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.
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....
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.
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.
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.
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 !
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.
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 :)
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.
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.
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. 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.
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.
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.
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.
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.
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?
The small-signal model used in this video (and described in ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-RDsM2jhbyY4.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.
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.
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.
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.
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
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!
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
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.
This analysis is fine as long as the output voltage swing is very small compared to Vcc , ie around one tenth of it . But once you try to have a bigger swing in the output, the gain will vary tremendously, and the output signal will get deeply distorted. These lectures may be good as a starter, but not in the real world. If someone wants to know what is in the real world, see this lecture and enjoy 🙂. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Q9HdYt3VGKU.html
You should provide simple Videos, based for poeple who have never studied on Universities. You are presenting only yourself. People who are familiar with it don't Need this Video anyway. The stars on RU-vid are People, who are able to explaing complex Things in a simple way.
Thanks for taking the time to comment but I don't really understand your points. 1. "You should provide simple Videos, based for poeple who have never studied on Universities." Can't we have a range of video types, aimed at people with different backgrounds? 2. "You are presenting only yourself." I don't know what this means. 3. "People who are familiar with it don't Need this Video anyway." This is true of every instructional video on RU-vid. 4. "The stars on RU-vid are People, who are able to explaing complex Things in a simple way." Um, I'm not trying to be a star. RU-vid is a convenient place to store material for my students so that's what I do. If other people find this material useful then that's great. I don't make people watch these.
What is the point of this? Transistors are inherently non-linear devices and there aren't many options for dealing with non-linearities. The mathematics rapidly becomes analytically intractable (that is, finding symbolic solutions with pen and paper) so for linear applications (such as amplifier design) we "linearise" the device by finding a range of operation such that the semiconductor devices behave approximately linearly. That's what is discussed in the video. The other option is to use circuit simulation software to numerically find answers to your questions, at the expense of a deeper and more nuanced understanding of the factors that control circuit response. Does anybody do this in practice? If you are an original circuit designer (developing new topologies, rather than using existing ones) then you absolutely do this AND perform numerical analyses as part of the design process. If you are a designer using existing topologies then you may do less of it, but the need to understand it is no less important. Without an understanding of small-signal analysis how would you know anything about the linear behaviour of an amplifier? The "fact" that the common-emitter amplifier input resistance is dominated by the small-signal base-emitter resistance of the transistor is revealed by small-signal analysis. The "fact" that the transconductance of the transistor is roughly 40 times the DC collector current at room temperature (a rule of thumb every electronics engineer knows) comes directly from the same theory. All these factors come into play when working as an analog design engineer and there are many other examples like this. Circuit simulation software uses the same theory (with admittedly much more sophisticated models) and every engineer should understand the tools he or she is using.
