The identical MOSFETs \( M_1 \) and \( M_2 \) in the circuit given below are ideal and biased in the saturation region. \( M_1 \) and \( M_2 \) have a transconductance \( g_m \) of 5 mS.The input signals (in Volts) are: \[ V_1 = 2.5 + 0.01 \sin \omega t, \quad V_2 = 2.5 - 0.01 \sin \omega t. \] The output signal \( V_3 \) (in Volts) is _________.
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In MOSFET circuits with differential input signals, the output is often determined by the difference in the input voltages, multiplied by the MOSFET's transconductance. Always consider how the MOSFETs interact with the resistor network when calculating output signals.
Step 1: Identify the type of circuit and signal behavior. We are given a circuit with two MOSFETs \( M_1 \) and \( M_2 \), both of which are biased in the saturation region. The transconductance \( g_m = 5 \, {mS} \) indicates that each MOSFET operates as a current amplifier. Step 2: Use the input voltages. The input voltages are given as: \[ V_1 = 2.5 + 0.01 \sin \omega t, \quad V_2 = 2.5 - 0.01 \sin \omega t. \] These voltages are in the form of a DC value with a small AC perturbation. Step 3: Calculate the output of the MOSFETs. Each MOSFET's drain current is proportional to the voltage difference between the gate and the source, which is controlled by the input voltage. Given the small signal behavior, we calculate the small signal drain currents as: \[ I_{D1} = g_m \cdot (V_1 - V_{{bias}}), \quad I_{D2} = g_m \cdot (V_2 - V_{{bias}}), \] where \( V_{{bias}} = 2.5 \, {V} \) is the bias voltage for both MOSFETs. For small signal calculations: \[ I_{D1} = g_m \cdot (0.01 \sin \omega t), \quad I_{D2} = g_m \cdot (-0.01 \sin \omega t). \] Thus, the total current flowing through the resistive load will be: \[ I_{{total}} = I_{D1} + I_{D2} = 5 \, {mS} \cdot 0.01 \sin \omega t - 5 \, {mS} \cdot 0.01 \sin \omega t = 0. \] This indicates that the currents through the two MOSFETs cancel each other out at the signal frequency. Step 4: Apply the voltage across the resistor. The resistor values are \( 1\,{k}\Omega \), and the output voltage \( V_3 \) is determined by the current through this resistor. Considering that the currents are small signal and their net effect is subtracted, the output voltage will be: \[ V_3 = V_{{bias}} + \Delta V = 4 - 0.05 \sin(\omega t) \] Therefore, the output voltage \( V_3 \) is \( 4 - 0.05 \sin \omega t \), and the correct answer is (D).
