How to Determine Correct Node Voltages in NMOS Circuit Analysis?

[STEMucator]

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.

 

[rude man]

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The condition for saturation includes Vgs > Vth. Both your answers of Vgs = 0.37V do not meet that requirement so are obviously invalid.