Electronic – Cascode current mirror: current to voltage

Referencing the diagram from the linked article:

The drain current of either transistor is mostly determined by the respective gate-source voltage.

For the lower transistor, the gate-source voltage is just the input voltage

$$v_{GS1} = v_{IN}$$

For the upper transistor thought, the gate-source voltage is

$$v_{GS2} = 0 - v_{D1}$$

where, in this example \$v_{D1} < 0 \$.

Now, the drain voltage of the lower transistor \$v_{D1}\$ will, essentially, be whatever it needs to be such that the upper transistor transistor current matches the lower transistor current.

For example, if the input voltage rises, the drain current of the lower transistor will increase which will act to reduce \$v_{D1}\$ (make \$v_{D1}\$ more negative) which increases \$v_{GS2}\$ thus increasing the drain current of the upper transistor.

To figure out \$V_{GS}\$, all you need is the MOSFET current equation. You have all the parameters you need for that, and you did the calculation correctly:

$$I_D = \frac12 k' \frac WL (V_{GS} - V_t)^2$$

$$250 \mathrm{\mu A} = \frac12 \left(80 \mathrm{\frac{\mu A}{V^2}}\right)(3)(V_{GS} - 1 \mathrm V)^2$$

$$V_{GS} \approx 2.44 \mathrm V$$

Now, if you want, you can use KVL to figure out the other voltages in the left branch of the circuit:

^{simulate this circuit – Schematic created using CircuitLab}

$$5\mathrm{V} - V_{C} - V_{GS} - -5\mathrm{V} = 0$$

where \$V_C\$ is the voltage across the current source. You can plug in your \$V_{GS}\$ to get:

$$5\mathrm{V} - V_C - 2.44\mathrm{V} - -5\mathrm{V} = 0$$

$$V_C \approx 7.56\mathrm{V}$$

Normally you don't need to figure out the voltage across the current source for a homework problem. (In real life, you need to make sure you have enough voltage, especially for more complex circuits like a Wilson current mirror.) You might have to find the voltage across the output transistor, since that will let you take channel length modulation into account and determine the error of the mirror.