Question

Which of the following statements is correct about the small-signal voltage gain of single-stage MOSFET amplifiers? Consider the statement: "Both common source and common gate amplifiers are inverting amplifiers."

Hint:

Remember the phase relationships for the three basic single-stage MOSFET amplifiers: \begin{itemize} \item \textbf{Common Source (CS):} Inverting (180° phase shift). \item \textbf{Common Gate (CG):} Non-inverting (0° phase shift). \item \textbf{Common Drain (CD) / Source Follower:} Non-inverting (0° phase shift). \end{itemize} Only the Common Source configuration inverts the signal.

Answer 1

Step 1: Understanding the Question:
The question asks to evaluate the correctness of the statement that both common source (CS) and common gate (CG) MOSFET amplifiers are inverting. An inverting amplifier produces an output signal that is 180 degrees out of phase with the input signal.
Step 2: Detailed Explanation:
1. Common Source (CS) Amplifier:

• Configuration: The input signal is applied to the gate, the output is taken from the drain, and the source is common (usually at AC ground).
• Operation: When the input voltage at the gate (\(V_{in}\)) increases, the gate-source voltage (\(V_{GS}\)) increases. This causes the drain current (\(I_D\)) to increase. The output voltage is given by \(V_{out} = V_{DD} - I_D R_D\). As \(I_D\) increases, the voltage drop across the drain resistor \(R_D\) increases, causing \(V_{out}\) to decrease.
• Conclusion: An increase in \(V_{in}\) leads to a decrease in \(V_{out}\). This signifies a 180-degree phase shift. Therefore, the **Common Source amplifier is an inverting amplifier**. Its voltage gain is given by \(A_v \approx -g_m R_D\), where the negative sign explicitly indicates inversion.

2. Common Gate (CG) Amplifier:

• Configuration: The input signal is applied to the source, the output is taken from the drain, and the gate is common (at AC ground).
• Operation: When the input voltage at the source (\(V_{in} = V_S\)) increases, the gate-source voltage \(V_{GS} = V_G - V_S = 0 - V_{in}\) becomes more negative (decreases). This causes the drain current (\(I_D\)) to decrease. The output voltage is given by \(V_{out} = V_{DD} - I_D R_D\). As \(I_D\) decreases, the voltage drop across \(R_D\) decreases, causing \(V_{out}\) to increase.
• Conclusion: An increase in \(V_{in}\) leads to an increase in \(V_{out}\). This signifies a 0-degree phase shift. Therefore, the **Common Gate amplifier is a non-inverting amplifier**. Its voltage gain is given by \(A_v \approx g_m R_D\), which is positive.

Step 3: Final Answer:
Since the Common Source amplifier is inverting but the Common Gate amplifier is non-inverting, the statement "Both common source and common gate amplifiers are inverting amplifiers" is incorrect.