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Problem 14: Given a JFET circuit with specified parameters and a diagram (Fig. 7.86), determine ID, VS, and V_DS.

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4 parts
a

3 mA\text{3 mA}

b

2.04 V\text{2.04 V}

9.96 V\text{9.96 V}

c

1.671 V\text{1.671 V}

-0.369 V\text{-0.369 V}

d

-0.952 V\text{-0.952 V}

Step-by-step solution

8 steps

Here's the solution for , referring to the network in Fig. 7.86:

Given parameters from Fig. 7.86:

  • Supply voltage VDD=18 VV_{DD} = 18 \text{ V}
  • Drain resistor RD=2 kΩR_D = 2 \text{ k}\Omega
  • Source resistor RS=0.68 kΩR_S = 0.68 \text{ k}\Omega
  • Gate resistor R1=680 kΩR_1 = 680 \text{ k}\Omega
  • Gate resistor R2=110 kΩR_2 = 110 \text{ k}\Omega
  • Gate bias voltage source VGG=12 VV_{GG} = 12 \text{ V}
  • Drain voltage VD=12 VV_D = 12 \text{ V}
  • JFET parameter IDSS=8 mAI_{DSS} = 8 \text{ mA}

Part a· 2 steps

Determine IDI_D

  1. Calculate the voltage drop across RDR_D.

    The voltage at the drain (VDV_D) is given as 12 V12 \text{ V}. The supply voltage is 18 V18 \text{ V}.

    VRD=VDD−VDV_{R_D} = V_{DD} - V_D VRD=18 V−12 V=6 VV_{R_D} = 18 \text{ V} - 12 \text{ V} = 6 \text{ V}

  2. Calculate the drain current IDI_D.

    The drain current flows through RDR_D.

    ID=VRDRDI_D = \frac{V_{R_D}}{R_D} ID=6 V2 kΩ=6 V2×103Ω=3×10−3 AI_D = \frac{6 \text{ V}}{2 \text{ k}\Omega} = \frac{6 \text{ V}}{2 \times 10^3 \Omega} = 3 \times 10^{-3} \text{ A} ID=3 mAI_D = \boxed{\text{3 mA}}

Part b

2.04 V\text{2.04 V}

Determine VSV_S and VDSV_{DS}

  1. Calculate the source voltage VSV_S.

    The source voltage is the voltage drop across RSR_S due to IDI_D.

    VS=ID×RSV_S = I_D \times R_S VS=3 mA×0.68 kΩ=(3×10−3 A)×(0.68×103Ω)V_S = 3 \text{ mA} \times 0.68 \text{ k}\Omega = (3 \times 10^{-3} \text{ A}) \times (0.68 \times 10^3 \Omega) VS=2.04 VV_S = \boxed{\text{2.04 V}}

  2. Calculate the drain-source voltage VDSV_{DS}.

    VDS=VD−VSV_{DS} = V_D - V_S VDS=12 V−2.04 VV_{DS} = 12 \text{ V} - 2.04 \text{ V} VDS=9.96 VV_{DS} = \boxed{\text{9.96 V}}

Part c

1.671 V\text{1.671 V}

Determine VGV_G and VGSV_{GS}

  1. Calculate the gate voltage VGV_G.

    VGV_G is determined by the voltage divider formed by R1R_1 and R2R_2 connected to the 12 V12 \text{ V} source.

    VG=VGG×R2R1+R2V_G = V_{GG} \times \frac{R_2}{R_1 + R_2} VG=12 V×110 kΩ680 kΩ+110 kΩV_G = 12 \text{ V} \times \frac{110 \text{ k}\Omega}{680 \text{ k}\Omega + 110 \text{ k}\Omega} VG=12 V×110 kΩ790 kΩ=12 V×1179V_G = 12 \text{ V} \times \frac{110 \text{ k}\Omega}{790 \text{ k}\Omega} = 12 \text{ V} \times \frac{11}{79} VG≈1.671 VV_G \approx \boxed{\text{1.671 V}}

  2. Calculate the gate-source voltage VGSV_{GS}.

    VGS=VG−VSV_{GS} = V_G - V_S VGS=1.670886 V−2.04 VV_{GS} = 1.670886 \text{ V} - 2.04 \text{ V} VGS≈-0.369 VV_{GS} \approx \boxed{\text{-0.369 V}}

Part d

-0.952 V\text{-0.952 V}

Determine VPV_P

  1. Use the Shockley equation to find VPV_P.

    The Shockley equation relates IDI_D, IDSSI_{DSS}, VGSV_{GS}, and VPV_P:

    ID=IDSS(1−VGSVP)2I_D = I_{DSS} \left(1 - \frac{V_{GS}}{V_P}\right)^2

    Rearrange the equation to solve for VPV_P:

    IDIDSS=1−VGSVP\sqrt{\frac{I_D}{I_{DSS}}} = 1 - \frac{V_{GS}}{V_P} VGSVP=1−IDIDSS\frac{V_{GS}}{V_P} = 1 - \sqrt{\frac{I_D}{I_{DSS}}} VP=VGS1−IDIDSSV_P = \frac{V_{GS}}{1 - \sqrt{\frac{I_D}{I_{DSS}}}}

  2. Substitute the known values.

    VP=−0.3691139 V1−3 mA8 mAV_P = \frac{-0.3691139 \text{ V}}{1 - \sqrt{\frac{3 \text{ mA}}{8 \text{ mA}}}} VP=−0.3691139 V1−0.375V_P = \frac{-0.3691139 \text{ V}}{1 - \sqrt{0.375}} VP=−0.3691139 V1−0.6123724V_P = \frac{-0.3691139 \text{ V}}{1 - 0.6123724} VP=−0.3691139 V0.3876276V_P = \frac{-0.3691139 \text{ V}}{0.3876276} VP≈-0.952 VV_P \approx \boxed{\text{-0.952 V}}
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