Physics 322, Lab 5: Diodes
Name:_______________
Key concepts
I vs. V for a diode, rectification, filtering, diode clippers, diode clamps
Text Material
Diefenderfer, Sections 5-1 to 5-9
Equipment
Resistor substitution box, function generator, digital oscilloscope
Components needed
(4) IN914 diodes
3:1 Isolation transformer
1 uF capacitor
C=__________
100 uF capacitor C=__________
1 nF capacitor
C=__________
(2) 1 kΩ resistor R=__________
10 kΩ resistor
R=__________
47 kΩ resistor
R=__________
3:1
The 3:1 isolation transformer. Output from the
secondary is isolated from signal ground.
Some background
Simple conductors obey Ohm's Law. Semiconductor devices like diodes and transistors, the building
blocks of modern electronics, do not obey Ohm's Law and in fact it is their high degree of non-linearity that
makes them so useful. In todays lab we look closely at the diode and a couple of its uses.
One of the main uses of diodes is in rectification of AC signals to produce a DC signal. It might be,
for example, that you are interested in measuring the average intensity of a sound wave with a data logging
program that can measure DC voltages. Microphones deliver an AC signal however, and so you would need
a circuit capable of converting this AC signal into a steady DC voltage that was proportional to the AC
amplitude. By the end of today's lab designing a simple circuit to accomplish this task will be easy for you.
The lab has five parts.
Part A:
Part B:
Part C:
Part D
Part E
Single Diode rectification
Filtering
The bridge rectifier
The Diode clipper
The Diode clamp
As in previous labs, in the exercises below you'll be asked to sketch what you see on the scope. Be sure
to draw both input and output traces in your sketch. Also please indicate scales and units on your
vertical and horizontal axes.
Part A: Single Diode Rectification
A rectifier is a circuit which lets through only one polarity of an input signal. For example say you have a
sine wave that oscillates between -5 and +5 volts. If you feed this into an ideal rectifier, you will get out just
positive peaks.
Question 1:
(a) Construct the above circuit using actual circuit
components and drive it with 5 Vpp, 1000 Hz sine wave
with no DC offset. Note that the vertical bar in the diode
symbol corresponds to the bar on the actual component.
Sketch the input and output voltages on the same graph
below.
(b) Reverse the orientation of the diode and describe
what happens. Why does this happen?
Question 2:
(a) Switch your generator over to a 5 Vpp, 1000 Hz square wave. Sketch the input and output voltages on
the same graph below.
(b) Are your observations consistent with the rule of thumb that the forward drop across a diode is about 0.6
V? If yes, explain, or mark on your sketches, where you see evidence of this drop.
Question 3:
Return to a sine wave and reduce the amplitude of the signal. At what input voltage does the output voltage
go to approximately zero (less than say 10 mV) volts over all time? Explain the significance of this
particular input voltage.
Vnon-conduction = ____________________
Question 4:
(a) Construct the following rectified differentiator circuit and drive it with a 10kHz square wave of 10Vpp.
Check with the instructor to make sure your output voltage looks reasonable. Sketch the input and output
signals below.
(b) Explain what your circuit is doing.
(c) Reverse the orientation of the diode. Describe what happens and explain why.
Part B: Filtering
The half-wave rectifier examined in part A produces a single polarity voltage, however one would not really
call it DC since the ripple is so large (see page 84 in the text for a precise definition of "ripple".). One way
to smooth out the ripple is to add a filter capacitor as shown in the circuit below.
Question 5:
(a) Construct the circuit shown using a filter capacitor of C=1 uF and an input sine wave voltage of 20 Vpp
and frequency 1 kHz. Sketch your input and output waveforms.
Question 6:
(a) Measure the AC ripple voltage , ∆V, as defined in Figure 5.12 of the text and compare it to the expected
value of ∆V ≈ I /(Cf ) , where I is the approximate DC current through the resistor. Also calculate the ripple
factor r = ∆V / VDC .
Measured values
∆V
VDC
C
f
r
Theoretical value ∆V=___________________
(b) Replace the filter capacitor with one of value C=100uF. Be careful to get the polarization of the
capacitor right (see instructor if you're not sure). Repeat the measurements above.
Measured values
∆V
VDC
C
f
r
Theoretical value ∆V=___________________
(c) Which capacitor does a better job of filtering? Explain why this is so by discussing the time constant of
the filtering circuit.
Question 7:
(a) Go back to C=1uF and try varying f through several decades. What happens to ∆V as f is increased?
Explain why this happens.
(b) Go back to f=1000 Hz and try varying R through several decades (use a resistor substitution box). What
happens to ∆V as R is varied? Explain why this happens.
Part C: The Bridge Rectifier
In general one wants ripple to be as small as possible. One way to reduce ripple is to use a full-wave
rectifier to convert both the positive and negative peaks in your AC signal to the same sign and then filter
the output. This has the effect of doubling the "f" in the theoretical ripple expression thus reducing ripple by
a factor of 2.
Question 8:
(a) Construct the bridge rectifier shown. The "isolation transformer" is used so that the signal generator's
ground is isolated from the circuit. This then allows us to look at the output voltage with the scope. Drive
the circuit with a 100 Hz sine wave. Sketch the output voltage, VAB, below.
(b) Explain how the circuit works.
(c) Use the scope to measure Vp of the transformer’s secondary. Then use the scope to measure Vp of the
voltage between A and B. (BEWARE: you cannot measure these voltages simultaneously with the scope
because of grounding issues! You can only have one grounded BNC cable connected to your circuit at
once.)
VpSecondary=__________________ VpAB=____________________. Are these two voltages equal? If not,
why?
Part D: The Diode clipper
Clipper circuits "clip" their input so that Vout does not exceed certain limits. In this exercise you will
construct simple circuits that clip the positive half of a signal, the negative half, or both halves.
Question 9:
(a)Construct the circuit shown and drive it with a 5 Vpp sine wave at 1000Hz. Sketch input and output signals on top
of one another.
(b) Explain how the circuit works.
Question 10:
(a) Modify the above circuit so that both positive and negative peaks are clipped to 0.7 volts at the output (Draw the
modifications on the schematic diagram below.) Drive the circuit with a 5 Vpp sine wave at 1000Hz. Sketch input and
output signals on top of one another.
(b) Explain how the circuit works. Suggest a possible use for the circuit.
Question 12:
(a)Construct the circuit shown and drive it with a 20 Vpp sine wave at 1000Hz. Sketch input and output signals on top
of one another. Use the +5VDC breadboard output as the 5 volt "battery". (As usual , the connection between ground
and the negative terminal of this “battery” is done inside the breadboard. There is no need for you to make this
connection! )
(b) Explain how the circuit works.
Part E: The Diode clamp
A common problem you may run into in science experiments is that your apparatus produces bipolar
signals (eg. An sine wave centered on zero volts.), and yet what you need is a signal that is entirely positive
(or negative). In other words you need some way to add an offset to your signal so that it oscillates between
0 and +2Vp. The Diode clamp circuit serves this function.
Question 13:
(a)Construct the circuit shown and drive it with a 10 Vpp sine wave at 10kHz. Sketch input and output signals on top
of one another. Use a 1 uF capacitor in the circuit.
(b)Is the value of the capacitor critical? Try switching it and report what you see.
(c)Design and sketch a diode clamp circuit that lowers the signal such that it is always negative. Sketch the circuit and
input and output voltages.