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## Homework Statement

Find the labelled node voltages. Assume ##k_n = 0.5 \frac{mA}{V^2}## and ##V_{tn} = 0.8V##. Neglect channel length modulation ##(\lambda = 0)##.

## Homework Equations

## The Attempt at a Solution

f) For this problem, I see ##V_D = V_G \Rightarrow V_G - V_D = 0V \Rightarrow V_{GD} = 0 V##.

Now ##V_{GD} = 0V < 0.8V = V_{tn}## which implies saturation mode.

The drain current is then given by:

$$I_D = \frac{1}{2} k_n V_{ov}^2$$

By using KVL, ##I_D = \frac{5V - V_D}{100k \Omega}##.

Now, ##V_{ov} = V_{GS} - V_{tn}##, but we know ##V_{GS} = V_G - V_S = V_G - 0V = V_G##. We also know that ##V_G = V_D##, so ##V_{GS} = V_G = V_D##. Hence we can write ##V_{ov} = V_D - V_{tn}##.

Subbing these into the drain current equation we obtain:

$$\frac{5V - V_D}{100k \Omega} = \frac{1}{2} k_n (V_D - V_{tn})^2$$

This yields a quadratic in ##V_D##, which has two solutions:

##V_D = 0.37V## and ##V_D = 1.2V##.

I am unsure how to exactly reason which of these is the proper solution.

h) For this problem, ##V_D = 5V## and ##V_G = 0V##. So ##V_{GD} = V_G - V_D = - 5V##.

Now ##V_{GD} < V_{tn}## which implies saturation operation.

The drain current is equal to the source current since ##I_G = 0##, so ##I_D = I_S = \frac{1}{2} k_n V_{ov}^2##.

Writing KVL we see: ##I_S = \frac{V_S + 5V}{100k}##.

Now, ##V_{ov} = V_{GS} - V_{tn}##, but we know ##V_{GS} = V_G - V_S = 0V - V_S = - V_S##. Hence we can write ##V_{ov} = - V_S - V_{tn}##.

Subbing these into the source current equation we obtain:

$$\frac{V_S + 5V}{100k} = \frac{1}{2} k_n (- V_S - V_{tn})^2$$

$$\frac{V_S + 5V}{100k} = \frac{1}{2} k_n (V_S + V_{tn})^2$$

This is another quadratic in ##V_S## that yields two solutions:

##V_S = -1.2V## or ##V_S = - 0.37V##.

Once again I would like to ensure my understanding of which solution is correct.

If someone could help me understand how to choose the right solution it would be very appreciated.

Thank you.