AMP I.12 – The Common Gate Amplifier


The last of the JFET classical amplifier configuration is the common gate amplifier, analogous to the common base BJT amplifier.  As usual, we will derive the voltage gain, current gain, input resistance, and output resistance of this configuration.

The Common Gate Amplifier

For the common gate amplifier, the input signal is applied at the source, while the output signal is sampled at the drain.  As you can see, the gate itself is grounded: the gate is “common”.

Voltage Gain

To derive the voltage gain, let’s draw the small signal equivalent circuit:

As usual, let’s express v_{out} and v_{in} to find the voltage gain A_v:

  • v_{out} = V_{DD} - R_D I_D = V_{DD} - R_D g_m v_{GS}
  • v_{GS} = v_G - v_S = -v_S = -v_{IN}, with v_G =0 because the gate is grounded

Substituting v_{GS} in the first equation gives:

    \[ v_{out} = V_{CC} + R_D g_m v_{in} \]

We can now calculate our voltage gain:

    \[ A_v = \frac{\Delta v_{out}}{\Delta v_{in}} = R_D g_m \]

If we add a load, this gain becomes A_v = (R_D // R_L) g_m

Current Gain

To find the current gain, we need to add a load, and determine the input and output current.

Let’s express i_{in}.  A node equation at the source terminal gives us:

    \[ i_{in} = g_m v_{GS} - \frac{v_{in}}{R_S} \]

Since v_{GS} = v_G - v_S = -v_S = -v_{in} (gate is grounded), we have:

    \[ i_{in} = -g_m v_{in} - \frac{v_{in}}{R_S} = -v_{in}(g_m + \frac{1}{R_S}) \]

Let’s express i_{out} next, using ohm’s law.  We already know the output voltage v_{out} since it was calculated in the previous section.

    \[ i_{out} =  \frac{v_{out}}{R_L} = \frac{g_m R_D // R_L v_{in}}{R_L} \]

We can now derive the current gain A_i:

    \[ A_i = \frac{i_{out}}{i_{in}} = \frac{g_m R_D // R_L}{R_L} \times \frac{1}{g_m + \frac{1}{R_S}} \]

    \[ A_i = \frac{g_m R_D // R_L}{R_M} \times \frac{R_S}{g_m R_S +1} \]

By approximating g_m R_S +1 \approx g_m R_S, we obtain:

    \[ A_i  = \frac{g_m R_D // R_L \ times R_S}{R_L g_m R_S} = \frac{R_D // R_L}{R_L} \]

As we can see, A_i < 1Our current gain is less than unity.  For R_L relatively low compared to R_D, we have A_i \approx 1.  This is why you will often see people say the common gate amplifier has unity current gain.

Input Resistance

To find the input resistance, we apply a test signal at the input and calculate R_{in} = \frac{v_{test}}{i_{test}}:

  • g_m v_{GS} = -g_m v_S = -g_m v_{test}
  • i_{test} = -g_m v_{test} - \frac{v_{test}}{R_S} = -v_{test}(g_m + \frac{1}{R_S}) , node equation at the source terminal

This gives us: R_{in} = \frac{1}{g_m + \frac{1}{R_S}} = \frac{R_S}{R_S g_m +1} \approx \frac{1}{g_m}.

The input resistance of the common gate amplifier is low.

Output resistance

To find our output resistance, we remove the load and ground the input.  Then, we apply a test signal at the output and calculate R_{out} =\frac{v_{test}}{i_{test}}.

Using a node current equation at the drain terminal, we have:

    \[ i_{test} = -\frac{v_{test}}{R_D} - g_m v_{GS} \]

However, since the gate and source are grounded (source is grounded because we grounded the input), we have v_{GS} = 0.  This gives us:

    \[ i_{test} = -\frac{v_{test}}{R_D} \]

And thus, our output resistance is R_D.  This is considered high.


To recap, the common gate amplifier has:

  • A_v = R_D g_m, a high voltage gain
  • A_i \approx 1, unity current gain
  • R_{in} \approx \frac{1}{g_m}, a low input resistance
  • R_{out} = R_D, a high output resistance

As expected, the common gate amplifier has very similar properties to the common base amplifier.

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