# NMOS Node Voltage Analysis

1. Mar 29, 2015

### Zondrina

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. Mar 29, 2015

### rude man

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