bias - FET Practical DC Analysis Problems

I just finished with theory understanding of FET transistor principles, etc.

Right after that I tried to make simple circuit Common-Source with FET and MOSFET the problems occurred one by another.

I could begin with JFET (n-channel) for start:

• Idss (when Vds = 15V) = 6mA


• Vgss(off) (when Vds = 15V & Id = 10nA) = from -0.5V to -8V



• Different admittances at Vgs = 0 and I don't know which to chose for calculating Id

All I wanted was to bias the FET, so that Vds = Vdd/2, so in the middle of the load line (for the start). Vdd would be 20V and Id would be 5mA > Vds would then be 10V, right?

I can't find appropriate Vgs and also don't know how to connect gate resistor (series or parallel) + I can't calculate its value since gate current cant be calculated and/or is not practically usable (neglected).

^{simulate this circuit – Schematic created using CircuitLab}

Transistor's datasheet: www.mouser.com/ds/2/149/bf244a-292510.pdf

Answer

If you have a one or a few JFET's in your disposal you can try to measure them.

You can use this setup:

AS you can see you can use a voltmeter to measure Vgs(off) and ammeter to measure Idss at once, all you need is to switch your multimeter between voltmeter/ammeter.

And you know Vgs(off) and Idss you can solve for Vgs:

$$V_{GS} = Vgs(off)* \left(1 - \sqrt{\frac{I_D}{I_{DSS}}} \right)$$

I have two JFET in my workbench First one is BF245A and I measure it and got

Vgs(off) = - 1.9V and Idss = 5mA

And BF245C:

Vgs(off)= -4.8V and Idss = 15mA

(the typical values given in datasheet are -5.6V/17mA).

So, If I want Id = 5mA I would use BF245C.

And Rd = 10V/5mA = 2.2kΩ

$$g_{m0} = \frac{2*I_{DSS}}{|V_{GS}(off)|} = \frac{30mA}{4.8V} = 6.25mS$$

Therefore the maximum gain passible for a given drain resistor is \$R_D = 2.2kΩ\$ is:

\$Av = g_{m0}R_D = 13.75 \; V/V\$ (without Rs)

For automatic bias you now need to select \$R_S\$ resistor

$$V_{GS} = Vgs(off)* \left(1 - \sqrt{\frac{I_D}{I_{DSS}}} \right) = -4.8V*\left(1 - \sqrt{\frac{5mA}{15mA}} \right) = -2V $$

$$R_S = \frac{2V}{5mA} = 400Ω = 390Ω$$

^{simulate this circuit – Schematic created using CircuitLab}

And now we can test the circuit on the workbench.