54. Cascode Amplifiers and the Miller Effect
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- Опубликовано: 30 июл 2024
- All inverting amplifiers suffer the Miller Effect, an amplified capacitance that reduces bandwidth. One cure is the Cascode Amplifier and it, along with the Mller Effect, are discussed here.
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I remember using nice shiny green capacitors like the ones you used in an active antenna circuit (PA0RDT Miniwhip) . The frequency response was limited to Long Wave and the lower frequency end of Medium Wave Broadcast Band. They were polycarbonate capacitors and, as the foils forming the plates were rolled up inside, they had appreciable inductance and a quite low self-resonant frequency. Replacing them with ceramic ones of the same value cured the problem and enabled HF Short Wave reception. Your video proved your point, but perhaps a different type of capacitor might have changed the response.
Great video, great instructor. Teaches you how to make quick practical EE estimations of resistances, gain, voltages and demonstrates theoretically and experimentally how to increase bandwidth using cascode amplifiers.
Thanks for creating this video. It is the most helpful video I have encountered on the cascode amplifier so far.
BEAUTIFUL
Excellent ❤
Excellent explanation and demonstration.
I *LOVE* these videos.
Wow! I hope I can continue to met those expectations.
Excellent video, needed a refresher on the details. Well explained.
Thank you. Much appreciated.
My only critique is that this is a very information dense video and hard to take it all in at once. Watched it in about four pieces I like that you usually don't droll on too long about obvious stuff, I like the level your targeted at, just don't rush too much and maybe smaller parts. Yeah, would be difficult to find a place to split this one...
Thanks for the feedback.
Very good video although I'd appreciate an explanation of how the different elements of the circuit were calculated.
At 16:40, Q2's base current can be estimated: R1's current is (15 V-Vbe)/680 k = 21 uA. Part of that flows through R2: Vbe/68 k = 9 uA; leaving 12 uA for the base current. Typically 200 gain gives collector current about 3 mA... probably a little larger because Vbe < 0.6 V for such small base currents. Emitter resistance is about
25 mV/ 3 mA ... about 6-8 Ohm. This all will fluctuate wildly with temperature, so a better design would put tens of Ohm between emitter and ground to stabilize the fluctuations.
Thanks a lot, everything is super clear now
Thank you for the helpful video
Thanks for taking the time to make these videos i really enjoy your teaching style very informative.
Thanks. I may not have the most engrossing "style" but I do try to maker them informative and accurate.
Thank you! Great explanation!
Scrumptious
i was actually thinking of doing the oposite, intentionaly making my jfet guitar preamp negative feedbacking itself to cut on some unwanted prequencies like rf interferences andhiss closer to 22khz
Great video. Is there any benefit in adding resistors (say 220 Ohm) to the E and C of Q2 to help 'swamp' the r'es and give a less distorted output, adjusting Q1 Rc if required for gain?
your work is fun and easy to understand. Thanks
Can I have Q2 be a lower Collector-Emitter (or in my case Drain-Source) max voltage than is supplied by the top rail? Because Q1 takes it...
Great, thank you much !
Glad it was helpful!
If the cascode load is a RFC then there is the posibility of just connecting the base of the top transistor to the rail as a form of biasing. The price is very limited output voltage swing being limited by the collector emitter saturation voltage.
There is also the folded cascode circuit.
Thanks for the video. @6:20 the result is charge (and not capacitance) so it should be in units of coulomb. Also the formula @7:48 does not make sense - why you add deltaV to (1+Av)*Cb to get a total charge?
Why we don't consider Miller effect in Common Gate configuration when we study high frequency response?
at around 3:00 the equations have an X that seems to denote multiplication in the first instance and then addition in the second instance. Could you explain the meaning of the X in the equations?
5:44 "We could just as easily have said 1 milli Volt", except that if you would have, the capacitance would have gone down... this is quite confusing, maybe the X was an addition operator here? Why would you use such notation in the first place???
Not a very good cascode amp. There is no need to capacitive couple between stages, and some emitter degeneration in the cascode stage enhances stability. You can get much better frequency response for the same gain by buffering the input and output with common collector stages. This "Miller Theorem" is really a simplification and nothing multiplies or divides the capacitance. What is really happening is the shunt resistance across those capacitance is much higher in a common emitter stage and lower with a well designed cascode stage. An estimate of the frequency response of a multistage transistor amplifier is the inverse sum of the open circuit time constants. So when the shunt resistance is reduced, the frequency response goes up. Common collector buffers reduce the shunt resistance, and the resistance looking into the emitter of the upper transistor is much lower than if there was just a load resistor there ( in the case of a common emitter amplifier). I have built cascode amplifiers, written out all the node equations and used a program to calculate the high frequency rool-offs. I also used the estimate I described above and calculated about the same value. Finally, by measuring the high frequency roll-off, I found all were in close agreement. Your Miller Theorem" concept is erroneous, and you should understand that all transistor stages have gain. The common collector and common base configurations have current gain. In any amplifier you want power gain, so both current and voltage gain are needed.
Idk if maybe i missed it which i probably did but ill ask anyways , what is the cap C2 for? Thats on Q1?/whats it do?/whats it called? Why is it going to ground?