04-03-2017, 09:50 AM
Mike,
You've picked an unfortunate circuit. I'm afraid that this is a non-trivial exercise.
The amplifier is a shunt feedback amplifier.
In theory, if the transistor offered infinity gain and Ra4 was bypassed with a large capacitor, the input impedance would be zero. But the gain isn't infinity, and that resistor isn't bypassed.
What about Ra2 and Ca1? They should be working outside the audio band, so we'll ignore them.
Always break a problem down.
If we ignore the effect of feedback first of all, the "raw" input impedance of the transistor is (re+RE) times hfe. "re" is the intrinsic emitter resistance (1/gm), and RE is Ra4.
("Little re" will be small, so we could ignore it. But we might as well work it out, as we'll need gm later.)
At room temperature, gm is Ic times 40. About 22. So re is 45 ohms. Let's assume the BC109 has a hfe of 200, so 200 times (45 + 2.2k) is 449k. Call it 500k.
So before the effect of negative feedback, Zin is 500k. Next would be to consider the effect of NFB on the value of the feedback resistor.
This is Miller effect. You put a small positive signal voltage on one end of the 5M6 and expect a certain current to flow. But because the other end of the 5M6 is not tied to ground - it has moved negative, so more current than you expect will flow, and you'll "see" a lower resistance than you think. It's like "anti-bootstrapping". Roughly, the resistance is reduced by the voltage gain of the amplifier.
The amplifier should have a voltage gain of approx 400 (gm times Rl). So that makes the 5M6 resistor look like 14k. This is in parallel with the 500k of the transistor.
However, that will be wrong because we don't know what comes next. When I worked out the gain, I used 18k for Rl, but Rl is in parallel with the next stage. Whatever that is, it'll reduce the open-loop gain of the input stage, and that'll mean that the 5M6 is reduced less by Miller effect. Post the whole diagram and I'll try to guess at it.
All of the above is approximate in a "back of the envelope" sense. However, the numbers aren't important - but trying to understand what's happening before doing the maths is. And there are many ways to approach this problem, but what I've said here is the most intuitive, least mathematical approach.
Of course, you can measure it really easily. Feed in a sine wave via a variable resistance in series with the input. Set resistance to zero and, monitoring the output, increase the resistance until the signal has fallen by half. Measure the resistance with a DMM. Viola
I'm afraid I couldn't follow your working - I got lost at the first line because I couldn't see where 10 came from. Sorry
You've picked an unfortunate circuit. I'm afraid that this is a non-trivial exercise.
The amplifier is a shunt feedback amplifier.
In theory, if the transistor offered infinity gain and Ra4 was bypassed with a large capacitor, the input impedance would be zero. But the gain isn't infinity, and that resistor isn't bypassed.
What about Ra2 and Ca1? They should be working outside the audio band, so we'll ignore them.
Always break a problem down.
If we ignore the effect of feedback first of all, the "raw" input impedance of the transistor is (re+RE) times hfe. "re" is the intrinsic emitter resistance (1/gm), and RE is Ra4.
("Little re" will be small, so we could ignore it. But we might as well work it out, as we'll need gm later.)
At room temperature, gm is Ic times 40. About 22. So re is 45 ohms. Let's assume the BC109 has a hfe of 200, so 200 times (45 + 2.2k) is 449k. Call it 500k.
So before the effect of negative feedback, Zin is 500k. Next would be to consider the effect of NFB on the value of the feedback resistor.
This is Miller effect. You put a small positive signal voltage on one end of the 5M6 and expect a certain current to flow. But because the other end of the 5M6 is not tied to ground - it has moved negative, so more current than you expect will flow, and you'll "see" a lower resistance than you think. It's like "anti-bootstrapping". Roughly, the resistance is reduced by the voltage gain of the amplifier.
The amplifier should have a voltage gain of approx 400 (gm times Rl). So that makes the 5M6 resistor look like 14k. This is in parallel with the 500k of the transistor.
However, that will be wrong because we don't know what comes next. When I worked out the gain, I used 18k for Rl, but Rl is in parallel with the next stage. Whatever that is, it'll reduce the open-loop gain of the input stage, and that'll mean that the 5M6 is reduced less by Miller effect. Post the whole diagram and I'll try to guess at it.
All of the above is approximate in a "back of the envelope" sense. However, the numbers aren't important - but trying to understand what's happening before doing the maths is. And there are many ways to approach this problem, but what I've said here is the most intuitive, least mathematical approach.
Of course, you can measure it really easily. Feed in a sine wave via a variable resistance in series with the input. Set resistance to zero and, monitoring the output, increase the resistance until the signal has fallen by half. Measure the resistance with a DMM. Viola
I'm afraid I couldn't follow your working - I got lost at the first line because I couldn't see where 10 came from. Sorry
