You are asking about feedback in this circuit, and how R3 fits into that:

First, note that the circuit is an emitter follower, and has a voltage gain less than 1. Second, R3 bypasses the transistor for most input voltages.

The block diagram you show is usually meant to be used when the forward open-loop gain is relatively large, and some negative feedback is supplied. The math still works whether the above is true or not, but the result is rarely of any use when G is small. In this case G is a bit less than 1, not 10⁵ like an opamp might have.

Another big gotcha is the DC offset from input to output. When the transistor is conducting, Vo will be one diode drop below Vi. Let's say the B-E voltage needs to be at least 600 mV to result in meaningful collector current. At 600 mV B-E, there will be 6 mA thru R3. That causes 6 V across R1.

For output voltages of 6 V or less (input voltages of 6.6 V or less), the transistor isn't really doing anything. The output is just the input thru the voltage divider of R3 and R1. There is no feedback anywhere.

Above about 6.6 V input, the transistor will contribute some of the output current. It will do so more as the input voltage goes higher, but that stops when the 10 V supply limit is reached. In this range, the circuit is a classic emitter follower with 100 Ω leakage around it for some reason. R3 is definitely not providing feedback. If anything, you could say it's providing some feed-forward.

You can think of an emitter follower as working on internal feedback in the transistor. There is surely much material on emitter followers out there.

posted about 3 years ago

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Olin Lathrop‭

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