# Overdrive Voltage Effect on Cascode Operation

Working on this exercise:

I understand that if all the transistors in a cascode have the same current and the same Vgs then they will have the same Vov. However in this case the Vgs do not seem to be the same and so I'm wondering how all Q's have the same Vov. Additionally, if all Q's have a Vov of 0.2 V then shouldn't the possible range of output voltages be this?: *Side note, when I initially tried the question I used Vds=Vov and worked from the top and bottom to get Vovs of Q1 = Q4 = 0.2 V and Q2 = Q3 = 0.3 V, which give the output voltage range given in the answers of the book.

The lowest value of $$\v_o\$$ where the output current starts to deviate from its nominal value is not $$\2*V_{ds,sat}\$$ as you calculate.

In that minimum $$\v_o\$$ situation, NMOS Q1 still needs to be in saturation mode as Q1 needs to set the output current. So Q1 does need one $$\V_{ds,sat}\$$ which has the same value as the overdrive voltage of 0.2 V.

But cascode NMOS Q2 can be in linear (or triode) mode so it will basically behave as a low-value resistor and the $$\V_{ds}\$$ of Q2 will be quite small.

The same is true for Q4 and Q3.

Edit:

The above is assuming $$\V_{g2}\$$ is chosen such that the lowest $$\v_o\$$ can be achieved. That's not the case here, $$\V_{g2}\$$ is lower here, at 1 V.

So when will Q2 shut off (and thus lower the output current)?

When Q2's $$\V_{gs}\$$ becomes lower than $$\V_t\$$.

That's at about $$\V_{s2}\$$ = $$\V_o\$$ = 0.5 V.

This is a bit "1st order approximation" as Q2 will need a bit more $$\V_{ds}\$$ than zero. In practice the real answer will be about 0.6 V to 1.2 V.

• So if only Q1 and Q2 need to be in saturation shouldnt the allowable range of Vo be somewhere around 0.2 to 1.6? This is still different from the range given in the exercise answer (0.5-1.3)
– Eric
Dec 12, 2018 at 17:08
• Oh my bad I meant Q1 and Q4
– Eric
Dec 12, 2018 at 17:18
• OK, my answer assumed optimal choice of Vg2 but that's not the case here, Vg2 is higher. I'll update my answer. Dec 12, 2018 at 17:23