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I have a question about an exercise from my electronic circuits class, number a): Find an expression for the large signal V_out at DC.

Since it is DC, we can ignore the capacitor, it will act as an open circuit. We are dealing with MOSFETs, so now current will flow into the base. That means I0 will flow up into from S to D. Now I am confused. If there was no battery, it would just flow to the right, divide 1:2 through M2 and M3 (since they are identical"), and recombine. Then, I_L would flow into the load, and I_0 - I_L would flow into R_0, which would give V_out = (I_0 - IL) * R0. Unfortunately, there is the battery V_bat. Some current will also flow in the battery right? So my way of solving it seems to be incorrect, but how else to I tackle the problem?

Thank you very much! :)

Exercise

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2 Answers 2

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If there was no battery, it would just flow to the right,

In this kind of circuit, the current source is not really meant to be an ideal current source. If there was no battery, the current source would be turned off too.

So my way of solving it seems to be incorrect, but how else to I tackle the problem?

The circuit is a current mirror. If you research, you will find numerous explanations of how it works and how to calculate the currents through M2 and M3.

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The current I0 can only flow through M1, the gates of M2 and M3 ideally have an infinite input resistance.

Since M2 and M3 have the same Vgs as V1 and the same dimensions as well, each of them will source a current I0, which results in a total current of 2*I0.

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