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Given the following p-type common-gate circuit, I'd like to try and calculate the voltage gain \$A_v=V_o/V_i\$. Given the following information: \$V^+=5V\$, \$V^-=-5V\$,\$V_{TP}=-0.8V\$,\$\lambda=0\$, \$K_p=0.001mA/V^2\$

Circuit

I began by doing a DC analysis. This means that the capacitors are seen as open circuits to DC and all AC sources are switched off, leaving only the independent DC sources. The simplified circuit therefore looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The drain current can be given by the equation \$I_{DQ}=K_p(V_{GS} +V_{TP})^2\$

The gate resistor is tied to ground directly and therefore no current flows into the gate. Then the gate to source voltage can be written with KVL as:

\$-V^+ + R_sI_D + V_{SG}=0 \implies V_{SG} = V^+-R_sI_D\$

Substituting this back into the equation for the drain current:

\$I_{DQ}=K_p(V^+-R_sI_D +VTP)^2\$

Solving for \$I_D\$ gives two values for \$I_{DQ}\$. \$I_{DQ} = 0.0008A\$ or \$I_{DQ} = 0.001339A\$. Choosing \$I_{DQ} = 0.001339A\$, the transconductance value \$g_m\$ can be solved for:

\$g_m = 2\sqrt{K_pI_{DQ}}\$

which gives \$2.314mA/V\$

The voltage gain \$A_v\$ for a common-gate configuration is given as:

\$A_v = \frac{gmR_D}{1+g_mR_{si}}\$

In this case \$R_{si}\$ was not given so I assume it simply falls away in the equation. I calculate \$A_v = 4.617 \frac{V}{V}\$

\$A_v = \frac{0.002314(2000)}{1+0.002314} = 4.617 \frac{V}{V}\$

The answer is given as \$A_v = 2.41\$. What did I forget to include or where were my assumptions wrong?

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  • \$\begingroup\$ I did not check the math, but isn't the gain be equal to gm*(RD||RL)? \$\endgroup\$
    – G36
    Apr 14, 2020 at 9:31
  • \$\begingroup\$ \$R_L\$ is in parallel with \$R_D\$ in AC analysis. \$\endgroup\$
    – Andy aka
    Apr 14, 2020 at 9:54

1 Answer 1

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Also, are you sure that \$K_P\$ is \$0.001\textrm{mA/V}^2\$? Because something is not quite right here.

If for example, I use \$K_P = 1\textrm{mA}/V^2\$ instead. I've got \$I_{DQ} = 0.823\textrm{mA}\$ and

\$g_m=2\sqrt{K_pI_{DQ}} \approx 1.81\textrm{mS}\$

And the voltage gain: $$A_V = g_m \times R_D||R_L = 1.81\textrm{mS} \times \frac{4\textrm{k}\Omega}{3} = 2.41\: V/V$$

And the input impedance \$R_{in} = \frac{1}{g_m }|| R_S = 485\Omega \$

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