#'s 10 and 11 **Transcription for Educational Website:**---### FIGURE 8–68This figure depicts a circuit diagram featuring a Junction Field-Effect Transistor (JFET). It includes:- A JFET symbol connected in series with: - A current meter (A) for measuring current. - A resistor (\( R_D \)), labeled as 4.7 kΩ. - A voltage meter (V) for measuring voltage across the JFET.- A voltage source (\( V_{DD} \)) is connected across the circuit.---### JFET Biasing Problems10. **Parameters from a JFET Datasheet:** - \( V_{GS(\text{off})} = -8 \, \text{V} \) - \( I_{DSS} = 5 \, \text{mA} \) - Determine \( I_D \) values for \( V_{GS} \) ranging from 0 V to -8 V in 1 V steps. - Plot the transfer characteristic curve from these data.11. **JFET in Problem 10:** - Determine \( V_{GS} \) required to set up a drain current (\( I_D \)) of 2.25 mA.12. **For a particular JFET:** - \( g_{mo} = 3200 \, \mu\text{S} \) - What is \( g_m \) when \( V_{GS} = -4 \, \text{V} \), given \( V_{GS(\text{off})} = -8 \, \text{V} \)?13. **Forward Transconductance:** - JFET biased at \( V_{GS} = -2 \, \text{V} \) - \( V_{GS(\text{off})} = -7 \, \text{V} \) - \( g_{mo} = 2000 \, \mu\text{S} \) at \( V_{GS} = 0 \, \text{V} \) - Determine forward transfer conductance, \( g_{fs} \).14. **P-Channel JFET Datasheet:** - \( I_{GSS} = 5 \, \text{nA} \) at \( V_{

As per Bartleby policy we can answer only one question at a time, I am solving question no 10 for you. Please repost for other questions.10. It is given that: VGSoff=Vp=-8 VIDSS=5 mA=5×10-3 ANow, the values of ID for each value of VGS ranging from 0 V to -8 V in 1 V steps can be calculated as: ID=IDSS1-VGSVp21.Taking VGS= 0 V. Then,ID=5×10-3×1-0-82=5×10-3 A= 5 mA2.Taking VGS= -1 V. Then,ID=5×10-3×1+-182=3.828×10-3 A= 3.828 mA3.Taking VGS= -2 V. Then,ID=5×10-3×1+-182=3.828×10-3 A= 3.828 mA4. Taking VGS= -3 V. Then,ID=5×10-3×1+-382=1.953×10-3 A= 1.953 mA5. Taking VGS= -4 V. Then,ID=5×10-3×1+-482=1.25×10-3 A= 1.25 mA6. Taking VGS= -5 V. Then,ID=5×10-3×1+-582=0.703×10-3 A= 0.703 mA7.Taking VGS= -6 V. Then, ID=5×10-3×1+-682=0.3125×10-3 A= 0.3125 mA8. Taking VGS= -7 V. Then,ID=5×10-3×1+-782=0.078×10-3 A= 0.078 mA9. Taking VGS= -8 V. Then,ID=5×10-3×1+-882=0 A The transfer characteristics curve can be drawn as shown below: