# Question over Coupling capacitors and Op-Amps

1. May 26, 2012

### KrNx2Oh7

So this is the op amp I am looking at and I have a few questions, not entirely sure how to ask them, but I'll ask more as I find out more

http://postimage.org/image/i228xrkj7/
http://postimage.org/image/i228xrkj7/

So the capacitor C1 and C3 are used to block DC, but when you add these do you not
get voltage drops in the AC signal? Like when the op amp outputs a voltage does it not drop some voltage across the capacitor before it reaches Vout.

My second question is over C2. Apparently it is suppose to eliminate power-supply noise from reachin ghte op amps noninverting input. I don't see how this works. How does a capacitor in parallel with R4 stop noise, a non-DC signal?

Finally my last question, which might sound silly, but why do capacitors not end up in the Gain. Instead of Gain= -R2/R1, why don't you do the current equations and include -jwC in the transfer function to have capacitors show up in the gain.

silly questions, but I want to know why everything is in this configuration and why it works

thanks

Last edited: May 26, 2012
2. May 26, 2012

### Staff: Mentor

The idea is to use coupling capacitors which are so large that even at the lowest frequency of interest their reactance is still comparatively much lower than other associated impedances in the circuit.

Last edited: May 26, 2012
3. May 26, 2012

### AlephZero

You haven't told us what frequency range the circuit is designed for, but the idea is the C2 provides a much smaller impedance from the op amp + input to ground than R3 and R4 in parallel.

You could get the same effect by reducing the values of R3 and R4, but then you would be wasting power because of the higher current flowing through them.

Similarly C1 will have a low impedance compared with R1, and C3 compared with RL, so for practical purposes they can be ignored when understanding how the circuit works.

Of course if you plotted the response of the circuit down to DC, the caps will affect the gain and phase shift of the amp at low frequencies.

4. May 26, 2012

### KrNx2Oh7

Cool that was VERY helpful. Let me see if I understand everything.

C1 and C3 block the DC signal and only drop AC voltage slightly since the impedance is low for the capacitors at higher frequencies (Xc= 1/-jwC)

For C2 it lowers the voltage seen at V+ because it's impedance is also small. And +V is a DC voltage which I assuming is superimposed with a small noisy AC signal. So for DC the capacitor does nothing, but for the noisy AC it reduces what is seen in the V+.

For gain I'm not sure about what you said when u mentioned "plotted the response of the circuit down to DC". When is the gain just -R2/R1 vs when I ...

V+ = V*(R4 || Xc)/ [(R4 || Xc)+R3]
V- = V+
-[Vo - (V-)]/R2 = Vin - (V-)/(Xc+R1)
solve for Vo/Vin

One more thing. Why disnt we just ground the pos terminal
thanks a million

Last edited: May 26, 2012
5. May 26, 2012

### truesearch

You would normally expect an op amp to have 2 symmetrical power supplies with a common 'ground' or 0Volts....usually +/-15V.
The + input is usually connected to the common ground to establish 0 volts.
Your amplifier is using a single supply which needs to be 'split', the 2 56k resistors act as a potential divider to 'split' the supply. The junction of the 56k resistors is connected to the + input of the amplifier.
This means that the output of the amplifier can increase and decrease symmetrically about an 'artificial' zero. The coupling capacitors play a role in this because
only AC (+/-) are transmitted by capacitors.