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