capacitive voltage dividers: applications and issues?

[gaussmarkov]

i was fooling around with AC math and accidentally discovered that the gain of an AC voltage divider made out of two capacitors does not depend on frequency.  i sure did not expect that.  i checked it two ways to make sure i had it right.  so that led to an immediate question:  "what is this useful for?"  because ideal capacitors do not dissipate power the way resistors do, there must be something.

i got a partial answer by googling and searching the forum.  R.G. has mentioned a cool way to measure the capacitance of cables.  and i read a quote by petemoore from sound.southwest.com that says "the capacitive divider is not frequency dependent, so long as the source impedance is low, and the load impedance is high compared to the capacitive reactance."  i guess this is a deviation from the ideal, but i don't understand it.  the best i can figure, the "source" and the "load" are not the two capacitors in the divider.  but that is as far as i get.

any help? 

[brett]

Do you mean that because the reactance of both caps changes with frequency (at the same rate), the division remains the same?
I guess that explains why one-pole filters use a cap and a resistor, rather than two caps (or two resistors) !

Overall losses would still be subject to the usual laws of impedance in and impedance out (Ohm's or Kirchoff's law or something). It's very hard to bend these laws. 
cheers

[brett]

Here's an idea:
Combine a regular cap with either (i) a resistor with voltage or current-dependent capacitance, or (ii) a capacitor with voltage or current-dependent resistance.
The result should have voltage or current-controlled gain.  (and phase-shift?)  Maybe that's sorta what happens in phasers already?  Oops. ??

[gaussmarkov]

Do you mean that because the reactance of both caps changes with frequency (at the same rate), the division remains the same?
I guess that explains why one-pole filters use a cap and a resistor, rather than two caps (or two resistors) !

i don't understand, really, how it works.  here's the math: the current through a capacitor is proportional to the rate of change of voltage or

I = C dV/dt

for two capacitors in series, C1 and C2, the current is the same through both so

I = C₁ dV₁/dt = C₂ dV₂/dt

and

dV₁/dt = C₂/C₁ dV₂/dt

that's the key thing.  the two capacitors' voltages are proportional no matter what.  kirchoff's voltage law says

dV_S/dt = dV₁/dt + dV₂/dt

and we can use this to solve for

dV₂/dt = C₁/(C₁ + C₂) dV_S/dt 

so that the divided voltage is proportional to the source.

it doesn't matter whether they are sine waves and it doesn't matter what their frequency is.  without a resistor in there to slow things down, i guess the capacitors don't charge and discharge in a lethargic way.  if you put in a square wave, it appears instantly at the junction of the voltage divider -- just like with resistors.   

Overall losses would still be subject to the usual laws of impedance in and impedance out (Ohm's or Kirchoff's law or something). It's very hard to bend these laws. 
cheers

except that resistors dissipate heat and capacitors don't.

[brett]

To me, it couldn't make any sense any other way.  To make either a resistor or a capacitor with double the Z, I just put 2 in series.  No need to think about frequency.

I guess this is because reactance is proportional to the inverse of frequency for the caps, individually *and* both combined. So the proportions stay constant.  The AC division by caps is similar to DC division for resistors - V x Z1/(Z1+Z2).  Or am I missing something? 

[gaussmarkov]

To me, it couldn't make any sense any other way.  To make either a resistor or a capacitor with double the Z, I just put 2 in series.  No need to think about frequency.

I guess this is because reactance is proportional to the inverse of frequency for the caps, individually *and* both combined. So the proportions stay constant.  The AC division by caps is similar to DC division for resistors - V x Z1/(Z1+Z2).  Or am I missing something? 

maybe. but if you are then you are not alone because i am too. 

here's what i understand to be correct.  impedance calculations are based on restricting the AC signal to be a sine wave.  this means that (strictly speaking) they are not a generalization of ohm's law to AC.  rather they are an extension that provides a useful guide, because guitar signals generally aren't sine waves.

and in that analysis  Z  is often frequency dependent.  for example, the noble capacitor has Z = 1/(j 2 pi f C) where f is the frequency of the sine wave AC.  that's why it surprised me that the capacitive voltage divider is not frequency dependent.  as you said, when you put two capacitors in series you get double the Z, but that is still frequency dependent.  which is great because that's how we get filters to work.

now this equation

dV₂/dt = C₁/(C₁ + C₂) dV_S/dt 

doesn't need a sine wave to be tractable.  the rate of change of the voltage at the junction of the divider is proportional to the rate of change of the supply voltage.  this means that, ignoring a possible DC offset (we could store a few extra electrons in between the two capacitors and they would just have to stay there), the divider voltage is proportional to the supply voltage as well.  no matter what the frequency and, more than that, no matter what the shape. 

if you put a square wave through an RC divider you don't get a square wave out.  LFOs use that phenomenon.  but with a CC divider (seems like a reasonable label  ) you put a square wave in and you get a square wave out.  and it's not a different phase either.  the jumps coincide.  i guess what this means is that for AC, Ohm's law for resistors holds for capacitors without any need to limit the signal to a sine wave.

[Edit] no, that's not right.  current is still not proportional to voltage.  scratch that last sentence.

unless, of course, i have made a mistake.  but that's how you learn sometimes.   

[Edit] prophetic, no?    i still think the rest is right. 

[gaussmarkov]

oops.  accidental double post.  (i'm inching my way up to 4 figures. )

[alanlan]

Cap potential divider (C1 top C2 bottom)?

Vout/Vin (jw) = (1/jwC2) / ( (1/jwC1) + (1/jwC2) )..............w = 2.pi.freq and j accounts for the fact that current through a capacitor is 90 degrees out of phase with the voltage across it.

multiply throughout by jw:

Vout/Vin = (1/C2) / ( (1/C1) + (1/C2) ) = C1 / ( C1 + C2 )

i.e. no dependence on jw (or frequency)

Also, there is no phase difference between Vout and Vin. 

BUT

This assumes that the input Vin has no source resistance.  So the above results may work reasonably well for the output of an op-amp circuit because the effective output resistance of an op-amp circuit is very low.  If you stick such a capacitive divider on the output of a single stage transistor amplifier (with higher output resistance) then the output of the capacitive divider will change with frequency because the circuit now has some significant resistance in series with "C1".

i.e. the effective potential divider is now:

Vout/Vin (jw) = (1/jwC2) / ( R + (1/jwC1) + (1/jwC2) ).......................R is the output resistance in series with Vin.

You can no longer cancel out the frequency term.

If you think about it you should see that this forms a simple low pass filter with R and the combination of C1 and C2, but the output at C2 will be reduced by the ratio C1 to (C1+C2).

...and this ignores any effect of loading by subsequent circuitry.

[gaussmarkov]

yes, i agree.   but i guess it's worth repeating that those impedance calcs are only for pure sine waves.  part of what i was noticing is that this result is not sine wave dependent.

it's good to remember about the likely presence of neighboring resistances, source and load.  thanks for chipping in.

i'm still hoping there's an application for capacitive voltage dividers in stompboxes.  perhaps it's revealing that i have never seen one.     i've found a lot of high voltage references ... not exactly in the ballpark. 

[gaussmarkov]

i got into this because i was working on understanding input and output impedance.  so i worked out the output impedance of a capacitive voltage divider and -- surprise -- it is frequency dependent:  the magnitude is 1/(sqrt(3) w (C₁ + C₂)).  for a resistive voltage divider, i figure it's the value of the resistors in parallel:  R₁ R₂/(R₁ + R₂).

[gaussmarkov]

so ... here's schem with what i think might qualify as a capacitive voltage divider:  jack orman's amz-fx dr. quack.  maybe this auto-wah needs such a divider?

[gez]

Well, this post has given me one idea.  If it works, I'll post the results. 

[alanlan]

yes, i agree.   but i guess it's worth repeating that those impedance calcs are only for pure sine waves.  part of what i was noticing is that this result is not sine wave dependent.

it's good to remember about the likely presence of neighboring resistances, source and load.  thanks for chipping in.

i'm still hoping there's an application for capacitive voltage dividers in stompboxes.  perhaps it's revealing that i have never seen one.     i've found a lot of high voltage references ... not exactly in the ballpark. 

All real life signals can be decomposed into a sum of sine waves (Fourier Transform).  The pure capacitive voltage divider has a transfer function which is independent of frequency so if you put a sqaure wave in, you will get a square wave with a reduced amplitude.

[R.G.]

Since capacitance is unavoidable, the oscilloscope people long ago learned to live with it.

The reason that high frequency scope probes are 10X dividers is so you can connect the probe to that "calibration" output on the scope and adjust the resulting high quality square wave on the display to actually look like a square wave. What you're adjusting is one of the caps in the resulting capacitive divider to match the precision 10:1 resistive divider. That makes the probe have a frequency independent square wave response and hence accurate high frequency representation on the display, not losing or gaining highs by capacitive gain or loss as frequency changes.

Capacitive dividers let you divide signals down *without* losing either highs or lows, well up into the many megahertz regions, in spite of the unavoidable losses in cables and stray wiring capacitance.

[gaussmarkov]

Well, this post has given me one idea.  If it works, I'll post the results. 

excellent!    and, gez, i hope you'll share your idea even if it doesn't work.  it would still be instructive. 

yes, i agree.   but i guess it's worth repeating that those impedance calcs are only for pure sine waves.  part of what i was noticing is that this result is not sine wave dependent.

All real life signals can be decomposed into a sum of sine waves (Fourier Transform).  The pure capacitive voltage divider has a transfer function which is independent of frequency so if you put a sqaure wave in, you will get a square wave with a reduced amplitude.

at first, i did not understand your point.    i was going to talk about phase shifts and how they might mess up an intuition based upon fourier decomposition.   that was when the force of your logic hit me:  phase shifts are frequency specific -- frequency independent implies no phase shifts.  or, in the math, if w cancels out then so does j.    thanks!

Since capacitance is unavoidable, the oscilloscope people long ago learned to live with it.

The reason that high frequency scope probes are 10X dividers is so you can connect the probe to that "calibration" output on the scope and adjust the resulting high quality square wave on the display to actually look like a square wave. What you're adjusting is one of the caps in the resulting capacitive divider to match the precision 10:1 resistive divider. That makes the probe have a frequency independent square wave response and hence accurate high frequency representation on the display, not losing or gaining highs by capacitive gain or loss as frequency changes.

Capacitive dividers let you divide signals down *without* losing either highs or lows, well up into the many megahertz regions, in spite of the unavoidable losses in cables and stray wiring capacitance.

nice. thanks, R.G.