# Can square waves cause a DC voltage

1. Jul 12, 2016

### Planobilly

I read this statement from Lenard Audio.

"When a large signal spike is created by un-plugging or plugging in signal leads, or when a pre-amp valve is driven hard into distortion (guitar amps), a large non-symmetrical square wave may cause a temporary DC Voltage to appear across a coupling capacitor, to the Grid of the following valve. This short DC pulse will cause the following valve to be shut off, or go into full conduction, for approx 1mS to 1 second, until the following Grid resistor discharges the coupling capicator."

First, is this statement correct. Second, if yes, how does this work?

Thanks,

Billy

2. Jul 12, 2016

### analogdesign

It isn't "technically" DC if you have a good quality capacitor, but a non-symmetrical square wave can easily change the effective DC level after the coupling cap due to bandwidth limitations on the following circuits. Since you probably have reasonably low bandwidth after the coupling capacitor, the circuit will average the spike and the level will rise and slowly fall as the pulse goes away.

The easiest way to visualize this is:

1. The coupling capacitor passes any spikes with little attenuation (the impedance is low at high frequency, and spikes have a lot of high frequency energy).
2. The signal path has limited bandwidth (for noise reasons, among others)
3. The high frequency spikes is then low-pass filtered.
4. A low-pass filtered square wave is just its average value (you can use a low-pass filter to measure duty cycle, for instance)
5. This appears as an effective DC shift across the coupling capacitor.

You can see this exactly thing by plugging in a scope probe on an oscilloscope. Do you see how the level jumps up and slides back down to baseline? This is the effect. Make sense?

To fight this kind of thing, you can put diodes in parallel with the coupling capacitor to limit the voltage excursion. This is sometimes done but it often is not due to cost/component count/space considerations.

3. Jul 12, 2016

### Planobilly

Thanks Analogdesign

As I am learning something new every day and I am far from knowledgeable about electronics, I hesitate to just say "that's not correct". So, when I read that statement I just did not think that this was DC. I had even less idea how it could be possible to create DC on a grid through a good capacitor by driving a preamp tube into distortion.

DC like effect, yes I can buy into that idea. I am pretty sure the guy writing the statement knew what he meant. Perhaps sometimes I read things in a "too literal"

Thanks for clearing this up for me.

LOL...every question answered brings up another question...lol Now I need to go investigate diode in parallel idea!!

Cheers,

Billy

Last edited: Jul 12, 2016
4. Jul 14, 2016

### tech99

DC can also arise if the amplifier is driven hard because the grid/cathode circuit acts as a diode rectifier.

5. Jul 14, 2016

### Staff: Mentor

Perhaps you are too literal. Any time varying signal whose average is not zero has a DC component. I think that is what they are describing.

6. Jul 14, 2016

### jim hardy

yes it's correct

here's a shot at how it works
it's a quirk of math that the circuit guys quashed for us.

It doesn't even have to be asymmetrical, a symmetrical square wave can have a DC component
but the coupling capacitors in your signal path soon force it to zero

If all is quiet,
then you switch on a nice square wave (or sine wave for that matter)
its DC component depends on where in the cycle you start it. Man oh man is that counterintuitive, though..

Thought experiment time

Let's think of DC as just the long term average, over several cycles

and also consider short term average, most recent half cycle

for the red square wave ,
gray is short term average note it builds from zero in first half cycle then heads back toward zero next half cycle, barely makes it there and turns right around again
brown is long term average which builds to average of gray wave, slower than i showed it though.

so that nice looking square wave has a DC component shown by the brown line,
and how much DC depends on when in its cycle it got switched on.

Same is true for a sinewave
and that's why power transformers have such huge inrush if you close on them at sinewave's zero crossing.

So,
the coupling capacitors in your signal path not only separate the high plate DC from the low grid DC,
they block any DC that's got your signal because of the random time somebody threw a switch.
More accurately they bleed it down to zero,
and thus spake Lenard .

old jim

Last edited: Jul 16, 2016
7. Jul 16, 2016

### Planobilly

Hi Jim,

I get where you are going with your post.

I guess this whole discussion should go back and define DC. I took this definition from a web site and it looks as good as any.

"DC (direct current) is the unidirectional flow or movement of electric charge carriers (which are usually electrons). The intensity of the current can vary with time, but the general direction of movement stays the same at all times. As an adjective, the term DC is used in reference to voltage whose polarity never reverses.

I ASSUME the following. Correctly??????

DC has a frequency of zero. If it does not have a frequency of zero it is not DC. A frequency of zero imply infinite impedance
A "ideal" capacitor will not pass DC
Pulsed DC, while everyone knows in general what that means...that is it alternates from from zero to some value over some period of time, it none the less alternates. It does not have a frequency of zero.

For a "spike" or call it a "pulse" a Fourier transform will clearly show frequencies other than zero for a pulse or spike.

I am saying this jokingly but I am beginning to wonder if at some level DC even exist...lol

I assume even a battery could have less than a perfect rate of discharge however small.

Cheers,

Billy

Last edited: Jul 16, 2016
8. Jul 16, 2016

### jim hardy

Ahhh now we're getting there

to get by in today's world we need to be able to think like a pure mathematician
and still get by as a practical technician.

If you redraw that square wave i drew up above
but instead of starting it at zero you start it at 1/4 cycle
go ahead, copy it into paint and do it
draw the short term average this time in say green not gray

observe the short term average now is centered about zero
so this symmetrical square wave has no DC component, but the only difference between the two is where in the first cycle they started

Your scope has FFT which breaks any waveform down into its component frequencies
one of them is 0 hz, the DC component ,,, often calleda0
first wave's 0 hz component is nonzero second one's is zero