# Open-circuited capacitor

1. Apr 1, 2015

### jaydnul

Say you are using an op-amp as a buffer, so the output is tied straight back to the inverting input. And in-between your voltage source and the non-inverting input you have a capacitor.

(Voltage Source)---------------||-------------(Non-inverting input of the op amp)

How would the op-amp know if there was an AC voltage source? There isn't a way for charge to accumulate on the other side of the capacitor, so no voltage will be detected at the input, right?

Or would the charge just become displaced in the second leg of the capacitor almost like an antenna? If so, what are the equations for such situation? Even though there is charge being displaced, I'd imagine the detected voltage at the input would still be considerably smaller than the voltage source, right?

2. Apr 1, 2015

### analogdesign

The op amp has no idea what is connected to it... it just responds to its inputs. In your case how the op amp responds depends entirely on the frequency of the ac voltage source. The capacitor appears as an open circuit at dc, so if it is a dc source the op amp will not respond. If the source is ac whether the op amp responds depends on the frequency (or, equivalently, the size of the capacitor. remember Z = 1/jwC for a capacitor).

The only equation you need here is Z = 1/jwC, then standard nodal analysis using your op amp. The op amp inputs will be dc biased somehow (depends on your circuit) so the only caveat is the breakdown voltage of your capacitor. If your ac voltage source output minus the op amp input dc value is greater than the breakdown voltage of your capacitor you will turn your cap into a resistor and everything will change (and you won't like what happens).

3. Apr 1, 2015

### jaydnul

With a DC source, what would be the voltage drop across the capacitor? It would be equal to the voltage source because there is no current, right? So the the charge on the plates should equal Q=CV.

But if there is a charge on the plates, then the capacitor lead connected to the op amp will be polarized (because the charge on the opposite plate would repel). Wouldn't an equal charge show up at the non-inverting input?

4. Apr 1, 2015

### Averagesupernova

Your approach here with a coupling capacitor tied to the non inverting input with NOTHING ELSE tied to it is asking for trouble. We can think of the input of the op-amp in this case as having very high or nearly infinite impedance. So it is possible for that input to float around unless it is tied to something through a resistor. A voltage across a capacitor only occurs due to a current through it at a previous point in time in order to charge it. Your setup will not really allow that as the input of the op-amp is such a high impedance.

5. Apr 1, 2015

### jim hardy

What a great way to think of it ! I never heard that before.
No current through it in the past = no voltage across it

So -
If OP proposes an ideal opamp with infinite Zin and no bias current,
Then
voltages on both sides of the capacitor are the same
........
so the opamp "feels" the applied voltage through the capacitor.

Equation would be voltage divider, Vout = Vin X Z2/Z1+Z2

finite reactance Z1 of capacitor in series with infinite impedance of opamp input Z2
makes the fraction
∞/( XC + ∞)
so take limit as Z2→∞ , and that = 1 .

is that right ?

That's theory for ideal opamp. Real world circuit will have to deal with this real world op-amp fact of life :

6. Apr 2, 2015

### sophiecentaur

I think the actual wording of the OP consists of what is called 'an ill conditioned equation'. There is no solution unless some more details are added, like the (finite) input resistance of the OpAmp and the voltage supply rails OR (more useful) the biasing arrangement at the input terminal of the OpAmp and the time since the connection was made. etc. etc.

7. Apr 2, 2015

### Averagesupernova

Well thank you Jim! That's an Averagesupernova original. Although I didn't think there was anything special about that statement when I posted it I will admit that it is a nicely condensed and simple approach.

8. Apr 8, 2015

### jaydnul

Sorry I'm coming back to this thread a little late.

Suppose the op amp input has infinite impedance. The battery will continue to work until the plate of the capicitor connected to it is at the same potential. Now that plate will have an electric field that effects the other plate which is connected to the op amp input. So wouldnt the charge on the op amp plate start displacing in response to this electric field? Meaning a negative charge on the plate and positive charge at the input?

9. Apr 8, 2015

### Averagesupernova

I am not sure I follow your last post. Simplify it by rereading what I posted before. Thought experiment: Connect one lead of a capacitor to the positive post of a battery. Connect a voltmeter with infinite input impedance between the unconnected plate and the negative post on the battery. Of course no such voltmeter exists in real life but in this thought experiment we know our ideal voltmeter will read battery voltage. It has to in order for the capacitor to behave in the way that we know they do.

10. Apr 8, 2015

### jaydnul

Ahh, ok. It just took a second explanation to stick. So the capacitor still blocks DC, but doesn't block a constant voltage?

11. Apr 8, 2015

### davenn

12. Apr 8, 2015

### Averagesupernova

Watch your wording. Notice in my thought experiment that it is not blocking the voltage. A capacitor cannot pass a steady current. In the real world our voltmeters will pass a small current and eventually a voltage will develop across the capacitor and it is said to be charged. Try this: Connect some capacitors up in parallel shooting for 50 to 100 uF. Try to use something with very low leakage such as tantalum capacitors. Hook the positive lead of the capacitor (assuming you will have caps that are polarized) to the positive terminal of a 12 volt DC source. Make sure the capacitors are discharged prior to doing this. Connect the negative lead of your voltmeter to the negative terminal of the battery. Touch the positive lead of the voltmeter to the negative lead of your capacitor bank and read the voltage. Remove the voltmeter lead. Come back an hour later. Do the same thing. If you don't spend a lot of time with the voltmeter hooked up you should be able to go for days doing this before the capacitor charges up through the voltmeter. Of course this assumes your voltmeter has an input impedance of about 1 Meg. Try leaving the voltmeter hooked up and watch the voltage fall. If you have a meter with a 1 Meg input and have 100 uF then it should take about 8 minutes for the voltage to fall from supply voltage to close to zero. Try the same thing with a cheaper analog meter. A meter rated at 20,000 ohms per volt should take a little less than half of that time if the meter is set to the 20 volt scale. If you had a magic voltmeter with infinite input impedance you could leave it hooked up indefinitely and the capacitor would never charge and the voltmeter would continue to read the battery voltage.
-
Edit: Actually there are some lab grade voltmeters out there that have INCREDIBLY high input impedance. They are slow reacting. It's been about 20 years since I have used one.

Last edited: Apr 8, 2015
13. Apr 9, 2015

### jaydnul

So using the circuit in my original post with ideal components, the cap WOULDN'T block a DC offset, correct? Since the charge has nowhere to sink from the second plate of the cap, whatever voltage is applied to the first plate will appear on the second plate, yeah?

14. Apr 9, 2015

### Averagesupernova

Correct! It is likely to give trouble with real-world components too. The inputs could float up, down, somewhere in between, etc.

15. Apr 9, 2015

### jaydnul

Nice! Thanks for the help Averagesupernova (and everyone else).

16. Apr 9, 2015

### sophiecentaur

This is 'how long is a piece of string?'. Every R and C will have an associated time constant. If they are both are big enough, then the time constant can be longer than the time taken to hook up the experiment, do some measurements, take it apart and go home.

17. Apr 10, 2015

### jim hardy

Great thought !

Try this thought experiment:

Make an inverting opmap, with 1 uf capacitor for Zin and 1uf capacitor for Zfeedback, instead of the usual resistors..
All parts are ideal, capacitors have infinite insulation resistance and amplifier has infinite Zin.....

Since gain = Zfeedback/Zin, this opamp has gain of one. But that's just a thought experiment. And since there's really no DC feedback, you have to set initial conditions.

I built one using polypropylene capacitors and OPA128 opamp(10^13ohms input impedance)
It would hold output constant within a millivolt overnight.
Makes a practical analog "track and hold" function block.

http://www.ti.com/lit/ds/symlink/opa128.pdf
http://www.ascapacitor.com/pdfs/products/X362.pdf

18. Apr 11, 2015

### Averagesupernova

That's pretty good Jim. I recall using poly caps in wien-bridge oscillators. They held tolerance pretty well but if I recall they could not take much heat.

19. Apr 11, 2015

### jim hardy

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