how to simulate small-signal transconductance (gm) of OTA?

[xiaoyanghqu]

According to definition of gm (gm=Io/Vin, Vout=0(gnd)),I add a large capacitor(1F) at the output to make it ac ground.Then do ac simulation,and plot the ac Io of the capacitor. But the result i think is not right . What is wrong with the simulation?
Thinks!

 

[LvW]

Re: how to simulate small-signal transconductance (gm) of OT

Because you want to measure/simulate the output current, the output must not remain open. However, don´t connect a capacitor at the OTA output because it introduces a frequency dependence which belongs not to the device under test (OTA). Instead, use a resistor which produces a maximum voltage in the milli-volts range. This is sufficient to realize the condition Vout ≈0 volts.

 

[xiaoyanghqu]

Re: how to simulate small-signal transconductance (gm) of OT

Because you want to measure/simulate the output current, the output must not remain open. However, don´t connect a capacitor at the OTA output because it introduces a frequency dependence which belongs not to the device under test (OTA). Instead, use a resistor which produces a maximum voltage in the milli-volts range. This is sufficient to realize the condition Vout ≈0 volts.

LvW，Thank you for your reply! What do you mean that the output must not remain open? Should the OTA be connected to closed loop by a unit feedback? If use a resistor as output load, how to set the DC point of the output transistors? The large capcitor can set the DC point and build the AC gnd, Am I right or wrong?
can you post the simulation setting? Thanks!

 

[LvW]

Re: how to simulate small-signal transconductance (gm) of OT

What do you mean that the output must not remain open?

When it is open (nothing connected), no current is possible.

Should the OTA be connected to closed loop by a unit feedback?

No, just loaded by a small resistor.

If use a resistor as output load, how to set the DC point of the output transistors? The large capcitor can set the DC point and build the AC gnd,

I was referring to an OTA with split supply voltages and symmetrical behaviour, which means: output current zero for input differential voltage also zero.
Is this the case?