How to Solve the Diode Circuits (Explained with Examples)

In this video, different methods for solving the diode circuits have been discussed.
There are two methods for solving/ analyzing the diode circuits.
1) Graphical Method
2) Diode Approximation
Both methods have been discussed in the video and using the diode approximation method, different circuit problems have been solved.
The timestamps for the different topics covered in the video:
1:02 Graphical Method (Using the Load Line)
3:19 Diode Approximations
4:29 How to Solve a circuit problem using diode approximation
7:46 Example 1 ( Series connection of Diode)
9:54 Example 2
12:11 Example 3 (Parallel Connection of Diode)
13:41 Example 4 (Parallel Connection of Diode with different diodes (Si and Ge))
16:11 Example 5 (Parallel connection of diode with different voltages)
Graphical Method:
In graphical method, the load line is drawn on the V-I curve of the diode.
The load line gives the possible values of voltage and current of the diode for the given circuit. The intersection of the V-I curve of the diode and the load line gives the operating voltage and current of the diode.
Diode Approximation:
In the diode approximation, the diode is replaced by the equivalent circuit. Although the values of voltage and current of the diode by this method is not exact, using the diode approximation the circuits can be solved/ analyzed very easily.
The link for the other useful videos related to diode:
1) Introduction to a diode and V-I characteristics of the diode
• Introduction to Diode:...
2) The diode resistance Explained:
• Diode Resistance Expla...
This video will be helpful to all students of science and engineering in learning how to solve the diode circuit problems.
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/ all_about.electronics
Music Credit:
www.bensound.com/
#HowToSolveDiodeCircuits
#DiodeExamples

[@ALLABOUTELECTRONICS]

The timestamps for the different topics covered in the video: 1:02 Graphical Method (Using the Load Line) 3:19 Diode Approximations 4:29 How to Solve a circuit problem using diode approximation 7:46 Example 1 ( Series connection of Diode) 9:54 Example 2 12:11 Example 3 (Parallel Connection of Diode) 13:41 Example 4 (Parallel Connection of Diode with different diodes (Si and Ge)) 16:11 Example 5 (Parallel connection of diode with different voltages) For more Solved Questions on the Diode Circuits, check out this playlist: kzbin.info/www/bejne/nGHbfn-MpdpjsNE

[@baljeerai3789]

In ex 3..why did u only subtract 0.7 ..there was 2 Si diode..so u should have to subtract 1.42??

[@rajugawala]

@@baljeerai3789 because only 0.7volt drop between common node will be there of two diodes suppose them as single element with 0.7 volts

[@vincentburdeos6152]

In example 5, how do you choose which diode is conducting?

[@mav1133]

In example 4 there are two diodes , what happens to the current , does it pass equally through two diodes?

[@orbhaisabkaisahoothik14]

Last one answer should be 4.3V

[@johangustafsson7396]

For every course I cannot understand, I can rest assured that there's and indian guy with a dedicated youtube channel for it somewhere.

[@clown12ify]

Not all heroes wear cape, some make educational diode circuits solving problems video

[@Honest-King]

Yeah they are everywhere ...

[@CarsonScholz]

Yeah literally man.

[@GStar1]

@@CarsonScholz being Indian I also know there'll some Indian to help me out 😂

[@azyle2104]

@@GStar1 lmao same

[@clown12ify]

Thank you indian guy, you're saving a student for doing this tutorial video

[@danielstoy171]

This is the most perfect video ever. Clear and concise with perfect explanations. Thank you!

[@Honest-King]

The accent though ...

[@danielstoy171]

man not really. I can still understand him pretty well with the accent

[@Astra20284]

yeah u r right

[@covertmediocrat]

@@Honest-King you probably are virgin because nobody around you looks like Kim Kardashian but you know with that face of yours you won't be able to impress bar dancers 😂

[@priyanshupreet8108]

@@covertmediocrat bro itna offend kyu ho gya?😂

[@yasminekudjoe7548]

I was so confused with this diodes stuff but you saved my llife. Thank you, you are the best

[@ahsanhabib8877]

I study in one of the top university in Bangladesh having one of the best physics teachers in South Asia. I admit you make it clearer than those over learned professors. Thank you so much sir.

[@harshalshah1058]

Before that video I was very much confused about diode solving but now I got confidence in this topic .Thanks sir for this amazing tutorial ✌️👍👍

[@SSRReloaded]

Thanks a lot, sir, my electronics exam is just 3 days away, and despite covering the syllabus, I was just blanked out of how to solve problems. Thanks a lot for making my concept clearer & helpin me on solving problems ❤️🙏

[@Scolar69]

I am here before 30 min lol

[@mnada72]

This is a great channel for studying circuits and electronics

[@pranavipranu607]

9:49 Current = 4 mA Voltage = 4 volt

[@neelimathakur2802]

9:49

[@neelimathakur2802]

I=4 ma V=4 volt

[@burraqamber663]

Right 👍

[@orbhaisabkaisahoothik14]

@MD TAIYAB 21PKBEA500 it should be 4.3V I guess the other terminal is at -1 hence total difference is of 4.3V

[@LousyPainter]

I can see an immediate use for the last circuit. Thanks. Excellent demonstration!

[@salmansabbir5692]

Thank's Sir. It's the best tutorial. After watching this tutorial, nobody else has any problem, I think.

[@kesusd]

The voltage across the 1k ohm resistor is 4 V and the current for the circuit is 4 mA for the example with one silicon diode and one germanium diode.

[@minepro1206]

Again you proved better than my professor.

[@aleemahmed9981]

Thanks have a test on diodes your way made everything more clear.

[@sarahjohansson957]

Hell clear now. Thank you sir!

[@chiraggupta8908]

best explanation I have ever seen.... Thank you Sir for making video like this....

[@shashanks7346]

the professor might be iitian or overlearned the thing which matters is "THE TEACHING STYLE" in which this guy is overlearned hatsoff !!!

[@akshatpanda9799]

Problem in 9:50 :- current = 4 mA voltage across Si diode = 0.7 V voltage across Ge diode = 0.3 V voltage across Load resistance = 4 V

[@gauravmaurya7755]

Nice

[@jhoanasuelo8474]

For item 9:38 the value of I= 4mA and Vo= 4V

[@KhalidAli-tr4nw]

How is that ??

[@user-de8os7yd6n]

@@KhalidAli-tr4nw 5V - (0.7V for Si + 0.3V for Ge)

[@ManojKumar-ho5mf]

I came here for concept used at 15:45 And it is cleared thank you 😊

[@wtfchazpwnt]

Perfect video. So clear and direct.

[@Honest-King]

I wouldn't say "clear"

[@vegas2287]

I was in a big trouble.This is a nice video.love it!Thank you very much

[@jaishreedagar]

ya same here thankyu sir

[@umairalii4737]

Thanks man ! Love from Pakistan 🇵🇰

[@drammeh1209]

Tamam... Your explanations are always Crystal Clear...

[@shikiwoo]

Thank you for the quick and simple explanation.

[@oppk4667]

Im 2022 neet aspirant...thank u for this vdo n u r perfect ❤️

[@Jared-Cruz]

Excellent video. Very clear and helpful.

[@altairishere2357]

You are life saver!!!!👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍

[@nguyentu-mc5nl]

At the example 5, 18:0, i think Vout should be (Vo= -1+Vr) because we have a -1V source at the bottom, therefore, it should be 4.4V right? pls i need explanation, someone helps me

[@kdu4249]

you just saved my life sir.. thank you so much for this.

[@vincentburdeos6152]

In example 5, how do you choose which diode is conducting?

[@kaleabgetahun4458]

i think there would be a 1.3V at the 4V source diode which results in the diode conducting because 4V is bigger than 1.3V , im not confident with my guess tho lol

[@diptishah3232]

very nicely Explained

[@codingfacts4835]

This video is very helpful guys.Thankyou ower of this channel

[@vanhoangtran8783]

The video is excellent with the clear explication ! Congratulation ! For me, I just have one question in the final example. In effect, what is the value of V0 if the first source is 3.5 V in stead of 2 V ? And the voltage of the first diode is 0.2 V in this case ? Thanks very much for your contribution !🎉🎉🎉

[@yarenkaya7872]

Thank you very much for the tutorial. Yet, I wish you had emphasized the importance of the directions and the signs. What could happen if your assumption were wrong and how can you check whether it is or not? If I am not mistaken, I have figured out this issue as follows: We always assign positive voltage to the entering port and negative to the leaving port of a diode. If we assume that the diode is ON, then after solving the circuit(maybe by the virtue of node analysis and substituting a corresponding voltage source whose potential difference is V0(threshold voltage), we should look at the direction of the current coming into the diode. If it is consistent(with the assigned voltage direction), we're right, if not the diode was actually OFF. If at first, we assume that the diode is OFF, it requires the potential difference on the diode is smaller than V0. After solving the eq. by assuming that the diode acts as an open circuit, the potential difference condition( must be smaller than V0) must be checked to see whether the assumption is correct or wrong.

[@DhruvPatel-jo9tn]

For example 5, can you not also make the assumption that the top diode is ON and the bottom diode is OFF? This assumption would still be correct, but v0 and a different value of 1.3 V instead?

[@aagmanarora2556]

same ques

[@ShreyaRajBEC]

Same doubt

[@lachgarnoureddine4458]

Indian are the best. From morocco ❤

[@persona_jam2458]

12:13 for output voltage why we minimize it from V2 not v1

[@bsbsb9068]

Your examples selection makes your teaching amazing sir👍👍👍

[@bilalaslam189]

KZbin University > Any University

[@bhpwner]

Curious why Vo is calculated different in example 2 and example 5? Why not care about the voltage drop across the 1k ohm resistor?

[@orbhaisabkaisahoothik14]

Yes the answer should be 3.3+1 = 4.3V

[@e-footballworld8828]

​@@orbhaisabkaisahoothik14 So current will be I= v/R which is 3.3-(-1)/1=4.3mA....so Vo will be -1+(I*R) = -1+(4.3*1) that is 3.3v 😅

[@e-footballworld8828]

In example 2 there is a resistor before the diode so there will be a voltage drop

[@NEERAJKUMAR-hc1wk]

Mast Sir.. I like it...this is recall my past study's

[@Bravo-xr9yr]

in the last example how about the -1V source and 1k resistor ? do they have any effect on the output ?

[@rejithretnan8365]

I too have that doubt

[@arsal16]

massive W for this whole series

[@bobbydreamsforu]

Thank you brother. Nice illustration and presentation.

[@shameelashaik6423]

Easy to understand the concept. Nice explanation.

[@coco-oc8qq]

For the first time i understood electronics ❤❤

[@vaishnavip2327]

Why In 2 example , u didn't consider V1?

[@sumitpundir9547]

Voltage across 1 kilo ohm resistance is equals to 4 volt

[@himanimalik6973]

In Eg. 2 for calculating V0 why not consider V1

[@ClaimItCrypto]

thankyou brother for guiding us🇵🇰

[@evelynfdz590]

In example 5, if D2 were changed to -4, what would happen to Vout?

[@gcjas1998]

At 12:04, why is the output voltage only V2-5V? What happened to 10V , V1, and the 0.7V of the diode?

[@ALLABOUTELECTRONICS]

see here we are applying the KVL. And we are interested in finding the voltage at a particular node in the circuit. You can start from that side too. In that case, Vo = 10 - V1 - 0.7 Or you can say that Vo = V2 - 5 V. Both ways you will get the same output.

[@gcjas1998]

@@ALLABOUTELECTRONICS Oh! So its a matter of choice between the two. Thank you.

[@6packabb]

@@ALLABOUTELECTRONICS But Vo=10 - 8.16 - 0.7= 1.14 Which is not equal when calculated from the other end

[@indhujag5920]

@@6packabb it's same only. Just .02 difference. It's not an issue.

[@jonesmath]

This really helped. Thanks!

[@VersatileAnthem]

great explanation !!! take love :)

[@hectormota6728]

I do not understand what I have to know in theory to calculate vo = v2 - 5v, what If I assume vo = 10v-v1-0.7? help! for 11:59

[@promitaghosh235]

I exactly don't know the theory...but.. suppose the -5v wasn't there...v0 would be v2 (I think that's clear)...now...we r adding this -5 v...adding v2 and -5v we get the result

[@theceltfan32]

Very well done!!!

[@shrutikadam4315]

Perfect explanation 👍🏻....but because of that subtitles the stuff u were writing on that part was not visible.....btw u don't need subtitles...ur English can be understood by everyone 👍🏻🔥

[@ALLABOUTELECTRONICS]

If required, you can manually turn off the subtitles in the video settings.

[@rohandevaki4349]

very nice explaination and tricky sums thankyou sir.

[@prathumnaradhakrishnan837]

In ex.5 you calculated that V0=4-0.7. but there is a negative voltage of -1v also. Then can i say v0=(4-0.7-0.1)volt

[@raghava8561]

Mera life save kar dhiya Thanks bhai

[@Harshavardhan-zr2ip]

Excellent...but please answer to it...as soon as.. While drawing equivalent ckt of diode we are taking opposite to voltage supply (but it is actually inside diode electric field is from higher potential+ to lower -) so same direction of supply...

[@righteousness185]

good teaching, thank you.

[@sohampanchal7710]

Can you please make a video where your doing example 2 except there is a diode before the 4kΩ and another diode before the 3kΩ

[@johnpautler4872]

The Calculations For Example #2 Are Incorrect. I = (10V - (-5V) - 0.7V) / (4K + 3K) = 614.29 uA The drop across R1 is 2.46 volts and R2 is 1.84 volts. Vo is therefore 6.9 volts.

[@ALLABOUTELECTRONICS]

I would recommend you to check your calculation once. (10 - (-5) - 0.7) / 7k is 2.04 mA Instead of adding 5V, you have subtracted it. And that is why you are getting incorrect result.

[@johnpautler4872]

@@ALLABOUTELECTRONICS Run a Spice sim and you get my result! The current is 633 uA.

[@ALLABOUTELECTRONICS]

I too checked it and I am getting around 2.04 mA. Please check the polarity of -5V. I think, you must have put +5V instead of -5V. That's the only case, where you will get 614 uA current.

[@britneygodwin]

@@johnpautler4872 yours is wrong

[@venkat.sairam]

great video , thanks for making

[@triptivashist4502]

Thnx for this video. It really helped a lot.

[@ravindragupta04]

An ac voltage of peak value 30V is connected in series with a germanium diode and a load resistor of 400 ohm. the forward resistance of the diode is 20 ohm and the barrier voltage is 0.3V. Find the peak current through the diode and peak voltage across the load,

[@ravindragupta04]

Please solve this Question........

[@kingkingsman5920]

How can u put the signs for these ciruits. I didn't understand. Plz xplain it

[@Gamingtax]

Aapna muja acha SA samja diya thanks,💖💖💗💗💗

[@gaurisharma2120]

very clear explaination..thank you...

[@sampansarkar3015]

Why in example 2 the v°(output voltage)=v²-5????

[@katsurakotaro552]

Very much appreciated! Thanks.

[@mayurshah9131]

Nice Narration

[@jakehealthx]

Wow, so helpful!

[@Peter_Andrews]

You are amazing man! Thank you

[@mohabubalam8068]

Sooo nice video sir...sooo nice.it is physics with beautyy.thanks for your good explanation.

[@hemanthadithya459]

Bro please use spot light while explaining the questions...🙏 Thank u for this amazing video...💖💖

[@akhilerla4770]

4mA current across each diode nd voltage across 1 kili ohm is 4 Because it is Seri es conection both

[@pensycoda6306]

THANK YOU!

[@winnieaguilarjr6737]

thankYou for the video. :) its very helpful.

[@oneeyedeagle549]

EX2 : why did we assume the current through v1 and v2 are same they are in parallel right?

[@ALLABOUTELECTRONICS]

Because they are connected in the series. when in doubt, just draw the supply voltages with + and - signs, (In this case, just draw the 10V and -5V voltage sources) and draw the complete circuit. You will get it.

[@anshumanpal9230]

Quite helpful for 12th boards

[@RafaqatAli-ow2hb]

Answer I=4mA VO=4v

[@Enigma758]

11:00 the -(-5) part is a little confusing.

[@rohithchalla1089]

Super explanation brother 👌. Keep it up!!

[@umatammisetti6483]

I got 4v for the question at 9:50. Does my answer is correct??

[@berylkeziahmonterona4844]

In example 5, why is it that the diode with the 2-volt source will be the one to stay off instead of the one with the 4-volt source? is it only based on assumption?

[@ALLABOUTELECTRONICS]

Let's say, for example, the diode which is connected with a 2V voltage source is ON. in that case, the voltage at the cathode of the diode is 2- 0.7 = 1.3V. The same voltage will also appear at the cathode of the second diode. (The connected with 4V source). With that assumption, the voltage at the anode of the second diode is 4V, and the cathode is 1.3V. That means the second diode should also conduct. And in that case, the voltage drop across the second diode is more than 0.7V (which is not possible) On the other end, if second diode is conduciton alone, then voltage at the cathode both diode is 4 - 0.7 = 3.3 V. And in that case, the first diode will remain off. (also there is no violation of KVL) I hope it will clear youir doubt, why first diode is OFF.

[@shatakshisingh2513]

I didnt understood the 5th example.....can someone please help me with this like what if the second diode remains OFF then too the voltage will remain same at both the nodes (1.3V)

[@shraddharastogi5176]

In the last example what if we assumed that the upper one silicon diode is on and the lower one if off . Then ? In that case the answer would have changed . So on what basis you made that assumption ???????????????

[@ALLABOUTELECTRONICS]

If you assume that, the first diode (the upper one) is ON, then the voltage drop across it will be 0.7V. Therefore, the voltage at the cathode of the lower diode will be 2 - 0.7 = 1.3V. The same voltage will also appear at the cathode of the second diode. So, for the second diode (the lower one), the voltage at the anode is 4V while the voltage at the cathode is 1.3V. And since the voltage at the anode is more than the cathode, the lower diode should also conduct. If the second diode is conducting then the voltage drop across it should be only 0.7V. But in this case, considering first diode is also conducting, it is 2.7V. (which is not possible for a silicon diode). Therefore, that assumption is not correct. And here only one diode is conducting. I hope, it will clear your doubt. In the video, I have already explained that. You go through the video if required.

[@TRICKSBYCHEMISTRYGURU]

how did u calculate vo in example 2

[@kewtomrao]

I think because they are parallel

[@student6830]

Vo = I * R

[@jhilmilgil4512]

9:55 Answer is I = 4mA and V = 4 V

[@PraveenKumar-zj9db]

Very good explanation. Thank yy

[@abuanas8569]

very fine explanation. Thank you

[@munsaffkhan617]

In the last example why you consider that first diode is off and 2nd is on but reverse can also be true that 1st is on and 2nd is off ?

[@sanskarkumar8674]

So helpful thank you Sir and keep going🖒

[@mostafaelgayar]

I think on 9:35, you meant 2.9 volts not 2.1 volts, am I right?

[@ritik839]

love from agra

[@saitejakothapalli663]

in example2 potential drop across silicon is 0.7 ,what we have to take if it is a normal diode

[@ALLABOUTELECTRONICS]

Unless it is specified, you can consider it as Si Diode and you can consider the 0.7V drop across it. In the question, if it is mentioned as the ideal diode, then the drop across the diode is 0. I hope, it will help you.