Chapter 3 problems

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3.9

Part A

  • Using KVL: 4=1.5 I_A+V_A
  • Thus the two points for the load line are V_A=4 and I_A=2.66
  • Overlay the above two points with the diode characteristics to find the answer.

Part B

  • Thevenin Equivalent: V_{oc}=200*.005=1V and R_{th}=200+200=400
  • Using KVL: -1+400*I_B+V_X=0, thus V=1 and I=0.0025 for the load line.
  • I_B can be read from the load line graph. We can then use this information to find the voltage over V_B.

Part C

  • KVL & KCL: I_C-V_C/500-V_C/500=0 and -0.5+V_C+V_X=0. Note that I_C is the same thing as I_X
  • Thus V_C=250I_C and V_X=1/2-250I_C. Using the load line to find the I & V of device X. Then plug into the second equation to find V_C

3.17

Part A

  • Guessing D1 is on, D2 and D3 are off. Looking at the voltage drops, this is very unlikely.
  • Guessing D1 off, D2 on, D3 off. I=7.5 mA and V=7.5.
  • Checking for positive current through presumed on diodes and negative voltage across the presumed off diodes.
  • D1 and D2 fail. D3 passes.
  • Guessing D1 and D2 on, D3 off.
  • I=0 and V=7.5. D1, D2, D3 pass.


Part B

  • V_{in}=0, V=5: D1, D2, D3, D4 on.
  • V_{in}=2, V=5: D1, D2, D3, D4 on.
  • V_{in}=6, V=5: D2, D3 on. D1, D4 off.
  • V_{in}=10, V=5: D2, D3 on. D1, D4 off.
  • V=-5 for -10 \le V_{in} \le -5
  • V= for -5 \le V_{in} \le 5
  • V=5 for 5 \le V_{in} \le 10

3.32

  • How does this circuit work?

3.33

P3.33.PNG

3.37

3.38