PHYS 536
DC Power Supply
Introduction
The process of changing AC to DC is investigated in this experiment. An integrated circuit regulator makes it easy to construct a high-performance voltage
source using only four parts: a transformer, full-wave bridge rectifier, capacitor filter, and regulator. A zener diode regulator is also included as a simple
illustration of regulator properties.
Rectifier
Figure 1: The bridge rectifier
The proper arrangement of the 1N914 diodes for full wave rectification is shown
in Fig. 1. The purpose of the rectifier is to convert the AC voltage from the
transformer in to DC voltage. This is accomplished using the diodes. An ideal
diode will pass current when forward bias, but no current when reverse biased.
A real diode will, however, pass current when the breakdown voltage is exceeded.
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However, this destroys the diode. A real diode will result in an 0.8 V drop across
the diode, while an ideal diode would not drop any voltage.
Because the diodes will only allow current flow in one direction, when the top
AC input is positive with respect to the bottom input, current will flow through
the top right diode, through the load attached to the right side of the rectifier
and exit the load on the left side of the rectifier, and finally pass through the
bottom left diode and leave through the bottom AC input. When the bottom
AC input is positive with respect to the top input, current will pass through the
bottom right diode, through the load, and out the top left diode. During both
sequences, current always flows through the load in the same direction, and so
the current provided is DC current.
Rectification
The peak voltage, Vp , from the rectifier is slightly less than the peak transformer
voltage, VT because voltage is dropped across it the diode, so
Vp = VT − 2Vd
(1)
Vd is the voltage drop across the diode, typically 0.8 V, and in the bridge rectifier
the current flows through two diodes. Since the voltage of the transformer
secondary is always specified as an effective or RMS voltage, Ve , and is related
through the peak voltage by the following relation
VT =
√
2Ve
Ripple Voltage
If the rectified wave is applied to an RC circuit, it will take time to discharge
the capacitor, and so smoothing of the rectified wave will occur, assuming that
the time constant of the RC circuit is chosen such that RC f1 , so that the
time constant is large compared to the period of the rectified wave.
For a capacitor,
i=C
dV
dt
dV =
i
dt
C
Hence,
2
The ripple voltage can be estimated, then, by
Vr =
ic
Δt
C
(2)
For half wave rectification, the capacitor will continue to charge throughout the
period (T = f1 ), of the original wave, so the ripple voltage can be estimated as
ic
fC
Vr =
For full wave rectification, the capacitor only discharges until the voltage begins
increases again, so the capacitor only discharges for half the period of the original
1
), so
wave (T = 2f
Vr =
ic
2f C
This relationship is, however, only approximate, since once the rectified voltage
becomes larger than the capacitor voltage, the capacitor will stop discharging
and instead increase its voltage as it gains more charge, so the capacitor is not
discharging over the entire half period, so a better approximation to the ripple
voltage is
Vr =
0.7ic
2f C
(3)
By choosing a large capacitance the ripple can be reduced to any desired level,
however this brute-force approach has several drawbacks. First of all, the capacitors may be prohibitively large and/or expensive. Another drawback is that
even though the ripple may be negligible line voltage variations are transferred
to the output voltage. A better approach is to regulate the output voltage.
The current drawn from the capacitor depends on the type of the regulator
used. We will consider two types of regulators: IC and zeners. Vo is the output
voltage of the regulator and RL is the external load resistor connected to the
regulator, i.e. it is the resistance of the device or load to which power is being
transferred. For an IC regulator
Vo
RL
(4)
Vc − Vz
VA − VZ
≈
R2
R2
(5)
Ic =
and for a zener
IC =
3
where Vc is the average capacitor voltage. Assuming Vr VA , this can be
assumed to be VA .
When the load resistor is attached directly to the capacitor without a regulator
the current is given by the relation
IC =
Vc VA
≈
RL
RL
(6)
Zener Regulator
Figure 2: Electrical characteristics of a zener diode
A zener diode is similar to a regular diode with one exception, when a reverse
bias is applied, the diode breaks down in a controlled fashion, so that the current
through the resistor, iZ , is greater than zero and the voltage across the resistor
is VZ ≈ VZK as shown in Fig. 2. While ideally the voltage would remain
unchanged, there is some small internal resistance in the zener, so the zener
under reverse bias is best described as an ideal diode in series with a voltage
source and small resistance rz as shown in Fig. 3.
This proves useful because, so long as the zener diode is reverse biased, it provides a nearly constant voltage. The output voltage of the zener is given by
Vo = Vz
R2
rz
+ VC
− I0 (rz R2 )
R2 + rz
R2 + rz
Note that for R2 rz and RL Rz ,
Vo ≈ Vz
4
(7)
Figure 3: Piecewise Linear model of a zener diode
Where rz is typically of the order of a few ohms.
Note that under eq. 7, a ripple in the capacitor voltage will result in a ripple in
the output voltage, and differentiating this equation,
dVo = dVc
rz
R2 + rz
(8)
The third term in eq. 7 also indicates that the output current affects the output
voltage, as given by
dVo = dIo (rz R2 )
If the load resistance draws too much current, the zener diode will drop out of
conduction and act as an open circuit, at which point the zener will cease to act
as a regulator. The maximum current that may be delivered is
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Io,max =
Vc,min − Vo
− Iz,min
R2
(9)
Where Vc,min is the minimum capacitor voltage (the peak voltage less the ripple
voltage) and Iz,min is the minimum current required by the zener diode to
maintain conduction.
IC Regulator
When using a voltage regulator, a minimum voltage, Vi,min must be maintained
across the input pin and ground pin. For the 7805 regulator, the required voltage
is 7.2 V. To insure that the voltage remains above the minimum required voltage,
the capacitor voltage must follow the condition
Vc,peak − ΔV ≥ Vi,min
(10)
IC Regulator Ripple Rejection
The ripple rejection for an IC regulator is specified in dB of attenuation.
RR(dB) = −20 log
ΔVo
ΔVi
(11)
Solving this for the output voltage ripple,
RR
ΔVo = ΔVi 10− 20
(12)
IC Regulator Temperature Drift
The power dissipated by the voltage regulator is given by
PD = Io (Vi − Vo ) + Vi IQ
(13)
where IQ is the quiescent current of the regulator, which is 6 mA for the 7805
regulator. The IC regulator’s junction temperature is a function of this power
dissipation. The junction temperature is
TJ = TA + θJA PD
6
(14)
◦
For the 7805 regulator, θJA = 27.3 WC . Changes in the junction temperature
will result in a shift in the regulator’s output voltage. The shift in output voltage
is given by
ΔVo =
The value of
1
ΔVo
ΔT
ΔVo
ΔT
ΔT
for a 7805 regulator is -0.8
(15)
mV
◦C .
Basic Power Supply
Figure 4: Initial circuit
Figure 5: Initial circuit with the capacitor
1. Begin by testing your 4 1N914 diodes. Do this using the multimeter.
On the multimeter, there should be a symbol which looks like a diode
symbol. Turn your multimeter to this setting. Attach the leads of the
multimeter to each diode, alternating the position of the leads. If the
diode is functioning properly, the multimeter should read 0.7 V in one
7
direction, and an overload in the other. If any diodes do not pass this
test, discard them and replace them with another diode.
2. The desired peak voltage across resistor R1 is 19.6 V. Calculate the necessary transformer voltage using Eq. 1.
3. Do not connect the provided variac and 5:1 step down transformer to the
circuit yet. Using the multimeter, set the voltage to the desired transformer voltage calculated in the previous step. Keep in mind that the
multimeter measures RMS voltage, not peak to peak voltage.
4. Before building the circuit, draw the shape of the expected voltage across
the resistor.
5. Build the circuit in Fig. 4. Use R1 = 1 kΩ using the variac and 5:1
transformer as the source AC voltage.
6. Measure the voltage across the resistor. Is it the expected 19.6 V? Is the
waveform as expected? If the voltage across the resistor is not 19.6 V,
adjust the voltage until it is.
7. Calculate the expected ripple voltage for the circuit in Fig. 5 if R1 = 1 kΩ
and C1 = 100 μF.
8. Build the circuit in Fig. 5 with R1 = 1 kΩ and C1 = 100 μF. Be careful
when inserting C1 , which is an electrolytic capacitor. For these capacitors, orientation matters, so be sure that the side connected to ground
corresponds to the side of the capacitor marked with a strip and negative
symbol.
9. While observing the voltage across the resistor, remove and replace the
capacitor in the circuit. What effect does the capacitor have on the output
voltage?
10. Measure the ripple voltage. How does it compare to the calculated value?
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Zener Voltage Regulator
A voltage regulator can also be built using a zener diode which is reverse biased.
Under reverse bias, the zener provides a nearly constant output voltage. The
zener diode does have a small resistance, however, rz . For the computations in
the section, assume rz = 10 Ω.
1. Initially, no load resistance is used. With no load resistance, VC = 17V ,
R2 = 1 kΩ, and the assumed rz = 10 Ω, calculate the capacitor ripple
voltage (ΔV ), and the output ripple voltage (ΔVo ). Because the ripple
voltage will be small, it can be assumed that VC ≈ VC,peak .
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Figure 6: Zener Voltage Regulator
2. Build the circuit in Fig. 6 using VC = 17V , R2 = 1 kΩ. Do no connect
any load resistor initially.
3. Measure the ripple voltage ΔV and ΔVo . From these measured values,
calculate rz .
4. Calculate the change in Vo that would result from an output current of
Io = 5.2 mA.
5. Use the multimeter to measure the output voltage Vo without a load resistor.Insert RL = 1.2 kΩ. Measure the output voltage again with the load
resistor in place. How did the output voltage change?
6. Calculate rz using the measured Io and ΔVo from the previous step.
7. Calculate the maximum Io that will leave a minimum of 2 mA of current
flowing through the zener diode. If the zener diode current falls below this
value, the diode will drop out of conduction.
8. If RL is replaced with a 470 Ω resistor, the zener diode will drop out of
conduction. Calculate the expected Vo with this load resistor.
9. Using RL = 470 Ω, measure Vo .
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Power Supply with an IC Voltage Regulator
Figure 7: DC power supply using an IC voltage regulator
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1. The 7805 IC voltage regulator requires an input voltage of 7.2 V to function correctly. If the voltage supplied by the capacitor drops below this
voltage, the regulator will not work correctly. Calculate the minimum
peak capacitor voltage that will keep the input voltage above 7.2 V. Assume C1 = 100 μF and RL = 100 Ω. The output voltage for the 7805
regulator is 5 V.
2. Build the circuit in Fig. 7 using C1 = 100 μF and RL = 100 Ω. Adjust the
variac until VC,peak is 2 V larger than the value calculated in the previous
step.
3. Measure the ripple voltage across the capacitor, ΔV .
4. With both VC and Vo shown on the oscilloscope, decrease the variac voltage until you see a ripple voltage appear in the output voltage. When the
ripple appears in the output voltage, the regulator has stopped functioning
as a regulator. Measure the VC at which this occurs.
5. Estimate the change in the Vo caused by changing the input voltage from
7 V to 25 V with a load resistance RL = 1 kΩ.
6. Replace the current load resistor with RL = 1 kΩ. Measure Vo with
an input voltage of 7 V, and with an input voltage of 25 V. Take both
measurements as quickly as possible to minimize temperature drift caused
by ohmic heating of the regulator. How does the change measured compare
to that calculated in the previous
Required Components
• Diodes: (4) 1N914, (1) 1N4375
• Capacitors: (1) 100 μF, (1) 0.1 μF
• Resistors: (1) 100 Ω, (1) 470 Ω, (1) 1 kΩ, (1) 1.2 kΩ
• ICs: (1) 7805 voltage regulator
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