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

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

3.32

• How does this circuit work?

3.33

3.37

3.38