# Help with JFET calculations

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#### I14R10

##### Full Member level 2 Hello

I'm new here but I have some experience in electronics. I'm currently building JFET amplifier for my antenna, for 14MHz radio. And I need some help with JFET calculations. I'm using BF244A JFET https://www.fairchildsemi.com/datasheets/BF/BF244C.pdf

And I made a few calculations and it seems that if I choose V(drain) close to V(power supply), and if I choose really small current, I would get higher amplification. For example:

Idss=5mA
Id=1mA
Vd=8V
Vgs=-0.5V
gm=4*10^(-3)

and with that I get amplification factor 32

But if I take for example
Id=4mA (higher than in 1st)

Than amplification factor is only 8

So what I want to ask, does my calculations make sense? I have run it in electronics workbench and I get similar results.

#### FvM

##### Super Moderator
Staff member does my calculations make sense?
Don't think so. You need to select a drain resistor, and you also must consider the connected load impedance. A gain of 32 would be only achieved with a drain resistor of 8k and no external load. But you need a respective higher supply voltage to get sufficient Vds. At 14 MHz, FET capacitances can't be ignored.

• I14R10

### I14R10

points: 2

#### erikl

##### Super Moderator
Staff member In first approximation, the gain (amplification factor) = gm*RL. Even if the transconductance increases with drain current (s. Transconductance vs Drain Current on p. 3 of the dataSheet), the output resistor RL has to decrease more, if Vsupply=const. Hence in such case the gain will decrease with increasing drain current.

• I14R10

### I14R10

points: 2

#### I14R10

##### Full Member level 2 I didn't write that, but I calculated with Rd=8k ohm and Rs= 470 ohm, I think, I can't find the paper now.

Than the answer is yes, with smaller drain current, the gain is larger.

Just to clarify this - I want to use high gain by using very small drain current, since the antenna produces very little voltage. Then the second FET would be biased to half the supply voltage and half the drain current. And then the BJT transistors would take over.

Here is my latest calculation

Vgs=-0.5V
Idss=5mA
Vdd=9V
Id=2.5mA
Rs=200 ohm
Rd=3200 ohm (with Vd=8V)
gm=4*10^(-3)

amplification is 12.8

This should make sense now

Last edited:

#### FvM

##### Super Moderator
Staff member With Id = 2.5 mA and Vdd = 9V, you get Vds of only 0.5 V, which is hardly suitable.

As another point, if you are designing real FET amplifiers, you should calculate it for the expectable Vgs,off and Idss range, not just assume a typical datasheet value.

#### LvW Did you bypass the source resistor Rs with a capacitor? If not - you have forgotten the feedback effect in your calculation.

#### I14R10

##### Full Member level 2 With Id = 2.5 mA and Vdd = 9V, you get Vds of only 0.5 V, which is hardly suitable.

As another point, if you are designing real FET amplifiers, you should calculate it for the expectable Vgs,off and Idss range, not just assume a typical datasheet value.
Why is Vds=0.5V not good?

Did you bypass the source resistor Rs with a capacitor? If not - you have forgotten the feedback effect in your calculation.
Yes I did.

#### FvM

##### Super Moderator
Staff member • I14R10

### I14R10

points: 2

#### I14R10

##### Full Member level 2 aah yes, now I understand. It needs to be in the saturation region

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