Wilson

The problem

Build a Wilson mirror from three NMOS. A 50 µA Current source from VDD feeds the input node: the drain of the first transistor M1 and the gate of the output transistor M3. M3’s source goes to a diode-connected transistor M2 (gate and drain together, source on ground), and M2’s gate also drives M1’s gate. M3’s drain is the output “out”.

If the output current rises, M2’s Vgs rises, M1 pulls more current from the input node, M3’s gate drops, and the current falls back: negative feedback. That multiplies the output resistance to about gm·ro/2, similar to a cascode, without any extra bias voltage. The cost is a larger minimum output voltage.

The test bench

• Supply: VDD = 1.8 V.
• Your reference is the Current source you place: set it to 50 µA.
• The test bench holds the net named “out” at 1.2 V with a voltage source, sweeps it from 0.1 V to 1.7 V, and measures the current your mirror exchanges with it.

Numbers to hit

• ◆Output current at 1.2 V between 46 µA and 54 µA
• ◆Output resistance (1.0 V to 1.6 V) at least 20 MΩ
• ◆Current within 2% above at most 1.1 V
• ◆Every transistor in saturation