Electronic – MOSFET threshold voltage and body effect

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I'm trying to simulate a common drain amplifier on LTSpice which can be used to shift the input voltage either up or down by the threshold voltage (\$V_{IN}+V_{TH}\$ or \$V_{IN}-V_{TH}\$, depending on PMOS or NMOS configuration). The bulk terminals of PMOS must be connected to the most positive rail while NMOS bulk terminal must be connected to the most negative rail for the transistors to work in the reverse bias hence, allowing control of current flow and \$V_{TH}\$.

My question is, would it be possible to adjust the threshold voltage by varying the connection of the bulk terminals? Or is this a bad practice?

Best Answer

The threshold voltage can be increased if the source is not connected to the body terminal. The threshold voltage is

$$V_T = V_{T0} + \gamma\sqrt{2\phi + V_{SB}} - \gamma\sqrt{2\phi}$$

where \$V_{T0}\$ is the threshold voltage when the source-to-body \$V_{SB} = 0\$, and \$\gamma\$ and \$\phi\$ are device parameters. If the NMOS source is connected to ground and so is the body then \$V_{SB} = 0\$ and \$V_T\$ is minimized (it's a similar argument for the PMOS).

So, yes, it is possible to adjust the threshold voltage by not connecting the NMOS body to the negative supply and the PMOS body to the positive supply.

However, this is usually not done intentionally. You typically want to minimize \$V_T\$ -- for example, this would allow you to use lower supply voltages.

The body effect is particularly undesirable for a common drain amplifier because it lowers the voltage gain. Without the body effect the unloaded voltage gain of a common drain amplifier is

$$\frac{v_o}{v_i} = \frac{g_m}{g_m + \frac{1}{r_o||r_{oc}}} \approx 1$$

where the approximation assumes the resistances are large. However, with the body effect the unloaded voltage gain is reduced:

$$\frac{v_o}{v_i} = \frac{g_m}{g_m + g_{mb} + \frac{1}{r_o||r_{oc}}} \approx \frac{g_m}{g_m + g_{mb}} < 1$$