JFET common source amplifier

[connect0715207]

voltage divider JFET common source amplifier with output voltage biased at the half of supply voltage.
what does the above sentence mean???

 

[schmitt trigger]

The sentence is incomplete, but I believe that the gate's bias must be set with a voltage divider such that at idling the drain voltage (of a JFET connected as common source amplifier, thus its output is the drain) is equal to half the total supply voltage.

 

[Audioguru]

But a Jfet is a depletion FET, its gate needs to be below its source voltage so a voltage divider is needed at the source, not at the gate. A single source resistor to ground will bias it and the gate can be at 0VDC. The source resistor can have a parallel bypass capacitor to increase the AC voltage gain.

 

[connect0715207]

want to design voltage divider JFET common source amplifier using load line analysis and the output voltage should be biased at the half of supply voltage.
Given:- Power supply = 15V, Power consumption < 5mw , voltage across drain and source = 7.5V
JFET device specification:- Idss = 12mA, Vgs(off) = -3V

 

[schmitt trigger]

You are correct, just like old fashioned vacuum tubes! Gate (grid) at 0v, source (cathode) self biases with a resistor.

 

[crutschow]

You are correct, just like old fashioned vacuum tubes! Gate (grid) at 0v, source (cathode) self biases with a resistor.

But the old grid-leak bias scheme won't work with FETs. ;-)

 

[FvM]

I don't see why a voltage divider would be used, a (possibly capacitively bypassed) source resistor is however necessary.

 

[Audioguru]

The Jfet and its source resistor make a voltage divider.