2
\$\begingroup\$

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?

\$\endgroup\$
  • 1
    \$\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
  • 2
    \$\begingroup\$ Equations 6.39 and 6.40 \$\endgroup\$ – Andy aka Jul 29 at 13:07
4
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Nice derivation. \$\endgroup\$ – Spehro Pefhany Aug 8 at 13:54
  • 1
    \$\begingroup\$ with thanks to who-ever edited the equations for legibility \$\endgroup\$ – analogsystemsrf Aug 10 at 16:56
0
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.