transfer function of a simple current mirror

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Hi all, I have a question about the transfer function of AC current. If we have a 1:1 current mirror, like the uploaded figure. Then if we inject an AC current at the input, what is the phase shift of the output AC current?
I think the gain would be 1 (0dB) below MHz level, but am not sure how to determine the phase shift between input and output. Is it 0 degree or 180 degree?
Thank you!
 

Below the effective cutoff frequency of the circuit, the phase shift should be 0 degrees (it's a current mirror, isn't it?).
 

erikl said:
Below the effective cutoff frequency of the circuit, the phase shift should be 0 degrees (it's a current mirror, isn't it?).
Click to expand...

Thank you erikl, I only knew it is a mirror in DC, but do not know if it is true in AC. Do you think is it correct to use MOS small signal mode to get your conclusion? I tried but did not figure out if that is the right way to go.
 

bhl3302 said:
Do you think is it correct to use MOS small signal mode to get your conclusion? I tried but did not figure out if that is the right way to go.
Click to expand...

Sure this should work: as stimulus use an ac current source in series with a DC current source, run an ac (or xf) analysis, and plot input and output current vs. frequency.

Here still an intuitive explanation for zero phase shift (far below the intrinsic cutoff frequencies of the transistors):

Increasing the control current of the mirror increases the Vgs of both transistors (0 phase shift), and this again increases the mirror current (0 phase shift again).
 
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