Can a capacitor let DC current through?

Discussion in 'Electronic Basics' started by vorange, Aug 20, 2007.

1. vorangeGuest

1st question : Till now, I believed that capacitors only let AC
signals through while blocking DC. But then, I saw a schematic whee
they put a capacitor on the output line of an opamp. The signal into
the opamp was a square wave signal (which I imagine is DC and not
AC). How then does the output of the opamp (presumably DC as well)
pass through the capacitor? This has confused the hell out of me.
Is there something I'm missing here?

2nd question : Is it fair to say that if a signal goes from say +5 to
-5 volts and then back to +5...etc that is is an AC signal because its
reversing its direction. But if it goes from +10 to 0 volts and then
back to +10 that it is a DC signal because its not reversing
direction?

I'm confused

2. Phil AllisonGuest

"vorange"

** That is quite wrong.

** A clue.

** Yes - that is pure AC.

** That is a signal with bath AC and DC components .

The DC component is 5 volts - ie the average value, as read on a DC meter.

The AC component is +/-5 volts peak, with the DC one removed.

Which is just what a series capacitor will do.

......... Phil

3. Stanislaw FlattoGuest

Don't be, dig into some 'very' serious books and read the definition of
capacitor.
You will find that the transfer function relates to dv/dt, meaning the
voltage change during time slice.
A digital wave has rising and falling edges, those are passed by the
capacitor, the DC component which does not change in time is blocked.

HTH

Stanislaw.

4. NobodyGuest

That's correct for a perfect capacitor. Real capacitors will "leak" some
DC, but it's normally negligible (if it isn't, you've probably chosen the
wrong sort of capacitor).
Any kind of "wave" is AC, although it may have a DC component.
The AC passes through, the DC doesn't.
No. The first one is a 5V-peak (10V peak-to-peak) AC signal with no DC
component (assuming that it's symmetrical, e.g. sine wave or 50% square
wave), the second is a 5V-peak AC signal with a 5V DC component.

In both cases, what will come "out" of the capacitor is a 5V-peak (10V
peak-to-peak) AC signal with no DC component.

5. Anonymous.Guest

You are describing an AC signal which is a square wave of +/- 5 volts
imposed upon a DC signal of +5 volts.

When you look at the output of the op amp, (assuming a gain of unity
and a good high frequency response) you will see only an AC signal
of +/- 5 volts.

1. The DC signal has not passed through the capacitor, but the AC signal
has.

2. The output of the capacitor is no longer referenced to 0V.

6. Anonymous.Guest

I'll rephrase that!

The voltage on the output of the capacitor no longer starts at 0v
and rises to 10V. It is equally dispose about 0V and is now just
the +/-5V AC signal and the DC signal has been lost.

Big mistake.

Graham

8. EeyoreGuest

And you think a rank beginner will even remotely understand those terms do you ?

Graham

9. Bob MyersGuest

Whether it was a mistake or not depends on the intent
of that particular design, the size of the cap, the frequency
of the square wave, etc., etc., etc....

Bob M.

12. Stanislaw FlattoGuest

From my failing memory, many times such chance remark stays with the
reader in a corner of brain to work for years as a whip to learn.
HTH.
_Understanding_ comes much later.

Stanislaw.

13. Michael A. TerrellGuest

Only if he actually listens to you, Donkey.

--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

14. Jon SlaughterGuest

square wave(or any wave) is composed of a series(which means a sum) of
simple sine waves. A capacitor passes these sine waves. It does attenuate
each frequency but as long as certain conditions are met the ouput will look
like the input.

A second way is to realize that the capacitor doesn't "block" except in
The capacitor will charge and discharge on each cycle of the square wave and
as long as this charging is fast enough the output will still look like a
square wave.

I would say this would technically be true. AC means alternating current. If
you have a square wave that goes from -5V to +5V then it must in some way
alternate the current when it changes from - to + or + to -. (even if its
just a fA)

If you take the square wave from 0 to +10 then the current never alternates
but just changes in magnitude(but not sign). -5 to 5 never changes
magnitude(for an ideal signal) but only sign.

Jon

16. JamieGuest

well, In the circuit you were looking at, it still is blocking DC how
ever, the change in levels on that DC will reflect on the out side of
the capacitor.
To break it down in simple terms.

Picture the capacitor as a battery. when a battery is discharged and
you connect a charger with an AMP meter on it. You'll see ampere's being
displayed on the meter until the battery fully absorbs all that its
going to take. At that point, the amp meter will show its lowest
reading, indicating the battery is charged.
When the source and absorbing device(Capacitor/battery) are at equal
voltages. No current flows.

Now lets translate that to an OP-AMP output.

The Op-Amp goes high to lets say 10 volts and you have a capacitor
connected to the output in series to a voltage meter for example.

While the capacitor is charging, current is being generated which will
allow the voltage meter to register a reading. When the capacitor
reaches it's full charge equal to what the op-amp voltage is, the
current will reduce to virtually zero. This will indicate on the voltage
meter no current which means no voltage will display. at this point, the
OP-AMP output can remain at the DC 10 volts and you'll see no voltage
at all on the meter because the 10 volts of the op-amp is equal to the
10 volts stored in the capacitor.
Just put in your mind 2 batteries connected in series back to back.
when you take a volt reading from the ends of these 2 in series. You'll
get 0 volts because the polarity of each are canceling each other.
this is what happens to a capacitor when it becomes fully charged to
the source in series..

Now, picture the op-amp going to ground or low after the capacitor has
fully charged on the + cycle. what we get now is something you may not
expect.
Think of taking a fully charged battery and reversing the polarity
connections.
In this case, the capacitor lead connected to the Op-Amp output was
fully charged to + volts, and now since the op-amp has shifted to
low/common, it has in effect, reversed the connections of that capacitor
so that the + side charge is now grounded.
since the output side of the capacitor is above the ground potential,
you will see the charge in the capacitor now discharging in reverse
which will give you a real (-) voltage from a system that only had 0 or
10 volts+
This is how below 0 volt AC signals are formed from a DC pulse for
example.
This voltage will be there until the capacitor has fully discharged
it's energy through the output load, in which case, would be your volt
meter.
And thus, starts the cycle of a fully discharged capacitor on the +
side again!

Hope you got something out of that.

18. JamieGuest

Hmm, You in a cocky mood these days John

19. JamieGuest

Old, I'd like to consider my self a young guy!

Gray hair are loved by many young ladies!

I don't know if it has anything to do with lack of
threat or what ever!

20. neon

1,325
0
Oct 21, 2006
a capacitor does not care what voltage the swing is as long there is a swing of voltage. [withing the cap specs] whatever happens on one side is transfered to the other side but not neccesseraly with the same levels. if one side is sitting at 10v and you input -5 volts it will spike to to 5v and ramp again to 10v with the paricular time constant as set.