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For BJT, the term "gm" refers to base-emitter conductance. But when we are talking about FETs and MOSFETs, where is this "gm"? Between gate-source junction for JFET? Drain to source conductance for MOSFET?

And what "gmb" means when talking about MOSFETs?

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2 Answers 2

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In general \$g_m\$ in simple term is a "gain" for any transconductance amplifier. And because transconductance amplifier is nothing more then a voltage controlled current source (VCCS) the gain expression is \$g_m = \frac{I_{out}}{V_{in}}\$.

For example if is \$g_m = 1\:Siemens\$ any change in the input voltage by \$1V\$ will change the output current by \$1A\$ (1 Ampere per Volt).

For BJT the transistor \$I_C\$ current is a controlled via input\$V_{BE}\$ voltage.

So, you plot \$I_C\$ vs \$V_{BE}\$

enter image description here

The \$g_m\$ is the slope of this curve

In MOSFET & JFET we have the same situation.

The output current \$I_D\$ is controlled via \$V_{GS}\$ voltage.

enter image description here

Hence again the slope of a \$I_D = f(V_{GS})\$ is a \$g_m =\frac{dI_D}{dV_{GS}} \$

The MOSFET in general is a 4 terminal device. The Gathe, Soure, Drain and the Body. And we can control the \$ I_D \$ via the gate terminal or via the Body. And this is why you have \$ g_{mb} \$

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  • \$\begingroup\$ Nice. From the first picture; the small vgs represents input AC (actually triangular) and Vgs represents DC biased quiescent point, right? \$\endgroup\$
    – MucaGinger
    May 1, 2017 at 19:35
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    \$\begingroup\$ Yep, you're right. \$\endgroup\$
    – G36
    May 2, 2017 at 5:14
  • \$\begingroup\$ So, input AC voltage vgs changes DC biased Vgs just a little bit (if we are looking onto the slope)? Even if vgs is ten times higher/bigger than Vgs, Vgs changes just a little bit with input signal? \$\endgroup\$
    – MucaGinger
    May 2, 2017 at 7:46
  • \$\begingroup\$ If Vin is lower then Vt the MOSFET is OFF and if Vin>> Vgs at DC. The MOSFET is in saturation. Hence $$V_{INmin} = V_T$$ and $$V_{INmax}= V_T+ \frac{-1+\sqrt{1 + 2KR_DV_{DD}}}{K*R_D}$$ In real life the range is even smaller. \$\endgroup\$
    – G36
    May 2, 2017 at 18:20
  • \$\begingroup\$ But if DC Vgs is biased above Vt, Vin does not need to be necessary above Vt. \$\endgroup\$
    – MucaGinger
    May 2, 2017 at 18:33
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In the bjt ic=gm*vbe refers to base-emitter conductance.

Similarly, in field effect transistors, and MOSFETs in particular, refers to gate-source conductance.Transconductance is the change in the drain current when we have small change in the gate/source voltage.id = gm*vgs

gmb: refers to transconductance which appears when we have the body effect.This occurs when VBS<>0(Voltage in the Bulk and Voltage in the source).

Notice:The terminal of bulk should usually be connected to the lowest potential of the circuit(ground or -VSS).This gmb provides a second current in the circuit with value ib=gmb*vbs

enter image description here

If body effect appears in your circuit your small signal model is the above

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  • \$\begingroup\$ But if bulk is connected to source, body effect cannot occur, right? Because if bulk and source are connected together there is no voltage difference, so no Vbs, right? \$\endgroup\$
    – MucaGinger
    May 1, 2017 at 19:16
  • \$\begingroup\$ All you refer is right \$\endgroup\$
    – elecV1
    May 2, 2017 at 9:17
  • \$\begingroup\$ But bulk is always connected to source, right? Or not? \$\endgroup\$
    – MucaGinger
    May 2, 2017 at 11:05
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    \$\begingroup\$ The terminal of bulk should usually be connected to the lowest potential of the circuit(ground or -VSS).In the most CMOS process bulk shouldn't connect in the source \$\endgroup\$
    – elecV1
    May 2, 2017 at 14:57

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