Cmoy Circuit
Jul 12, 2005 at 3:21 AM Thread Starter

#### TURBO

Hello. I have a doubt about this circuit. Could someone explain to me the path of the electrons in this amp section from the start? Why do I need R2?. R2 doesn’t make sense to me, why is there? Thanks.

Jul 12, 2005 at 3:30 AM

#### nikongod

##### DIY-ku
it forms the high-pass filter with the cap.

it also serves to reduce input offset current diferences.

it also helps to avoid a hardcore pop, and possible large dc offset at output if the wiper of the pot "skips" off of the pot..

Jul 12, 2005 at 3:44 AM

#### TURBO

Quote:

 Originally Posted by nikongod it forms the high-pass filter with the cap. it also serves to reduce input offset current diferences. it also helps to avoid a hardcore pop, and possible large dc offset at output if the wiper of the pot "skips" off of the pot..

But, I understand that the cap (C2) will regulate the offset current differences, but R2??????. What is the relation between the cap C2 and resistance R2 in more detail?.
Thanks a lot.

Jul 12, 2005 at 4:48 AM

#### nikongod

##### DIY-ku
c2 has nothing to do with current differences. that is r2. for most chips it is nto terribly important to have a specfic value.

c2 compared to r2 sets the "-3db" point. tangent wrote about it here:
link to applicable article on tangent's site
the basic equation is : f(-3db)=1/2(pi)R2*c2 (r in ohms, c in ferads) it looks prettier on tangents writeup. be sure to get unit/decimal parts correct. units/decimals will kick your arse hard.

Jul 12, 2005 at 6:30 AM

#### drewd

##### Banned
Whoa there, fellas - R2 sets the input impedance of the amplifier. Its primary purpose is to present a constant, high impedance load to the source. Without it, the source will see a varying load from the potentiometer and, as the volume is changed, the sound quality of the amplifier will change.

While the exact value of R2 is not critical, a good rule of thumb is for it to be 5 to 10 times the value of the potentiometer.

C2 serves to block DC offset voltage from the source. A byproduct of the RC network is that it also forms a high pass circuit, but with a properly selected capacitor value, the effect is minimal - you just want to keep DC out of the signal path.

Turbo, to (sort of) answer your question, the signal flow is through the pot to C1, which appears as an open to DC. The AC signal reaches the non-inverting (+)output of the opamp. R1 provides a high impedance signal reference to ground. Since an opamp (in a CMoy) is a voltage amplifier, it doesn't matter that there is current flowing through the resistor - all the opamp cares about is voltage.

The opamp will try to keep the voltages at the inverting (-) and non-inverting inputs equal, through the feedback loop. So, as the voltage changes at the non-inverting input, the output varies in proportion to that change. Some of the output voltage goes to the inverting input, through the feedback resistor. The resistor divider made up of R3 and R4 sets the amount of output voltage that is necessary to match the two input voltages. If R4 increases or R3 decreases, then more voltage will be necessary to match the two inputs. That also means that more voltage will be available on the output of the amplifier.

-Drew

[EDIT: Oy vey, I shouldn't post late at night! PeterR is spot on about R2 in his post right below.]

Jul 12, 2005 at 7:26 AM

#### PeterR

Quote:

 Originally Posted by drewd Whoa there, fellas - R2 sets the input impedance of the amplifier. Its primary purpose is to present a constant, high impedance load to the source. Without it, the source will see a varying load from the potentiometer and, as the volume is changed, the sound quality of the amplifier will change. While the exact value of R2 is not critical, a good rule of thumb is for it to be 5 to 10 times the value of the potentiometer.

A pot without a load will look like a constant load to the source regardless of wiper position. You'd get a varying imput impedance if R2 was smaller.

Jul 12, 2005 at 10:48 AM

#### TURBO

Hey fellas, thanks. Now, everything make more sense.

Drewd, thank you for explaining the flow from the start point.

Jul 12, 2005 at 3:03 PM

#### 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.

Jul 12, 2005 at 4:02 PM

#### PeterR

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.

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

Jul 12, 2005 at 4:46 PM