How do I measure the small signal output impedance (ro) of a modified wilson current mirror?

example Q4:

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  • \$\begingroup\$ Are you looking for a sophisticated Test Engineer's approach? Do you have some acceptance criteria? or just idle curiosity? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 26 at 0:21
  • \$\begingroup\$ At this point just doing some side research for a project \$\endgroup\$ – Roo Apr 26 at 0:23
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    \$\begingroup\$ If you apply a square wave to DC at V1 with a current sense resistor of suitable value, what do you expect for \$R_O = \frac{\Delta V}{\Delta I}=10^7?, 10^{10}?\$? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 26 at 0:28
  • \$\begingroup\$ I found at gain of 6.3G ohm at 1 Hz, however, it gradually decreases with frequency. I'm not sure why the resistance decreases as the frequency increases. \$\endgroup\$ – Roo Apr 26 at 0:53
  • \$\begingroup\$ What do you expect Coss to be? 1pF? 100pF? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 26 at 0:58

\$R_O = \frac{\Delta V}{\Delta I}\$

In other words, vary the output voltage and measure the corresponding change in current.

If you're asking something more subtle than that, please clarify by editing your question.

  • \$\begingroup\$ updated the question \$\endgroup\$ – Roo Apr 26 at 0:13
  • \$\begingroup\$ Same answer. Apply an AC voltage and measure the AC current. You can make it as large or small as you like. You can even vary the frequency. \$\endgroup\$ – Dave Tweed Apr 26 at 0:16
  • \$\begingroup\$ where is the ideal place to place the AC source? \$\endgroup\$ – Roo Apr 26 at 0:20
  • \$\begingroup\$ In series with V1. \$\endgroup\$ – Dave Tweed Apr 26 at 0:36
  • \$\begingroup\$ I found at resistance of 6.3G at 1 Hz \$\endgroup\$ – Roo Apr 26 at 0:54

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