Which one of the following statements is true about the small signal voltage gain of a MOSFET based single stage amplifier?

MOSFET Amplifier Small Signal Voltage Gain Analysis

This question requires understanding the polarity of the small signal voltage gain for two common single-stage MOSFET amplifier configurations: the Common Source (CS) and the Common Gate (CG).

Common Source (CS) Amplifier Characteristics

In a CS amplifier, the input signal voltage ($v_{in}$) is applied between the gate and source ($v_{gs}=v_{in}$), and the output voltage ($v_{out}$) is taken at the drain.

- An increase in $v_{gs}$ leads to an increase in drain current ($i_d$).
- This increased $i_d$ flows through the drain resistor ($R_D$) and the MOSFET's output resistance ($r_o$).
- The output voltage ($v_{out}$) is typically related to the drain current by $v_{out}\approx -i_d\times (R_D||r_o)$ (assuming the source is AC grounded and $V_{DD}$ is the supply voltage).
- Therefore, an increase in $v_{in}$ (which increases $i_d$) causes a decrease in $v_{out}$.
- This indicates a 180-degree phase shift between the input and output signals.
- Conclusion: The Common Source amplifier exhibits an inverting voltage gain. The approximate gain is $A_v=\frac{v_{out}}{v_{in}}\approx -g_m(R_D||r_o)$, where $g_m$ is the transconductance.

Common Gate (CG) Amplifier Characteristics

In a CG amplifier, the input signal voltage ($v_{in}$) is applied to the source (with the gate typically held at a fixed DC voltage, acting as an AC ground), and the output is taken from the drain.

- The gate-source voltage is $v_{gs}=v_{gate}-v_{source}$. Since the gate is at AC ground ($v_{gate}=0$) and the input signal is applied to the source ($v_{source}=v_{in}$), we have $v_{gs}=0-v_{in}=-v_{in}$.
- The drain current change is $i_d=g_m{v}_{gs}=g_m(-v_{in})=-g_m{v}_{in}$.
- The output voltage change is $v_{out}=-i_d\times (R_D||r_o)$.
- Substituting $i_d$, we get $v_{out}=-(-g_m{v}_{in})\times (R_D||r_o)=g_m(R_D||r_o)v_{in}$.
- The voltage gain is $A_v=\frac{v_{out}}{v_{in}}=g_m(R_D||r_o)$.
- Since $g_m$, $R_D$, and $r_o$ are typically positive values, the voltage gain $A_v$ is positive.
- Conclusion: The Common Gate amplifier exhibits a non-inverting voltage gain.

Summary Statement

The Common Source amplifier is inverting, while the Common Gate amplifier is non-inverting.