vorange said:
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
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.