This video shows a way to calculate transistor bias and the values of the actual circuit. (This technique only works with a higher impedance load.)
Пікірлер: 101
@BarryMcGowan6 ай бұрын
Finally, some who explains, how to calculate transistor biasing without mystifying it, and even goes on to show it in operation. Brilliant, and thanks!!!
@user-io3jz9nb2b Жыл бұрын
Thank you so much for this video. I'm studying for my finals and no matter how hard I try to go through the NIDA lectures it simply doesn't explain it as well as you do. Short, simple explanation even a dumb college student can understand. Thank you again!
@keithnoneya8 жыл бұрын
One of the best simple videos on this subject I've seen. Thanks for teaching us. Best Wishes n Blessings Keith
@DeeegerD8 жыл бұрын
A better title for this would be how to analyze a transistor circuit. To figure out the bias from scratch you would have some requirements like swing voltage on the output, current limits on the output and through the transistor, etc... I've watched many of these videos now and have yet to see anyone take the "requirements" approach and show how to determine the bias and current components. Still a nice explanation of a transistor circuit.
@DeeegerD8 жыл бұрын
This is showing how to analyze a transistor circuit - not bias one? To demonstrate how to bias a transistor you first need requirements of the circuit and the type of transistor to be used. Then you can figure out what resistors, etc... will be required for the input and output of the circuit. This has to be the umpteenth video I have watched and they all do the same thing - analyze a circuit but never show how to create one from scratch given only the input and output requirements.
@Speleorat18 жыл бұрын
Totally agree digger. It's useful for for fault diagnosis on an existing circuit. It's no good for calculating biasing etc. for initial design.
@JustFun-iz9rf4 жыл бұрын
I thought the same thing because I’m trying to understand how to design a simple circuit. Thank you
@arjunsachar92288 жыл бұрын
How u find resistor value R1 and R2. Did you pick it randomly
@knowall57926 жыл бұрын
Very good analysis for new electronic enthusiasts.
@michaelwilson189 жыл бұрын
Great video.. Good job explaining this process...
@preciseinstruments96109 жыл бұрын
is the biasing same for common base and common emmiter and common collector configration???
@pragill9 жыл бұрын
Excellent video, very simple explanation.
@mrdouble8 жыл бұрын
Short, fast and to the point. Ty
@juanitoandrade54808 жыл бұрын
Ok. Then I only need to calculate the transistor to get close to the middle point between cutoff and saturation considering the gain and the amplitude of the signal imput. It will work kinda class A amplifier (All the time in resistive/active mode) Right?
@INCYTER Жыл бұрын
How do you find the base current? Why is it that you didn't need to account for this current splitting at that node, but only had to account for the voltage drop V_BE?
@edilmarlulab59648 жыл бұрын
is the diode in emitter is silicon?
@bingosunnoon93415 жыл бұрын
Why do you start with the 39k resistor? Why not 10k or 100k?
@ronaldrosete40862 жыл бұрын
That's a good content. You earned a subscriber, bro.
@darwenmint19 жыл бұрын
Great, simple, clear!
@Jgc812 жыл бұрын
No doubt video super,pls make clear video so easy to learn
@alchemy12 жыл бұрын
Why do you calculate VR2 and not VR1 and then use that to calculate voltage drop between the base and the emitter, when the base is supplied from the positive rail R1 side. What would be the voltage drop be between the base and the emitter if R2 was not there.
@q8legeng19986 жыл бұрын
You helped me so much much thanks to you sir 💯
@billwilliams63384 жыл бұрын
HEILMAN HACKATRONICS Does the AC/DC load line mean the maximum output wattage power of the transistor or does the AC/DC load line mean the power dissipation of the transistor? Because the AC/DC load line is selected by the EE designer of the operating voltage and operating current for the transistor and the DC bias Q point which will determine the maximum output wattage power out of the transistor and the power dissipation of the transistor. I'm not sure what the LOAD means in the AC load line and DC load line what they are calling the LOAD.
@NeilCaulfield8 жыл бұрын
best explanation and workflow I have found! Thanks! Will certainly help on my test tomorrow!
@markschmitt55508 жыл бұрын
Now for the tricky part... Input and Output Impedance... DC Switching Applications are easy... try throwing an AC signal through them like Audio... you use a common voltage divider value... say... with a 9 vdc supply and the collector to V+ and a 10k value on the emitter (a common collector NPN buffer situation), a base divide of 100k/15k, and it sounds good with a low impedance signal but try feeding it with something high like a guitar and it sounds muted and weak... then you add a small value resistor of 47k (or larger) between the divider and the base of the transistor and ka pow... the impedance of the transistor input is fooled into seeing a high impedance and the guitar sounds much better... impedance is everything when it comes to Audio and the difficult aspect to grasp when it comes to gain... And I did enjoy your video... simple and effective... what can you tell us about saturation?
@RexxSchneider2 жыл бұрын
You just use base dividers of 100K/100K and your problem is solved without any extra resistors. Simple.
@nunonavega10438 жыл бұрын
thank you very clear explanations!
@user-wk6gr1cs3o5 жыл бұрын
Thank you for the explanation of the beautiful .. I am a Transistor who did not work in Daira BLY89
@TheEtbetween5 жыл бұрын
Question, how is it you make this so easy to understand? And Professor make it so difficult. Thanks, greatly appreciated.
@sebidsouza40293 жыл бұрын
How u got R1 and R2
@mattkan28766 жыл бұрын
That is a very nice tutorial
@Jerkwaad7 жыл бұрын
Great video!
@edilmarlulab59648 жыл бұрын
hello sir i'm beginner, can you explain why you subtract 0.7v to find the VRE. is the diode is silicon type or what?
@HeilmanHackatronics8 жыл бұрын
There is approximately 0.7v dropped across the base emitter junction, when the transistor is turned on. This transistor is a N-P-N: Collector - N, Base -P, Emitter - N. A silicon diode is P-N: Anode-P and Cathode-N. in both cases there is a 0.7v drop across the P-N junction.
@siyabongandlovu48929 жыл бұрын
Mr Heilman you deserve a hand shake #TheBest
@ashiq32264 жыл бұрын
How to calculate pnp transistor bias
@jeffreymorgan84428 жыл бұрын
In the explanation you just took the current through r1 and r2. What about the base current that splits at the junction of r1 and r2? Should it be ignored?
@Lortagreb4 жыл бұрын
The motivation is probably that its very small if the beta factor of the transistor is high and such can be ignored if that is the case.
@RexxSchneider2 жыл бұрын
When analysing these sort of circuits, a simple tip is to look at the value of the collector resistor (Rc) and the lower base bias resistor (R2). As long as R2 is not a lot bigger than Rc, you are safe to assume that the base current will be negligible compared to the bias current through R1 and R2 - as long as the transistor has a β in excess of about 50, which will be true almost all of the time.
@tasnimgolwala46817 жыл бұрын
how to determine the transitior is whether in saturation or in cut off or in active region? Pls help
@321reh7 жыл бұрын
Vce=.2volts The Transistor is in Saturation Mode.When Vce=Vcc The Transistor Is In Cutoff Mode And When Vce=1/2 x Vcc The Transistor Is In The Normal Operation Mode. Vcc Is The Supply Voltage... We Have 2 KVL Loops Vcc=Rb1+Rb2 and Vcc= Vrc+Vce+Vre Voltage Gain= -Rc/Re I Hope This Information Helps!! Best Regards MR.OHM1970
@royrogers76448 жыл бұрын
Nice video, but how do i calculate the resistance of R1,R2, RC and RE?
@Honest-King4 жыл бұрын
Ohm's law ? V/ I
@martinmartinmartin29963 жыл бұрын
Heilman has done good work in showing the various DC voltages when biasing a bipolar transistor. The trouble with this approach is the accuracy and the application of this design. - the hfe of a bipolar transistor is SPECIFIED with an variation of (say) 3:1. -A further variation depends upon TEMPERATURE due to the self heating of the transistor -A complication of the design arises due to the magnitude of LOW FREQUENCY AC output signals . Many factors must be considered and juggled..., too many to quickly discuss.
@RexxSchneider2 жыл бұрын
Because the current through the bias resistors is significantly larger than the base current, the hfe of the transistor has little effect on the bias points. Because the collector current is fairly low, the dissipation in the transistor is small (about 15mW), so the self-heating will be no more than about 3°C at the junction. That means a reduction in Vbe of about 6mV, but that is swamped out by the voltage across the emitter resistor, which provides negative feedback and stabilises the bias points. Because the emitter resistor (470R) is very much larger than the intrinsic emitter resistance (25mV/Ic = 9R), the non-linearity of the intrinsic resistance is swamped put and has virtually no effect on the large signal response. Now, of course, all of that depends on a sensibly designed circuit, which the one shown is, so the points made are valid considerations if you have transistors with very low hfe, or you have base bias resistors that are too large, or you have a circuit with high current and voltage, or you have a very small emitter resistor.
@PabloGonzalez-wi5gq9 жыл бұрын
Funny how your piece of paper says the project is biasing a transistor when in reality all you're doing is calculating parameters given the bias.
@0dbm3 жыл бұрын
That’s the first time I have ever understood transistor biasing
@daviddavud567810 жыл бұрын
Matt Heilman at 2.30 you say it is between saturation and cut off. BUT: Vce is much greater than Vre +0.7V so it is definitely operating in saturation right??
@HeilmanHackatronics10 жыл бұрын
A BJT is current dependent so difficult to determine by voltage.
@321reh6 жыл бұрын
For Normal Operation Vce=1/2 x Vcc, Saturation Vce=.2 volts and Cutoff Vce=Vcc Vce must be>> Vbe for Proper Operation Rb Controls Ib Ib controls Ic Ic Controls Vrc Vrc Controls Vce.
@L1Christopher9 жыл бұрын
very clear and concise, many thanks. There was no mention of beta??
@RexxSchneider2 жыл бұрын
Because the circuit was properly designed to minimise the effect of β on the bias points, so it didn't really matter.
@roadeycarl8 жыл бұрын
Really useful video. Thanks! Is there any way to calculate the Beta using this method?
@LetsPlayMC20138 жыл бұрын
B = Ice / Ib current through the collector (or through emitter) divided by current through the base.
@roadeycarl8 жыл бұрын
Ah yes! thank you.
@sooohaib9210 жыл бұрын
very useful ! thank you
@HeilmanHackatronics10 жыл бұрын
Glad it helped.
@juanalvares115 жыл бұрын
can we calculate it with the kirkof's law?
@Honest-King4 жыл бұрын
I am a student so I think it is possible but to do so one must ignore the transistor because he didn't used it in the upper loop to find the current
@primalitopanaguiton27854 жыл бұрын
ehere did you get the .7v
@RexxSchneider2 жыл бұрын
Any silicon transistor passing a few milliamps of current will have a base-emitter voltage drop around 0.6V to 0.7V. It's an inherent characteristic of all silicon P-N junctions.
@mjkirk1210 жыл бұрын
I always get confused on the current thru R1 and R2 in the first step. Can you assume the current is equal in both R1 and R2 to compute the voltage at the base? Won't some of the R1 current flow to the base and the balance flow thru R2? It would not be a true voltage divider. It seems as if you need a set of simultaneous equations (in a spreadsheet?) to get the correct answer. Or maybe your disclaimer "this only works with a high impedance load" is the key? This makes the current flowing to the base from the R1-R2 junction negligible? But the emitter resistor is much smaller (470 ohm) - is that considered a low or high impedance? I guess it must work - since your circuit measurements agree with the maths. Thanks for posting.
@SeattlePioneer5 жыл бұрын
That seems plausible to me....
@RexxSchneider2 жыл бұрын
You start by assuming that the collector current (Ic) is about half the supply voltage divided by the collector resistor - in this case 6V/2K = 3mA. Then you make a worst case estimate of the current gain β of the transistor, say 50. That tells you that the base current (Ib) can't be more than about 3mA/50 = 60μA (and probably a lot less). So when you calculate that the current through the bias resistors is 300μA, you can see that it's much bigger than Ib and the effect of Ib on the bias point will be small. if you want to compute the actual effect, then you can use Thevenin's theorem to treat the voltage divider as a voltage source of 2.05V (as calculated) with an "internal" impedance equal to the parallel combination of 33K and 6.8K = 5.6K. If you draw a base current of 60μA through 5.6K, you drop 340mV. so the actual voltage at the base would be 2.05V - 0.34V = 1.71V. Of course that is worst case when β=50, so in most real cases where β might be 200 perhaps, then you'd have 1/4 of the base current and therefore 1/4 of the voltage drop, so the base bias point would be more like 2.05V - 0.09V = 1.96V. Remember that in a real circuit, resistor tolerances can often have a bigger effect than that.
@shipwrech9096 жыл бұрын
thank you so much
@tgee74857 жыл бұрын
all well and good until your input is 20 mV pk-pk, -0.7v gives you a negative value... then what ? im so confused
@RexxSchneider2 жыл бұрын
You couple in your ac signal through a capacitor to the base of the transistor, so that the 20mV pk-pk stays centred around the +2V that the base is biased to. The circuit as shown will accept input signals up to around 2V pk-pk.
@SeattlePioneer5 жыл бұрын
I don't see where the .7 volts for the base/emitter voltage drop comes from....
@RexxSchneider2 жыл бұрын
Any silicon transistor passing a few milliamps of current will have a base-emitter voltage drop around 0.6V to 0.7V. It's an inherent characteristic of all silicon P-N junctions.
@ArunKumar-sp7ts5 жыл бұрын
nice video
@johnchin37349 жыл бұрын
I thought it is a tutorial on how to bias a transistor. But what your doing is trying to compute current voltages for a given bias. You should have shown us how to compute for the resistance of those resistors so that the transistor is in bias condition. Yours is the resistance is already given. But the problem is we dont know how to determine the value of those resistors. Hope you can make another video showing how to compute for the resistance of the resistors to stabilize the transistor.
@RexxSchneider2 жыл бұрын
That's the difference between analysing an existing circuit (which is what this video set out to do, and does well) and designing a circuit to meet a particular specification (which is not the intent of this video). The design process can be complex and requires rather more than a basic knowledge, so it isn't the same target audience.
@nick663338 жыл бұрын
thanks bro
@zee0337 жыл бұрын
not the " bias" I was looking for😂😂
@Jerkwaad7 жыл бұрын
You were looking for liberal bias.
@prashanths12537 жыл бұрын
awesome
@jjay11a9 жыл бұрын
i think the calculations were off because of the beta of your transistor, either way great video.
@321reh9 жыл бұрын
+Jose Alfonso And the Resistors have Tolerances of + - 5% or + - 10% This amplifier is Independent of Beta
@dicko195 Жыл бұрын
39.8 divided by 12 = 3.316
@jonsanford010 жыл бұрын
Is it the calculations that cause the circuit to work?
@HeilmanHackatronics10 жыл бұрын
Of course, just like the sun didn't rise until we determined the earth rotated :p
@ElectronicSupersonic10 жыл бұрын
My com teacher wrote this to help us bias bjt's www.uwplatt.edu/~drury/bjt_bias.pdf
@jonsanford010 жыл бұрын
Jesse Robinson All that is impossible for me to remember.
@ElectronicSupersonic10 жыл бұрын
I like the name of the file. "How to bias bjts for fun and profit"
@wolflahti4128 жыл бұрын
You don't need to hold the entire process in your head at once. (I certainly couldn't.) You just look at the circuit and decide what you need to figure out, step by step.
@sangtranvan34587 жыл бұрын
thankS
@bacho74105 жыл бұрын
This guy went from black to white real quick
@B1r3a4n5i6a7c88 жыл бұрын
Ha! beautiful!
@tarekmostafa87278 жыл бұрын
R1 and R2 shouldn't have same current why u plus the resistance ?
@HeilmanHackatronics8 жыл бұрын
to determine the total resistance
@tarekmostafa87278 жыл бұрын
so they are in series ?
@HeilmanHackatronics8 жыл бұрын
This method is an approximation technique, which as you can see at the end of the video works.
@tarekmostafa87278 жыл бұрын
ohhh I see ok ) thanks much
@scubamaster15046 жыл бұрын
Tarek Mostafa ... the current through the base is very small and negligible (uA) compared to that through the resistors.
@woodwright21278 жыл бұрын
If there is voltage drop of 0.7 volts across the BE junction in the transistor, shouldn't the impedance of that junction be taken into account as a series "resistance" with RE? And shouldn't the combination of those two be taken into consideration as a parallel resistance to R2? All of the above before you can calculate the current through R1? Just asking here, because I'm no expert on this. Perhaps you are, but it seems you are ignoring the part of the circuit that is parallel to R2 in your initial calculation. Please explain.
@juanitoandrade54808 жыл бұрын
+Wood Wright The resistance of the PN junction is not linear, so is better take it out as voltage.
@RexxSchneider2 жыл бұрын
When calculating the dc bias points, the dynamic emitter resistance of the BE junction will be so small that we can assume the voltage Vbe is always between 0.6V and 0.7V for all practical collector currents. In most cases, it's simplest to just assume Vbe is fixed at 0.7V. When you're calculating the gain of the stage, the dynamic impedance of the BE junction will need to be considered. For a bipolar transistor, that is the slope of the Vbe vs Ie curve, and is given by 25mV/Ie, which turns out to be around 9 ohms in the given circuit. because Re (470R) is so much larger than that, we can ignore the effect of the junction impedance. Because the current into the base is a factor of β smaller than the collector current, it means that resistances seen in the emitter are effectively multiplied by β when considering their effect on the base. In this case, there is indeed an effective resistance of β x (470R + 9R) in parallel with R2. However, if β > 50, then that resistance looking into the base turns out to be at least 24K (and probably a lot more), so it only has a small effect on the 6.8K resistor R2.