Hi,
I'm trying to simulate the power supply jection of OPAMP.
But, I'm not sure my simulation method is correct or not.
I designed the simple  OPAMP with 60dB gain
And I'd like to simulate the power supply jection.
So, I give AC signal at power source (VDD) and I tied the output singal to one of inputs of OPAMP.
Is it correct?

And I have another question.
If OPAMP have resistor feedback, how to simulate the PSRR ?
should I connect the output and other input not to be connect with feedback node, and simulate AC analysis?

Thanks

Won

The PSRR is simply the ratio of the gain of your circuit from its normal input to the gain of your circuit for a signal injected on the power supply.

PSRR = A_in / A_supply

where

A_in = v_out / v_in with v_supply = 0
A_supply = v_out / v_supply with v_in = 0

Thus, you should configure your opamp as you expect it to be used, and then perform two AC analyses, one where the normal input is driven and one where the supply is driven. With Spectre, you can get the same information using one XP analysis.

- MM -

The XF analysis is very similar to the AC analysis, except that while AC allows you to specify a specific source and see the response at any possible observation point; XF allows you to identify a particular observation point, and compute the transfer function from every souce to that point.  (This should all be explained by running "spectre -h xf".)

To use XF, you must identify the observation point, which is typically the output of your circuit. This is generally done by specifying a pair of nodes. Then you would specify the start and stop frequencies just as you would with AC. Then run the analysis. Then the transfer function from every source in the circuit to the observation point is available in the analysis results. To find the PSRR  compute Vin/Vdd if Vin is the name of your input source and Vdd is the name of your supply.

- MM -

You should probably not set the xfmag. If you leave it alone it defaults to 1, meaning that it will directly compute the transfer function.

I cannot understand why the computed result is -6.4kdB. That indicates that there is no path from the supplies to output of the circuit. Did you properly specify the circuit output to the XF analysis?

- MM -