Test: MOSFET Amplifier with CS Configuration - Electronics and Communication Engineering (ECE) MCQ

We construct the rπ model and find that the voltage gain from the gate to the drain of the MOSFET is g_m * R₁. Since Channel Length Modulation is neglected, the voltage gain won’t be g_m * R₁ || R_O.

Ideally, the input impedance while looking into the gate of the MOSFET is infinite. This is because of the SiO₂ layer which behaves as an insulator.

To find the output impedance, we perform a small signal analysis with the help of our r_π model. After placing every voltage source (V_cc and V₁) to ground, we connect a simple voltage source at the drain node. Thereafter, the ratio of applied voltage to input current gives us the impedance looking into the drain which is ro. But this voltage will be applied in parallel to R₁. Hence the total output impedance is R₁ || r_o.

This is easily observable by performing a small signal analysis at the output of the C.S. stage. We need to turn off all voltage sources, Vcc mainly, and give a small input at the gate. Thereby, the voltage gain becomes g_m * R₁||r_o. The presence of early effect reduces the gain a bit and deviates M1 from its ideal characteristics.

By performing a simple small signal analysis, we find that the input resistance is simply R₃. The impedance is not infinite since we have a resistor between the gate and the input voltage.

The voltage gain for a degenerated CS stage is  Hence, after putting the values, we get 5/4 and hence the answer becomes 1.25. R_d is the total resistance connected to the drain of the M₁ while R_s is the total resistance connected to the source of the M₁.

If we perform a small signal analysis at node S, we will find that three resistors are connected from node S to ground. They are R₁, and the resistances appearing between source and drain of the MOSFET’s due to channel length modulation ie r_o1 and r_o2. Hence, the output resistance is R₁ || r_o1 || r_o2.

By performing a small signal analysis of the following circuit, we find that the output impedance of the circuit is simply (1 + (g_m * r_o)) * R_s + r_o. For doing this analysis, we have to short V₁ and V₂ to ground. Thereafter, we place a voltage source at the input node and measure current. The impedance measured will be the output impedance which is (1 + (g_m * r_o)) * R_s + r_o.

The transconductance is the ratio of a small change in the output current due to a small change in the input voltage. By differentiating the equation relating the current to the input voltage of a MOSFET with respect to the input voltage, we’ll get 

The input impedance of the C.S. stage, i.e. the impedance looking into the gate of M₁ is always infinite. Hence, in presence of early effect, the input impedance remains infinite.

If the early effect is neglected, r_o –> ∞ and hence, the output impedance is only R₁. This is inferred by performing the small signal analysis at the output node or the drain of the M₁.

The above shown stage is a degenerated CS stage. This stage is called so because the current source connected at the source of M₁ reduces the total gain of the CS stage. The current source provides a finite output impedance which is connected the source. Thereby, the overall gain decreases.

We calculate the output impedance by shorting the two voltage sources to ground. Thereafter, as we apply a simple step input at the output node, i.e. the collector node, we’ll find that the total impedance at connected to the drain of M1 is nothing but (1 + gm * (R₁ || R₂)) * R_i + (R₁ || R₂) where gm is the transconductance of M₁, R₁ || R₂ is the total resistance connected at the drain and R_i is the total resistance connected at the source. The output impedance would’ve been R₁ || R₂ if the current source was absent.

Since M₁ and M₂ receive the same bias voltage V₁, the current generated by both the MOSFET’s are same i.e.  Both the currents enter node S and hence the voltage at node S is C_ox (W/L) * (V₁ - V_th)².

The dependent current source has a variable resistance. If K doubles, the magnitude of current provided by the current source doubles, and thus, the total resistance connected to the source of M₁ reduces by 2. By using the expression of voltage gain, , we find that a decrease in Rs leads to an increase in the voltage gain.