Quick Guide: Designing A BJT Common Emitter Amplifier

[@ctonew6155]

After watching over 50 videos this is one of the best explanations of how bjt transistors works as an audio amplifier. It was well presented, clear from beginning to the end.

[@lawrencejelsma8118]

That was a wonderful review of the Class A (Preamplifier BJT from your intro) small signal inverting amplifier. I liked how you just did a great component configuration review from theory and coupling AC capacitor filter designs making anyone into designing with basic PNP transistors will turn a small signal input to a gain of 10 amplified inverted output signal. Well done!

[@KewlTo]

After 5 years of regret that I failed my design as a student I'll finally learn again. You explained this well. I realized the flaws of my old design.

[@G4KDXlive]

At last I have found a video that explains what I need to know. Many explanations on KZbin go into far to much detail … or … are just not very good ( one calculates the supply voltage you need as the final step!! ) Bedankt Trevortje!

[@wlewis8336]

Wow... great vid. Exactly what I was looking for.. thanks trevor

[@Yorumcu63]

Best explanations on youtube videos also most simple formulas.Thank you for video.I understand al details Only I want ask a question I must check transistor's dataseeet for cutoff graphic?

[@basiljackson9318]

Thanks for the explanation keep explaining how to use jfets and mosfets.

[@sameedansar3552]

Thank you it really helped me in making the project that was assigned to me

[@eugenepohjola258]

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.

[@omarramirez5764]

You are de Big Man!... I dont know too much english but I underdtood everyting, thanks.

[@andresmendez6870]

I was going to give you a like, but then I saw your video already has the perfect number of likes

[@socksygen]

Amazing video, exactly what I was looking for!

[@PacificNatureTV]

immensely helpful!! keep it up!!

[@Headbutter-Lettuce90]

So, ive been doing treble boosters that are just simply common emitter amplifiers, been trying different transistors too

[@shardulbhosale819]

For a maximum power gain and transfer, the load impedance should be equal to R3 (10K)

[@jackdaniel8763]

Hi first time I found real course of electronics straight to the point, no out of range theories.I wish you make complete guide about Transistors.Are you German?:) Thank you

[@trevortjes]

Glad you like the stuff I make, I'm Dutch tho! And I am not sure what the future brings for this channel. I've got a lot of interests and time just keeps ticking. Who knows what the future brings!

[@jackdaniel8763]

Thumb up

[@user-ub5su9vj6d]

@@trevortjes You did a good job. I like Dutch Cookies👍👍👍🙏🙏🙏

[@davidluther3955]

YOU SHOULD USE A BYPASS CAPACITOR ACROSS EMITTOR RESISTOR AND,YOU SHOULD GET BETTER GAIN.

[@Aaron-nq9rh]

Can you use this type of amplifier for AM Modulation and Demodulation?

[@SecondaryLoop]

Upon building the preamplifier, do you simply connect it to your typical common-emitter amplifier and then to the speaker? No additional circuits required?

[@trevortjes]

As the name suggests, the PREamplifier merely boosts up the voltage and prepares the signal for the next stage. Driving a speaker requires a circuit that can also deliver current because a speaker needs power. (Power=voltage x current, we only accounted for the voltage with this preamplifier). The common emitter amp is not meant to do this. It can barely deliver current and thus not drive your speaker to good levels. Follow up the preamplifier with a power amplifier like a push pull configuration or something different. Then attach that to the speaker.

[@SecondaryLoop]

@@trevortjes Do the technical parameters of the amplifier itself matter much if the voltage output of the pre-amp matches with the input range of the amplifier?

[@materialdialectics]

4:43 Where is this 90k number coming from? EDIT: nevermind I got it

[@rodericksibelius8472]

What is the 'PHYSICS' behind 'why' there is 'INVERSION / 180 degreess OUT OF PHASE' of the 'signal input' at the 'output"?

[@trevortjes]

To understand I'm gonna make some bold assumptions. Let's exaggerate! We put 0v on the base of the transistor. In this schematic not possible. But this would shut off the transistor and the voltage on the collector would be equal to the power supply voltage. So low input = high output. Now exaggerate the base voltage to something way higher than 0v. The transistor would be saturated/fully turned on and the collector voltage would roughly be equal to the voltage between the divider of the resistor on the collector and emitter. So input high = output low. Now our amplifier operates somewhere in between but you can see how high input equals low output, and low input equals high output. This is the behaviour which explains the 180 degrees shifted/inverted output.

[@darrylgodfrey9604]

A great guide! I was following very well until the point where you calculated R2 (at about 4:15). Can you please explain how you came up with that equation? Many thanks in advance.

[@trevortjes]

Fairly simple, don't be afraid of the wall of text. We want to make sure the voltage divider creates a stiff voltage on the base. But when you put another resistance across one of the voltage divider resistors (R1 or R2), you create parallel resistors which in turn alters the voltage divider ratio and thus the base voltage, which is undesirable. The resistor I'm talking about that could ruin the voltage divider, is the input impedance/resistance of the transistor. If you would model that input impedance as a resistor, it would sit in parallel with R2. But what is the input impedance of the transistor? Simply beta * R4. There are sources as to why we can assume it is this simple formula, one I recommend is a book called "Designing Audio Power Amplifiers by Bob Cordell". In this video I satisfy all conditions as to why this assumption can be made. So now we do another assumption that happens a lot in analog electronics design. To make sure one variable doesn't influence another variable too much, we take a factor of more or less of 10. Sure there still be an influence and you will probably never really have 0 influence, but it is good enough for rock and roll. When putting a resistor in parallel with another, we call that loading. And loading in context of current means that a substantial amount of current branches off to the resistance that is loading our intended resistor. So in this case, more current goes into the base down R4 than down R2. We do not want this to happen, so we need to create a path for the current to travel which is more attractive. We simply do this by making the resistance of R2 10 times smaller than the input impedance of the transistor. Hence beta * R4 / 10. Hope this helped :)

[@darrylgodfrey9604]

@@trevortjes Thanks for the reply - very much appreciated. All the best, -darryl

[@BetterCallHardik]

Positive swing estimation -> 3.72V for ic=0 assuming swing of 409uA*9.09 kOhm Negative swing estimation -> vce=0 -> 4.51V

[@eugenepohjola258]

Howdy. Very nice. However. JohnAudioTech has shown that the output impedance is 2 x the collector resistor. I have verified that this is true. My test circuit: B+ = 9 V, Rc = 10 k, Re = 1 k, Rb1 = 47 k, Rb2 = 12 k. Loading with a 22 k resistor the output signal dropped to a value of 1 / sqrt2 of the unloaded value. This is the most energy transfer efficient design, assuming the Rc is fixed. If the circuit is designed for a lower output impedace the load power will increase. Yes. But the total current consumption is way larger than the increase in load power. Best Regards.

[@MaintenanceOfficer-n3w]

I think there is an error : At 4:20 into the video, R2 is being computed incorrectly. The equation given ignores the base current through VBE. If that is accounted for, R2 should be about 26 K ohms (and R1 will be about 176K)

[@trevortjes]

I indeed ignore base current on purpose. As stated at around 4:08 I mention how we want to neglect the base current by making sure the current through R2 is at least 10x bigger.

[@MaintenanceOfficer-n3w]

@@trevortjes If I take you equation at time 4:15 and multiply both sides by 10*IB, I get 10* ib* R2 = (Beta* ib) * R4. Now the left side is the same as the voltage across R2. Call it VR2. Looking at the right side, beta*ib = ic and ic*R4 is the voltage across R4. So what you are saying is that VR2 = VR4, which of course is not true. What I suggest instead in that we use the values you have for VR2 = 1.06 V and ic = 409 uA and then now say that we want the current through R2 to be 10 * ib. Note that 10 * Ib = 10 * (ic/beta) , which is the same as ic / 10, which equals .0409 mA. Now we know the current through R2 and we know the voltage across R2, so we can compute the resistance R2 = VR2 / 0.0409 = 26 K ohm. So your estimate of 10 K ohm is a bit different! I would like to know where this logic falls apart if I am wrong. I've tried looking at many sources to see what I might be missing. Also ,note the obvious;: if we change R2 to be 26 K ohms, we obviously need to adjust R1 to get the proper voltage divide. Luckily your way of doing that is the same as my way of doing it.

[@MaintenanceOfficer-n3w]

@@trevortjes PS - I really did love your video. Well done and was instructive to me. I just think we need to account for VBE.

[@trevortjes]

@@MaintenanceOfficer-n3w Hey! This video uses a lot of simplifications and does not account for some deviations in transistors. All this to get a result which is "in the ballpark" yet very usable and satisfying in real life situations. Approaches like this can be a breath of fresh air for amateurs or other electronic enthusiasts who want to dip their toes in transistors without having to follow hours of lectures or dive into books. Obviously, one can introduce any detail they can to get closer to a realistic/scientific result. In my electronics journey I often came across problems which required me to do these more in depth stuff and put a lot of time and effort in it. My goal with the video is to give people interested somewhat of a shortcut to the knowledge, yet a bit of theory to also understand it. Videos like these also serve as my way to remember some theory cause even after 3 years of not having dipped my own toes in the matter, I also sadly forget how most of these things work. But thanks for your insights, I always support people who keep analog theory alive in an age where it slowly becomes obsolete.

[@MaintenanceOfficer-n3w]

@@trevortjes I agree that you have done a really good job at simplifying the design process and appreciate your efforts. We certainly do want to encourage others to dip their toes in these transistors. Thanks for taking the time to put together the video!

[@KiariaJohnson]

In the simulation example for LT Spice, where did the value of R1 come from? could you explain? does this work in PSPICE or orcas?

[@trevortjes]

Hi, R1 is the assumed load. In this case the output impedance of the amplifier is designed to be 10k. To not load the amplifier too much, I use the rule of thumb to use a load of at least 10x the output impedance. In this case that results in 10k * 10 = 100k. I have sadly no experience with PSPICE or orcas.

[@kiariajohnson4374]

Thanks so much!

[@richardandrews573]

Superb video.

[@KiariaJohnson]

could you do an LTSpice tutorial for this circuit?

[@KiariaJohnson]

I am confused on the SINE(0 0.1 1k) is this an LTSpice command from the directive or just a comment?

[@yousef1419]

Brilliant video

[@joem3581]

Really useful video. :-)

[@U812-k7j]

I can't seem to get 900nf no matter how I try using 1 / 6.28 x 8823 x 20 I'm doing something wrong on the calculator.

[@trevortjes]

1/(6.28x8823x20) = 9.02x10^-7 = 902x10^-9 x10^-9 is nano So 902 nanofarad! :)

[@U812-k7j]

Thanks got it!@@trevortjes

[@TOmaynardOL]

Any recommended reading for amplifier design?

[@trevortjes]

Designing Audio Power Amplifiers by Bob Cordell

[@arashyusefi1889]

Thanks 🙏👍💯😊

[@anurasenarathna1703]

Very good. Thank you

[@justchary]

Nice, thank you!

[@wegi9621]

Epic!

[@RylanVlawgs]

I love you

[@RexxSchneider]

I disagree that this is not an exact science. It certainly is, if done properly, and you should get the expected results that you design for, within component tolerances. if this were not so, it would be impossible to manufacture products consistently. Your requirements are not mutually independent, since the supply voltage determines the maximum possible gain from this design, which is Vcc/50mV if you don't care about distortion. That can be lowered to Vcc/500mV if you want to keep distortion to reasonable levels. So the 9V supply voltage determines a maximum useful gain of x18. Specifying x10 is therefore acceptable. There is no need to complicate the process by drawing load lines. Biasing to the "middle of the load line" is exactly the same as setting the "Q-point" (the quiescent voltage at the collector) at (Vcc + Vce(sat) + Ve) / 2. Since the voltage across R4 is the voltage across R3 divided by the gain and the voltage across R3 is near to half the supply voltage, that shows that Ve (the voltage at the emitter) is approximately Vcc/20 in this case. That means the collector voltage should be about (9V + 0.2V + 0.45V) / 2 = 4.825V and so the collector current should be (9V - 4.825V) / 10K = 418μA. Your method ignored the transistor saturation voltage, which is okay, but that's the reason for the slight difference. The rest of your method is fine, although the output coupling capacitor should be ten times bigger as the worst case loading impedance from the next stage won't be lower than 10K, so you design for that. Now we know the collector current, we can calculate the intrinsic emitter resistance (re), which is 25mV/Ic = 60 ohms. Since the gain is actually R3 / (R4+re), using 1K for R4 will give a calculated gain of 10k / (1k + 0.06K) = 9.4, which is close to the design requirement. The gain you found in your simulation was, of course, the open circuit gain reduced by the loading of the 100K from the next stage, i.e. by a factor of 100K/(10K+100k) = 0.91. If you multiply the calculated gain of 9.4 by 0.91, you get 8.55. That's less than 2% difference from the gain you found in the simulator and well within component tolerances. This really is an exact science.

[@trevortjes]

You can make it as exact and complicated as you want but that wasn't the point of the video. The idea of the video was to give people a way, a roadmap, to quickly get a decent idea on how a transistor amplifier works and how they could be arranged to do what you want them to do with some tips and tricks here and there. I don't think many people care for designing an amplifier with x gain at a very low tolerance, we got opamps for that. I do like your explanation, transistor theory is kinda dying. I'd invite you to use your knowledge to make your own videos or lectures or whatever to keep education on this topic relevant.

[@RexxSchneider]

​@@trevortjes I absolutely appreciate that you did a fine job of making a comprehensive introductory video for the subject that will appeal to a lot of folks who are just starting to dip their toes into the water. Nevertheless, one of the strengths of the KZbin format is that it allows folks to expand on those basics in the comments section, thus providing further information without in any way detracting from the original video. I hope you don't mind me taking that liberty. The problem with opamps is that they are either low-noise or cheap, but never both, and a cheap single stage common emitter front-end with a modest gain can beat all but the most expensive of opamps. I'm afraid I don't make videos, but I do have a PowerPoint presentation that I've used to teach common emitter circuits. You (or anyone else) are welcome to make use of it, should you wish. I've put the latest version on GitHub at github dot com /RexxS1 /Common-emitter if you're interested.