Chapter 3 problems

3.9

Part A

• Using KVL: $4=1.5I_A+V_A$
• 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

• Thevenin Equivalent: $R_{th}=500$ and $V_{oc}=$

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. $I=7.5mA$ 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 $-10\le V_{in}\le -5$
• $V=-5$ 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

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