I'm trying to figure out how cascode current mirror works.
I've serached it all over the web and haven't found an answer.

The diode connection sets: V_gs1 = V_gs3 = V_ds1 = V_ds3
(given that all the transistors are the same).

Given the current source's value, I can calculate V_gs1 = V_gs3 = V_ds1 = V_ds3 = V_gs2.

How can I now determine the current in the right branch (through N2, N4) ?
I've read that I can assume the voltage over N2 is constant. Why is that? how does that assist me to find the current in the right branch?

Thanks !!!

Aaron, thank you for your reply.

I'm not following you.
What am I missing here?

Correct me if I'm wrong -
let's assume V_DS1 = V_DS3 = V_GS1 = V_GS3 = 2V.
Take V_D2 = 1V.
All I know now is that V_DS2 + V_GS4 = 4V, since V_G4 = 4V.

What can I infer about the the current through N2, N4 (which is the same)?

Hi Gil,


following your math, lets let VD2=1V. Also, VD4 is determined by the load of the current mirror, so lets say for example, that VD4 = 4V. Now VGS4=3V, and VDS4=3V while VGS2=2V, and VDS2=1V. Clearly this is an impossible situation since based on the voltage, N4 should be passing a much larger current than N2. The only way this would work is if VD2 increases to 2V. Then VGS2=VGS4=2V, and VDS2=VDS4=2V. Also, in this situation, the same current flows through N2 as N1.

That was obvioulsy a simply example, but what if VD,n4 is equal to 3V? Well, again, the current through N4= that through N2, so in order to get the same current, the drain of N4 might reduce to 1.9V so that VGS2=2V, VDS2=1.9V, VGS4=2.1V, and VDS4=1.1V. Remember that a MOSFET's current is less sensitive to VDS than VGS. So in this situation, the VDS of N2 decreased from 2V to 1.9V. So the current is quite close to that through N1.

Take the case without the cascode, if VDS1=2V, and VDS2=3V, the difference in current could be quite great.


hope it helps,
Aaron