- #1
jegues
- 1,097
- 3
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 [tex]V_{CMmin}[/tex] 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 [tex]V_{CMmin}[/tex] by writing a KVL from the common mode input down to the -Vss.
Like so,
[tex]V_{CMmin} = V_{GS} + V_{CS} - V_{SS}[/tex]
Where [tex]V_{CS}[/tex] 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 [tex]"V_{CS}"[/tex] is going to be replaced by the voltage [tex]V_{DS3}.[/tex]
Computing [tex]V_{DS3}[/tex],
First note that,
[tex]V_{GS} = -V_{S}[/tex]
Where [tex]V_{S}[/tex] is the voltage at the source of Q1 and Q2.
[tex]-V_{S} - V_{t} = V_{ov}[/tex]
[tex]V_{S} = -V_{ov} - V_{t} = -0.7V[/tex]
Thus,
[tex]V_{DS3} = V_{S} + V_{SS} = 0.5V[/tex]
Now writing my KVL,
[tex]V_{CMmin} = V_{GS} + V_{DS3} - V_{SS} = 0V[/tex]
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!