Does anyone know the formula to calculate the Vgs of a JFET used as a variable resistor? Say, I have a Vds of 24v and I require a current of 20mA (just arbitrary values) I would need a 1.2k resistance, how would I calculate the Vgs to get that resistance value?

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    \$\begingroup\$ It does not work like that, the JFET behaves similar to a current source (not a resistor). The drain current of the JFET (assuming \$V_{DS}\$ is high enough) depends on the model JFET and its \$V_{GS}\$. Often the drain current is only specified when \$V_{GS}\$ = 0 V. \$\endgroup\$ – Bimpelrekkie Jul 29 at 13:04
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    \$\begingroup\$ Equations 6.39 and 6.40 \$\endgroup\$ – Andy aka Jul 29 at 13:07

Following up on Andy's link, we differentiate the equation # 6.39, with respect to \$V_{DS}\$.

We want the resistance. That requires incremental \$V_\mathrm{Drain}/I_\mathrm{Drain}\$.

We end up with a constant portion: $$ \frac{\mathrm{d}I_D}{\mathrm{d} V_{DS}} = 2 K (V_{GS} - V_P) $$ and a variable portion: $$ \begin{split} \frac{\mathrm{d}I_D}{\mathrm{d} V_{DS}} &= K 2 (2 V_{DS}) \\ &= 4 K V_{DS} \end{split} $$ The constant portion is the transconductance, or the inverse of channel resistance.

The variable portion is the DISTORTION.

  • \$\begingroup\$ Nice derivation. \$\endgroup\$ – Spehro Pefhany Aug 8 at 13:54
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    \$\begingroup\$ with thanks to who-ever edited the equations for legibility \$\endgroup\$ – analogsystemsrf Aug 10 at 16:56

Yes - the JFET can be used as a controllable "quasi-linear" resistor (between the nodes D and S) within the following limitations:

1.) The drain-source voltage VDS must be small enough. In practice: VDS<(VGS-VP) and for "good" linearity: VDS<<(VGS-VP),

2.) The source node must be at ground potential or at "virtual ground" (inverting opamp input). Therefore, the gate voltage VG is the control voltage

3.) The value of RDS=f(VGS) is as given in analogsystemsrfs answer.


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