**Q1. Fig. 1 shows the transfer characteristic curve of a JFET. Write the equation for drain current.**

Fig.1

**Solution.** Referring to the transfer characteristic curve in Fig. 1, we have,

**Q2. A JFET has the following parameters: IDSS = 32 mA ; VGS (off) = – 8V ; VGS = – 4.5 V. Find the value of drain current.**

**Solution :**

**Q3. A JFET has a drain current of 5 mA. If IDSS = 10 mA and VGS (off) = – 6 V, find the value of (i) VGS and (ii) VP.**

**Q4. For the JFET in Fig. 2, VGS (off) = – 4V and IDSS = 12 mA. Determine the minimum value of VDD required to put the device in the constant-current region of operation.**

Fig.2

**Q5. Determine the value of drain current for the circuit shown in Fig. 3.**

Fig.3

**Solution. **It is clear from Fig. 3 that VGS = – 2V. The drain current for the circuit is given by;

**Q6. When a reverse gate voltage of 15 V is applied to a JFET, the gate current is 10−3 μA. Find the resistance between gate and source.**

**Solution.**

**Q7. When VGS of JFET changes from –3.1 V to –3 V, the drain current changes from 1 mA to 1.3 mA. What is the value of transconductance?**

**Q8. A JFET has a value of gmo = 4000 μS. Determine the value of gm at VGS = – 3V. Given that VGS (off) = – 8V.**

**Q9. The datasheet of a JFET gives the following information: IDSS = 3 mA, VGS (off) = – 6V and gm (max) = 5000 μS. Determine the transconductance for VGS = – 4V and find drain current ID at this point.**

**Q10. A JFET in Fig. 4 has values of VGS (off) = – 8V and IDSS = 16 mA. Determine the values of VGS, ID and VDS for the circuit.**

Fig. 4

**Solution. **

Since there is no gate current, there will be no voltage drop across RG.

**Q11. Find VDS and VGS in Fig. 5, given that ID = 5 mA.**

Fig. 5

**Q12. The transfer characteristic of a JFET reveals that when VGS = – 5V, ID = 6.25 mA. Determine the value of RS required.**

**Solution.**

**Q13. Determine the value of RS required to self-bias a p-channel JFET with IDSS = 25 mA, VGS (off) = 15 V and VGS = 5V.**

**Solution.**

**Q14. Select resistor values in Fig. 6 to set up an approximate midpoint bias. The JFET parameters are : IDSS = 15 mA and VGS (off) = – 8V. The voltage VD should be 6V (one-half of VDD).**

**Q15. In a self-bias n-channel JFET, the operating point is to be set at ID = 1.5 mA and VDS =10 V. The JFET parameters are IDSS = 5 mA and VGS (off) = − 2 V. Find the values of RS and RD. Given that VDD = 20 V.**

**Solution. **Fig. 7 shows the circuit arrangement.

**Q16. In the JFET circuit shown in Fig. 8, find (i) VDS and (ii) VGS.**

Fig.8

**Q17. Determine ID and VGS for the JFET with voltage-divider bias in Fig. 9, given that VD = 7V.**

Fig.9

**Solution.**

**Q18. In an n-channel JFET biased by potential divider method, it is desired to set the operating point at ID = 2.5 ****mA and VDS = 8V. If VDD = 30 V, R1 = 1 MΩ and R2 = 500 kΩ, find the value of RS. The parameters of JFET are IDSS = 10 mA and VGS (off) = – 5 V.**

**Solution.** Fig. 10 shows the conditions of the problem.

Fig.10

**Q19. Draw the d.c. load line for the JFET amplifier shown in Fig. 11.**

Fig.11

**Solution.**

Fig. 12 shows the d.c. load line AB.

Fig.12

**Q20. The JFET in the amplifier of Fig. 13 has a transconductance gm = 1 mA/V. If the source resistance RS is very small as compared to RG, find the voltage gain of the amplifier.**

Fig.13

**Solution.**

Transconductance of JFET, gm= 1 mA/V

**Q21. The transconductance of a JFET used as a voltage amplifier is 3000 μmho and drain resistance is 10 kΩ. Calculate the voltage gain of the amplifier.**

**Solution.**

**Q22. What is the r.m.s. output voltage of the unloaded amplifier in Fig. 14? The IDSS = 8 mA, VGS (off) = – 10V and ID = 1.9 mA.**

Fig.14

**Solution.**

**Q23. If a 4.7 kΩ load resistor is a.c. coupled to the output of the amplifier in Fig. 15 , what is the resulting r.m.s. output voltage?**

Fig.15

**Solution. **

The value of gm remains the same. However, the value of total a.c. drain resistance RAC changes due to the connection of load RL (= 4.7 kΩ).

Total a.c. drain resistance, RAC = RD || RL

**Q24. In a JFET amplifier, the source resistance RS is unbypassed. Find the voltage gain of the amplifier. Given gm = 4 mS; RD = 1.5 kΩ and RS = 560Ω.**

Thus with unbypassed RS, the gain = 1.85 whereas with RS bypassed by a capacitor, the gain is 6. Therefore, voltage gain is reduced when RS is unbypassed.

**Q25. For the JFET amplifier circuit shown in Fig. 16, calculate the voltage gain with (i) RS bypassed by a capacitor (ii) RS unbypassed.**

Fig.16

**Q26. For a certain D-MOSFET, IDSS = 10 mA and VGS (off) = – 8V.**

**(i) Is this an n-channel or a p-channel ?**

**(ii) Calculate ID at VGS = – 3V.**

**(iii) Calculate ID at VGS = + 3V.**

**Solution. **

**Q27. A D-MOSFET has parameters of VGS (off) = – 6V and IDSS = 1 mA. How will you plot the transconductance curve for the device?**

**Solution. **

When VGS = 0 V, ID = IDSS = 1 mA and when VGS = VGS (off), ID = 0A. This locates two points viz IDSS and VGS (off) on the transconductance curve. We can locate more points of the curve by changing VGS values.

Thus we have a number of VGS – ID readings so that transconductance curve for the device can be readily plotted.

**Q28. Determine the drain-to-source voltage (VDS) in the circuit shown in Fig. 17 if VDD = +18V and RD = 620Ω. The MOSFET data sheet gives VGS (off) = – 8V and IDSS = 12 mA.**

Fig.17

**Q29. The D-MOSFET used in the amplifier of Fig. 18 has an IDSS = 12 mA and gm = 3.2 mS. Determine (i) d.c. drain-to-source voltage VDS and (ii) a.c. output voltage. Given vin = 500 mV.**

Fig.18

**Solution.**

**Q30. The data sheet for an E-MOSFET gives ID(on) = 500 mA at VGS = 10V and VGS (th) = 1V. Determine the drain ****current for VGS = 5V.**

**Q31. The data sheet for an E-MOSFET gives ID (on) = 3 mA at VGS = 10V and VGS (th)= 3V. Determine the resulting value of K for the device. How will you plot the transconductance curve for this MOSFET ?**

**Q31. Determine VGS and VDS for the EMOSFET circuit in Fig. 19. The data sheet for this particular MOSFET gives ID (on) = 500 mA at VGS = 10V and VGS (th) = 1V.**

Fig.19

**Solution.** Referring to the circuit shown in Fig. 19, we have,

The value of K can be determined from the following equation :

**Q32. Determine the values of ID and VDS for the circuit shown in Fig. 20. The data sheet for this particular MOSFET gives ID (on) = 10 mA when VGS = VDS.**

Fig.20