Semiconductor Devices: Common Emitter Amplifier

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The basics of a small signal common emitter transistor amplifier stage.
Reference: Chapter 7 sections 2 and 3 of Semiconductor Devices.
My free texts and lab manuals are available for download at my college web site www.mvcc.edu/jfiore and at my personal site www.dissidents.com
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25 июл 2024

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Комментарии : 55

@davidluther3955 26 дней назад

EXCELLENT PRESENTATION!ONE OF BEST EXPLANATIONS OF CE CONFIGURATIONS I HAVE EVER HAD.I WISH I HAD THESE GIFTS WHEN I WAS A COLLEGE STUDENT STRUGLING THROUGH TO UNDERSTAND ALL OF THIS.

@ElectronicswithProfessorFiore 26 дней назад

Glad you liked it. Tell your pals!

@nilsmclellan868 Год назад

I have been waiting for someone to explain it the way you did. Thank you so much. Please do more!!

@luisramos6676 3 года назад

Thank you!! Taking an Electronics Technician course and just went thru this yesterday and I could tell the entire class was confused. Going to show this video to classmates so hopefully it helps them as much as it helped me!

@ElectronicswithProfessorFiore 3 года назад

Glad you liked it, Luis.

@simonyoungglostog Год назад

I just watched this again and understood all of it. Your methods are working Prof!! :)

@0451Deus Год назад

Perhaps one of the most comprehensive and well-explained videos on this topic in all of RU-vid. Thank you! I was going through your videos and was wondering if you had one (or will make one) about designing a multistage amplifier given a certain set of parameters? Such as a specific voltage gain, a specific Vcc, a specific input resistance, etc.

@ElectronicswithProfessorFiore Год назад

I don't have one of those out yet, but you will find all of the requisite bits either in the Semiconductor Devices playlist or in the (free) text book of the same name. If your interest is in making audio amplifiers and associated equipment, I strongly suggest taking a look at the Op Amps playlist and associated (free) text book. There's tons of applicable material there, and, at least for the line level stages, designing with modern op amps tends to be a more successful endeavor for the average individual (as in quicker and less expensive). Modern op amps can be of excellent quality, and a smart use of them with some discrete design elements can accomplish most anything you want.

@0451Deus Год назад

@@ElectronicswithProfessorFiore Oh I only asked because I'm actually an EE student right now and one of our lab assignments is to design a multistage amplifier using BJT transistors. I find it interesting how the R(in) of one stage becomes the load of another, and having to take that into your calculations for each stage. It was and still is slightly confusing to me but still a lot of fun.

@janaksodha9582 2 года назад

Pls can you consider a cascode circuit of 2 transistors and use the same t-model used in this video.

@RexxSchneider Год назад

That's one of the best descriptions of the common emitter stage with fixed biasing that I've seen on RU-vid. I would add a few points: (1) At 0:32 the other two things to look at are frequency response and distortion. For audio use, the input capacitor needs a reactance equal to the sum of the source and input impendences at the -3dB point, usually 20Hz. Similarly, the output capacitor needs a reactance equal to the sum of the output and load impedances at the -3dB point. Finally, the emitter bypass capacitor needs a reactance equal to the sum of r'e and Re1 at the -3dB point. It's a common mistake to assume that it needs to be calculated from Re2, which often then tanks the low frequency response. Distortion in BJTs in common emitter mode is mainly caused by the dependence of r'e on collector/emitter current. It means that large input signals have greater amplification of the positive going half-cycle than of the negative going half-cycle, and it's pretty nasty. It is reduced by making Re1 an order of magnitude greater than r'e, at the expense of gain. (2) The maximum unloaded voltage gain is given by maximising Rc/(r'e + Re1). that occurs when Re1 = 0 and so is equal to Rc/r'e. Since Rc will normally drop around half the supply voltage under quiescent conditions it is equal to Vcc/2 divided by Ic. So the maximum gain Av can be written as (Vcc/2.Ic) / (26mV/Ic) = Vcc / 50mV. This show the maximum achievable voltage gain depends just on the supply voltage. However, the distortion in such a stage is likely to be unacceptable for anything other than minute input signals. So we might pick Re1 to be 9x bigger than r'e, reducing the distortion significantly, but lowering the gain by a factor of 10. The maximum realistic gain is then Vcc / 500mV. The result is that you always drop 250mV quiescent across Re1 at a maximum usable gain. And those are good design rules-of-thumb to remember. (3) Since the output impedance is Rc, setting the output impedance (Zout) then sets the collector current to Vcc / (2.Zout) approximately. So for a 10V supply with a 5K output impedance, you will set a collector current of 1mA. You want the current through the base bias resistors to be significantly bigger than Ib to reduce the variation in operating point with variations in transistor β. So the parallel resistance of R1 and R2 (what you called Rb) is going to be significantly less than β x Re1. But for gain Av, Re1 = Rc / Av = Zout /Av. That gives Rb

@ElectronicswithProfessorFiore Год назад

All good stuff, but a bit much to squeeze into an introductory video! You will find this and more in my associated text. Free for the download-see the links in the video's description.

@RexxSchneider Год назад

@@ElectronicswithProfessorFiore Well, of course it always depends on your target audience, but if you look at other similar videos on RU-vid, you'll see that the points I raise above are very common questions in the comments. The common emitter amplifier stage can be "designed" by simply copying other working designs, and that's certainly one route for the beginner. But as soon as you examine what is needed to design the circuit from a set of specifications, you'll realise that there's a core of information that is essential to be able to master that task. You certainly have to show how to calculate capacitor values, and it's useful to demonstrate how the choice of collector current determines the input and output impedances, as well as how the choice of supply voltage determines the maximum gain. Anyone who aspires to become a designer needs that core at least.

@ElectronicswithProfessorFiore Год назад

@@RexxSchneider Right, and the goal of this video is as an introduction, with more to follow. It is not intended to cover the topic in its entirety. This video represents what I would present typically in our first lecture on common emitter amplifiers in a classroom setting. All of the videos I create for this channel are modeled on what I would cover in a single lecture, and are designed to support my free textbooks. They are not designed in any way to compete with what anyone else is doing on RU-vid.

@simonyoungglostog Год назад

@@ElectronicswithProfessorFiore I'm finding your lectures perfectly paced. Trying to give 'all' of the relevant information in one go is overwhelming and poor teaching. Your teaching pace is great, for me. Thank you. We could also mention Miller effects and frequency response but this is better done in a separate lecture. My poor brain can only absorb so much at once. I'm really enjoying your teaching style.

@torikenyon Год назад

This video is beautiful (although it took me a full month to understand it)

@eugenepohjola258 7 месяцев назад

Howdy again. I was wrong. You and JohnAudioTech are correct. Rc is the output impedance. Then the output signal is ½ the unloaded and most power is transferred. I remembered JAT's reply wrongly. Regards.

@ElectronicswithProfessorFiore 7 месяцев назад

Glad you got that cleared up!

@yousifali1737 2 года назад

I wish i found this channel long time ago

@ElectronicswithProfessorFiore 2 года назад

Better late than never...

@torikenyon Год назад

I’m confused about the impedance looking into the base (Zin base). With the AC model, it looks like it should just be r’e + rE, as the current source should be considered an open when calculating impedance?

@ElectronicswithProfessorFiore Год назад

Yes, it is an open when looking out toward the collector. The issue is that the current source is a dependent source whose value is beta times larger than the incoming base current. When you multiply this current by the emitter resistance, you get a voltage much greater than the base current could create alone. Thus, the emitter resistance appears to be beta times larger when seen from the base. The reverse also happens. When you look into the emitter, whatever you see in the base appears to be beta times smaller.

@simonyoungglostog Год назад

I watched it again, I want my memory back! Would you calculate the loaded input voltage by including the voltage-divider resistors or just use the V-in at the base and V-out at the collector to calculate the Av? I think that as long as we state our assumptions it doesn't really matter but was wondering if there is a standard method?

@ElectronicswithProfessorFiore Год назад

You must include the divider resistors in order to get a proper Zin value. This may or may not affect the overall gain- that depends on the size of the source impedance.

@simonyoungglostog Год назад

@@ElectronicswithProfessorFiore Thank you.

@Chris-hi2hn Год назад

I was confused at the point where you talked about the unloaded versus loaded output impedance. You showed that you can use a voltage divider to determine The output voltage with an added load. But wouldn't the load that's added be imperial with Rc? Not in series. I'm sure i'm missing something here

@ElectronicswithProfessorFiore Год назад

The load is indeed in parallel with Rc. The point is that you can do the analysis two ways: First way is to use the loaded calc, including Rload in parallel with Rc. Second way is to calc the unloaded gain, and then multiply that by the loss created between Zout and the load. What may be confusing you is that we start with a model of the transistor as a current source with Rc in parallel with it, and then you do a source conversion, turning it into a voltage source with Rc in series with it. This makes the voltage divider effect with the load obvious. (You could also skip that conversion and do a current divider between Rc and the load, but my experience is that students tend to see the voltage divider effect more clearly).

@lukerandolph900 25 дней назад

Unfortunately the top of you page is outside the camera frame - v hard to guess what’s going on there. Are you able to repost in a better position?

@ElectronicswithProfessorFiore 25 дней назад

Go to 21:00. There is a still frame of the entire page. It turned out that very little was chopped off, but I understand that it can be disconcerting.

@eveinah7552 Год назад

Hi Professor Fiore.. Can you help me figure out the the AC and DC loadlines of this specific circuit, pls? Thank u so much.

@ElectronicswithProfessorFiore Год назад

Let's start with this: Here is a video that includes the DC load line for a very similar circuit: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-TZbgD-uZ69U.html The obvious difference is that, in the circuit above, the emitter resistance is split into two parts (which are combined into one for DC as the bypass cap is open).

@eveinah7552 Год назад

@@ElectronicswithProfessorFiore thank you so much! appreciate the response, Professor..

@Pooja-zz9tu Год назад

Hi, Professor Fiore can you please explain what is an ac ground and why it is needed to short dc source, I tried to look for an explanation but cannot get one. Thanks

@ElectronicswithProfessorFiore Год назад

This is a standard circuit analysis technique. We are looking at an AC equivalent circuit. In other words, "How does the DC source 'behave' in the AC equivalent?" Think in terms of Thevenin's Theorem. We are replacing the DC source with its ideal internal resistance, which is a short. Thus, we say that the DC source "goes to AC ground". That is, the power supply node is the same as ground when performing an AC analysis. So "AC ground" is not the same as the circuit ground. It is a mathematical abstraction, a place of zero potential for AC signals.

@Pooja-zz9tu Год назад

Thank you, ....when we are figuring out the collector current through an exponential equation and we consider the dc input bias and ac varying small input signal and assume a linear model while calculating the gain due to ac input the dc battery source does not appear in the ac gain equation whereas in dc gain it appears So mathematical abstraction is modeled in ac circuit without the need of dc source...am I understanding correctly ?@@ElectronicswithProfessorFiore

@ElectronicswithProfessorFiore Год назад

@@Pooja-zz9tu More or less, but remember that the DC source is required to help set up the AC equivalent in the first place!

@eugenepohjola258 8 месяцев назад

Howdy. Objection. JohnAudioTech has shown that the output impedance is 2 x the collector resistor. I have verified this. Loading with 2 x Rc the signal drops to 1/sqrt2. And. If the emitter resistor has no bypass capacitor the transistor input impedance is very high and may be omitted. Only Rb1IIRb2 remains. If the emitter resistor is bypassed with a large enough capacitor the Rb1IIEb2 may be omitted leaving only the transistor input impedance. How to determine the transistor input impedance ? Check data sheet for variation of base-emitter junction threshold voltage with varying base currents. Zin is roughly dVbe / dIbe. Zin will vary with varying drive levels though. Regards.

@ElectronicswithProfessorFiore 8 месяцев назад

Really? That's pretty interesting because Rc winds up in parallel with the BJT output, so how do we get a value that's larger than Rc? (e.g., what do you put in parallel with a 1k to get 2k?) Now, regarding your "proof", given Rc

@BartKus Год назад

Slight correction: r'e = 26mV / Ic

@ElectronicswithProfessorFiore Год назад

Yes (the derivation can be found in chapter 7 of my semiconductor devices text), but we're splitting hairs here. From a practical standpoint, there is no real difference. With typical betas for small signal BJTs, the difference between Ic and Ie is under 1%, and perhaps more importantly in this case, the circuit schematic shows a swamped configuration. That means that the precise value of r'e is not all that important because of rE.

@BartKus Год назад

@@ElectronicswithProfessorFiore Absolutely right, and I'm glad this is documented here now for anyone else watching the presentation in future. I was feeling particularly sensitive to r'e when I left the comment, since I was seeing a discrepancy in LTspice simulation results vs theory when measuring an un-degenerated CE amp. r'e was 17.3333 computed, but was measuring ~18.8679 (1.5mA quiescent Ic, 5k RL, 10V supply). Predicted gain of 288 but measured gain of 265, an 8% difference! Gain was predicted from the typical RL/r'e. So why an 8% error? In a simulator of all things? Took me the rest of the day to realize I needed to also consider the *bulk* resistance of the collector, which was defined as 1 ohm in the SPICE model. 1 ohm becomes a significant part of r'e when r'e is only 17.333 to begin with. With the improved denominator of 18.333 (r'e+Rc) the gain calculation (Av=273) is much closer to what is measured (265), but still not quite right. I'm not sure where the remaining error is coming from. 😞 And yes I realize in a typical circuit this difference is insignificant because you'd typically degenerate that CE amp, but I'm trying to understand the physics of the transistor alone more precisely.

@ElectronicswithProfessorFiore Год назад

@@BartKus Don't forget, the model we're using here is simplified. It ignores a bunch of things (that are usually fine to ignore in many cases). For example, what is the equivalent internal resistance of the collector current source in the model (we assume infinite)? Ultimately, that will lower the gain and may be noticeable with large load and biasing resistances. What about r'b (an effective base resistance)? Usually, too small to bother with but it can reduce the input voltage (i.e., the signal that effectively appears across r'e) which lowers the load voltage. Models are always a balancing act between computational efficiency and accuracy. The thing I always tried to reinforce in lecture is not to get too caught up in this. As long as you know the limits of your model, you should be OK. What is the point in chasing down deviations that are a fraction of a percent in your computations when your production circuit uses 5 or 10% component tolerances that will create much larger deviations? Better to spend your time doing an analysis of min/max performance so you can quantify the variability coming off the production line. A basic Monte Carlo sim can shed some light on this very quickly.

@RexxSchneider Год назад

Actually, Vtherm = kT/q, so for an ideal transistor, Vtherm/T = 86.17μV/K That gives you Vtherm = 25mV at 290K = 17°C and Vtherm = 26mV at 302K = 29°C That 12 degree range is more or less what we consider room temperature, so neither value is any more correct than the other.

@RexxSchneider Год назад

@@BartKus Another complicating factor is that the output impedance of small signal BJTs is only of the order of a few hundred kilohms under normal conditions and that could account for maybe 2% reduction in gain below predicted. Similarly the bulk base resistance is just a wild guess at 1 ohm. In a real transistor it is just as likely to be 0.5 ohm or 2 ohm. That can account for another few percent variation. Finally, the Miller effect due to collector-base capacitance can be significant at high gains and can easily account for another few percent. Try varying your test frequency and see if the gain changes in the real circuit.

@gammersunity4117 13 дней назад

Doesn't voltage divider between a base of a transistor cause voltage drop to around 600mv

@ElectronicswithProfessorFiore 13 дней назад

0.6 volts as the approximate value of Vbe, the base-to-emitter voltage when forward biased. The divider establishes Vb, the base voltage. Not the same things here as the emitter is not at ground.

@gammersunity4117 12 дней назад

@@ElectronicswithProfessorFiore sir I been trying to learn to design/build a class AB power amplifier, where should I start to learn this, I wasn't expecting you to reply. There isn't a lot of information on anywhere. VAS stage is the most difficult to understand.

@ElectronicswithProfessorFiore 12 дней назад

@@gammersunity4117 What's your background? Do you have any formal training in engineering or electronics? That will tell us where to begin. I will state at the outset that designing a quality power amplifier is not a minor undertaking. Building an existing design is much less work, but still not trivial.

@gammersunity4117 12 дней назад

@@ElectronicswithProfessorFiore I don't have a training infact I'm mechanical engineering student, I am just big fan of speakers and amplifier from very young age, my knowledge is kind of dispersed.I just know known the very basics.

@ElectronicswithProfessorFiore 12 дней назад

@@gammersunity4117 OK, as an ME student, you would have the requisite math and physics, and possibly an intro electricity course, so that's good. I suggest that you download my AC Circuit Analysis text along with my Semiconductor Devices text if haven't done so already (links in the video description above). There are matching video playlists for both. You'll want to work your way through the first four or five chapters of the AC text and then start at the beginning of the Semi Dev text. Yes, this will take some time but you can either do it fast or you can do it right. Trying to skip through material to "get to the good stuff" will only mean trouble later. Besides, if you enjoy the material, you might opt for a double major in ME and EE, and that would be a pretty cool combination. This recipe is probably not what you want to hear but it is what you need to hear. I should add that there are some things that you can skip over in the Semi Dev text such as the chapter on diode applications and specialized types (clippers, clampers, LEDs, varactors, and so forth) as well as some of the material on power supplies, assuming you're just going to buy an off-the-shelf power supply versus designing your own. Even with that, you're still looking at the equivalent material of a 4 credit hour course, assuming you've taken an electricity/circuit analysis course with your ME studies. (If not, grab the DC Circuit Analysis text and run through the first five or six chapters before doing anything else.) At my college, most engineering students took an intro circuit analysis course in their second year.