# JFET I-V questions, SedraSmith

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#### peterpops

##### Junior Member level 3
Hi!

I have the following problem
"The figure a) shows the circuit symbol of a device known as the p-channel junction field-effect transistor (JFET). As indicated, the JFET has three terminals. When the gate terminal G is connected to the source terminal S, the two-terminal device shown in b) is obtained. Its i-v characteristic is given by

$i = {I}_{DSS}\left[2\frac{v}{{V}_{p}}-{\left(\frac{v}{{V}_{p}}\right)}^{2}\right]$ for v <= Vp
and
i = IDSS for v >= Vp

where IDSS and Vp are positive constants for the particular JFET. Now consider the circuit shown in c) and let Vp = 2 V and IDSS = 2 mA. For V+ = 10 V show that the JFET is operating in the constant-current mode and find the voltage across it. What is the minimum value of V+ for which this mode of operation is maintained? For V+ = 2V find the values of I and V."

My solution:
Question 1.
I do a Norton equivalent circuit with In = 10/2.5k = 4 mA and Rn = 2.5k // 2.5k = 1.25k. Since I know it is in constant current mode 2 mA is going through the JFET and the rest through the Rn resistor which gives me the voltage = 2 mA * 1.25k = 2.5 V, hence it is in the constant-current mode.

Question 2.
If I do it backwards with 2 V, I get current through Rn resistor is 2/1.25k = 1.6 mA + 2 mA (through JFET) = 3.6 mA * 2.5k = 9 V is the min V+ value for which the mode of operation is maintained.

Question 3.
Here I'm confused... Now the i-v characteristic is not linear anymore, can I still use the Norton theroem? One attempt is that a voltage divider would give me V = 1 V and if I substitute that into the current equation I get I = 1.5mA, is that the correct way?

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