Min. Common Mode Input MOS Diff Pair with Current Mirror

[jegues]

Homework Statement

See first figure attached for problem statement.

Homework Equations

The Attempt at a Solution

I've solved everything in this circuit, I just have some confusion about solving $$V_{CMmin}$$ when I have a current mirror acting as my current source connected to the sources of Q1 and Q2.

Usually when I have this setup with an ideal current source instead of a current mirror I can solve for $$V_{CMmin}$$ by writing a KVL from the common mode input down to the -Vss.

Like so,

$$V_{CMmin} = V_{GS} + V_{CS} - V_{SS}$$

Where $$V_{CS}$$ is the minimum voltage required across the current source.



(See 2nd figure attached for example)

How do I do this now with my current mirror in place?

It looks as though my $$"V_{CS}"$$ is going to be replaced by the voltage $$V_{DS3}.$$

Computing $$V_{DS3}$$,

First note that,

$$V_{GS} = -V_{S}$$

Where $$V_{S}$$ is the voltage at the source of Q1 and Q2.

$$-V_{S} - V_{t} = V_{ov}$$

$$V_{S} = -V_{ov} - V_{t} = -0.7V$$

Thus,

$$V_{DS3} = V_{S} + V_{SS} = 0.5V$$

Now writing my KVL,

$$V_{CMmin} = V_{GS} + V_{DS3} - V_{SS} = 0V$$

I'm not entirely sure if my reasoning is correct so if somebody could point out any mistakes or confusions I'm having it would be greatly appreciated.

Thanks again!

 

[KataKoniK]

Your approach to solving for V_{CMmin} with a current mirror in place is correct. The voltage V_{CS} is indeed replaced by the voltage V_{DS3}, which is the voltage across the current source of the current mirror. Your computation of V_{DS3} is also correct. Your KVL equation is also correct, and you have correctly solved for V_{CMmin} as 0V.

One thing to note is that the voltage V_{DS3} is not necessarily equal to V_{S} + V_{SS} as you have written. V_{S} is the voltage at the source of Q1 and Q2, while V_{DS3} is the voltage across the current source of the current mirror. These two voltages may not be equal, as the current mirror may have a different output impedance than the ideal current source. However, in this specific circuit, the output impedance of the current mirror is large enough that V_{DS3} can be considered equal to V_{S} + V_{SS}.

Overall, your solution is correct and your reasoning is sound. Keep up the good work!