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Resistive loads vs capacitive loads. what are the differences?

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Syafiq

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Normal op-amps drives resistive loads while Operational Transconductance op-amps drive capacitive loads. What are the contributing factors to these loads? Also, what are the differences between VCVS and VCCS? Does the 'S' means source output? Thanks!
 
Hello,

By VCVS do you mean Voltage Controlled Voltage Source, and
by VCCS do you mean Voltage Controlled Current Source?

If so, then using a resistive load will work for either of these but capacitive loading has to be minimized with the VCVS simply because it is a VS (Voltage Source).

The capacitor has equation:
dv/dt=i/C

and solving for 'i' we get:
i=C*dv/dt

Now dv/dt is just the rate of change of the voltage source, which is a statement about how fast the source voltage rises. For a ramp this can be slow or faster, but for a step change this means it rises very very fast, meaning dv/dt is actually infinite. So we have more or less something like this:
i=C*(infinity)

and since anything times infinity is still infinity, we end up with:
i=infinity

which means it takes an infinite current to drive the capacitor.

This is obviously not possible. What actually happens in a real circuit is the driver internal resistance limits the actual charge current, but that requires power dissipation within itself which could cause overheating. One time during turn on of the circuit might not hurt anything, but often these circuits have to do this many times per second depending on the frequency, so it ends up overheating the IC chip that does the driving.

So driving a capacitor with a voltage source that steps from some value to another value in zero time is not a good idea. However, using a current source is another story.

Using a current source the cap charges up slowly, a little at a time, and so there are no infinities to deal with.
Going back to the equation for the capacitor:
dv/dt=i/C

and now we see that with a small current of 0.001 amp and capacitor value C=0.001 Farad we have:
dv/dt=0.001/0.001
which of course equals:
dv/dt=1

so the rate of rise of voltage is 1 volt per second, which is not harmful usually.

For an op amp, the difference is that if it puts out a voltage then we have to be mindful of what we are driving and what the wave shapes are, but for a current output a capacitive load is easier to deal with.

This doesnt mean you'll never see a regular op amp driving a capacitor, but when you do it will probably have a resistor in series with it too in order to prevent over current.

There are specialized IC's made for driving large capacitor loads. They are designed to be able to put out a very fast rising voltage even into a capacitive load. But these are made that way by design so that the applications designer can create a circuit that can quickly charge capacitive loads such as a MOSFET.
 
The previous poster has answered the original question pretty completely.

I would just like to add that most op amps operate with closed loop feedback to stabilize and establish the operating parameters.

Driving a heavy capacitive load may add enough phase lag at the output that can, and often does reduce phase margin, if not produce actual oscillation.
 
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