University of California, San Diego
Department of Electrical and Computer Engineering
ECE65, Spring 2006
Lab 5, Diodes
Experiment 1: Measurement of Diode iv Characteristics:
This circuit allows one to measure the diode
iv characteristics by using a current-to-voltage
converter circuit as is shown below.
+
vi
+ vD −
R2
−
+
−
+
vo
−
Circuit Analysis: Show that vi = vD and vo = −iD R2 .
Lab Exercise: Set up the circuit on your breadboard using a diode and R2 = 1 kΩ. Apply a
square wave at 1 kHz with no DC offset to the input. Use the 20dB attenuation output
of the function generator and set the AC amplitude to zero before connecting the
function generator to the circuit. Display both vi and vo on the scope and center both
signals so that zero voltage is in the middle of the scope of display. Slowly increase the
amplitude of the input until the output square wave has a an amplitude of about 10 V
peak-to-peak. Because the input is a square wave, we really applying only two distinct
voltages to the input, call them high and low. For example, if the input has a peak-to-peak
amplitude of 1 V, we are applying +0.5 (high) and −0.5 V (low) to the circuit. The output,
therefore, will be a square wave, the high (low) voltage of output is proportional to diode
current when the input is high (low). By measuring high and low voltages of vi and vo ,
we will get two iv pairs for the diode.
1) Explain why the output square wave is not symmetric around zero volt.
2) Measure high and low voltages of vo and high and low voltages of vi . Compute the two
corresponding iv pairs for the diode. Lower the input and repeat the experiment. Take 8
measurements down to about 10 mV on vo . Tabulate your results, and make a linear and
a semi-log plot of iD (Y axis) versus vD . From the semi-log plot, estimate Vt of the diode.
3) Set the function generator to provide a triangular wave at 1 kHz with no DC offset
to the input. Use the 20dB attenuation output of the function generator and
set the AC amplitude to zero before connecting the function generator to the circuit.
Display both vi and vo on the scope and center both signals so that zero voltage is near
the bottom of the display on one grid line. Set the scope of show (x vs y). Slowly increase
the amplitude of the input. The scope shows the i − v characteristics of the diode. Print
out the scope output and mark and label the axis.
ECE65, Lab 4 Exercises
Page 2
Experiment 2: Rectifier and Peak Detector:
In this circuit, the input is a sinusoidal wave
with an amplitude of 5 V (no DC offset) and a
frequency of 1 kHz, R = 1 kΩ, and the diode
is a 1N4148 general purpose diode.
+
−
VS
C
+
R
Vo
−
PSpice Simulation: Simulate the circuit with PSpice for the following capacitor values: no
capacitor, 10 nF, 1 µF, and 100 µF and plot vo as a function of time. Note that it takes a
few periods before the transients die away and you get steady-state signal (or alternatively,
you tell PSpice to ignore initial transients). Explain PSpice Results.
Lab Exercise: a) Assemble the circuit (with no capacitor). Attach Scope channel A to
the input and Scope channel B to the output. Printout your input and output traces and
compare with your simulations.
b) Repeat part a for capacitor values of 10 nF, 1 µF, and 100 µF. Printout your input and
output traces and compare with your simulations. Note that electrolyte capacitors
should be placed on the circuit according to their polarity
Experiment 3: Zener Diode Power Supply
Set up the circuit below with a 1N5232B Zener diode (vZ = 5.6 V). In this circuit, the
9-V supply represents the “unregulated voltage.” The elements in the box (1 kΩ resistor
and the zener diode) form the regulator circuit. The combination of the variable resistor
(potentiometer) and 100 Ω resistor, represents the “load” in this circuit (call their combination RL ). With varying the resistance of the potentiometer, we can draw different
amount of current from the regulator circuit. What is the purpose of the 100 Ω resistor?
1k
9V
+
−
IL
+
+
V in
VL
0−10 k
100
−
−
Volage Regulator
Circuit Analysis: Calculate the load voltage (VL ) as a function of the load current (IL ).
Estimate the maximum load current for the circuit to act as a voltage regulator.
PSpice Simulation: Simulate the circuit with PSpice with RL as a parameter with a range
of 100 Ω to 10 kΩ (do NOT include the 100 Ω resistor in your simulation!). Plot V L versus
IL and compare with your analytical results.
ECE65, Lab 4 Exercises
Page 3
Lab Exercise: Assemble the circuit. Start with the potentiometer set at maximum resistance (i.e., about 10 kΩ). Measure the load current and the load voltage. Then, vary the
potentiometer resistance and measure the load voltage for a range of load currents. Plot
VL , versus IL . Compare with your circuit analysis and PSpice simulation.
Experiment 4: Improved Power Supply(Do this experiment for an extra credit of 30
points)
Set up the circuit below with 2N3904 transistor and 1N5232B Zener diode (v Z = 5.6 V).
Use the potentiometer in series with a 100 Ω resistor as RL .
Circuit Analysis: Calculate the load voltage (VL ) as a function of the load current (IL ).
Estimate the maximum load current for the circuit to act as a voltage regulator. Hint:
Use your solution of the experiment 3, iL in the circuit of experiment 3 is the same as iB
in this experiment.
PSpice Simulation: Simulate the circuit with PSpice with RL as a parameter with a range
of 10 Ω to 10 kΩ (do NOT include the 100 Ω resistor in your simulation!). Plot V L versus
IL and compare with your analytical results.
Lab Exercise: Assemble the circuit. Start with the potentiometer set at maximum resistance (i.e., about 10 kΩ). Measure the load current and the load voltage. Then, vary the
potentiometer resistance and measure the load voltage for a range of load currents. Plot
the load voltage, VL , versus the load current, IL . Compare with your circuit analysis and
PSpice simulation.
IL
+
9V
+
−
+
0−10k
V in
1000
−
VL
100
−
RL