- #1

- 199

- 1

## Homework Statement

## Homework Equations

## The Attempt at a Solution

My instructor gave us some sample problems he said might be on the exam, I tried working them out but I was having difficulty with 3 of them:

For the 1st problem, when there is positive value for voltage, I got the 1st diode as off (open circuit) and the 2nd diode as on (short circuit). This would mean Vout is always 0 for positive voltage. When the voltage becomes negative, the 1st diode becomes a short and the 2nd becomes open circuit. So 20||20=10 and Vo=(0.5)Vin using voltage division. Graph for Vout would be 0 when the 5V peak sine wave is positive curve, and a curve of half the amplitude when it is a negative curve (min=-2.5).

For the 2nd problem, I was used to working with 1K ohm resistors so the 500 ohm kinda confused me. But if the value of resistor doesn't matter, I find that diode starts conducting when Vi=2.5 and conducts fully when Vi=2.7 (+2V). This would mean graph is a linear line (Vo=Vi) until Vi=2.7 when it becomes a horizontal line. For the -3V, diode starts conducting when Vi=-2.5 and conducts fully when Vi=-2.3. Graph would look similar to the +2V one.

For the 3rd one, I had a zener diode for another problem so I am assuming there is positive value at the negative end. Also VZ=VZ0 because rz is very small. For +20V, I got current flowing through diodes B, C, D. the voltage is 20-2Vd-Vz=10.4V. For +5V, I got a negative voltage after the voltage drop across zener diode so I found no current would flow through the loop. For -5V, I got a positive voltage after the voltage gain in zener diode, so zero current again. For the last one (-20V), I found current flowing through doides A, C, E if I assume zener diodes can conduct voltage in both ways. Then the voltage is -20+Vz+2Vd=-10.4V.