the resistance to the LEDs on the right of the diagram is going to be much more than the one little LED with the resistor on the left. Therefore the best way to get those LEDs to light up is to set fire to the page and ignore the current they draw.

[@userPrehistoricman]

You're damn right. I've never seen a working schematic. It's just symbols and lines!

[@erikbraun2867]

Yep we learned Kirchhoffs law in the first few months, that and ohms law is the first thing that is learned

[@johnwilliamson467]

Agree it would be a fantastic employment question . Designed to see if the one person questioned know the difference from real world and text book . Enjoyed the video.

[@treelineresearch3387]

I don't know what it is with the "educator" class and coming up with really goofy contrived scenarios (that are often "trick questions" to some extent) when there are plenty of actually useful real life design patterns to analyze and you'll regularly encounter in the future. Goes for really any type of educator.

[@absurdengineering]

These questions are written by people who would have a very hard time putting this circuit actually together and explaining what it does. I’ve had a similar question on an EE exam, but it was explicit about using non-ideal components (at least to the extent we were taught), expecting the student to provide assumptions and simplifications, and use nonlinear modeling. The junctions weren’t in LEDs but in BJTs, and the circuit looked like it should just go up in smoke or something. The answers most of us got were within 50% of measurements on an actual circuit with jellybean parts. People were stunned. The professor just said “you know there’s a reason I pushed hard for you to learn all this arguably dense theory - it works quite well even if you pull some numbers out your ass”. It was one of the most memorable classes I ever took.

[@ProfessorV.]

I was a professor of electrical / electronic technology for thirteen years while still running my engineering firm. Based on my years in public education, questions like this hardly surprise me and are more a commentary on the competency of the instructor than the student's lack of understanding. It's unfortunate that students are subjected to these kinds of confusion but sadly, it is not that uncommon to find instructors over their heads teaching courses that they have no expertise in, largely because administrators simply need a body at the head of the class to keep the financial wheels turning, irrespective of whether or not the chosen instructor is the best person suited to teach that class. At a time when we need STEM fields to be rigorously taught more than ever, here in North America (Canada and US), we're seeing a rapid decline in the quality of public education at all levels, including colleges and universities. This is attributable in large part to cultural shifts and political forces that no longer value true knowledge or skills and their importance to society. The physics of the universe remains nevertheless immutably complex and we either rise to its level or seek employment elsewhere.

[@franktechniek]

It's not such a bad question, as it invites to express all sorts of concern, giving you a chance to show your understanding of leds. A possible (quick) answer could start with confirming the left hand side is perfectly ok, and the right hand side is 20ma through the bottom, and 10ma through each individual led. That said, you could elaborate a bit, about leds being non-linear. So, in real life, the current through the bottom led will be more than 20mA (perhaps 22mA) and through each individual led also a bit more, perhaps 11mA. An precise calculation is not possible, with the data provided. But I agree it is a silly circuit to be forgotten quickly.

[@skuula]

It's not about theory vs practice... Those who insist that the two are in opposition have not much idea about either. It's flawed theory vs good theory

[@GertvandenBerg]

The other method is A = 20 mA, B > 20mA, B < 40 mA (if B reaches 40 mA, C is 20 mA, which only leaves 2V over that LED at the bottom (which can only sustain 20 mA), which means the current needs to be less than that), C=B/2 (the exact value of B (and C) in that range would need more information)

[@besenyeim]

Fun fact: semiconductors are sensitive to all kinds of energy, not just heat. Theoretically, the LED is reacting to incoming light too. Obviously, the effect is negligible, probably unmeasurable.

[@janami-dharmam]

It is well known that a common LED can be used as a photodiode; that it is rarely used in this mode is simply because the active surface of the LED is too small to be useful as a photodiode.

[@sdgelectronics]

I suspect xrunner also had some measurement error in the parallel LED pair as a result of the shunt resistor of the multimeter.

[@danman32]

Reminds me of a question in electronic music class in high-school: Considering perfect human hearing can hear from 20hz to 20khz, is it worth the extra money to get an ideal amplifier (no sloped roll-off) with a frequency response 20hz to 30khz compared to one 20hz to 20khz? Expected answer was yes as you'd want to pass harmonics. But I always argued you can't hear any harmonics past 20khz. I couldn't find a way to articulate that the ear itself would be a bandpass filter just like the amps.

[@PainterVierax]

Yeah that's how non-educated people are pushed to buy newer and more expensive audio equipment. and you were still assuming a perfect human hearing freq range. In reality audition falls in the young age and most of us can't perceive sounds past 15kHz. Perceived sound level is heavily non-linear though so even in hi-fi there is no need to give a shit about high freq distortion. Also, a well-built hi-fi system will have a low pass filter to cut those >20kHz just to preserve the speakers. And finally, the sound characteristics is made primarily by the room, secondarily by the speakers/microphone. The amp, DAC or audio source have a marginal impact, especially in today's world.

[@danman32]

@@PainterVierax that's true, but you missed the point. The argument I had with the teacher was that with the ears having a bandpass of 20-20k, no amount of high frequency harmonics output by the amp would make it past the ear. Probably best way to explain it would have been to switch from frequency spectrum to time spectrum. The ear would miss any change in the waveform shorter than 50uS. I suppose another demonstration would be a square wave output from a 30k amp passed through a 20k low pass filter (to emulate the ear) would end up the same as a 20k amp.

[@PainterVierax]

@@danman32 I didn't miss it, it's just that I didn't react to it since that teacher clearly didn't understand his topic. What I pointed out is that this all amp bw question he asked is even more irrelevant in real life situation than the younger you supposed during that time. You can make that LP filter even narrower in your demo, still the perceived result won't be altered. That's why we hear more differences in sq wave v/s sine in the lower freqs rather than the higher.

[@cubbucca]

the ear can't hear it, but maybe the soul can feel it.

[@PainterVierax]

@@cubbucca that's why a double-slit experiment is required. With a proper methodology, many cognitive biases and pseudosciences are teared apart.

[@denismilic1878]

I made just glance over this and do calculations In the head left side A = 20mA (4V, 200ohm), right side B=26 mA (4V, 150ohm) C=B/2= 13mA if you ignore LED U/I nonlinearity. Left LED normal power, the right bottom LED overdrive, right up led pair 60% power. I'm just an amateur who likes math and electronics. In this type of exam is usually stated that you ignore nonlinearity, heat dissipation, radiation, etc. This is just a way to learn simple circuit logic. It's like calculations of force, acceleration, power, work without friction, and efficiency.

[@lachlansmith6746]

Assuming ideal diodes RHS will burn out (either the leds or wires) and become an open circuit leaving you with just the LHS with the led and current limiting resisitor. easy q 🙃

[@ongunkanat]

Exactly my thoughts.

[@vihai]

If the leds are modeled as real, matched, components they cannot burn because 40 mA is an upper limit for the current of the bottom led.

[@smunaut]

Huh ... no ... An ideal diode is one that obeys Shockley ideal diode equation ... and given the Vf=2V @ 20mA data point you can be _sure_ that the current is less than 40 mA in the RHS.

[@lachlansmith6746]

@@smunaut An ideal diode just experices a voltage drop across it with 0 internal resistance. I think you might be confused.

[@ongunkanat]

@@lachlansmith6746 I was going to write the same thing but I thought on the subject a little bit. The thing is "ideality" of a diode is a spectrum. Schockley models non-ideal diodes with ideal *p-n junctions*. Since the diode in the question has a specific amperage at a specific Vf, it must be non-ideal. A truly mathematically ideal diode will pass infinite amount of current for any positive voltage, as you said.

[@f-s-r]

They shouldn't use this kind of circuit in an exam, because that would teach the student to expect something that doesn't hold in real life, and a circuit that is a very bad idea to design like that. At least do not simply ask for the current to be calculated, which would be an impossible task with real-world components. Instead ask if the current can reliably be calculated on each of the test points, and justify the answer. Just by looking at the datasheet on the video (which the exam doesn't even provide), it says that a the LED has a typical Vf of 2v, and a max Vf 3v (if i understood the mess they made to the columns right). And what's the point of using ideal (or better call them magical) components? Just use resistors instead of leds on the right branch, and nobody's hurt. On the right branch, assuming identical leds (and good luck finding that!! :P), then the current thru the upper leds has to be half of that on the lower led, which means that the Vf on the upper leds is going to be less than 2v, and on the lower led it's going to be higher than 2v, and the current thru the lower led is going to be higher than 20 mA, but exactly by how much? damn if i know! By the way, spanish speakers always call them led instead of L.E.D., as the pronunciation of any letter combination is obvious in spanish, and a lot shorter to speak "led" than "L.E.D." In fact, we always use letter sounds in acronyms, instead of letter names, probably because letter names are never shorter than letter sounds, usually around 2 or 3 times the lenght, and most letter names include E, so the result of using all that letter names put togheter would sound funny. For example: "ele.e.de.", instead of "led".

[@davidjohnston4240]

In practice, those LEDs have non linear V/I curves and with three of them in circuit, the only rational approach is to do an iterative approximation to the stable point, with Monty Carlo parameters to show the distribution from which you can estimate the yield. Of course that's pretty much what we do for real analog circuits on real silicon chips. The math gets you 1/3 of the way there. The simulations get you the rest of the way.

[@rarbiart]

from the question I would assume the follwing tasks are expected from the student: 1) looking up at least one matching diode datasheet 2) calculating the currents by the the appropriate differential equations. 3) and only finally (after doing the work of 2) explaining why it's a bad design by calculating the effect if temperature drift on b/c 4) finally go on the topic of binning and production/reliability issues if used in mass production

[@JessicaTranaker]

OMG! I remember a very similar question from my students' days. (I am from Sweden)

[@electricreflex]

If you had the exact diode equation, it is not a solvable equation by hand. You could either use iteration or a computer solving method but the intuition is likely spot on: the parallel LEDs will each draw less current/voltage than the ideal diode and the following diode in series will make up for it and draw more current/voltage. Those exact values will have an equilibrium point where kirchoff’s current law is valid AND the exact diode equations are valid for each diode, but the solution will require solving something like 3 exponential equations with 3 unknowns. An absolute nightmare so solve by hand and not what the author intended, but if we can build it, there WILL be a solution.

[@ColinTimmins]

"The poor bastards on the right will have to sort them selves out..." Haha, loved that one.

[@KJ6EAD]

Long ago I calculated the effective series resistance of a common red, orange or yellow LED to be 100𝛀 at 20mA. It's been a useful bit of knowledge whenever I've needed to quickly approximate component values for a circuit.

[@esecallum]

v= zero across the series leds as vf=2 plus vf=2 4vcc-4vf =0. in real world you will get some current

[@space_engineer17]

Step 1:Build the circuit Step 2: Probe the multimeter wherever you want

[@dtiydr]

We got this question in the electronics higher class once. While the majority started to calculate and stuff and got all kind of answers I just wrote that it was an instant short. I was one of very few that had a right answer which was really surprising to say the least.

[@JimGriffOne]

Oh no, parallel diodes and no current limiting resistor on one side! Thermal runaway, here we come! How can anyone calculate what's going to happen here, or is it that you calculate "ideal" values?

[@lightningmcqueen1577]

Ideal values obviously, these tests are made to test math skills not electronics

[@rjrussell63]

Since there is no limiting resistor in the second branch, the LEDs will draw a current level that develops an ~ 2vdc voltage drop, so the two LEDS in parallel will draw ~20mA each and the bottom LED will have ~ 40mA assuming they are kept at 20C and don't run away.

[@vincei4252]

You know what's worse? Being penalized for questioning such a stupid question in the margin. Years later I am still being thrown these types of idiotic questions even at work where all answers are wrong and you have no recourse to point this out, however, you have to complete the multiple choice "training" in something not related to my actual job because compliance says so. It is so infuriating.

[@EEVblog]

I remember failing a digital exam because I used IEC logic symbols instead of ANSI which was the only one the lecturer knew.

[@jkobain]

I'm glad that I took a glance at the thumbnail of the video before listening to you, Dave. I'm especially glad that I was able to roughly estimate the values, but for me it was even more important that I could realize the stupidity and uselessness of such a circuit on an exam. If only it wasn't an intentional trap, of course. And thank you so much for another great video!

[@noobian458]

I like you calling it a led (rhymes with dead) Dave! Growing my dad would always say it like that, and I cringe every time I hear L-E-D... It's a led!

[@lightningmcqueen1577]

The Converse is also true we have been hearing L E D separately from childhood and thus cringe at the alternate pronounciation, English is a funny language after all

[@ferrumignis]

And once you've passed too much current through it, it becomes a ded.

[@copernicofelinis]

It all depends on the model used in that course. An unrealistic but plausible (from the point of view of evaluating circuit theory skills) piecewise linear model where the diode is an open for negative V and a simple resistor for V>=0, the circuit can be solved without any inconsistencies. Once it has been determined that all diodes are forward biased, just substitute them with the resistor (what value? You are given vf and Id, so 2V/20 mA is 100 ohms.) Then you can find all the currents you want. The circuit is (R+R) // ( (R//R) + R ) Edit: and the currents will be: iA = 20 mA iB = 26.7 mA iC = 13.3 mA

[@Paxmax]

At best; mildly infuriating. EEV: "These diodes on the right side they just gonna hafta sort 'em selves out... the poor ba**ards" Douglas Adams: Marvin trudged on down the corridor, still moaning. "I've got this terrible pain in all the diodes down my left hand side..." Me: *hard chuckle* while eating, checks myself... ok, ok... *swallows again* ...I'm good =o)

[@youtuuba]

Dave, I would have to disagree that the right branch (lower half) has 20mA through it; the "Vf = 2V @ 20mA" does not mean that the LED regulates the current to 20mA. Without a current limiting resistor (or active current regulator) in that right branch, those right LEDs will have a huge current, limited by whatever the power supply can source, and soon the current in the right branch will be 0 after then burn out and open.

[@flymypg]

I need to spend more time in the forum....

[@EEVblog]

No responsibility taken if it consumes your life!

[@vincei4252]

@@EEVblog exactly why I don't spend more time in the forum.

[@cdw4255]

On a civil service exam a question was presented (bear with me, it was 18 years ago.) Describe the condition of the circuit, including any explanations and voltages in the diagrammed circuit if R (whatever) failed as a short.

[@srmofoable]

oh dave. You silly man, thinking university teaches anything practical. Its pretty unbelievable I have senior EE undergrads coming to me for an internship and they have never even shopped digikey let alone looked at a datasheet.

[@lunstee]

What constitutes an correct answer to this question depends on the context in which it's being asked. In many situations, Dave's take on it, that it's nonsense, is reasonable. However, at a certain level, it is certainly a reasonable question, calling on the examinee to identify what, if any, additional information would be necessary to solve, and what assumptions can be made to guess at it. Simply saying that the Diode is nonlinear and giving up on going further is literally saying to an interviewer that you know enough to make a certain amount of money and don't care about being worth more than that. If I were using this question in an interview, it would be because I would be looking for whether the candidate understands that the nonlinearity of a diode's IV curve is exponential, and how one works with exponential functions. Tomorokoshi gives the best reasonable final result, but still doesn't quite get full marks, relying too much on equations rather than a good intuition to get there. A solid A, but not quite A+. The answer I would be looking for is: 1) Assume the diodes are all matched, and there is no thermal runaway between the two parallel diodes. 2) In the absence of more detailed information of the LED's IV characteristic, assume it's in the exponential region of it's IV curve (e.g. series resistance isn't yet significant) From 1), we have right away that the top diodes each pass an identical amount of current, and the bottom diode, passes the sum of that, or twice the individual top current: Ib = 2*Ic If we hold Vcc at 4V and the node where all three diodes are connected together at exactly 2V, then from the diode spec given, all three diodes would be passing 20mA. There's an imbalance of 20mA that needs to be provided to make this happen. If you don't provide (sink) that 20mA, the voltage of that node will be pulled to a higher voltage by the two diodes being stronger at pulling the voltage up than the bottom diode trying to hold the voltage down. As we let the intermediate voltage deviate from 2V, the currents respond exponentially. The top diodes will see their current reduce by a factor of k=e^((Vmid-2V)/nVt) where Vt is the thermal voltage (~26mV), but n isn't given. Conversely, the bottom diode will see its current increase by the same factor, assuming it's an identical LED (same colour) to the others. We're not given the wavelength, hence n, so can't figure what Vmid ends up being, but do have that the two top diodes pass 2*Ic=2*20mA/k, while the bottom diode passes 20mA*k, and that these are the same. Taking out the common 20mA factor, we have 2/k=k, or k=sqrt(2). From this, Ic = 20mA/sqrt(2) ~= 14.1mA and Ib ~= 20mA*sqrt(2) * 28.2mA

[@kc4cvh]

The current at Test Point A is indeed 20 mA. However, the current at TP-B and TP-C would require the conductance curve of the LEDs to calculate, as the function is non-linear. Also, whether all three LEDs are of the same type is not specified, making an answer conditional.

[@paulharrow7897]

The most correct and complete answer that it is possible to give is... A = 20 mA, B > 20 mA, C < 20 mA (and C = B/2). Without the LED characteristic curve, we cannot be any more accurate than that, but we do know B will be > 20 mA and C < 20 mA because the voltage across the top two diodes will be less than the voltage across the lower diode.

[@demindor]

You can check that if the voltage drop is 2 V for 20 mA, then 100 Ohms resistor provides the same drop, and the current at point A is 20 mA. But all you really can tell about points B and C is that current at point B is twice the current at point C. Hence, the voltage at point B is higher than 2 V, and the real current depends on how those LEDs conduct at voltages above and below 2 V. All you can say is that the current at point B is higher than 20 mA, and the current at point C is higher than 10 mA, and they relate to each other as 2:1.

[@ambassadorkees]

Knowing the typical LED Vf curve, the *only possible correct answer* is: A: 20mA B: >20mA 10mA

[@drcyb3r]

In school I had a physics teacher that was a physics doctor (don't know if that's the right term) and he would have loved that problem. He wasn't a very good teacher (didn't know how to tell the stuff to the students) but he was really good at math and physics. And he was also interested in questions that were a bit off-topic or only topic-related. So if you didn't understand something or had a question to another topic, he took the time in his break or even after school to calculate the answer or to explain it with really good drawings on the board. If he didn't know something, he told us he didn't know instead of trying to find an easy answer to not "look bad" in front of the class. But he then took the time to look the question up and explain it in detail in the next lesson.