# Wilson BJT current mirror output impedance

I am trying to have a small signal analysis of a Wilson BJT current mirror to find out what is the output impedance. Already I know that the answer is $$R_o=\beta r_o/2$$ In the first step, I need to draw the small signal model. My model is different from what is depicted here: More confusing about this model is the direction of $$\I_{C3}\$$. I find another model in my textbook which is somehow the one that I expected, but still not quite the same: I expect to see $$\r_{\pi 1}\$$ in the model and I cannot understand where $$\1/g_{m2}\$$ comes from.

• Q2 is a diode-connected BJT, The Rout for diode-connected BJT is equal to: $r_{out} = \frac{1}{gm}||r_\pi||r_o \approx \frac{1}{gm}$ Also notice that 1/gm2 is much smaller then r_pi1, and this is why they omitted r_pi1 from the diagram.
– G36
Aug 19, 2022 at 18:52
• @G36 thank you. I got it. Is there another assumption that I can take and make KCL and KVL easier? I am not able to reach to the answer. Ro=Bro/2 Aug 19, 2022 at 20:40
• Which small-signal mode did you decide to use?
– G36
Aug 20, 2022 at 6:37
• @G36 second one! It seems easier to follow! Aug 21, 2022 at 7:23

The simplified small-signal model Wilson current mirror will look like this:

simulate this circuit – Schematic created using CircuitLab

Now we can write KCL:

$$V_X = (I_X - I_{C3})r_{o3} + V_{BE}$$

Also notice that $$\V_{BE1} = V_{BE2} = V_{BE}\$$ and that: $$\I_X = g_{m2}V_{BE} + g_{m1}V_{BE}\$$

And from this we have -> $$\V_{BE} = \frac{I_X}{ g_{m1}+ g_{m2}}\$$

Next $$\I_{C3}\$$ current

$$I_{C3} = V_{BE3}*g_{m3} =- V_{BE}*g_{m2}*r_{\pi3}*g_{m3} =$$ $$= -\frac{I_X}{ g_{m1}+ g_{m2}}*g_{m2}*r_{\pi3}*g_{m3} = -\frac{g_{m2}\: g_{m3}\: r_{\pi3}}{g_{m1}+ g_{m2}}I_X$$

Therefore:

$$V_X = \left(I_X - (-\frac{g_{m2}\: g_{m3}\: r_{\pi3}}{g_{m1}+ g_{m2}}I_X )\right)r_{o3} + \frac{I_X}{ g_{m1}+ g_{m2}} =$$

$$= \left(I_X + \frac{g_{m2}\: g_{m3}\: r_{\pi3}}{g_{m1}+ g_{m2}}\:I_X \right)r_{o3} + \frac{I_X}{ g_{m1}+ g_{m2}}$$

$$V_X = I_X \left(\frac{g_{m2}\: g_{m3}\: r_{\pi3} }{g_{m1}+ g_{m2}} r_{o3} + r_{o3} + \frac{1}{g_{m1}+ g_{m2}} \right)$$

$$V_X = I_X \left( \left(\frac{g_{m2}\: g_{m3}\:r_{\pi3} }{g_{m1}+ g_{m2}} +1\right)r_{o3}+\frac{1}{g_{m1}+ g_{m2}} \right)$$

And finally, we have the output impedance value:

$$R_{OUT} = \frac{V_X}{I_X} = \left(\frac{g_{m2}\: g_{m3}\:r_{\pi3} }{g_{m1}+ g_{m2}} +1\right)r_{o3}+\frac{1}{g_{m1}+ g_{m2}}$$

Additional notice that: $$\\:\large g_{m3}\:r_{\pi3} = \beta_3 \$$

and $$\ \large g_{m1} = g_{m2}\$$ thus

$$\ \large \frac{g_{m2}}{g_{m1}+ g_{m2}} = \frac{1}{2}\$$

So we can simplify the equation

$$R_{OUT} = \left(\frac{g_{m2}\: \beta_3}{g_{m1}+ g_{m2}} + 1\right)r_{o3}+\frac{1}{g_{m1}+ g_{m2}}$$

$$\large R_{OUT} \approx \left(\frac{\beta_3}{2}+ 1\right)r_{o3}$$