feeling dumb... single supply op amp question.

[armstrom]

ok... I'm suffering from a bit of an understanding gap here so I'm hoping someone can help me out... If you have a simple single-supply op amp circuit that is capacitor coupled on both the input and the output. will the voltage swing after the coupling capacitor be centered around ground (0V) or around your bias voltage? (4.5V for a 9V supply)

I had always been under the impression that the coupling cap was there to simply block the constant 4.5V DC that would be on the output, however, when I simulate this circuit in LTSpice the value is centered around 0V.  Meaning the voltage is pulled below ground.. where does this negative potential come from if there is only a positive supply rail? I seem to be missing something fundamental here...

sorry for the dumb question.... I'm sure this is obvious to everyone

[petemoore]

  Supply for the amp is 9vdc.
  To swing +/- from a fixed voltage, something between 9v allows more room before one or the other phases hits a supply rail [gnd or V+].
  The outputs should follow the difference between the 2 inputs.
  The input 'idle' voltage [before any AC signal 'twiddles' it] is biased between the rails, say 4.5v.
  Referring to many an opamp circuit [TS / DIST+ / Microamp] there is a voltage divider string [of two equal value resistors], one end of this string is V+, the other is grounded. When equal R's are seriesed between a voltage, the divide [where the R's are connected] 'in the middle' will be a reference: voltage of 1/2 of supply. In the center of the available swing room between the 2 supply rail voltages [+/Gnd.].
  With a symmetric signal input, the greatest output swing before hitting one of the rails is when swing headroom is equal for + and - signal output swing excursions from Vbias [= 1/2v = Vb =....usually, occasionally the divider resistor values aren't made equal, thus creating a bias voltage that is 'offset'.
  Sometimes 3 resistors are used instead of just 2 'biggies' to provide the bias at low current/low feedback.
  Two 'smaller' resistors voltage divide, then one large resistor is used to between Vbias and input to 'further separate' the power supply from the input, yet still provide a reference voltage [see DOD250].
 

[brett]

Hi

will the voltage swing after the coupling capacitor be centered around ground (0V) or around your bias voltage?

All capacitors block DC, so there is 0V DC plus the AC signal after the cap.
cheers 

[Gurner]

ok... I'm suffering from a bit of an understanding gap here so I'm hoping someone can help me out... If you have a simple single-supply op amp circuit that is capacitor coupled on both the input and the output. will the voltage swing after the coupling capacitor be centered around ground (0V) or around your bias voltage? (4.5V for a 9V supply)

I had always been under the impression that the coupling cap was there to simply block the constant 4.5V DC that would be on the output, however, when I simulate this circuit in LTSpice the value is centered around 0V.  Meaning the voltage is pulled below ground.. where does this negative potential come from if there is only a positive supply rail? I seem to be missing something fundamental here...

sorry for the dumb question.... I'm sure this is obvious to everyone

The negative voltage comes from the fact a capacitor can't change it's charge instantly ...so if there's 0VDC  sitting on the right hand side of your DC blocking output cap (ie the side connected to your output cable), and then a 'negative heading'  AC signal is presented on the left hand side of the same cap  - then if we know the cap can't change it's charge instantly - the right hand side must move down! The result is negative AC voltage.

[PRR]

> when I simulate this circuit in LTSpice the value is centered around 0V. 

You have a pull-to-ground resistor after the cap?

In real life, the cap starts with zero V DC across it. You turn-on the power and the amp node rises to say 4.5V. Instantly the output (after the cap) rises to 4.5V. But the pull-down resistor bleeds this off. In theory it never decays to dead-zero V DC. In practice, with audio R-C ratios (sub-Second), if we wait several seconds after turn-on we find very low V DC declining to very-very-low.

The typical SPICE skips turn-on. It starts by opening all caps and shorting all coils. Now it computes the DC conditions which would exist after infinite time. Then it notes the DC voltage on each side of each cap, and charges that cap to that voltage. (And coil currents.) So the TRANSient analysis starts from "infinite time after turn-on".

Either way, after that initial 4.5V turn-on thump had bled-down close-enough to zero V DC:

Your amp idles at 4.5V. The cap has a 4.5V charge. The output is at zero V DC.

You put in a signal and the amp output quickly moves to 3.5V DC. The cap voltage can not change quickly. The output shifts to negative 1V.

If you kept it there, that -1V would bleed-down toward zero V.

If you kept it there, it would not be "music". Music is vibrations. If your amp swings from 4.5V idle to 3.5V, it will surely swing back through 4.5V to 5.5V in much less than a second. And a quick swing to 5.5V shifts the after-cap output to +1V DC.

As long as you keep up the vibrations (play), the cap stays charged near 4.5V steady, amplifier swings above/below 4.5V idle push the other end of the cap up and down from zero V DC.

[armstrom]

Thanks for the replies guys. I think I realize now why I was having issues understanding this. I was not taking into account that a capacitor (at least an "ideal" one) is a barrier to electron flow. As voltage decreases on the op amp side of the coupling cap electrons pile up inside the capacitor, causing an increased negative charge. This additional charge acts as if it were a "pressure". This repels the electrons on the "other" side of the capacitor, causing a negative voltage in the output conductors. When the output voltage of the opamp moves in a positive direction it begins to pull electrons back out of the capacitor which then causes a sort of charge "suction" pulling electrons back into the output side.

So it's not so much a case of a negative voltage potential being switched into the output, but more accurately the capacitor compels a reverse or "negative" current flow in the output circuit, which develops a negative voltage potential across the load.

ok, I knew there was something I wasn't thinking of... ugh.. having a brain drain kind of day anyway.

[DougH]

Here's a simple way to think about this stuff: the sig coming from your guitar swings around 0v. At an instant of time that is some voltage- we'll call it 'x'. On the opamp input that value is x + 4.5 . After the output cap, that 4.5 goes to 0 and the sig is x (* the gain of the circuit).