AC-DC Converter Design
By
Guoan Chen & Alex Soto
Course: ECE 345
TA: In Seop Lee
Date: 5/4/1999
Project Number: 19
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ABSTRACT
The goal of a AC-DC converter is to simply function as a voltage source. The converter is to provide
any voltage between 5-15 volts. The output voltage must not have an output voltage ripple greater than
1% for it’s given output. The AC-DC will implement a Boost converter that should provide a rather
high efficiency. These conditions will be observed when a load is attached.
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TABLE OF CONTENTS
Page:
1. Introduction ( by Guoan Chen) .......................................................................... 1
2. Design Procedure and Details( by Guoan Chen & Alex Soto)………………… 1
Transformer ............................................................................................................. 2
Rectifier..................................................................................................................... 2
Voltage Regulators................................................................................................... 3
Gate Drive Controller............................................................................................... 4
Boost Converter......................................................................................................... 4
3. Design Verification ( by Alex Soto)......................................................................... 5
4. Costs ( by Guoan Chen)............................................................................................ 7
5. Conclusions ( by Guoan Chen)... ............................................................................. 8
6. Figures and Plots ……………………………………………………………………9
7. Equations (by Alex Soto)……………….………………………………………….. 15
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AC-DC CONVERTER
Introduction:
In this project, the goal is to design and build a AC-DC Converter. Most electronic devices are sold in
the stories that require fixed AC-DC conversion. This design will be able to rectify the AC voltage to a
DC voltage anywhere between 5 to15 volts. An adjustable DC-DC converter allow this design to
convert the AC voltage to the desired DC voltage. The whole design can de divided into three
subprojects. The tree subprojects are the AC rectification, the DC-DC converter and the gate drive of the
switching device.
Design Procedure and Details:
AC
Input
Transformer
Rectifier
Voltage
Regulator
DC-DC
Converter
Output
Gate
Drive
To begin the design of a AC-DC converter, there is an obvious need to convert the standard utility 120
ac voltage to a useful DC voltage which would power the components within the converter. To use any
other voltage source other than the AC voltage for the AC-DC converter would defeat its purpose.
Having chosen a single phase transformer the electronics shop, it became necessary to construct a
rectifier circuit, filter and voltage regulator for the assembly of the power source for the AC-DC
converter.
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This AC-DC converter basically composed five main design blocks: they are the transformer, rectifier,
Voltage regulators, Gate drives controlled chip, and the boost converter. Now, let’s go to the detail of
every single building blocks.
The Transformer:
The first few blocks of the block diagram are composed of the building blocks of the power supply,
specifically the transformer, rectifier, filter and voltage regulator blocks. The transformer, the first
building block of the converter required the ac input. A basic two pin ac plug connector was connected
to the transformer allowing it to be plugged into any standard wall plug. For safety reasons we affixed
the transformer in a metal case and wrapped the outside of the box in a layer of electric tape. Also,
instead of plugging the transformer directly into a wall plug, a surge protector with an on/off switch was
used. (A.S)
The transformer is used for this project which is a step down control transformer.
Here are some data specifications of the transformer:
Style
Secondary
In Parallel
In Series
Each winding
LHV
12V at 4-8A
12V at 16A 25V at 8A
Height
Width
Depth
Mounting
(cm)
(cm)
(cm)
(cm)
4(3/16)
3(1/2)
3(5/8)
2(3/4)*3
The transformer is designed to connect in series for AC-DC converter design. It is needed to input at 120
Vac, and output at 25 Vac. It is about a 4:1 step down rate. An isolation box is also designed for the
use of this transformer. It is not only providing electrical isolation between the powerline circuit and
the electronic circuits, but also it biased the rectifier circuit. This isolation can totally reduces the risk
of electrical shock (GC)
Rectifier:
A full bridge rectifier is used for the rectification of the step down AC voltage from the transformer. The
rectifier design is composed 4 diodes in a full-wave bridge configuration. D1and D2 are forward biased,
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D3 and D4 are reverse biased. Since Vac is a sinusoidal input voltage, during the positive half of the
input voltage cycle, Vac is positive, D1 and D2 are forward biased, have the same current direction with
Vac, but since D3 and D4 are reverse biased, so they are cut off. At the negative half of the input
voltage cycle, D3 and D4 are reverse biased, have the same current direction with Vac, but since D1 and
D2 are forward biased, they are turned off. By using the full bridge rectifier, both of positive and
negative cycle of Vac are totally rectified. Thus, this full bridge rectifier design can minimize the lost of
current during the Vac to Vdc transition between the transform and the rectifier.(G.C)
The use of this rectifier will reduce the size of the capacitor then would otherwise be necessary with
half-wave rectification. The reason being that the sinusoidal peaks would occur at every half period in
full wave rectification as opposed to peaks occurring every full period in half wave rectification,
lessening the discharging time of the filter capacitor. (A.S)
Voltage Regulators:
For the sake of simplicity and debugging simple Motorola three pin voltage regulators will be used
instead of Zener Diodes. The two voltage regulators, 7812CT and 7805CT, used will limit the voltage to
twelve and five volts for the integrated chip and Boost converter voltage inputs, respectively. The input
to the five-volt regulator was taken from the twelve-volt regulator output to reduce the power dissipation
of the voltage regulator. (A.S)
The 25 volt Vac from the transform becomes Vdc after it pass through the rectifier, but however, the
magnitude of the voltage are still the same. Thus, voltage regulators are needed to regulate the 25 Vdc to
some useful levels. There are two different voltage regulators used in this AC-DC circuit design, a12
volt regulator which supplies voltage to the pulse width modulation chip, and a 5 volt regulator is acted
as the voltage source for the boost converter.(GC)
The voltage regulators, when loaded, produced an unanticipated rippling output voltage. As the case
with most undesired voltage ripples, the solution was to simply place a capacitor at output and ground,
eliminating the voltage ripples. The capacitance’s were 1 uF, which happened to be on hand, avoiding
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the need to calculate any other values. With the necessary DC voltages needed for the Boost converter
and the gate drive, thus concludes the power assembly of the AC-DC (A.S)
Gate Drive Controller:
The third subproject was to assemble a gate drive for the switching device, the mosfet. This gate drive
would have to generate a square wave with an adjustable duty ratio. The high voltage of the square wave
must be safely above the 2.7 volts necessary to turn on the mosfet. There are several gate drives that can
be used in this low power application. The gate drive considered in this converter, was suggested the
Prof. Krein, is shown in Fig.2. Similar gate drives can be referred to in the ECE 369 lab notes.(A.S)
A current mode pulse width modulation control chip, UC 3842 is used for the gate drive control. Please
look at the PWM Control diagram for detail of the circuit design. The UC 3842 generate a square wave
signal to the switching device , a MOSFET gate. The square wave has an up and down cycle. The up
cycle is 5 volt and the down cycle is –5 volt. The MOSFET gate need 2.7 volt to turn on, and only if the
gate is on, the current can go through the MOSFET to the boost converter. Thus, at the up cycle , the
gate is on and at the down cycle, the gate is off. And the duty ratio of the MOSFET is controlled by a
potentiometer which is connected with Pin 8 and Pin 1. The frequency of the switching is set at 10
kHz.(GC)
Boost Converter:
The actual manipulation of the output voltage requires a DC-DC converter. The DC-DC converter used
was a Boost converter that has an output voltage greater than the input voltage. The simple description
on the boosting principles can be seen in the Boosting Principles diagram, Fig.3. The characteristics of a
Boost converter are to take a voltage input and output a voltage equal greater than the input voltage
depending on the duty ratio of the switching device that allow current to flow for a certain time.
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A simulation of the circuit shown in Fig.4 yielded a voltage output show in Fig.5. Pspice does not
simulate these kinds of converters very well. In reality the output is a DC voltage. The calculated values
of this converter can found in the Equations chapter of this report. (A.S)
It basically is a step up voltage converter. The boosting principle behind this converter is to draw a
current through the inductor and in a short burst deliver the current across a load. The relationship of the
boost is :
Vout = Vin/[1-D]
Where D1 is the duty ratio of the mosfet.
However a boost converter needs to have a minimum voltage to start with. Then the starting voltage can
be boosted up to the level as desired. How much range the boost converter can go, that depends two
major factors : the inductor and the pulse width modulator chip. A bigger inductor is harder to have it
go to saturation, and more current can be draw to the output resistor of the boost converter, thus, the
voltage output of the converter can go to higher range. Other important factor is the pulse width
modulation chip. As it was introduced in the previous section, it is the gate drive controller for the
mosfet. And the mosfet totally control how much current to the inductor by turning ON or OFF. So if a
square wave which generate by the chip have a longer up cycle width, the gate would stay open for a
longer period of time, and more current can go to the boost converter , the output voltage would be
bigger. Thus, a higher performance PWM chip can definitely improve the output range of the boost
converter. (GC)
Design verification
Initially the Boost converter exhibited characteristics of a Buck converter. That is to say that the output
voltage could be varied from five volts to a smaller voltage. Apparently the inductor was at fault. The
inductor was a physically small device with very thin wire. This may have caused saturation in the
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inductor. When saturation occurs the magnetic flux, the energy stored in the inductor, leaks from the
core to the air surrounding it. To avoid saturation in an inductor we consider the magnetic relation:
N i < Bsat R A
The physical parameters of the inductor can’t be changed unless different inductors are used. The
inductors in the electronics shop were of the same type is thin wound wire so using a single different
inductor would not work, however, if there were a number of inductors, each supporting a fraction of the
current, saturation could be avoided. Using four 10 mH inductors in parallel, the Boost converter
assumed it’s expected characteristics. The unloaded output voltages can be seen in Fig.6. For all output
voltages, the output voltage ripple was far less than 1%. Using a multimeter the observed fluctuation for
the 15-volt output voltage was only 60 mV which results in ripple of 0.4% in the worst case.
As a voltage source, it is expected to be able to support an external load connected in parallel to
the internal resistor. Using a ten kilo-ohm potentiometer we measured that minimum load, which would
draw the most power before the output voltage varied more than the specified tolerance of 1%. The
minimum load was 8 kbefore the 15-volt output voltage varied 1% yielding a maximum power yield
of:
P = V I = V ( V/R) = V² / R = 15² / 8 k
P = 0.0281 W
The efficiency of the converter:
Vin = 5 V
Vout = 16.6 V
Iin = Iout / [1-D1]
Iout = 16.6 V / 1000
Iin = 0.4040 A
Iout = .0166 A
Pin = 0.2023 W
Pout = 0.1759 W
Efficiency =  = Pout / Pin = 0.869 = 86.9 %
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Cost:
Parts
Quantity
Price/per-piece
Cost
Transformer
1
$20
$ 20
US 3842
1
$3.3
$ 3.3
Resistors
6
$ 0.20-0.7
$2
Diodes
6
$0.2-0.8
$ 2.4
Capacitors
6
$0.5 – 1.50
$ 12
Mosfet
1
$0.7
$0.7
Inductors
4
$0.5
$2.0
AC Plug connector
1
$2.8
$2.8
Voltage regulators
2
$1.1
$2.2
LED
2
$0.5
$1.0
Subtotal:
$42.3
Labor
Dollars/ per hr
Cost ( 6hr/per week, 7 weeks)*2.5
Donald
$25/per hr
$2625
Alex
$25/per hr
$2625
Total
$5250
The total cost of this project : $42.30+ $5250.00 = $5292.30
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Conclusion:
The AC-DC converter design has been tested with computer simulation and analysis and real resistance
load. It has not only achieved the voltage output range requirement ( 5V-15 V) but also it have operated
at a great efficiency, 86%, which is compatible with the industry standard.. But however, the testing
results also show some uncertainties about the power output of this AC-DC converter, which is below
our expectation. The problem is most likely is caused by the old transformer which we obtained from the
electronics shop and became inoperable after the demonstration.
Figures and Plots
( they are attached with hard copy report)
Equations
To calculate the values of the components of the DC-DC converter, the following observations must be
taken. When the switching device is on, the mosefet, the inductor voltage is simply Vin = 5 V. The
current will rise linearly as:
5 V = L di/dt = L i / t
where i acceptable current change related to the output voltage ripple by
V=IR
Vout = Iout * 1000  = 15
Iout = 0.015 A
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and,
Iin = Iout / [ 1 - D1] = 0.015 / [ 1 - 2/3 ] = 0.045 A
and t is the on time of the mosfet.
Vout = Vin / [ 1 - D1 ]
15 = 5 / [1- D1 ]
D1 = 2/3
The on time is calculated as follows:
t = T * D1 = (1 / 16000Hz) * 2/3 = 41.6 µs
Thus, the inductor value is:
L = 5 * t / i = 5 * t / Iin
L = 4.55 mH
In the case of the output the capacitor must maintain the output voltage. Remember that the
capacitance equation is:
Ic = C dv / dt = C v / t
Thus,
C = Ic * t / v
C = (15/1000) * (1/16000 * 1/3) / (12 * 0.01)
C = 2.08 uF
This is an unloaded case where the output duty ratio is at it’s smallest for the “off” mosfet state. To
insure that the voltage doesn’t vary for the loaded case and different voltage levels, the capacitance
should be set much higher. The specific capacitance would depend on the load current and voltage.
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