Quote:
Originally Posted by TURBO
So basically R2 set the gain on the amp (one channel) and R3 / R4 controls the voltage in the ouput of the amp. Am I correct?. Why do I want to control the voltage of the output? (Just, to not to blow up the headphones and set the impedance of the amp?). Thanks.
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R3/R4 set the gain of the amp. An opamp takes the difference between its two inputs and amplifies it by a very large amount, like 100dB (x 100,000) and more. This would be quite useless, but here comes the trick: feedback. You take the output signal and feed it to the negative input. Let's say you have a positive input signal p, a negative input signal n, an output signal o and your amp has a gain of g:
o=(p-n)*g
The difference of the inputs is multiplied by the gain.
Now we connect the output to the negative input, i.e. n=o, and get:
o=(p-o)*g
=> p=o*(1+1/g)
Now if g is very large (and that's point of an opamp, ideally it would be infinite), 1/g becomes very small. Therefore we can disregard that term and get:
p=o
That means the output of the opamp is identical to the input, i.e. we've reduced the 100,000x gain to 1. Usually we want some gain however, so there's another trick. We don't feed back all of the output signal. Let's say we take e.g. only half of it, i.e. n=o/2:
o=(p-o/2)*g
=>p=o*(1/2 + 1/g)
again, we neglect 1/g and get:
o=p*2
Voila, by using only half the output signal for feedback, we have got a voltage gain of 2. That's exactly what the feedback network R4/R3 in the schematic above does. The two resistors form a voltage divider, so that the negative input sees only a portion of the output signal: n=o*(r3/(r3+r4)). As we've seen above, the gain is the reciprocal value of the factor by which the output signal is reduced, i.e. g=1+r4/r3