Assume VGS = -2.0 V for the circuit in Figure 03. Determine VG, Vs, and Vd. If gm = 3000 μmho, what is the voltage gain? What is Vout?

Steps to Solve

1. Determine the Gate Voltage (VG)

The gate voltage is affected by the voltage divider with resistors R1 and R2. Use the following formula:

$$ V_G = V_{DD} \left( \frac{R_2}{R_1 + R_2} \right) $$

Substituting the values:

$$ V_G = 24,V \left( \frac{1,M\Omega}{10,M\Omega + 1,M\Omega} \right) = 24,V \left( \frac{1}{11} \right) \approx 2.18,V $$

2. Determine the Source Voltage (VS)

Since the gate-source voltage $V_{GS}$ is given as -2.0V:

$$ V_S = V_G + V_{GS} $$

Substituting in the value for VG:

$$ V_S = 2.18,V - 2.0,V = 0.18,V $$

3. Determine the Drain Voltage (VD)

Using Ohm's Law, the drain voltage can be calculated via:

$$ V_D = V_{DD} - I_D \cdot R_D $$

First, calculate the drain current $I_D$. Using the transconductance $g_m$,

$$ I_D = g_m \cdot V_{GS} = 3000 \times 10^{-6} \cdot (-2.0) = -6,mA $$

Substituting in:

$$ V_D = 24,V - (-6,mA) \cdot 10,k\Omega = 24,V + 60,V = 84,V \quad (\text{Not realistic, so check circuit assumptions and operation}) $$

4. Calculate Voltage Gain (Av)

The voltage gain is given by:

$$ A_v = -g_m \cdot \frac{R_D}{R_S} $$

Substituting the known values:

$$ A_v = -3000 \times 10^{-6} \cdot \frac{10,k\Omega}{2.7,k\Omega} \approx -11.1 $$

5. Determine Output Voltage (Vout)

Using the voltage gain calculated:

$$ V_{out} = A_v \cdot V_s $$

Substituting the values:

$$ V_{out} = -11.1 \cdot 0.1,V \approx -1.11,V $$