0
\$\begingroup\$

Here, in this question, I am referring to 3 stage Lin topology amplifier (differential stage - voltage amplification stage - output power stage + negative feedback).

enter image description here

I am wondering about portion of negative feedback being fed from the output to T3 via resistor network.

Here, in this case, voltage gain equals Av = Rf1/Rf2 + 1 = 23. So, I am wondering if the difference in amplitudes between two base-to-emitter voltages of TR2 and TR3 is increased if we increase the Rf1? And vice versa if we decrease the resistance of Rf1?

\$\endgroup\$
  • \$\begingroup\$ But isn't this difference equal to Vdiff = Vb2 - Vb3 = Vo/Aol ? Where Aol is open loop gain (the amplifier gain without the feedback ) and Vo is an output voltage? \$\endgroup\$ – G36 Feb 10 '18 at 9:52
  • \$\begingroup\$ \$R_{f1},\: R_{f2}, \: C_2\$ form a low pass filter with corner frequency \$\frac{1}{C_2R_{f1}R_{f2}}\$. Low frequency gain is unity, high frequency gain is \$\frac{R_{f2}}{R_{f1}+R_{f2}}\$ \$\endgroup\$ – Chu Feb 10 '18 at 9:59
  • \$\begingroup\$ @G36 I don't understand what you are trying to say here. \$\endgroup\$ – Keno Feb 10 '18 at 10:26
  • \$\begingroup\$ If for example output voltage is 1V and the open loop gain is 500V/V the voltage difference between Vb2-Vb3 is equal to 1V/500 = 2mV \$\endgroup\$ – G36 Feb 10 '18 at 10:31
  • \$\begingroup\$ @G36 Okay, this seems true, I guess. \$\endgroup\$ – Keno Feb 10 '18 at 10:37
2
\$\begingroup\$

I am wondering if the difference in amplitudes between two base-to-emitter voltages of TR2 and TR3 is increased if we increase the Rf1?

The open loop gain of this amplifier is finite so, to drive a bigger output signal for the same input signal, something has to increase and that will be the difference in signal amplitudes between TR2 and TR3 bases.

It won't be much because at 1 kHz (for instance), the open loop gain will be 1,000 to 10,000. With this sort of gain, the base signal difference to drive 10 Vp-p at the output is in the range 1 mV to 10 mV.

So, with a closed-loop gain of 23, to drive 10 Vp-p at the output, the voltage at TR2's base will be 435 mVp-p and the voltage at TR3's base will be 1 to 10 mV lower. If closed-loop gain doubled, there would still be this voltage difference but TR2's input voltage would halve to achieve 10 Vp-p at the output.

\$\endgroup\$
  • \$\begingroup\$ But lets say that we leave Rf1 and Rf2 as they are and we increase the input signal amplitude (Vpp) - what increases the output voltage swing of an amplifier? The difference between two bases probably stays unchanged, right? \$\endgroup\$ – Keno Feb 10 '18 at 10:31
  • \$\begingroup\$ If you want a bigger output signal then differences between the two base signals MUST be equally bigger for a given open-loop gain. \$\endgroup\$ – Andy aka Feb 10 '18 at 10:33
  • \$\begingroup\$ Yes, this seems true indeed. But if we increase input signal amplitude of TR2 then the signal amplitude at the base of TR3 also increases - which points to the fact that difference between two base voltage amplitudes stays unchanged? \$\endgroup\$ – Keno Feb 10 '18 at 10:37
  • \$\begingroup\$ Not quite - if the TR2 base signal doubles you don't quite get double the output signal amplitude - that small difference in what you get (versus what you expected) drives the "system" to produce maybe a 2 mV to 20 mV base signal difference. \$\endgroup\$ – Andy aka Feb 10 '18 at 10:42
  • 1
    \$\begingroup\$ Correct - the output falls-short by a small amount in order to drive a bigger difference voltage between the bases for a given open-loop gain. \$\endgroup\$ – Andy aka Feb 10 '18 at 11:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.