What you say makes sense. Breadboard up a circuit and see what your idle
point winds up being. That's really the only way you will find out for
sure.

tim ab0wr

David wrote:



Tim,

Say we look at VG1s = 0.1V as per your example.

The graph for BF998 shows that if VG2s = 4V and VDs = 8V then ID approx
= 12.5mA

This would mean that unless I applied a negative voltage on the source I
would need to apply 0.1V forward bias to G1 and 4V to G2 ?
As Rs is creating a negative self bias voltage ?

If I set the bias point lower - say 5mA then VG1s is approx. -0.2V
according to the graph.

I can achieve this by using a resistor in the source of 0.2/5mA (40
Ohms) and then set VG1 = 0 (so that VG1s = -0.2V) and then 4.2V on G2
so that VG2s = 4V.

Is this correct ?

The transfer characteristic curve shows that for say 10mA. If VG2s = 4V
then gm = around 24mS and if VG2s is reduced to 0V the gm reduces to
about 7mS.

Thanks

regards

David



tim gorman wrote:
David wrote:


Tim,


Thanks for the info.

So, say I wanted to set the bias up at 10mA.

I find the Vd/Id curve on the datasheet.

If I have say a 8V power supply and decide to use a choke in the drain.
This gives me Vd of around 8V. The curve indicates that VGs1 should be
0V and VGs2 = 4V.

This means Rs would be 8/10mA = 800R.

Where I am now confused is that VGS voltages are the Gate to source
voltage. If the source voltage is 8V from the example above then to get
VGS2 of 4V then the bias on VG2 would need to be 12V ?
Also if VGs1 = 0V then the actual voltage on G1 should be 4V ?

Is this correct or am I misinterpreting something here ?

Thanks.

Regards

David

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



I'm sorry, I should have picked up on the fact that you are using a
dual-gate mosfet.

A dual-gate mosfet is a lot like 2 fet's in series. Gate2 is usually used
with an external bias to set the dynamic range of the device. The signal
is usually associated with Gate1. You can apply a fixed bias to Gate2 or
tie in something like an AGC signal to vary the device amplification.

For this type of device you probably would be better off looking at the
graph of the Transfer Characteristics. The graph will show the change in
Id for changes in Vgs1 with Vgs2 at a fixed value.

For your device I would probably run Vgs2 at 3v to 4v. Looking at the
transfer graph, you would want Vgs2 to be around 0.1v to get in the
middle of the linear curve. That would put your standing Id at about 9-11
mA.

This would make your source resistor 0.1v/10mA = 10R.

Remember that you'll want to breadboard the circuit and try this out
before actually including it in a production unit. Use a fixed voltage
divider to get the 4v for Vgs2 and a source resistor of 10R and see how
the circuit works. You can always change the source resistor to get what
you need.

tim ab0wr