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Question

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

The correct answer is
Common source amplifier is inverting and common gate amplifier is non-inverting 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.

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Important Questions from Field Effect Transistors

  1. FET is like a switched on condition when it operates in ______ mode.

  2. Which of the following is the characteristic of Field-effect transistor?

  3. When VDS is the drain voltage and VDS(max) is the maximum drain voltage, the JFET will breakdown if:

  4. Which of the following is true about the transconductance of a MOSFET in saturation (I Dis the Drain Current)?

  5. Transconductance in an FET indicates how effectively the input voltage controls the

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