Which parameter primarily determines the maximum possible drain current (IDSS) in a JFET’s transfer characteristic?
The physical properties of the channel and VGS = 0 V
IDSS is the drain current of a JFET with the gate shorted to the source (VGS = 0 V) and VDS beyond pinch-off. It is the maximum drain current on the transfer curve.
With VGS = 0 V the depletion regions are narrowest, so the current depends only on the channel itself — its doping, width, length and carrier mobility. Shockley's equation \(I_D = I_{DSS}\left(1 - \frac{V_{GS}}{V_P}\right)^2\) shows ID = IDSS at VGS = 0.
Raising VDS above pinch-off hardly changes the current, because the JFET is in its saturation (constant-current) region.
Threshold voltage is a MOSFET parameter; a JFET is characterised by its pinch-off voltage VP.
Resistance between source and ground (self-bias) sets the operating point and lowers ID below IDSS; it does not set IDSS itself.
Hence, the answer is The physical properties of the channel and VGS = 0 V.
Similar Questions
If a JFET is operating in the ohmic region, how does its behavior compare to its operation in the saturation region?
Important Questions from Types of FET
The transconductance g mof a JFET is equal to:
\(\frac{ I _{ DSS }}{ V _{ P }}\left(1-\frac{ V _{ GS }}{ V _{ P }}\right)\)
The CMOS inverter can be used as an amplifier when:
Which of the following is NOT a minority carrier semiconductor device ?
The metal gate that is insulated from the channel is also referred to as:
If the output of a CMOS inverter gets accidently shorted to ground
