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enter image description here

Vtn = 0.4 V, W = 1um, L1=0.25um, L2=0.5um.

For a question I got the circuit above, and the solution said that M1 is in triode and M2 is active. Also, V1 is unknown so I can't figure out how they determined this. Any help?

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  • \$\begingroup\$ Wherever you read that, throw it away and get a better book. "Active" isn't even a standard term for an operating mode of a MOSFET. \$\endgroup\$ – The Photon Apr 14 '17 at 17:44
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When facing these kind of exercises your should first of all write down the typical MOS operating region conditions. A good, simple (and very approximated) start could be the following:

Is M2 ON?

3V - V1 > Vth

Assuming it is conducting, then V1 should have a maximum value of 2.6V

Is M2 working in saturation?

3V - V1 > 3V - V1 - Vth 3V > 3V - Vth

Yes, M2 is working in SATURATION

Is M1 ON?

3V > Vth

Yes, M1 is ON

Is M1 working in saturation?

V1 > 3V - Vth

V1 > 2.6 V

In the "worst case" (V1 = 2.6V) M1 would be ALMOST in saturation. But here the key is the Length of M1 and M2. As you can observe, L1 is smaller than L2. Lets now suppose both M1 and M2 are working in saturation and obtain the value of V1 using the drain-source current equation without taking into account the body effect.

Ids1 = K'/2 * (2*W/L2) * (3 - 0.4)^2

Ids2 = K'/2 * (W/L2) * (3 - V1 - 0.4)^2

Ids1 = Ids2

Solving this equation we obtain V1 = 6.27 V and -1V, which are not possible values. Thus both transistors cannot be working in saturation, and we can conclude saying that:

M2 is working in saturation and M1 is working in ohmic region

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The top FET should be in region where the IV characteristic provides straight and nearly flat lines. The bottom FET will be in triode (acting as a resistor).

This circuit will, for processes where longer channels have lower Vthreshold, achieve very stiff current sources ---- at least from what I recall.

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For M2 to be on: Gate must be > Source Therefore: Vin must be > V1

For M1 to be in saturation: Drain must be > Gate Therefore: V1 must be > Vin

The two conditions are opposite to each other. If both transistors are on, this means that Vin > V1, therefore Gate(1) > Drain (1) ==> M1 must be in triode region.

PS: I took Vth out of the picture for simplicity. By the way, if Vth1 is NOT equal to Vth2, then both transistors can be saturation!

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