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ELEC 350L
Electronics I Laboratory
Fall 2002
Lab 8: Simple DC Power Supply
Introduction
In almost every DC power supply a rectifier circuit and a filter capacitor are used to convert an
AC voltage to a pulsating DC voltage and then to smooth out the pulsations over time. Because
most electronic circuits require DC power, rectifier circuits are almost always found in
equipment that operates from 60-Hz AC power lines. Although sophisticated voltage regulation
circuits are usually employed in practice to maintain DC output voltages at specified levels, the
simple circuit that you will study in this experiment forms the basis of most commercial DC
power supplies. In fact, this basic circuit, with a few minor modifications, is sufficient for many
applications requiring DC power. In this lab experiment you will investigate the operation of a
common type of rectifier circuit and the use of filter capacitors to smooth out power supply
“ripple.”
Theoretical Background
A very basic power supply circuit is shown in Figure 1. A source of AC (such as the 120 Vrms
available at a wall socket), represented by vin, drives a transformer, which might step up or step
down the voltage. If no load resistor (RL) is present, the capacitor charges during the first halfcycle that the diode is forward biased (a positive half-cycle). During the next (negative) halfcycle the capacitor remains charged because the diode is reverse biased, and there is no path
available through which charge can leave the capacitor. The capacitor retains its charge (and the
voltage across it stays constant) thereafter because the diode remains reverse-biased at all times.
No additional current can flow into the capacitor, but no current flows out either. The voltage
across the capacitor remains at the peak positive value of the voltage across the secondary
winding, minus the value of the diode’s turn-on voltage (approximately 0.8–1 V for common
rectifier diodes).
iL
+
vin
C
RL
vout
_
Figure 1. Basic power supply circuit using half-wave rectifier.
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Of course, a load (represented by RL) must be connected to this circuit if it is to do any useful
work. If a load is present, vout is no longer constant because the capacitor discharges though the
load whenever the voltage across the secondary winding falls below the value necessary to keep
the diode forward-biased. This circuit is called a rectifier, because it allows current to flow in
only one direction though the load. The current varies considerably, however, as the capacitor
voltages rises and falls.
The major disadvantage of the circuit shown in Figure 1 is that the diode conducts during only
one of the input AC voltage’s half-cycles, and for this reason the circuit is more specifically
called a half-wave rectifier. During the half-cycle when the diode does not conduct, the
capacitor continues to discharge through the load. The rate of discharge depends upon the value
of RL. If RL is small enough, the current flowing out of the capacitor is significant. As a result,
the voltage across the capacitor can drop to an intolerably low level. The continual charging and
discharging of the capacitor causes the output voltage of the power supply to fluctuate, creating
what is called a ripple voltage that is superimposed on the average DC output voltage. This
effect is shown graphically in Figure 2. Since most applications require a nearly steady supply
of DC, the presence of significant ripple is undesirable. The ripple voltage is often expressed as
a percentage and is calculated using the formula [1]
% ripple  100
peak - to - peak ripple voltage
.
average DC voltage
vout
ripple
voltage
(p-p)
average DC
value
t
Figure 2. Ripple voltage in the output of a DC power supply. (adapted from [1])
This problem is mitigated somewhat by using a full-wave rectifier circuit like the one shown in
Figure 3. In this circuit, diodes D2 and D4 conduct during positive half-cycles (when the top end
of the transformer secondary is positive), and diodes D1 and D3 conduct during negative halfcycles. Thus, during both half-cycles current flows downward through the load resistor, so this
circuit is a rectifier. The full-wave rectifier is an improvement over the half-wave circuit
because the capacitor is charged during both half-cycles. Consequently, the capacitor does not
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have as much time to discharge, and the output voltage does not drop as much. For a given
capacitor value, the output voltage remains more stable than with the half-wave circuit, and the
ripple voltage is much smaller in magnitude.
D2
D1
vin
D4
+
D3
C
ic
+
RL
vo
_
Figure 3. Full-wave bridge rectifier circuit with filter capacitor. The diode symbols would
normally be filled in but are not in this case because of the limitations of the drawing software
used to produce the diagram. The “+” symbol next to the capacitor indicates that it is an
electrolytic type. Since they are available in large values, electrolytic capacitors are usually used
in power supply circuits,. Electrolytic capacitors are polarized; that is, the voltage across them
must have the indicated polarity for proper operation.
The capacitors shown in Figures 1 and 3 are often called filter capacitors because their purpose
is to reduce the ripple voltage to a tolerable level. As the sinusoidal input voltage rises to its
peak, the capacitor voltage rises with it because the Thévenin equivalent resistance of the source
is typically very small. When the input voltage begins to fall again, the capacitor begins to
discharge. However, if the capacitor is large enough (and if the load resistance isn’t too small),
the rate of discharge is much less than the rate at which the input voltage falls. Thus, there will
be a period of time during each half-cycle when none of the diodes conduct; that is, all of the
diodes are reverse-biased (off). During this time the capacitor and resistor are effectively
isolated from the rest of the power supply circuit, and the capacitor voltage drops exponentially
with the time constant RLC as the capacitor discharges through the load. Therefore, in order to
keep the ripple voltage down to a tiny fraction of the average (target) DC output voltage level,
the time constant is made much larger than the length of one half-cycle. The smaller the load
resistance, the larger the value C must be for minimal ripple.
The required value of C is easily determined. Recall that the current ic that flows into the top of
the filter capacitor is given by
ic  C
3
dv o
,
dt
where vo is the output voltage. (vo is also the capacitor voltage.) If the ripple is a small
percentage of the total output voltage, then the voltage across the capacitor decays almost
linearly. (See, for example, Figure 2 above or, for a more accurate depiction, see Figure 4.44 in
the text.) Let vmax be the maximum value of the output voltage and vmin be the minimum value.
If the ripple is small, then the decay is very slow, and the capacitor will not start charging again
until almost the next time the input voltage reaches a peak. For a full-wave rectifier, the next
input voltage peak occurs one half-period (T/2, where T is the period) later. The time derivative
of the output voltage over the half-cycle can therefore be approximated as
dv o
v  v min
,
  max
dt
0.5T
where the negative sign indicates that the output voltage falls as the capacitor discharges.
Substituting this result into the i-v relationship for the capacitor given earlier yields
ic  C
dvo
v  v min
 C max
.
dt
0.5T
The diodes are off (reverse-biased) when the capacitor discharges, so the only place the capacitor
current can flow is through the load resistor. Since ic is defined as flowing into the top of the
capacitor, this means that ic = –vo/RL. Substituting this result into the equation for ic above leads
to the relationship
 ic 
vo
v  v min
 C max
,
RL
0.5T
where vo is now understood to be the nominal (average) value of the output voltage. The ripple
voltage Δvr is defined as the peak-to-peak value of the fluctuation of the output voltage, so Δvr =
vmax – vmin. Thus,
vo,avg
RL
C
v r
,
0.5T
where vo,avg is the average DC output voltage. This can be rearranged to give
v r
0.5T
.

vo,avg RL C
Note that the left-hand side is essentially the percentage ripple. (In this case, the ripple is
expressed as a fraction rather than as a percentage. If the percentage ripple is 5%, then
Δvr/vo,avg = 0.05). The frequency of the input voltage is typically known, and the load resistance is
either also known or can be estimated with a fair degree of accuracy. If this is the case, then the
equation can be used to find the required filter capacitor value to achieve a target percentage
ripple. Using the relationship f = 1/T,
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C
0.5T
1
1
1
.

RL vr vo,avg  2 f RL vr vo,avg 
The output voltage of the simple DC power supply discussed here is very close to the peak
magnitude of the sinusoidal voltage that appears across the secondary winding of the
transformer. If a certain DC output voltage level is required for a given application, then a
transformer must be selected that has the appropriate secondary voltage rating.
Experimental Procedure
Construct the full-wave bridge rectifier shown in Figure 3. For safety reasons, the transformer is
enclosed in a large box with a power cord coming out of it. There are three jacks on the box that
are connected to the secondary winding of the transformer. The two red jacks connect to the
ends of the winding, and the green jack connects to a center tap on the winding. For this lab
experiment, take the AC voltage for your power supply from the two red jacks. Before
connecting the transformer to the rectifier, sketch or plot the AC voltage measured across the
secondary winding. Is it sinusoidal?
Now build the rectifier circuit using four 1N4007 diodes, and use a 1 k resistor for the load.
Without a filter capacitor in place, observe the output waveform (measured across RL) on the
oscilloscope. Sketch or plot the waveform, and compare it to the plot of the voltage across the
secondary winding. Is it what you expected? Why or why not?
Warning: Do not attempt to measure both the AC voltage on the secondary winding and
the output voltage across the load resistor at the same time. You could create a ground
loop (a short between the two ground leads) that could destroy one or more components.
Calculate the capacitance value needed to obtain a 4% ripple rating for the output voltage. From
the stock of available capacitors, select one with a value as close as possible but greater than the
value you calculated. (Why should it be greater than the value you calculated?) Turn off the
power, and add the capacitor to the circuit.
Warning: Electrolytic capacitors are polarized. Failure to pay attention to their polarity
could result in their spectacular destruction and a trip to the hospital for anyone nearby.
Reapply power to the circuit, sketch or plot the output waveform, and record the nominal DC
output voltage and the ripple voltage. Calculate the percentage ripple. You may need to change
the oscilloscope’s coupling setting from DC to AC in order to observe the ripple clearly.
Comment on the difference (or closeness) between your measured ripple value and the value you
predicted. Also, is the nominal DC output voltage value what you expected to obtain?
A data sheet for the 1N4007 is available at http://www.fairchildsemi.com/ds/1N/1N4007.pdf
Reference
1. M. Wasserman, Laboratory Manual for Microelectronic Circuits and Devices, 2nd ed., by M.
N. Horenstein, Prentice-Hall, Inc., Upper Saddle River, NJ, 1996.
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