Re: Biasing of Dual Gate Fets

David wrote:



Hi,

Pretty fundamental I know but can someone please explain the steps for
setting up bias for a Dual Gate MOSFET.

I know I could place a pot on the gate and source for each circuit and
play with values but I would like a method that enables me to calculate
the values.

The main issue is how to determine values for Rs and Gate 2 Voltage.

I am using BF998 and want to have a "play" at 5V and 8V supply.

The formulae for Id is Id = Idss(1-Vgs/Vp) ^ 2
But Idss is stated as 2-18mA
Vp Gate 1 is given as a range from 1-2V
Vp gate 2 is given as range from 0.5 to 1.5V

If I apply say 4V to G1 and 0V to G2, how do I calculate the voltage at
the source to determine Vgs ?

Any help much appreciated.

Regards

David


One way of doing this is to get the datasheet for the FET you are using.
There should be a graph that shows the operating characteristic curves. The
x-axis will be Vds and the Y-axis will be the drain current Id. The
characteristic curves will be for various levels of Vgs. Pick an operating
point based on the type of amplifier you want. Let's suppose it will be
Class A. Assume the FET has a power supply voltage of 40v and an Idss of
10ma. Let's say that you pick a point in the middle of the operating curves
that gives an Id of 6ma and a Vds of 20v in order to get the maximum swing
out of the amplifer. Looking at the characteristic curves shows that this
will require a Vgs of about -1v. Now you have everything you need.

If Vgs needs to be -1v and Id is 6ma (assume Id and Is will be the same) you
need a resistor of Vd/Id (R = V/I) or about 166 ohms.

The gate resistor you see in FET amps is not really there for biasing but
more to set the input impedance of the amplifier. As long as the leakage
current from the gate to the source is small, Vgs is set by the bias
resistor in the source lead.

tim ab0wr


.

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