### Capacitor Self-Resonance

```Experiment No. 3.
Power Supplies and Linear Regulators
By:
Prof. Gabriel M. Rebeiz
The University of Michigan
EECS Dept.
Ann Arbor, Michigan
All electronic systems which operate on DC voltages (+5 V, +12 V, etc.) and are plugged to
the AC (120 V, 60 Hz) system require a power supply. This includes all audio components (CD
player, tape player, receiver, etc.) and most home/office electronic units (computers, printers,
alarm clocks, microwave ovens, and dishwashers/fridges/washing machines if they have
microprocessor controls). A power supply schematic is shown in Figure 1. The 120 V AC
voltage is first stepped down to ~12 V AC using a transformer, then passes by a diode bridge
rectifier and a low-pass “filtering” capacitor. The resulting DC voltage has a ripple of around
+10% Vp (for example 12V + 1 V).
The main focus of Experiment No. 3 is on the input section of the power supply, mainly the
transformer, diode rectifier and the low-pass capacitor. The relevant equations for a full-wave
or half-wave bridge rectifier are given in the course notes. Notice that these equations are
accurate to +25% since they are derived from very simple assumptions. They are intended to
give a conservative estimate of the output voltage ripple, the peak diode current and the peak
inverse voltage.
0.5 A Fuse
D1
+
+
-
Vi
–
AC 120 Vrms
AC 10 Vrms
+
D3
D4
C = 220 µF
R L=
470 Ω
Vo
–
Figure 1: A full-wave bridge rectifier.
Dual (Positive and Negative) Output Power Supplies:
In order to get a dual output power supply, a center-tap transformer must be used. The center
tap is usually connected to ground. Relative to the center tap, one of the output transformer
lines is at a positive voltage and the other is at a negative voltage. A full-wave rectifier
delivering dual-output voltages is shown below:
1
Fuse
+
+
-
D2
D1
+
Vi
–
D3
D4 C
Vo (positive)
RL
–
+
C
Vo (negative)
RL
–
A full-wave single-output rectifier can also be obtained using a center-tap transformer as
indicated below:
Fuse
+
C
+
-
R L Vo
–
There is basically no difference in Vpk, VR, and Id between this topology and the four-diode
bridge, except that this topology requires a center-tap transformer and results in a peakinverse voltage across the diodes of (2 Vpk – VD).
Linear Regulators:
The full-wave bridge rectifier has terrible voltage regulation! This means that if RL decreases
by a factor of 2 (thus the output current is doubled), the output voltage ripple doubles! Also, if
the input voltage (called line voltage) changes by +10% (which is quite typical in 120 V AC
systems), then the output DC voltage also changes by +10%! Therefore, a power supply
composed of only a transformer, a bridge rectifier and a capacitor is quite bad for high
performance applications, and linear regulators must be used to ensure very low voltage ripple
and excellent load and line voltage regulation.
Linear (or series) regulated power supplies were introduced many years ago and are still used
extensively today. The basic design technique consists of placing a control element in series
with the rectifier and load device. Figure 2 shows a simplified schematic of a series regulated
supply with the series element depicted as a variable resistor. Feedback control circuits
continuously monitor the output and adjust the series resistance to maintain a constant output
voltage. Because the variable resistance (series element) of Fig. 2 is actually one or more
power transistor operating in the linear (class A) mode, supplies with this type of regulator are
often called linear power supplies.
2
Variable
Tap
Rectifier
Linear Regulator
Series
Element
AC Input
VoDC
Power
Transformer
Filter
Feedback
Control
IL
RL
Figure 2: The Agilent E3631A power supply.
In terms of performance, linear regulated supplies have a very precise regulating properties
and respond quickly to variations of the line and load. Hence, their line and load regulation and
transient recovery time are superior to supplies using other regulation techniques (such as
switching mode power supplies). These supplies also exhibit low ripple and noise, are tolerant
of ambient temperature changes, and with their circuit simplicity, have a high reliability.
Series regulated power supply are not efficient and in some cases dissipate a lot of power. Let
us take the example of a bridge rectifier designed to give 15 V + 1 V DC and connected to a
(series) linear regulator with an output regulated voltage between 5 and 12 V. If the output
voltage is 12 V, then the voltage drop across the series regulator is 3 V. For an output current
of 1 A, the output power is 12 W (Po), the power consumed by the regulator is 3 W and the
bridge rectifier therefore delivers 15 W (Pin). Thus the regulator efficiency (Po/Pin) is 12 W/15
W = 80% which is quite good. However, if the output voltage is 5 V, the series regulator
voltage drop is 10 V and the power consumed by the regulator is 10 W. This results in a poor
regular efficiency (Po/Pin) of 5 W/15 W = 33%! For high output currents/powers a poor
efficiency means that a huge amount of power is consumed in the linear regulator!
One way to solve this problem is to always make sure that the rectified bridge voltage is only a
few volts (1-2 V) above the required output voltage, thereby resulting in a low power
dissipation in the regulator (and high efficiency).
For a high-power variable voltage power supply such as the Agilent E3631A, the feedback
controls the transformer tap so as to change the input DC voltage to the regulator. This is
expensive and requires a DC motor or SCR switches to move the tap, but results in a very
efficient power supply.
3
Multiple Output Power Supplies:
If several output voltages are needed such as +5, +12 and +24 V, then it is best to use a
multiple-tapped transformer with 3 different rectifiers and regulators as shown in Fig. 3a.
High efficiency
(a)
+
+
–
Bridge +
Series
Rectifier –6 V Regulator
Vi1
+
5V
–
110 V
High efficiency
+
Bridge +
Series
Rectifier –14 V Regulator
Vi2 +
–
–
12 V
–
High efficiency
Vi3 +
–
+
Bridge +
Series
Rectifier –26 V Regulator
24 V
–
Very poor efficiency
(b)
+
+ Series
26 V – Regulator
+
+ Series
26 V – Regulator
+
+ Series
26 V – Regulator
+
5V
–
Poor efficiency
110 V
Bridge
Rectifier
12 V
–
–
High efficiency
–
24 V
Figure 3: An efficient multiple output power supply with multiple-tapped transformer (a) and a
simple but very inefficient implementation (b).
This is much more power efficient than using a simple transformer/rectifier and three regulators
(Fig. 3b).
Linear Regulator Specs:
Linear regulators specifications are easy to understand and their performance is summarized
in a few specifications:
Regulated Output Voltage ≡ Vo
(~1 V to Vline - 2 V)
Output Voltage Ripple ≡ VR
(mV levels)
ΔVo
Δ Vin
ΔVo
Δ IL
Line Regulation ≡
Change in output voltage
(<mV/V)
Change in input voltage
Change in output voltage
(~mV/A)
Change in output current
Short Circuit Current ≡ Isc
(10 mA to 10 A)
The data sheet of the LM 105/205/305/376 linear regulator family is in the Appendix.
4
Switching Power Supplies:
As seen above, linear regulators are not efficient especially if a large range of output voltages
are needed. For high power systems a switching power supply is used. The theory of these
supplies are beyond this course but it suffices to say that they are very efficient (85-95%) and
can deliver large amounts of power (50-1000 W). However, they do not offer excellent line and
load regulation, low ripple and low noise as linear power supplies. For this reason, all
analog/instrumentation/audio systems use linear regulators and all digital/computer/printer
systems use switching power supplies.
5
Diodes & Bridge Rectifiers
Full-Wave Rectifier
Half-Wave Rectifier
+
+
+
Vi
~
Vo
~
C
–
Vopk
Diode Off
Vo
(Vi)
+
Vo
–
RL
C
–
Vi
Diodes On
Vr
Diode On
RL
–
Diodes Off
Vo
Vr
(Vi)
Δt
Δt
t
1/120 sec
1/60 sec
Vopk
Vipk VD
Voav
Vopk
Vr
2
Vopk
f RLC
Vr
t
IL
6
(VD 0.7V)
(f
2Vr
Vopk
60 Hz)
2
2Vopk
ID av
IL 1
ID
IL 1 2
IC av
IL
IC
IL
pk
Vipk 2VD
Voav
Vopk
Vr
2 Vopk
Vr
2Vopk
Vr
2
2Vopk
Vr
Vr
2
(f
2 f RLC
t
f
(VD 0.7V)
Vopk
Vr
IL
Vopk RL
pk
Vopk
2Vr
Vopk
2
Vopk RL
Vopk
2Vr
ID av
IL 1
ID
IL 1 2
pk
IC av
IL
IC
IL
pk
60 Hz)
Vopk
2 Vr
Vopk
2Vr
2
Vopk
2Vr
f
Experiment No. 3.
Diodes and Bridge Rectifiers
Goal: To learn about diodes and to use a diode bridge to build a 14V full-wave rectifying
DC power supply.
Equipment: •
Agilent E3631A Triple output DC power supply
•
Agilent 33120A Function Generator
(Replacement model: Agilent 33220A Function / Arbitrary Waveform
Generator)
•
Agilent 34401A Multimeter
•
Agilent 54645A Oscilloscope
(Replacement model: Agilent DSO5012A 5000 Series Oscilloscope)
1.0 Diode Test Circuit:
1. Set the Agilent E3631A power supply to give a maximum current of 0.6A for the
+6V supply. This is important so as not to burn the diode while testing it.
2. Connect the MUR 105 silicon rectifier diode as follows:
3.
Measure the diode current for V = 0.3, 0.4, 0.5, 0.7, 0.8, 0.9, 1.0, 1.2 and 1.3
up to a diode current of 0.5A.
Determine the diode voltage for Id = 1mA, 10 mA and 100 mA.
2.0 Full-wave Bridge Rectifier with a Single Output:
1.
Measure RL and the 0.5 Ω resistor accurately using the multimeter.
Measure C using the capacitance meter in the lab.
2. Connect the full-wave bridge rectifier as shown below and make sure that the
diode polarity and the capacitor polarity are correct. There is a small (0.5 Ω)
resistor in series with the 220 µF capacitor. This resistor is needed so as to
measure the current in the 220 µF capacitor (IC = VC/R).
TURN OFF THE POWER SUPPLY AND THE MULTIMETER. DISCONNECT ALL
CABLES FROM THESE UNITS AND RETURN THE CABLES TO THEIR ORIGINAL
PLACE. THIS IS VERY IMPORTANT SO AS NOT TO HAVE GROUND LOOPS AND
BURN THE FUSE! YOU WILL NOT USE THESE UNITS AGAIN.
ALSO, ALWAYS OPERATE WITH A SINGLE COAX TO BANANA-PLUG
OSCILLOSCOPE CABLE!
7
0.5 A Fuse
C
+
+
-
Vi
–
AC 120 Vrms
D AC 10 Vrms
D2
D1
A
B
D3
+
220 µF
D4
+
V1
–
0.5 Ω
R L=
470 Ω
Vo
–
B
After you have checked the circuit, connect the transformer to the 120 V AC wall
outlet. If there is a short-circuit, the fuse will blow and Vi = 0!
a.
Draw the circuit in your lab notebook.
b.
Measure Vippk and Virms using the scope. At this time, connect the
oscilloscope ground to point D. It should be 10-11 Vrms.
c.
Measure the spectrum of Vi in dB (fo ,3fo and 5fo). You will notice that the
waveform is not perfectly sinusoidal due to the cheapie wall tranformer.
The magnetic case is saturating at high voltages and distorting the
transformer output signal (Vi)!
BE CAREFUL WHEN YOU ARE DOING VOLTAGE MEASUREMENTS ACROSS
THE DIODE BRIDGE ACROSS RL, ETC. MAKE SURE THAT THE SCOPE
GROUND IS NOW CONNECTED TO NODE B!
3.
Remove the capacitor and measure Vo. Make sure that you have a full-wave
rectified voltage. (Do not worry if you see a distorted sinewave.) Measure
Vopk and determine the diode voltage drop (per diode).
4. a.
Connect the capacitor to the circuit and measure Vopk, Vo(av) and VR
(the ripple voltage) and compare with your pre-lab calculations. Plot Vo.
Measure the frequency of the output voltage ripple.
From the output voltage waveform, measure the charging and
discharging time of the capacitor.
From the output voltage waveform, measure the slope (∆V/∆T) of the
discharge portion at the center of the linear portion. As discussed in
class, this results in the discharge capacitor current (I = C ∆V/∆T)
which is equal to the load current.
b.
Measure Vo in the frequency domain and write the fundamental
harmonics in dB (120, 240, 360 and 480 Hz).
5. On the scope, measure the voltage (V1) across the 0.5 Ω resistor. This
measurement results in the charging and discharging capacitor current (I = V1/R ,
R = 0.5 Ω). Again, the discharge capacitor current is equal to the load current.
Draw accurately the current waveform (I = V1/R) on your lab notebook and
label the peak positive (charging) current and its duration, the average
negative (discharge) current and its duration (the average discharge current is
found at the center of the linear portion of V1). Label the charging and
discharging periods of the capacitor.
3.0 Half-Wave Power Supply With a Single Output:
Assume that one of the diodes blew up and became an open circuit (D2 for example).
the circuit becomes a half-wave rectifier.
Take D2 out of the circuit now. Draw the circuit in your lab notebook.
Repeat parts 3, 4 and 5 of section #2.
8
Experiment No. 3.
Diodes and Bridge Rectifiers
Pre-Lab Assignment
1. For the full-wave bridge rectifier with C = 220 µF, RL = 470 Ω and Vi = 10 Vrms:
a. Calculate the ripple frequency, Vpk, VR, IL, ID(av) and ID(pk).
b. Calculate the time in msec (per period) that the diode(s) is turned on.
c. Plot the current in the 220 µF capacitor for one period.
2. Repeat (1) but for a full-wave bridge rectifier when one of the diodes is blown. This is
equivalent to a half-wave rectifer but with two diodes in series. Therefore, in this special
case, you should take into account the voltage drop of two diodes and Vopk = Vipk - 2 VD
in the half-wave rectifier equations.
3. Regulators are essential for low output ripple voltages and this problem will show why! For
the full-wave rectifier with RL = 470 Ω,
a. Choose C to result in VR = 100 mV and VR = 10 mV.
b. Calculate ID(av) and ID(pk) in each case.
c. Calculate the time in msec or µsec (per period) that the diode(s) is turned on.
You will find that in order to get a low ripple voltage from a bridge rectifier, the time
each diode is “on” is very short and the peak diode current is extremely large! This is
why most bridge rectifiers have a ripple voltage around +10% and are followed by
linear regulators.
4. Calculate the efficiencies of the multiple output linear regulators of Fig. 3a and 3b if each
output is delivering the same current I (the answer is independent of I).
5. Assume that the Agilent E3631A power supply with a variable output from 0–25 V does
not have a controlled-tap transformer, and V= 26 V (fixed) after the bridge rectifier/lowpass filter.
a. Calculate the efficiencies of the linear regulator when the output voltage is set at 2 V,
then at 25 V, and delivers 1 A to a load in both cases.
b. What is the power consumed by the linear regulator in each case?
c. Repeat a and b above with a controlled tap transformer (take a 1 V drop across the
regulator).
9
Experiment No. 3
Diodes and Bridge Rectifiers
Lab Report
1. Report all your measured values in the full-wave bridge rectifier C = 220 µF, RL = 470 Ω
(voltages and capacitor current times, input voltage, component values, etc.) and compare
with calculations done using your measured Virms (or Vippk), and assuming an ideal
diode model (VD = 0.7 V). Put the measured data and calculations in table form. Comment
on the discrepancies, especially for the peak current in the capacitor.
2. Report all your measured values in the half-wave bridge rectifier C = 220 µF, RL = 470 Ω
(voltages and capacitor current). Put the measured data and calculations in table form.
Compare with calculations and comment on the discrepancies.
3. The load current can be obtained in three separate methods:
– Using Vo (av)/RL = IL(av)
– Using the slope of the discharge portion of Vo (I = C ∆V/∆T).
– Using V1/R in the middle of discharge period of the capacitor.
Using your measurements and the exact values of RL, C and the 0.5 Ω resistor,
determine IL(av) using the above methods and comment on any discrepancies. You will
find that the “slope” method is the least accurate. Why?
4. A high performance regulated power supply for audio systems with the following outputs is
needed:
Power Supply #1:
+5 V (Imax = 500 mA):
Microprocessor Circuitry
+12 V (Imax = 1 A):
Analog Small-Signal Audio Circuitry
Power Supply #2:
+24 V (Imax = 6 A):
Output Audio Power Amplifier to drive the speakers
a. First of all, ideally how much audio power can this system give into an 8 Ω speaker
without clipping? The #2 supply can deliver a ± 24 V max. and ± 6A current max.
Therefore, the output of the power amplifier can go to ± 24 V (assume no voltage drop
in the output amplifier).
b. Design the power supplies (#1 and #2) for high efficiency. You can use a multiple-tap
transformer and several rectifiers/regulators. Choose appropriate capacitors for each
supply. The unregulated output of each rectifier should have a ±15% (or 30% total)
ripple voltage at peak current load.
c.
Assume that the voltage drop is 0.5 V across the 5 V linear regulator and 1 V across
each of the 12 V and 24 V linear regulators. Calculate the overall regulator efficiency.
These experiments have been submitted by third parties and Agilent has not tested any of the experiments. You will undertake any of
the experiments solely at your own risk. Agilent is providing these experiments solely as an informational facility and without review.
10
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FOR ANY DIRECT, INDIRECT, GENERAL, INCIDENTAL, SPECIAL OR CONSEQUENTIAL DAMAGES IN CONNECTION
WITH THE USE OF ANY OF THE EXPERIMENTS.
```