I am analyzing the schematic for an audio amp:


I am interested only in the top left op-amp (JP2 on the left) where it runs into the op-amp, pictured below.

enter image description here

A few simple things that I don't understand.

  1. Why is the audio fed initially through a capacitor? Won't the filter out any audio signal?

  2. If Vcc is much higher than the audio signal, say 12V, won't if feed back into the lower-power audio source and cause problems?

  3. Similar to my first question, why is the output of the op-amp then fed through another capacitor? Again, won't this filter the signal out?


3 Answers 3


C9 is placed in the audio input path to block the DC that would otherwise flow because of the DC bias (~\$V_{CC} / 2\$) that will develop on the inverting input. A capacitor passes AC current and blocks DC current. The size of the capacitor determines how much AC at what frequency will pass through, that's why C9 is large-ish.

\$V_\text{CC}\$ doesn't come into it here. It powers the op amp, but doesn't connect directly to the op amp output. The DC portion of it that develops on the inverting input is blocked by C9.

The output capacitor blocks the DC in the output.

An ideal capacitor has infinite impedance at DC, which you can think of as AC having a frequency of zero. The capacitor's impedance (which is like resistance, except for AC) decreases as the frequency increases. Also measured in ohms, a capacitor's impedance is:

$$\frac{1}{2\pi fC}$$

Where \$f\$ is the frequency in Hertz (Hz) and \$C\$ is the capacitance in farads. 0.47uF seems a bit on the small side for low audio frequencies, so it might not have terrific low frequency response. The capacitor would have an impedance of ~17kΩ at 20Hz, for example, 1.7kΩ at 200Hz, etc. So about 1/3 of the input signal is dropped at 20Hz, about 1/30th at 200Hz. More may be dropped on the output depending on the impedance of the following stage.

The LM358 is a so-called "single supply" op amp, meaning it operates effectively with only a single voltage supply, \$V_{CC}\$ in this case. To do that, it "centers" the signal around half of the supply voltage. That's what the resistor divider on the non-inverting terminal is there for. So the output signal is AC superimposed on a DC value of say 6V for a 12V supply. That's why the output coupling capacitor is needed, to block the DC component, leaving only the amplified AC signal.

  • 1
    \$\begingroup\$ The really important point about the input capacitor is that the bias on the inverting input (set at Vcc/2) isn't taking dc current through R2 (or R4) and thus saturating the amplifier. \$\endgroup\$
    – Andy aka
    Commented Oct 7, 2015 at 8:40
  • \$\begingroup\$ Ah, interesting, I see what you're saying, that certainly makes sense of the input capacitor. I was wondering under what conditions DC would be on the input signal and was thinking C9 was just belts and suspenders. But of course, you're right, the cap is protecting against DC coming from inside the circuit rather than arriving through the input. Thanks @Andyaka, I've updated my answer to reflect this :) \$\endgroup\$
    – scanny
    Commented Oct 8, 2015 at 3:27

All three of your questions are related.

  1. This serves a dual (triple? see #2) purpose of blocking any DC voltage on the line, and re-referencing the voltage to something the circuit can use.

  2. Note that the voltage on the + pin of U1G1 is connected to Vcc/2. If you apply your laws of op amps, you know that in negative feedback mode (where the output is connected to the - pin through feedback) the + and the - port will be equal voltage. This MIGHT cause current to flow into the output of whatever you've plugged in to JP2, but the capacitor prevents that.

  3. The output capacitor is an easy way to prevent DC voltages from appearing on downstream amplifiers (which would amplify the error voltage, causing bigger offset problems), as well as preventing current flow in speaker coils that are attached which at best burns energy and at worst burns up the speaker coil!

DC error/offset in the output is an inevitable effect of reality in most amplifiers. There are some designs (called DC servo designs) that have a feedback loop to cancel this output DC offset. This reduces the need for output capacitors in power amplifiers, but most of the authors I've read are pretty down on the idea as a practical solution to the offset problem.

A note on filtering: You're probably looking at this like an RC circuit, but at the frequencies we're looking at, the source impedance is low enough and the capacitance is high enough that the effect is minimal. Generally the larger the capacitor, the better the bass response and lower distortion.


For question 1&2: This intentional capacitor removie any DC off set of that input signal. This offset voltage may change the Q point of circuit which can cause unexpected values.Audio signal is mainly an oscillation which is not cancelled out through this capacitor (capacitor act as short circuit for small signal ac).

Q 3: What I understnad from the diagram, your VCC is higher then input. For op amp, if VCC is not hhigher,output will be lesser then what it should be and for electrical IC this also be true.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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