Morning Sir. I'm struggling a bit with this one. I hope you don't mind me asking questions of you. When it comes to calculating the re values of each transistor, wouldn't they have vastly different emitter currents? That is, the first transistor might have a base current of 100 times less than the second transistor, therefore the emitter currents would also be about 100 times different. If we say r'e = 26mV/Ie, if the second transistor is about 3mA, the first would be about 3uA. Am I making a silly mistake somewhere? Further to yesterday's video on Common-Collector (emitter follower), where we could have used a second stage with a PNP, could we do the same here and return the emitter voltage to the same as the original DC base voltage? Is there a reason to use 2 x NPN? Thanks again, I'm finding your videos perfectly paced (even if I've only just found them) and I hope that my questions aren't too daft.
Yes, the two transistors have vastly different emitter currents, and thus, each transistor's individual r'e is vastly different from the other. The trick is in how the pair is perceived by the rest of the circuit. Consider a pair with beta = 100 for both devices and r'e1 is 200 ohms, meaning that r'e2 is 2 ohms due to its higher current. If you look into the base of the first transistor and consider calculating Zinbase, what do you see? r'e1 gets multiplied by beta1 for 20k ohms. OTOH, r'e2 gets multiplied by beta1 and beta2 which also produces 20k. Those are in series and add to 40k. The total beta is 100*100 or 10k, thus, you can model this as a single transistor with a beta of 10k and an r'e double that of the second transistor (and double the usual 0.7V Vbe).
11:00 I’m confused, wouldn’t re be just 150 ohm because cap acts like a short for ac?? Also the 150 ohm looks like a swamping resistor but it comes after cap instead of before, can you help please?
Yes, the cap acts like a short at these frequencies and that's why the 150 is in parallel with the 3.3k (remember, the DC power supply is an AC ground). That combo is pretty close to 150, though. And yes, this resistance does act like swamping in that it will lower distortion compared to a typical CE amp.
You don't want DC going to a loudspeaker, so in that regard, yes, if the output is not naturally sitting at 0VDC. It would be at 0VDC if you were using a symmetrical output stage with dual (+/-) DC power supplies, and thus, no cap needed. Feeding DC into a loudspeaker will just waste power and increase distortion.
@@ElectronicswithProfessorFiore yes I know what DC does to a speaker is guess what im trying to learn is what kind of design doesn't require DC blocking caps. So are you saying that if I were to use a + and - DC power supply then I wouldn't have to use DC blocking capacitors? Ps I appreciate your help and all of your videos
@@SheikhN-bible-syndrome Yep. There are many variations on using symmetrical bipolar power supplies. For example, instead of using a single +40 VDC supply for a class B power amp, you'd use +20 VDC and -20 VDC. The operation of the output section is pretty much unchanged, however, as the output would be sitting at 0 VDC, no output coupling cap would be needed. Check out the videos on class B power amps. It's easy to configure them in this manner. Class A (like the one in this video) is a bit more involved to achieve that goal (you can't just substitute the supplies). Try this: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-6gLo_eaB2cc.html If you haven't seen the videos on class A power amps, it would make more sense to watch those first, like this one: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-0akOGTXwJWY.html For the record, the circuit in this video would require further modification to eliminate the output cap. Imagine that you had a voltage divider on the leftmost base of the pair and it set Vb to 1.4 VDC. That would place the rightmost emitter at 0VDC, and thus, no cap would be needed. (In a practical circuit you'd need some adjustment/calibration because the Vbe drops would not be exactly 0.7 V each). That's a bit of a moot point though, because class A amps are notoriously inefficient and class B (or its variants) would be preferred (and those designs are easy to implement with dual supplies).
