Finally I really get it, thank you. We put together a circuit like this in class yesterday and we really didn't quite understand what we were doing, despite many weeks of prestudies. So many teachers are just all over the place and wants to teach everything at once, making it incomprehensable.
This was so well done! Other videos do not cover concepts that are essential for a clear understanding. Your teaching is truly unique and does not even allow for confusion to arise. I would suggest a next video for an emitter follower however with watching this video I can figure it out on my own.
Thank you very much sir for your generosity in sharing this very needed knowledge and delivering it in the most effective way. From now on wards I'm a subscriber. Thank you again. God bless you sir.
I'm studying these topics for the doctoral program interview. I studied from the book during my undergraduate education. Your expression is very nice and fluent. Thank you from Turkey :)
I'm so glad I've found this video, you don't know how this helped me. Not only I learnt how to re-draw the circuit into equivalent one, but I also revised the circuit solving. Thank you.
I learned and understood more from this than I ever did in my electronic devices class. It's too bad that many people assume this is the level of instruction they will receive when attending a university - if this level of clarity and explanation was the standard, tuition prices may be a little more justified.
Thank you James. I'm sure your tuition fees weren't wasted. Maybe this video just pulled a few more of the pieces together for you. I'm really pleased it helped you.
Very clear, concise, and easily applicable to other amplifier circuits....much easier to understand and apply than the electrical circuits courses I took in school....
It was a very clear and nice explanation. I have seen a lot of videos but it was really good considering you have showed the logic behind the analysis instead of random formulas. Thank you so much, sir.
This is THE MOST beautifully presented and explained video on Small signal analysis of BJT! Excellent, excellent job! A small request, can you please make a video of small signal analysis of Mosfet?
The MOSFET uses medium circuit analysis! Large or small don’t work. It can also be referred to as “Goldilocks” analysis. You probably won’t learn about this at your state school but here at MIT, it’s a closely guarded secret and only taught to upper class men.
Tried other videos but hands down, this is the best out there. Thank you very much for explaining in a very simplified way. Hpoe you will make video on frequency response as well.
This is amazing!! I had been looking for so many videos, but this is the actual video that I could understand !! Very detailed explanation, love it. Hope you could keep making video with such good content and excellent presentation.
thank you very much, sir. watching your video i understood the small signal concept very well thanks to you. and i got to say you really are a good tutor please keep up the good work.
Really good and understandable presentation of the subject and the mathematical background. There is a lot of substitutions work to do. But anyway, very clear and easy to follow step by step for those who are fit in mathmatics. Thanks !
For the question, Ic goes through Rc cause if you look back through dc perspective again, ic is the ib at base times beta, in which current is drawn form a dc source, in case of ac analysis, dc is basically a/c ground.
You could make life a lot easier by making use of the concept of the dynamic emitter resistance, that is the slope of the Vbe vs Ie curve. It is equal to the reciprocal of gm and depends only only the emitter current and the absolute temperature. At room temperature, it is simply 26mV/Ie (sometimes taken as 25mV/Ic), so should be immediately calculable from the dc analysis. Now see how it simplifies the calculations and makes them more practical: At 15:22 "So we feel pretty happy with this result", referring to Av = -β.Rc/rπ, but that looks to all the world as if the voltage gain depends on transistor β, which it doesn't. In fact rπ is just equal to β.re. That is logical since rπ is just the slope resistance of the junction seen from the base, and the base current is β times smaller than Ie (for any modern small signal transistor, β >> 1 and Ic = Ie). So Av = -β.Rc/rπ can be simplified to Av = - (β.Rc) / (β.re) = - Rc/re which is a much more usable result and shows that the gain is independent of the transistor parameters. In fact, Av = - Rc/(25mV/Ic) = -Ic.Rc/25mV. It should be obvious that as Ic.Rc is just the quiescent voltage across the collector resistor, the maximum voltage gain of a common emitter stage is necessarily limited by the supply voltage alone. If you want the output to be able to swing symmetrically without clipping, then the collector bias point will be about half the supply voltage (Vs) giving a maximum possible gain of Vs/50mV. Similarly, at 30:28, you have Av = -(gm - 1/Rf).(Rc || Rf). If we write the ratio Rc/Rf as FB - a feedback ratio which gets smaller as Rf gets larger - we can simplify Av as follows: Av = -(gm - 1/Rf).(Rc || Rf) = -(1/re - 1/Rf).(Rc.Rf/(Rc + Rf) = -(Rc/re - Rc/Rf).(Rf/(Rc + Rf) = - (Rc/re - FB).(1/(FB + 1)) So if feedback is very small (Rf >> Rc) then FB = 0, and Av approximates to - Rc/re as is the case without feedback. If feedback is larger, so that Rf = Rc, then Av = - (Rc/Re -1)/2 or about half of the gain without feedback. Those are useful results.
The low-frequency small-signal model of the transistor looks like a simple resistor r_\pi between base and emitter, and a dependent current source \beta i_b from collector to emitter. Since the emitter is connected to ground in the original circuit, one end of the current source must therefore also be connected to ground.
at 13:40 on the video, for Voltage gain, The resistor R pi is basically (Beta)x(re), so the beta's will cancel and you will get Av= -RC/re re = 26mv/IE
In the first exercise (from 5:47) R1 and Rc were had a common wire and when we simplify them as R1 and R2 are put together into a parallel resistance it separates totally from Rc. Won't that affect anything? A little bit lost here
Many thanks for these lucid presentations. Where can I buy your book!? One thing still hurting my head: you say at the start 'the dc power supply looks like an infinite capacitor', so you short-circuit it. But the a.c. input signal is never travelling north towards Vcc, so never going to ground via the power supply cap, so...??
Wow, this was great! But I've some naive doubts. • how can the transistor run on a small signal, without any biasing? Or Do we primarily assume that the transistor is running at some voltage ?
In order to function as an amplifier the transistor must be biased into the active region and this is achieved by the DC power supply and the various resistors connected to the transistor terminals. Depending on the circuit we may also need to level-shift the input signal (via the input coupling capacitor) so that the transistor remains in the active region as input signal swings from its most negative to most positive values. This is why we perform a DC, or bias, analysis -- to ensure that the operating point (or quiescent point) of the device is appropriate. The input signal is treated as a deviation about the bias level in the circuit, and the AC, or small-signal, analysis is solely concerned with calculating how these deviations propagate from the input to the output. We are able to separate our analysis into two parts -- DC and AC -- because we are restricting the size of the input signal so that the transistor appears to be linear (note that the small-signal models contain only linear components). However, at all times the circuit has a DC state and an AC state, even if we consider them independently while doing our full analysis.
@@markandrews5167 wow, thank you very much. I got this doubt because, during small signal analysis, we short the sources right? Once we do that we lose the bias, don't we? How does that current source and resistor(small signal model of BJT) count in the bias. Lemme tell u what I think,correct me if I'm wrong; we are able to account the bias because, we have the collector and base currents in the small signal model (a Res and dependent current source) as those currents found during large signal analysis. Hence shorting the sources doesn't change the bias, as we consider the bias currents, and further look at the changes of those currents.
@@aravindhvasu195 This is basically correct but your reasoning can be made even simpler. Recall the superposition theorem? The response of a circuit to a combination of stimuli is the sum of the responses to each stimulus, provided the circuit is linear. As long as we keep our AC excitation amplitude small enough the transistor responds linearly to it. We now treat the complete transistor circuit as a linear network responding to two separate stimuli: the DC bias "stimulus" and the AC small-signal. Work out the response to each stimulus and add the responses -- this will be the complete (bias + small-signal) response of the circuit. One minor point: You mention "large signal analysis", but perhaps you mean "bias analysis"? The large signal analysis is usually rather messy and non-linear.
@@markandrews5167 Thanks for the question and the careful response. When you solve for Rout, you short the input to the ground. The input is the sum of a DC supply signal and an AC small signal. When you short the input to ground, essentially both the DC and AC signals are shut off. I understand the part that shuts off DC supply, but what about that AC small signal? Is there some kind of approximation here? Sorry for pressing on this point, but something I feel is not clicking in my mind, although I know your solution is correct.
That is awesome Sir. You made things simple and vivid clear. But I have been desperately looking for current gain(hfe) expression in this hybrid pi model(the second one with feedback resistance rf, in my case it would be Rbc). Could you pliz explain that part too? Thank you.
If you take the expression for Vout = Vin you got from the nodal analysis and apply it to the second expression you got from the mathematical tools. would that still be correct?
Determining r_pi is discussed in the companion video "Transistor Small Signal Model". It turns out that the AC parameter r_pi depends critically on the quiescent (DC) state of the circuit!
The precise values of the small-signal device parameters (AC part) depend on the bias point of the transistor (DC part). Check out the accompanying video on the Small-signal Transistor Model. It explains where the AC parameters come from and why they depend on the DC state of the circuit.
There is a slight ambiguity in what you have asked because you haven't written an equation. If you mean "can we write i1 = ic + if ?" then the answer is no, because all of the currents are leaving the output node. If you mean "can we write i1 = - (ic + if) ?" then the answer is yes, as is shown in the video. KCL does indeed allow you to write this equation, but nodal analysis aims to express the currents in terms of the voltage at the node in question. I hope I have understood your question correctly.
@@MarkTheEngineer thanks for the reply. I understood where I was wrong. But now i applied kvl from Vin to 0(ground). Vin +iFrF +(ic+iF)rC=0 and i substituted this value of iF in eqn of Rin but i get a different answer :/ eqn at 31:08
@@MarkTheEngineer yes and i1 is -(ic+iF) and if you substitute this value in Rin equation you get different formula. Maybe a solved example on collector feedback bias will solve my doubt. Anyway thanks sir
The small-signal model used in this video (and described in th-cam.com/video/RDsM2jhbyY4/w-d-xo.html) assumes the collector current is independent of the collector-emitter voltage and in this case r_o is infinite. In reality i_C increases slightly, and linearly, as a function of v_CE (the so-called Early effect) but with a shallow slope depending on the specific device. The small-signal output resistance r_o is the reciprocal of the slope of the i_C vs. v_CE curves in the active region. The simplified model used in this video also ignores h_re, the small-signal feedback path from the collector to the base, and the capacitive effects between the base, emitter, and collector. All of these refinements can be included to provide a more accurate model (just look at a Spice model for the transistor to see what can be included!) but as this is an introductory lecture on small-signal analysis piling on more parameters won't make the process any clearer.
I didn't know the true direction of i_f. The cool thing is that you can draw the currents going in any direction you like and, providing you write down your circuit equations correctly, the mathematics will take care of the details. For example, if the actual current really does flow in the direction I have drawn it then i_f will be positive. If the actual current flows in the opposite direction to the way I have drawn it then i_f will be negative. All you have to do is ensure your starting equations are correct, and the mathematics does the rest for you.
If someone can help me understand the first AC equivalent circuit I would greatly appreciate it. I'm a little confused as to how RC and RE can be said to be in parallel. To me, it would appear that from an AC perspective, shorting the power supply appears to shunt everything out of circuit except for R1 as the entire side of the circuit up to what was the collector side of R1 would have to be at the same voltage potential. To my limited understanding, this AC circuit would simply look like R1 is one branch, and the other branch is a parallel connection between R2 and RE if the transistor is on. It just looks like the series combo of RC and RL would have the same potential at both ends so there could be no current flow through them
I'd contend that the transistor symbol should not appear in the ac circuit but be replaced by its ac model immediately. That'll prevent the mistaken belief that a 0.7 V still exists across the base-emitter junction.
There are two "new" things happening here as far as students are concerned: (a) shorting the capacitors and DC voltage sources (and thereby apparently changing the circuit topology), and (b) replacing the transistors with their small-signal models. In my experience separating these two things in time results in fewer mistakes being made.
Outstanding teaching; clear, concise and neat. The best explanation on TH-cam.
Finally I really get it, thank you. We put together a circuit like this in class yesterday and we really didn't quite understand what we were doing, despite many weeks of prestudies. So many teachers are just all over the place and wants to teach everything at once, making it incomprehensable.
Best video on Small signal as far as i have seen. Wish I had this while i had the course.
Thanks bro. I wish you taught my analog electronics class. My professor is known as Prof. Power Point.
hahahahah, same bro
Same here too bro
I usually don't leave comments on TH-cam videos but this video really helped me. Thank You!
This has to be the best video on the topic in all of TH-cam. Thank you for making this!
This was so well done! Other videos do not cover concepts that are essential for a clear understanding. Your teaching is truly unique and does not even allow for confusion to arise. I would suggest a next video for an emitter follower however with watching this video I can figure it out on my own.
Thank you for the kind words. I'm really pleased this was helpful.
This is by far the best video I've watched on youtube about any topic on EEE. So so good and nicely explained.
Thank you to make me understand in a half-hour while uni. prof couldn't in weeks.
Thank you very much sir for your generosity in sharing this very needed knowledge and delivering it in the most effective way. From now on wards I'm a subscriber. Thank you again. God bless you sir.
I'm studying these topics for the doctoral program interview. I studied from the book during my undergraduate education. Your expression is very nice and fluent. Thank you from Turkey :)
I'm pleased this video was helpful. Good luck with the interview!
I wish I could give more than 1 like. Thank you for the effort that you've put in to explain clearly.
I'm very pleased you found it useful. Thanks for the feedback.
Excellent teaching.
Thanks ❤
I'm so glad I've found this video, you don't know how this helped me. Not only I learnt how to re-draw the circuit into equivalent one, but I also revised the circuit solving. Thank you.
I learned and understood more from this than I ever did in my electronic devices class. It's too bad that many people assume this is the level of instruction they will receive when attending a university - if this level of clarity and explanation was the standard, tuition prices may be a little more justified.
Thank you James. I'm sure your tuition fees weren't wasted. Maybe this video just pulled a few more of the pieces together for you. I'm really pleased it helped you.
Very elegant explanation. Great video.
Very great appreciation for an "every effort made" thorough tutorial.
Thanks Buck. I am pleased you found it helpful.
Very clear, concise, and easily applicable to other amplifier circuits....much easier to understand and apply than the electrical circuits courses I took in school....
I am literally paying $27k per semester (less than 4 months) to learn this. And my class isn't as clear as this.
Immensely valuable video ! A video to be seen many times and studied thoroughly ! Enormously helpful !!!
Test tomorrow. Great refresher! Very clear and concise.
Good luck!
It was a very clear and nice explanation. I have seen a lot of videos but it was really good considering you have showed the logic behind the analysis instead of random formulas. Thank you so much, sir.
Fantastic explanation!
Best lecture ever ! Thank u
Amazing explanation. Thank you
Fantastic job! Well thought out examples. I really appreciate the thoroughness put into this video.
Thanks Joshua. I’m very pleased you found it useful.
This is THE MOST beautifully presented and explained video on Small signal analysis of BJT! Excellent, excellent job!
A small request, can you please make a video of small signal analysis of Mosfet?
The MOSFET uses medium circuit analysis! Large or small don’t work. It can also be referred to as “Goldilocks” analysis. You probably won’t learn about this at your state school but here at MIT, it’s a closely guarded secret and only taught to upper class men.
Tried other videos but hands down, this is the best out there. Thank you very much for explaining in a very simplified way. Hpoe you will make video on frequency response as well.
Thanks for the feedback. I'm really pleased this has helped you!
The way he explained is sweet and simple ☺️
This video has helped me a lot. You are a great teacher. Keep up!
This is amazing!! I had been looking for so many videos, but this is the actual video that I could understand !! Very detailed explanation, love it. Hope you could keep making video with such good content and excellent presentation.
Fantastic explanation… thanks so much!
thank you very much, sir. watching your video i understood the small signal concept very well thanks to you. and i got to say you really are a good tutor please keep up the good work.
Thanks. I'm glad you found it useful.
Nice video sir could you please make one on mosfet?
This vid is helped me alot in EE class
Out standing video sir. I am subscribing your chanel right now
You just saved my ass. Thanks from Italy!
I'm pleased this was useful!
Really good and understandable presentation of the subject and the mathematical background. There is a lot of substitutions work to do. But anyway, very clear and easy to follow step by step for those who are fit in mathmatics. Thanks !
best video that explains every step
Great video, thanks a lot!!
What I dont understand is why does the current i_c flow through R_C and not directly to ground at 11:55 ?
This guy used π model for his analysis which is quite hard to catch, T model is rather easier for beginners.
For the question, Ic goes through Rc cause if you look back through dc perspective again, ic is the ib at base times beta, in which current is drawn form a dc source, in case of ac analysis, dc is basically a/c ground.
Great Work!!
I don't have word this is great
Amazing video, very good explanation!
I'm pleased you found it useful.
Excellent job!
Thanks so much :)
Excellent work.
Great stuff, really helpful!
Mr. Mark : Why are the input signals shown with a +/- in the a/c analysis circuit? I have seen them in some text books also. Regards.
Hello Mark, Your videos are very usefull. We miss you and your lectures. Would it be possible to get more please? Regards
This was a great video. Thank you so much, I'm finally understanding this topic.
Excellent explanation of bjt small signal analysis. I really loved the way you teaching
A small request Sir
Please make a video on Mosfet
I'm pleased you found the video helpful. I certainly will get around to covering MOSFETs but I don't have a delivery date just yet, I'm afraid.
Really great video. Hope you can add more video about Circuit!!!!
Thank you, I'm really pleased this was helpful. There will certainly be more videos.
You could make life a lot easier by making use of the concept of the dynamic emitter resistance, that is the slope of the Vbe vs Ie curve. It is equal to the reciprocal of gm and depends only only the emitter current and the absolute temperature. At room temperature, it is simply 26mV/Ie (sometimes taken as 25mV/Ic), so should be immediately calculable from the dc analysis. Now see how it simplifies the calculations and makes them more practical:
At 15:22 "So we feel pretty happy with this result", referring to Av = -β.Rc/rπ, but that looks to all the world as if the voltage gain depends on transistor β, which it doesn't. In fact rπ is just equal to β.re. That is logical since rπ is just the slope resistance of the junction seen from the base, and the base current is β times smaller than Ie (for any modern small signal transistor, β >> 1 and Ic = Ie).
So Av = -β.Rc/rπ can be simplified to Av = - (β.Rc) / (β.re) = - Rc/re which is a much more usable result and shows that the gain is independent of the transistor parameters.
In fact, Av = - Rc/(25mV/Ic) = -Ic.Rc/25mV. It should be obvious that as Ic.Rc is just the quiescent voltage across the collector resistor, the maximum voltage gain of a common emitter stage is necessarily limited by the supply voltage alone. If you want the output to be able to swing symmetrically without clipping, then the collector bias point will be about half the supply voltage (Vs) giving a maximum possible gain of Vs/50mV.
Similarly, at 30:28, you have Av = -(gm - 1/Rf).(Rc || Rf). If we write the ratio Rc/Rf as FB - a feedback ratio which gets smaller as Rf gets larger - we can simplify Av as follows:
Av = -(gm - 1/Rf).(Rc || Rf) = -(1/re - 1/Rf).(Rc.Rf/(Rc + Rf) = -(Rc/re - Rc/Rf).(Rf/(Rc + Rf) = - (Rc/re - FB).(1/(FB + 1))
So if feedback is very small (Rf >> Rc) then FB = 0, and Av approximates to - Rc/re as is the case without feedback. If feedback is larger, so that Rf = Rc, then Av = - (Rc/Re -1)/2 or about half of the gain without feedback. Those are useful results.
This video is very clear! But why is the current source connected to ground at 23:08?
The low-frequency small-signal model of the transistor looks like a simple resistor r_\pi between base and emitter, and a dependent current source \beta i_b from collector to emitter. Since the emitter is connected to ground in the original circuit, one end of the current source must therefore also be connected to ground.
Thank you 💓
at 13:40 on the video, for Voltage gain, The resistor R pi is basically (Beta)x(re), so the beta's will cancel and you will get Av= -RC/re
re = 26mv/IE
In the first exercise (from 5:47) R1 and Rc were had a common wire and when we simplify them as R1 and R2 are put together into a parallel resistance it separates totally from Rc. Won't that affect anything? A little bit lost here
thank u so much Mark
You are welcome. I hope it was useful.
Best video out there
35:35 If the circuit has L parellel R --> Rout=RL? Right?
Many thanks for these lucid presentations. Where can I buy your book!? One thing still hurting my head: you say at the start 'the dc power supply looks like an infinite capacitor', so you short-circuit it. But the a.c. input signal is never travelling north towards Vcc, so never going to ground via the power supply cap, so...??
I love you MARK!!!!!!!!!!!!!!!!!
great content good sir, thank you.
I'm very pleased you found it useful.
26:40 how did u find( vout-vin)/Rf
This is amazing 👏🙌.
May you please do the same with MOSFETs as well.
You are the 🐐
Thank you. 🙂
As perfect as they get 👌✌️
Thank uu soo much ♥️♥️♥️♥️♥️♥️♥️♥️
Wow, this was great! But I've some naive doubts.
• how can the transistor run on a small signal, without any biasing? Or Do we primarily assume that the transistor is running at some voltage ?
In order to function as an amplifier the transistor must be biased into the active region and this is achieved by the DC power supply and the various resistors connected to the transistor terminals. Depending on the circuit we may also need to level-shift the input signal (via the input coupling capacitor) so that the transistor remains in the active region as input signal swings from its most negative to most positive values. This is why we perform a DC, or bias, analysis -- to ensure that the operating point (or quiescent point) of the device is appropriate. The input signal is treated as a deviation about the bias level in the circuit, and the AC, or small-signal, analysis is solely concerned with calculating how these deviations propagate from the input to the output. We are able to separate our analysis into two parts -- DC and AC -- because we are restricting the size of the input signal so that the transistor appears to be linear (note that the small-signal models contain only linear components). However, at all times the circuit has a DC state and an AC state, even if we consider them independently while doing our full analysis.
@@markandrews5167 wow, thank you very much. I got this doubt because, during small signal analysis, we short the sources right? Once we do that we lose the bias, don't we? How does that current source and resistor(small signal model of BJT) count in the bias. Lemme tell u what I think,correct me if I'm wrong; we are able to account the bias because, we have the collector and base currents in the small signal model (a Res and dependent current source) as those currents found during large signal analysis. Hence shorting the sources doesn't change the bias, as we consider the bias currents, and further look at the changes of those currents.
@@aravindhvasu195 This is basically correct but your reasoning can be made even simpler. Recall the superposition theorem? The response of a circuit to a combination of stimuli is the sum of the responses to each stimulus, provided the circuit is linear. As long as we keep our AC excitation amplitude small enough the transistor responds linearly to it. We now treat the complete transistor circuit as a linear network responding to two separate stimuli: the DC bias "stimulus" and the AC small-signal. Work out the response to each stimulus and add the responses -- this will be the complete (bias + small-signal) response of the circuit.
One minor point: You mention "large signal analysis", but perhaps you mean "bias analysis"? The large signal analysis is usually rather messy and non-linear.
@@markandrews5167 Thanks for the question and the careful response. When you solve for Rout, you short the input to the ground. The input is the sum of a DC supply signal and an AC small signal. When you short the input to ground, essentially both the DC and AC signals are shut off. I understand the part that shuts off DC supply, but what about that AC small signal? Is there some kind of approximation here? Sorry for pressing on this point, but something I feel is not clicking in my mind, although I know your solution is correct.
Great cool
27:40 Did you mean to say that 1/R_c + 1/R_F is R_c//R_F instead of 1/( R_c//R_F) ?
may i ask on the last circuit why do you ignore the fact that r pi is in series with rf?
Can you clarify what you mean? r_pi is not in series with R_F in the last amplifier circuit.
4:18 "the ac sauce that's driving the circuit"
鸟小
We short circuit Vin for measuring Rout as there is a voltage source of Av*Vi in internal composition of the right part of the two port :)
hi do you have common emitter for pnp ? im really struggling with my assigment as so many good tutorial are for npn and not pnp :(
That is awesome Sir. You made things simple and vivid clear. But I have been desperately looking for current gain(hfe) expression in this hybrid pi model(the second one with feedback resistance rf, in my case it would be Rbc). Could you pliz explain that part too? Thank you.
Thankyou so much for this
At 21:12, when the current source iC is = to 0, why is it not replaced by a short circuit?
Because then you would have a short circuit across v_x, giving a very large current instead of no current in the current source branch.
much worth then any other transister lecture.
Thanks for the feedback. Much appreciated.
Thanks a lot!
If you take the expression for Vout = Vin you got from the nodal analysis and apply it to the second expression you got from the mathematical tools. would that still be correct?
Paul, I don't quite follow your question. Can you explain, perhaps with a time-stamp?
@@markandrews5167 can we use the voltage gain expression at 29:01 and substitute it into the expression you got at 33:55 and still be correct?
LOVELY , THANK YOU
How can i find the value of the r π of a BJT transistor? Is it written in the datasheet?
Determining r_pi is discussed in the companion video "Transistor Small Signal Model". It turns out that the AC parameter r_pi depends critically on the quiescent (DC) state of the circuit!
@@MarkTheEngineer thanks !
fancy making one on mosfets??
tindo tare There almost certainly will be some videos on MOSFETs but at the moment I’m not sure when. Thanks for the suggestion.
excellent !
My professor was talking about calculating DC Bias before we get to the AC part. could you explain what exactly that means?
The precise values of the small-signal device parameters (AC part) depend on the bias point of the transistor (DC part). Check out the accompanying video on the Small-signal Transistor Model. It explains where the AC parameters come from and why they depend on the DC state of the circuit.
Can you it same with h-parameter equivalent model plss
Nice video sir but i have a small doubt. Can't we write i1 as (ic+iF) 25:35 using KCL?
There is a slight ambiguity in what you have asked because you haven't written an equation. If you mean "can we write i1 = ic + if ?" then the answer is no, because all of the currents are leaving the output node. If you mean "can we write i1 = - (ic + if) ?" then the answer is yes, as is shown in the video. KCL does indeed allow you to write this equation, but nodal analysis aims to express the currents in terms of the voltage at the node in question. I hope I have understood your question correctly.
@@MarkTheEngineer thanks for the reply. I understood where I was wrong. But now i applied kvl from Vin to 0(ground). Vin +iFrF +(ic+iF)rC=0 and i substituted this value of iF in eqn of Rin but i get a different answer :/ eqn at 31:08
@@HDgaming345 KVL requires that vin = -if RF + i1 RC.
Oops, sorry this is what you have stated. I suspect your error is a simple algebraic one.
@@MarkTheEngineer yes and i1 is -(ic+iF) and if you substitute this value in Rin equation you get different formula. Maybe a solved example on collector feedback bias will solve my doubt. Anyway thanks sir
is this pi model?
thanks sir you are the best
I'm pleased the information was useful.
What happened to the r_o in the hybrid pi model?
The small-signal model used in this video (and described in th-cam.com/video/RDsM2jhbyY4/w-d-xo.html) assumes the collector current is independent of the collector-emitter voltage and in this case r_o is infinite. In reality i_C increases slightly, and linearly, as a function of v_CE (the so-called Early effect) but with a shallow slope depending on the specific device. The small-signal output resistance r_o is the reciprocal of the slope of the i_C vs. v_CE curves in the active region. The simplified model used in this video also ignores h_re, the small-signal feedback path from the collector to the base, and the capacitive effects between the base, emitter, and collector. All of these refinements can be included to provide a more accurate model (just look at a Spice model for the transistor to see what can be included!) but as this is an introductory lecture on small-signal analysis piling on more parameters won't make the process any clearer.
Btw perfect lesson, thanks a lot!
thanks
Amazing video, very good explanation! but am quite naive on how did you know the direction of If in 24.42. Thanks if anyone can be of help also
I didn't know the true direction of i_f. The cool thing is that you can draw the currents going in any direction you like and, providing you write down your circuit equations correctly, the mathematics will take care of the details. For example, if the actual current really does flow in the direction I have drawn it then i_f will be positive. If the actual current flows in the opposite direction to the way I have drawn it then i_f will be negative. All you have to do is ensure your starting equations are correct, and the mathematics does the rest for you.
If someone can help me understand the first AC equivalent circuit I would greatly appreciate it. I'm a little confused as to how RC and RE can be said to be in parallel. To me, it would appear that from an AC perspective, shorting the power supply appears to shunt everything out of circuit except for R1 as the entire side of the circuit up to what was the collector side of R1 would have to be at the same voltage potential. To my limited understanding, this AC circuit would simply look like R1 is one branch, and the other branch is a parallel connection between R2 and RE if the transistor is on.
It just looks like the series combo of RC and RL would have the same potential at both ends so there could be no current flow through them
i love you
What's "Small-Signal"?
I'd contend that the transistor symbol should not appear in the ac circuit but be replaced by its ac model immediately. That'll prevent the mistaken belief that a 0.7 V still exists across the base-emitter junction.
There are two "new" things happening here as far as students are concerned: (a) shorting the capacitors and DC voltage sources (and thereby apparently changing the circuit topology), and (b) replacing the transistors with their small-signal models. In my experience separating these two things in time results in fewer mistakes being made.
why 1/Rc + 1/Rf = 1/(Rc//Rf ) ?
22:05 all the time, every time.