OK, I think I got it.

Let's say I want Vdd = 8V. Say I have an RF chock in the Drain then Vds
is almost 8V.

Say I want 5mA bias current.

I look at the Vds vs Id and find that VG1-s = -0.2V for this condition.
(So actual VDs is 7.8V)

Next I use a source resistor of VG1-s/ Id = 40 Ohms
As VG1 = VG1-s + VG1 - IDxRS the Gate voltage applied is 0V.

The graph on the datasheet mentions that this is the output when VG2-s
is 4V.

So now I calculate VG2 = VG2-s + ID x RS = 4.2V

I suppose that I can make Rs smaller to the point where VG1-s is 0V on
the graph.(Occurs at 10mA for BF998).

If I want higher Q point I then add positive bias to V1. I also
understand I can get around 10dB attenuation by lowering VG2-s from 4.2V
towards 0V but need to make it negative if I want to switch the signal
off altogether (If I was to use the FET as say a ASK or OOK modulator).

Thanks



tim gorman wrote:
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