#### peterpops

##### Junior Member level 3

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 <= V

_{p}

and

i = I

_{DSS}for v >= V

_{p}

where I

_{DSS}and V

_{p}are positive constants for the particular JFET. Now consider the circuit shown in c) and let V

_{p}= 2 V and I

_{DSS}= 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 I

_{n}= 10/2.5k = 4 mA and R

_{n}= 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 R

_{n}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 R

_{n}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?

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?