# Equivalent resistance problem on cascode mirror circuit

I'm trying to analyze the cascode mirror, specifically its sensitivity to VCC variations.

Before facing with complicated calculations with the equivalent circuit, the tutorial (I'm following) advices to find the equivalent of the circuit block on the right of point A, giving only the final result of Req :

$$R_{eq} = \frac{v_{A}}{i_{B}} = \frac{2(1+\beta)}{g_{m}}$$

but without explaining the line of reasoning used to find it, with the only exception to consider equal the transconductances of Q1 and Q3.

By using different approaches, I tried different times to find the same result for the equivalent resistance, but I was not able to find it, and to find an expression for the voltage on point B, too.

At this point, I guess that the circuit block on the right of point A cannot be solved, and that the tutorial is wrong, unless I'm missing something important...

Could some of you help me to find a line of reasoning in order to find the above expression for Req?

• So you want to find the equivalent resistance seen from Q1 base into the BJT? Because Req seen at point A (between point A and GND) is equal to 1/gm2 + 1/gm4. Due to the fact that r_pi is (beta +1) times large than 1/gm, thus we can ignore r_pi influence.
– G36
Commented Dec 24, 2020 at 18:47
• @G36 - Could you please detail where the resistances 1/gm2 and 1/gm4 can be found on the equivalent circuit, in order to be able to compare them to r_pi? Commented Dec 24, 2020 at 20:16
• Q2 and Q4 work as diode-connected BJT's. This is why you have 1/gm2 and 1/gm4 on the equivalent diagram. i.sstatic.net/QkMTY.png
– G36
Commented Dec 24, 2020 at 20:30
• @G36 - I see. But as explained in my post, I'm looking for finding the Req looking from A to the right side, where there are no 1/gm2 and no 1/gm4... Commented Dec 24, 2020 at 20:45
• But I've already shown you (in my answer) how to find the Req.
– G36
Commented Dec 24, 2020 at 20:48

At first glance, we can assume that:

$$V_{\pi3} \approx \frac{v_A}{2}$$

Do you know why we can do that?

So, we know the $$\Q_1\$$ emitter current value:

$$i_E \approx g_{m3}\frac{v_A}{2}$$

And $$\Q1\$$ base current will be equal to:

$$i_B = \frac{i_E}{\beta +1} = \frac{g_{m3}\frac{v_A}{2}}{\beta +1} = \frac{g_{m3} v_A}{2(\beta+1)} = v_A \frac{g_{m3}}{2(\beta+1)}$$

$$\frac{i_B}{v_A} = \frac{g_{m3}}{2(\beta+1)}$$

Therefore:

$$R_{eq} = \frac{v_{A}}{i_{B}} = \frac{v_A}{\frac{g_{m3} v_A}{2(\beta+1)}} = \frac{2 (\beta +1)}{g_{m3}}$$

And this is the end.

• I suppose that your first assumption is valid for a DC regime, but not for small signals...otherwise, I do not know where it comes from. Commented Dec 24, 2020 at 20:39
• Why do you think that V_pi3 = VA/2 does not hold in the small-signal analysis?
– G36
Commented Dec 24, 2020 at 20:52
• From the complete circuit I can see that, approximatively, the Q1 base voltage (point A) is 2 times that of Q3 base, but the same relation is not so evident from the incremental circuit. On the other hand, I do not think that what is valid for DC is also valid for small signals. Commented Dec 24, 2020 at 20:59
• But in this circuit Ic2 ≈ Ic4 which means that gm2 ≈ gm2 and V_pi3 = Va/2.
– G36
Commented Dec 24, 2020 at 21:08
• I see it now, even though this is only valid if r_pi3 >> 1/gm4, does it? Commented Dec 24, 2020 at 21:20