Coupling capacitors

[Kevin Weddle]

I believe that a coupling capacitor is directly affected by DC. I am talking about low value capacitors that see the signal as discreet DC values because the signal rate of change is low. I can prove this theoretically. Whenever the signal goes DC, the other side of the capacitor will go to it's DC bias. It is often taken for granted, but when the signal stops changing, the signal on the other side of the capacitor jumps to it's DC bias and this is a distortion of the input.

Also, I have stated before that there exists a problem with using very small capacitors. The RC time constant is so short that capacitor completely charges with every value of the signal. It is desireable to have a capacitor charge in a few time constants so that you don't witness the charging curve of a capacitor. When the capacitor is sitting there at 3v, and then the signal goes 4v, you have a capacitor charging curve. Now if you do it in 3 time constants instead of 5 then the signal is better reproduced.

There is a linear protion to the capacitor charging curve. By linear we mean that we introduce a function between voltage and time that follows a straight line. So although the angle of the function can be anything, it is at least predictable. As the capacitor charges for every discrete value of the signal, it inserts a slightly different rate of change than the input because of the charging curve of the capacitor.

 

[audioguru2]

when the (input?) signal stops changing, the signal on the other side of the capacitor (its output?) jumps to it's DC bias and this is a distortion of the input.

No, the capacitor's output signal doesn't jump, and certainly not when the input signal stops changing.
For high frequencies, the capacitor's impedance is low and its output follows its input. If the input suddenly changes its DC voltage, its output will follow then slowly charge or discharge to the bias voltage.

Coupling capacitors are frequently chosen to have a low value for a highpass filter affect and they don't cause distortion.

 

[Kevin Weddle]

I am sorry Audioguru but I just don't agree. You have to admit to some kind of distortion if even just a little. And what about the jumping to a DC. If the signal is 1vpp and the output of the capacitor is biased at 5 volts then the output changes from 4.5 to 5.5. This only occurs when the rate of change is there to pass the signal. We'll assume that it passes the entire signal. If the input goes to 4.5 volts with a zero rate of change, the other side of the capacitor will go to 5 volts. This 5 volts does not indicate the input of 4.5 volts. This is what I was talking about. As far as the time constants go, you can use the example of a low value capacitor as sort of a high pass filter. Remember that there is a critical frequency where the impedance of the capacitor and it's resistor lends itself to voltage dividing. You then get a 20db per decade attenuation between the capacitors impedance and the resistor. But there is to account for the small capacitor which will have a high critical frequency because you need a high frequency to get the signal to pass. Below that frequency you get the attenuation. Now for the frequency of the signal which is low. The capacitor will attenuate this frequency very well and I am just guessing that this will lead to complete charging of the capacitor. It is most desireable to not completely charge the capacitor as 5 time constants leads to the nonlinear portion of the charging curve. The linear charging is what you want and it takes place in the 1 to 3 time constant range.

By the way it is easy to do. All you have to make sure of is that the circuits time constant is high compared to the frequency of the signal. If the signal is 25Khz, then the time of the signal is 40 us. The time constant of the circuit needs to greater than 40 us. So all you need is to take care of the resistance and the capacitor will not completely charge in 40 us.

 

[Kevin Weddle]

Hey Audioguru. I want you to investigate this article where it talks about the time constant and the signal time which must be smaller to achieve less distortion.

http://www.kpsec.freeuk.com/capacit.htm

 

[audioguru2]

Kevin,
A coupling cap with a value that is too low causes visual distortion to a square-wave or any other kind of waveform with sharp edges. It really isn't harmonic distortion, it is just the too-small coupling cap acting like a high-pass filter, passing the sharp edges but attenuating the broad low-frequencies.
Maybe you could say that poor low frequency response caused by the high-pass action of having coupling caps too small is some kind of distortion, but not harmonic distortion that adds harmonics to the signal.

The charging curve of a capacitor has nothing to do with harmonic distortion. If you feed a pure sine-wave into a good (no electrolytic coupling caps) high-pass filter, the output won't have any harmonic distortion at any frequency.