# Chapter 3 problems

## Contents

### 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?