Download ECE 250 Homework #5: Circuit Analysis with Zeners and Diodes - Prof. Marc E. Herniter and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
ECE 250 Homework # 5
Due 10/7/
Problem 1: a) Plot Vo(t) and Vin(t) for the circuit below by hand. Do not assume ideal diodes. Indicate numerical values for all breakpoints. Let Vin be a ± 15 volt triangle wave. The breakdown voltage of the Zeners is 5.1 volts. Assume a diode voltage drop of 0.7 volts. D
R1 Vo D
Vin
b) Verify your answer with PSpice. Use the D1n4733 zener.
Problem 2: An I-V plot for a Zener is shown below:
Create a separate model for the Zener in each of the three regions shown above.
Problem 3 : a) Plot Vo(t) and Vin(t) for the circuit below by hand. Indicate numerical values for all breakpoints. Let Vin be a ± 15 volt triangle wave.
Region 2
Region 3
Region 1
Vz
I z
Vo
D
R 3k
R 1k
D
R 3k
D
R
Vin
b) Verify your answer with PSpice.
Problem 4 : The donor concentration in the n-region of a silicon pn diode at 300 °K is Nd = 10^17 /cm^3. The minority carrier concentration in the p-region is npo=3× 103 /cm^3. The zero bias junction capacitance for this diode is Cjo=2 pF. Find the junction capacitance with a reverse bias of 10 volts. Assume that m= 0.5.
Problem 5:
Plot VR(t), VZ(t), VO(t), and Vin(t) for the circuit below. Do not assume ideal diodes. Indicate numerical values for all breakpoints. Let Vin be a ± 15 volt triangle wave. The breakdown voltage of the Zeners is 5 volts. Assume a diode voltage drop of 0.7 volts.
-^ -
Vin
VZ
D
R
0
VR
D
Vo
b) Verify your answer with PSpice.
Problem 6
The donor concentration in the n-region of a silicon pn diode at 300 °K is Nd = 10^17 /cm^3. The minority carrier concentration in the p-region is npo=3× 103 /cm^3. The zero bias junction capacitance for this diode is Cjo=2 pF. Find the conductivity in the n region and the conductivity in the p region.
Some constants:
- μp=480 cm^2 /volt-sec,
- The charge on an electron = 1.602× 10 -19^ coulombs,
- μn=1350 cm^2 /volt-sec,
- Boltzmann’s constant = 86×10-6^ eV/ºK = 1.378×10-23^ J/ºK,
- Silicon band gap energy = 1.1 eV,
- Silicon material constant B = 5.23× 1015 ( ) ( ) (^)
( # ^ cm^3 ⋅ o^ K^32.
