Output caps and DC offset

[ic]

I have a hobby background in electronics, robotics, and microcontrollers, but I'm new to audio. I thought I would start off by breadboarding a few of Jack's buffers from http://www.muzique.com/lab/buffers.htm

I breadboarded the transistor and op amp buffers, and have been using the breadboard oscilloscope/AWG from http://www.gabotronics.com/development-boards/xmega-xprotolab.htm to feed in a .125V 770Hz sine wave and to see the result.

When using the 10 microF electrolytic output cap shown in the schematic the output initially has a marked +ve offset, and takes a minute or so to come down to 0V. If I reduce the output cap to 1 microF it takes a few seconds to normalise, and is almost instantaneous with a .1 microF greencap. If I go too low on the output cap the trace gets a low frequency "wobble".

Can anyone explain what's happening? Is it real or an artifact? Would this behaviour have consequences in a stompbox?

The scope has a 1M input impedance if that matters. I've noticed a lot of other circuits use a much smaller output cap - how critical is this value?

Thanks for your help!

Ian.

[tubegeek]

The very high impedance of the scope is giving you a much slower discharge time for the cap than a typical input would - a 100K input would be 10 times faster, for example.

[R.G.]

When using the 10 microF electrolytic output cap shown in the schematic the output initially has a marked +ve offset, and takes a minute or so to come down to 0V. If I reduce the output cap to 1 microF it takes a few seconds to normalise, and is almost instantaneous with a .1 microF greencap. If I go too low on the output cap the trace gets a low frequency "wobble".

Can anyone explain what's happening? Is it real or an artifact? Would this behaviour have consequences in a stompbox?

When a circuit has been un-powered for a while, all the voltages dissipate more or less quickly down to 0V across each part. Output caps in single-power-supply circuits start with 0V across them, and when the power comes on, the end attached to the circuit is yanked up to the quiescent DC bias voltage in the circuits. Capacitors being capacitors, there has to be current flowing to make the outboard end go down to 0V, so the outboard end is pulled up to the DC bias voltage too - which is another way of saying that the voltage across just the output cap stays at 0V until something forces it to change.

Those circuits do not show a "pull down" resistor to ground on the outboard side of the output capacitors, so the capacitors stay with 0V across them and with both capacitor terminals at the internal DC bias voltage on both terminals until you connect your scope probe. This connects the equivalent of the probe's 1M to ground on the outboard side of the cap, and starts pulling the outboard terminal to ground.

The time constant of 10uF and 1M is 10 seconds, so it drops by 0.639 of the remaining voltage towards its final voltage every time constant. 10uF and 1M takes a long time. 1uF and 1M has a 1 second time constant, so in five seconds, it's pretty nearly at ground. 0.1uF and 1M is only 0.1S time constant, so it's pretty much at final value in 0.5S.

To answer your questions, then:
1. Yes, it's real.
2. Yes, it has consequences, both AC and DC. The "DC" effect is a loud pop when the power is turned on, as the cap sucks current from the amplifier following it to charge to it's final value. The AC consequence is that different values of output cap change the amount of low frequency that gets through the output cap into whatever follows it. Notice that this depends on the perhaps unspecified impedance of the thing that follows it, which is not stated, and is probably unknown. 10uF and 1M (if the whatever after the buffer has an input impedance of 1M) has a low frequency rolloff of F = 1/(2*pi*R*C)= 0.0159Hz, or about 16millihertz. That's lots of lows. 1u and 1M is 159mHz, and 0.1uF and 1M is 1.59Hz.

By now you should be appreciating how much the perhaps-unknown impedance which might connect to your buffer affects frequency response.

I've noticed a lot of other circuits use a much smaller output cap - how critical is this value?

How critical is your bass rolloff frequency, and what is this buffer going to be driving?

One mistake that's very common to electronicists of all kinds is to not appreciate that circuits are not isolated in their use, the way they're taught. There is always something providing your circuit a signal, and the signal source has its own source impedance. And circuits will in general drive something after their outputs. That thing has an input impedance. These "phantom" impedances are always there when you use a circuit in the real world. You have to learn to anticipate that they're there, and to estimate their effect on what your circuit is trying to do.

[ic]

Thank you Gentlemen for a very important lesson in practical electronics!

I connected a 100K resistor to ground and, as expected, things settled much more quickly. From reading various commentaries on different circuits I had somehow reached the conclusion that high input impedances were desirable and output cap values weren't critical. I realise now that it depends on the application, and that there aren't any simple answers. I shall spend some time boarding some RC circuits for my own education.

I also rather naively thought I could patch together fragments of circuits and that things would work nicely, but I can see that I'm going to have to bust out the calculator and do some theory. For me the joy is in learning about ths stuff, if I get something that sounds cool afterwards, its a bonus.