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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: Wilson BJT SMALL SIGNAL MODEL 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: Wilson current mirror SMALL SIGNAL MODEL I expect to see \$r_{\pi 1}\$ in the model and I cannot understand where \$1/g_{m2}\$ comes from.

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    \$\begingroup\$ 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. \$\endgroup\$
    – G36
    Aug 19, 2022 at 18:52
  • \$\begingroup\$ @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 \$\endgroup\$
    – chami
    Aug 19, 2022 at 20:40
  • \$\begingroup\$ Which small-signal mode did you decide to use? \$\endgroup\$
    – G36
    Aug 20, 2022 at 6:37
  • \$\begingroup\$ @G36 second one! It seems easier to follow! \$\endgroup\$
    – chami
    Aug 21, 2022 at 7:23

1 Answer 1

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

schematic

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

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

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

I hope this will help you.

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