A video showing how to do AC analysis of a common emitter amplifier. AC analysis involves figuring out the voltage gain, the input impedance and the output impedance of the amplifier.
If anyone is wondering why the output impedance is found by treating Rc as if its in series with the load when its clearly in parallel in the diagram, its because in the simplified amplifier model (the triangle) we model the output as a thevinin equivalent circuit with the output impedance being the thevinin equivalent resistance. Its kind of unexplained in the video but him opening up the current source is just the usual step you take to finding thevinen resistance, which ends up being just Rc. And so when you put together the output as a thevinin equivalent circuit in the final model you have Zout as Rc in series with the load. Very helpful video by the way i just wanted to clear this up because it stumped me for a while
you what's even more treacherous, the fact that a forward baised diode on a voltage divider circuit gives of a .637mV out put that means the whole time Ve is just .637mV, since base to emitter junction is a forward biased.
Good video, confirmed a different method I learned for solving these beasts. Highly satisfying when you make your calculations, build it, and your in a close ballpark to actual values.
To those asking where the RE went in the AC analysis, it was shorted by the bypass capacitor. Capacitors are shorted in AC and therefore current passe through the shorted component to avoid the resistance.
+David Williams I did not understand the Rin part, why not do Vin/Iin for the 24.4k ohm resistor at the input too but do it for the 20.36k ohm??? what is going on here dude?
@@user-ww2lc1yo9c It's a mistake. The ac input impedance of the transistor is β x re or 150 x 20.36 = 3K. Put that in parallel with the 24K and you'll get sensible answers.
BRO!!! I have learned a lot more from you than from my whole electronics semester at school. I FUCKING LOVE YOU! So clear and you explain every detail... keep it up man, I wish I could take back that tuition money and give it to you LOL
You've got a great voice for tutorials. My circuit professors are Indian, and they always sound like they're shouting and angry when they explain these things.
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.
The quick answer is that at high enough frequencies a capacitor looks like a short circuit. At low enough frequencies, a capacitor looks like an open circuit. The deeper reason is that a capacitor presents a reactance to the signal which is an opposition to a change in voltage. The reactance of a capacitor is equal to 1/(2*pi*f) where f is the frequency. As the frequency increases, the reactance of the capacitor decreases.
Output swing refers to the maximum and minimum voltage that the output reaches. I have a list of video ideas that I would like to create. I'll add clampers to the list, but don't expect it too soon.
Hello David ...great great video as usual..... when you reached to the part about finding the gain , input and output impedance , i got lost since i am missing basic theory about this matter . can you guide me to a link/links that i could learn the basic theory about how to calculate the gain , input and output impedance of any circuit ....thx alot
Thank you for your efforts .. If we applied this circuit c emmiter to the radio frequency via a resonant circuit, we know that the signal will be inverted 180 degrees. Can we receive the signal in the reception circuit and be mirrored, or should it be the same as the phase ?
Nicely done! I would prefer to get out a general-purpose equation for gain in terms of source and load impedances, and the resistors and source voltages being used, but perhaps that’s one of those exercises better left to the viewer. In any case, CE-amplifiers are a lot more demystified as far as I’m concerned.
Sure. If Rs is the source impedance, Rl is the load impedance, Zin is the amplifier input impedance, Zout is the amplifier output impedance, and Av is the amplifier open-loop gain, then the overall gain is given by: Zin/(Zin+Rs) x Av x (Rl/(Rl + Zout). For a well-designed common emitter stage, Av will equal -Rc/Re; Zin will approximately equal the lower base bias resistance, and Zout will equal the collector resistance. Unfortunately, the circuit in the video doesn't fit my definition of "well-designed".
Dear David,I really like this video, as well as your video on the emitter follower circuit. I just had one question on this circuit: Could you please tell me how to determine the peak AC current leaving the amplifier? The reason I am asking is I am trying to combine an emitter follower with a common emitter. I see now how to find the output impedance of the amplifier, and how to get Vout. Would the outward current just be Vout/amplifier output impedance?Thanks,Tim
Load impedances for audio systems are usually 4 or 8 ohms, so common emitter amplifiers are not used for driving speakers. Amplifiers are used in many different types of applications that have higher load impedances. Also common emitter amplifiers are single stage amplifiers and are often found as part of a multistage amplifier system.
On the contrary, a pure Class-A amplifier will have a common emitter stage as its output, and many high-quality audio amplifiers use that, although they may use an active collector load to improve the open loop gain. There are plenty of examples on the web.
+Sylwester Kogowski. The beta of the transistor is the current gain while the gain of 193 is the voltage gain of the amplifier circuit that uses the transistor
The midband AC gain of a common emitter amplifier is equal to about -Rc/(re+RE). When you bypass the emitter resistor (RE) with a capacitor, you are making RE invisible from an AC point of view (i.e you are bypassing it). This makes the gain equal to about -Rc/re
That's only true when Xc (the reactance of the capacitor) is negligible compared to re. For slightly lower frequencies, the ac gain is -Rc/(re + Xc) as long as Xc
Thanks for the video. I got a doubt in between which goes..why can't we use a hybrid pi model in the AC equivalent of the transistor instead of the T model you have used? (using a Hie in b/w E and B and current source of Hfe(Ib) btw C and E..here the emitter is grounded due to the bipass. ) PS : can you make any videos on MOS amplifiers?? please, it'd be of a lot of help to guys like me :)
Hi, thank you for the tutorial i really like it. just a question. when you found Av gain which is equal to Vout / Vin. i believe that Vout should be Rc || RL because they are in parallel. am i right ? thank again sir
This video was somewhat helpful however I think it will fall short for most students studying the concepts. In the future could you make a video that describes the circuit in much more detail. For instance, the effects of the capacitors? What are the poles and zero's of each capacitor. How can one find the gain when taking them into consideration?
I wanna build a small practice guitar amp for me, but I did not know it was going to be hard so hard, damn! However I am enjoying research and learn...
@David Williams why is the current source replaced by an open ?? If the idea is to obtain the output impedance by using a thevenin equivalent, dependent sources shouldn't be removed if I remember correctly ...
