Opps, I see my finger trouble.

The input was stated as Y = 0.06mS +j 2mS . This would be almost 17K in
parallel with 2.1pF NOT 1k7

The output was Y = 0.08mS +j 0.8mS. This should be 12.5k in par with
just under 1pF. Not 1250 Ohms

OK, I am getting some understanding I just need some gaps filled in now.

I see that when the recommended gate voltage is applied to G2 that Rs is
adjusted for the required current (say 10mA for lowest noise figure).

So, If I put 4V of G2, grounded G1 via and RFC and RFC on drain to Vcc
then placed a 100R in the source, I could adjust the 100R till 10mA
flowed to set the bias point.

I suppose then you would match the 12k drain impedance to a tank
circuit. Or place the tank circuit as the load.


Is the gain of the amp just related to Gm and the load impedance on the
drain.
So that if the DC bias was set for specific VDS and VGS then there would
be a specific transfer admittance that would produce a current through
the load. With RFC there would be max. gain or gain could be fixed by
specific load less than this?

Thanks

Regards

David




Leon wrote:
David wrote:
I am going around in circles attempting to understand the datsheet for
the BF998 Dual Gate FET.

From what I can gather, with Vcc = 8V and 4V on G1 and ID of 10mA, at
150 MHz
Input appears to be 1700 Ohms with 2.1pF par
Output appears to be 1250 Ohms with 1pF par

(I thought Fets had hundreds of K ohm input impedance - or is this only
at low Freq ?)

The typology I have is.

L/C tuned circuit to G1
100K+100K providing 4V to G2 (with 100n Bypass cap close to gate)
The source has 100R with 100n Par to ground.
The Drain has the Inductor of the tuned circuit to Vcc (and bypassing on
Vcc side as well as 47R series resistor for additional decoupling).

The tuning cap goes from Drain to ground.

I have about 250mV on the source (only 2.5mV bias current).

How do I obtain 10mA drain current ? Or are they only referring to AC
current ?

Would someone mind please giving me a bit of overview of these devices.

MOSFETs have a very high input resistance, but the actual input
impedance at high frequencies is a lot lower.

Leon