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 :(
"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
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.
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.
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.
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
And you think a rank beginner will even remotely understand those terms do you ?
Graham
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.
-- If you do I don\'t see why he should have any difficulty with them.
-- Yes, it is, but its output swings from 5V more positive than 0V to 5V more negative than 0V.
?
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.
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
There are a few ways to think about this. One way is to realize that a 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 steady state conditions(and this is where your confusion is comming from). 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
-- Yup. What we old fuckers used to call "pulsating DC"
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.
-- "I\'m never wrong, once i thought i was, but was mistaken" Real Programmers Do things like this. http://webpages.charter.net/jamie_5
-- I suggest that since there\'s no galvanic connection between the windings of a capacitor, its secondary can be biased at any convenient voltage and its output will swing about that point.
Hmm, You in a cocky mood these days John :)
-- "I\'m never wrong, once i thought i was, but was mistaken" Real Programmers Do things like this. http://webpages.charter.net/jamie_5
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! :)
-- "I\'m never wrong, once i thought i was, but was mistaken" Real Programmers Do things like this. http://webpages.charter.net/jamie_5
What does gray hair have to do with age? Mine started to turn while I was still in high school, and I was completely gray by the time I was
20.-- 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
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