Amplifier output with capacitive loading

Hi guys
can I ask a question?
I am designing a most basic amplifier: nmos differential input, pmos current mirror loading. The output terminal is only planning to drive a cap.

My simulation output is in the picture attached. The red is the output without a cap and the green is with a cap. We can see the green curve is way less ideal compare to red. My analysis is as follow: I believe that the green are curves, instead of lines, suggests that it is not slew rate limited (also agree with my hand calculation). So I suspect that the reason of of the bad performance was due to the RC constant: huge output resistance make the RC constant really bad.

I have read that if your amplifier only need to drive a cap, no additional low-impedance output stage is needed. Is this statement contradict my simulation result?
Thanks
Allen

If you need to charge/discharge a capacitor very fast and you don't want huge operating current you will need a low output impedance driving stage. But capacitance load cannot cause DC error, I think this is the reason why you read no additional stage is required.
A big load capacitor slows down the circuit, but big time constant is not a "really bad" thing, sometimes it is necessary, especially if you want to implement feedback and make a stable circuit without oscillation.

A Differenatial Pair has pretty high Output Impedance therefore it cannot drive a large capacitive/inductor loads.
That's why there is an Output Stage ( AB-Class or Totem-Pole ) in OpAmps.An Source Follower will also work well ( try and see )

A differential pair with capacitive load can be used as a Gm-C filter because its output resistance is quite high. So it can drive. Point is the charge/discharge speed matters for somebody or its used for something else.

Hi frankrose
Thanks for your reply.
I was referring to that the difference between an OTA and an opamp was the output stage. Opamp=OTA+output stage. An opamp is used to drive a small resistive load because the output stage have a low output resistance. And OTA is used in, for example, switched capacitor circuit, is only to drive a capacitive load and no resistive load. (so the high output resistance won't be reduced by paralleling a small resistor )

Then how does the OTA with a capacitive load situation handle the big output time constant problem, which lead to the slow charging time?
Thanks
Allen

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Hi BigBoss
Thanks for your reply.
I was referring to that the difference between an OTA and an opamp was the output stage. Opamp=OTA+output stage. An opamp is used to drive a small resistive load because the output stage have a low output resistance. And OTA is used in, for example, switched capacitor circuit, is only to drive a capacitive load and no resistive load. (so the high output resistance won't be reduced by paralleling a small resistor )

Then how does the OTA with a capacitive load situation handle the big output time constant problem, which lead to the slow charging time?
Thanks
Allen

In OTA the Gm determines the speed of the circuit, output resistance is infinite, which is not for a real circuit, and not just the big time constant leads to slow charging time.
The finite Gm of the diff-pair causes slewing, the output resistance causes an exponential settling with the load capacitor, on the transient plot you can see the effect of both. With infinite output resistance (infinite time constant) the chargeing time won't be higher, but won't be 0 either, due to slewing.
If you want a fast OTA you can increase the Gm of the differential pair with extra current.