Elenco Electronics XP-620K User guide

POWER SUPPLY KIT
MODEL XP-620K
ElencoTM Electronics, Inc.
Assembly and Instruction Manual
Copyright © 2003, 1989 ElencoTM Electronics, Inc. Revised 2003 REV-O 75320K

PARTS LIST
If any parts are missing or damaged, see instructor or bookstore. DO NOT contact your place of purchase as
they will not be able to help you. Contact ElencoTM Electronics (address/phone/e-mail is at the back of this
manual) for additional assistance, if needed.
RESISTORS
Qty. Symbol Value Color Code Part #
1 R5 .18W5% 3W 101804
2 R3, R4 2.7W5% 1/2W red-violet-gold-gold 112701
2 R1, R2 150W5% 1/4W brown-green-brown-gold 131500
2 VR1, VR2 2kWPotentiometer 192422
CAPACITORS
Qty. Symbol Value Description Part #
5 C1 - C4, C8 10mF Electrolytic 271045
2 C5, C6 2200mF Electrolytic 292226
1 C7 4700mF Electrolytic 294744
SEMICONDUCTORS
Qty. Symbol Value Description Part #
4 D1 - D4 1N4001 Diode 314001
4 D5 - D8 1N5400 Diode 315400
1 Q1 A70 Transistor PNP 320070
1 Q2 2N6124 Transistor PNP 326124
1 IC1 LM-317 Integrated Circuit 330317
1 IC2 LM-337 Integrated Circuit 330337
1 IC3 LM-7805 Integrated Circuit 337805
MISCELLANEOUS
-1-
Qty. Description Part#
1 Transformer 440720
1 PC Board 512013
1 Fuse 1.25A 530125
1 Rocker Switch 541204
1 Cover 611120
1 Chassis 612020
1 Heat Sink 615010
2 Knob 622009
1 Strain Relief 624003
5 Insulating Washer 624007
4 PC Board Stand-off 625001
2 Black Binding Post 625031
5 Lockwasher, Binding Post 625031LW
5 Nut, Binding Post 625031HN
2 Red Binding Post 625032
1 Yellow Binding Post 625034
5 Screw 6-32 x 3/8” Phillips 641640
2 Screw 8-32 x 3/8” Phillips 641840
4 Screw #6 x 3/8” Black Truss Head 642652
2 Screw #6 x 3/8” black AB Phillips 642660
2 Nut 7mm 644101
Qty. Description Part#
1 Nut 6-32 Large 644600
4 Nut 6-32 Small 644601
2 Nut 8-32 644800
2 Flat Washer 8mm x 14mm 645101
2 Lockwasher 5/16” 646101
2 Lockwasher #8 646828
4 Rubber Feet 662003
1 Fuse Holder Body 663005LB
1 Fuse Holder Nut 663005N
1 Fuse Holder Cap 663005UB
1 Fuse Holder Washer 663005W
3 Mica Insulator 780002
1 Silicon Grease 790005
6” 20 Ga. Red Wire 813210
48” 22 Ga. Red Wire 814201
48” 22 Ga. Orange Wire 814310
48” 22 Ga. Blue Wire 814610
1 Line Cord 862105
2” Shrink Tubing 1/2” Dia. 891101
1.5” Shrink Tubing 3/4” Dia. 899110
1 Solder Tube 9ST4

-2-
PARTS IDENTIFICATION
Resistor 2kW
WPotentiometer Capacitor Diode Transistor
Integrated Circuit Transformer PC Board Fuse
Heatsink
Switch Knob Strain Relief
PC Board Stand-off
Binding Post Assembly Fuse Assembly
Flat Washer Lockwashers
MicaNuts
7mm 6-32 / 8-32
#8 5/16”
Binding Post
Lockwasher
Nut
.18W3W Resistor

-3-
IDENTIFYING RESISTOR VALUES
Use the following information as a guide in properly identifying the value of resistors.
BAND 1
1st Digit
Color Digit
Black 0
Brown 1
Red 2
Orange 3
Yellow 4
Green 5
Blue 6
Violet 7
Gray 8
White 9
BAND 2
2nd Digit
Color Digit
Black 0
Brown 1
Red 2
Orange 3
Yellow 4
Green 5
Blue 6
Violet 7
Gray 8
White 9
Multiplier
Color Multiplier
Black 1
Brown 10
Red 100
Orange 1,000
Yellow 10,000
Green 100,000
Blue 1,000,000
Silver 0.01
Gold 0.1
Resistance
Tolerance
Color Tolerance
Silver +10%
Gold +5%
Brown +1%
Red +2%
Orange +3%
Green +0.5%
Blue +0.25%
Violet +0.1%
BANDS
12 Multiplier Tolerance
IDENTIFYING CAPACITOR VALUES
Capacitors will be identified by their capacitance value in pF (picofarads), nF (nanofarads), or mF (microfarads). Most
capacitors will have their actual value printed on them. Some capacitors may have their value printed in the following
manner. The maximum operating voltage may also be printed on the capacitor.
Second Digit
First Digit
Multiplier
Tolerance The letter M indicates a tolerance of +20%
The letter K indicates a tolerance of +10%
The letter J indicates a tolerance of +5%
FortheNo.01234589
Multiply By 1 10 100 1k 10k 100k 0.01 0.1
Multiplier
Note: The letter “R” may be used at times
to signify a decimal point; as in 3R3 = 3.3
103K
100V
Maximum Working Voltage
The value is 10 x 1,000 = 10,000pF or .01mF 100V
10mF 16V
METRIC UNITS AND CONVERSIONS
Abbreviation Means Multiply Unit By Or
p Pico .000000000001 10-12
n nano .000000001 10-9
mmicro .000001 10-6
m milli .001 10-3
– unit 1 100
k kilo 1,000 103
M mega 1,000,000 106
1. 1,000 pico units = 1 nano unit
2. 1,000 nano units = 1 micro unit
3. 1,000 micro units= 1 milli unit
4. 1,000 milli units = 1 unit
5. 1,000 units = 1 kilo unit
6. 1,000 kilo units = 1 mega unit

-4-
Introduction
The most important factor in assembling your XP-620K Power Supply Kit is good soldering techniques. Using
the proper soldering iron is of prime importance. A small pencil type soldering iron of 25 - 40 watts is
recommended. The tip of the iron must be kept clean at all times and well tinned.
Safety Procedures
• Wear eye protection when soldering.
•
Locate soldering iron in an area where you do not have to go around it or reach over it.
•Do not hold solder in your mouth. Solder contains lead and is a toxic substance. Wash your hands
thoroughly after handling solder.
• Be sure that there is adequate ventilation present.
Assemble Components
In all of the following assembly steps, the components must be installed on the top side of the PC board unless
otherwise indicated. The top legend shows where each component goes. The leads pass through the
corresponding holes in the board and are soldered on the foil side.
Use only rosin core solder of 63/37 alloy.
DO NOT USE ACID CORE SOLDER!
CONSTRUCTION
Solder Soldering Iron
Foil
Solder
Soldering Iron
Foil
Component Lead
Soldering Iron
Circuit Board
Foil
Rosin
Soldering iron positioned
incorrectly.
Solder
Gap
Component Lead
Solder
Soldering Iron
Drag
Foil
1. Solder all components from
the copper foil side only.
Push the soldering iron tip
against both the lead and
the circuit board foil.
2. Apply a small amount of
solder to the iron tip. This
allows the heat to leave the
iron and onto the foil.
Immediately apply solder to
the opposite side of the
connection, away from the
iron. Allow the heated
component and the circuit
foil to melt the solder.
1. Insufficient heat - the
solder will not flow onto the
lead as shown.
3. Allow the solder to flow
around the connection.
Then, remove the solder
and the iron and let the
connection cool. The
solder should have flowed
smoothly and not lump
around the wire lead.
4. Here is what a good solder
connection looks like.
2. Insufficient solder - let the
solder flow over the
connection until it is
covered. Use just enough
solder to cover the
connection.
3. Excessive solder - could
make connections that you
did not intend to between
adjacent foil areas or
terminals.
4. Solder bridges - occur
when solder runs between
circuit paths and creates a
short circuit. This is usually
caused by using too much
solder. To correct this,
simply drag your soldering
iron across the solder
bridge as shown.
What Good Soldering Looks Like
A good solder connection should be bright, shiny,
smooth, and uniformly flowed over all surfaces.
Types of Poor Soldering Connections

-5-
ASSEMBLE COMPONENTS TO PC BOARD
Figure A
Diodes have polarity. Be sure
that the band is in the correct
direction.
Figure B
Electrolytics have a polarity
marking indicating the (–)
lead. The PC board is marked
to show the lead position.
Figure C
Mount the transistor with the
flat side as shown on the top
legend. Leave 1/4” between
the part and PC board.
Band
D5 - 1N5400 Diode
D6 - 1N5400 Diode
D7 - 1N5400 Diode
D8 - 1N5400 Diode
(see Figure A)
C1 - 10mF Electrolytic
C2 - 10mF Electrolytic
C3 - 10mF Electrolytic
C4 - 10mF Electrolytic
C8 - 10mF Electrolytic
(see Figure B)
C7 - 4700mF Electrolytic
(see Figure B)
Q1 - A70 Transistor
(see Figure C)
D4 - 1N4001 Diode
D3 - 1N4001 Diode
D2 - 1N4001 Diode
D1 - 1N4001 Diode
(see Figure A)
R2 - 150WResistor
R1 - 150WResistor
(brn-green-brn-gold)
C6 - 2200mF Electrolytic
C5 - 2200mF Electrolytic
(see Figure B)
R3 - 2.7WResistor
R4 - 2.7WResistor
(red-violet-gold-gold)
R5 - .18WResistor
Flat
1/4”
(–) (+)
Polarity
Mark

-6-
PC BOARD WIRING
Cut the wires to the required length. Strip 1/4” of insulation off of both ends. Insert the lead into the hole and
solder it to the foil side.
0123
Use this ruler to measure the wires when cutting them to their required lengths.
4567
4” Red
Hole K
4” Orange
Hole J
3” Red
Hole I
3” Orange
Hole H
3” Blue
Hole G
4” Blue
Hole F
4” Red
Hole E
6” Blue
Hole D
6” Red
Hole C
3 1/2” Red
Hole W
3” Orange
Hole V
3 1/2” Blue
Hole U
3” Blue
Hole T
3 1/2” Orange
Hole S
3” Red
Hole R
4 1/2” Blue
Hole Q
5” Orange
Hole P
4 1/2” Red
Hole O
5” Blue
Hole N
4 1/2” Orange
Hole M
5” Red
Hole L

-7-
Install the binding posts with the colors in order, as shown in Figure D. Remove the large nut and splitwasher.
Insert the post into the hole and fasten it with the nut and splitwasher. Tighten down the nut with pliers.
Cut off the tabs on the two potentiometers and install them with the lugs up, as shown in Figure D. Secure
in place with a 5/16” lockwasher, 8mm flat washer and 7mm nut.
Turn both potentiometer shafts all the way counter-clockwise. Line up the line on the
knobs with the first line on the voltage scale. Press knobs onto the shaft of the
potentiometers.
Note the lug configuration on the rocker switch. Push the switch into the hole in the
chassis with lug 1 on top as shown in Figure D.
PANEL ASSEMBLY
Figure D
Lug 1
Lockwasher
Nut
Red
Black
Yellow
Rocker Switch
Rear View of Rocker Switch
7mm Nuts
8mm Washers
1
5/16” Lockwashers
Potentiometers * Cut off tabs
2
3
Red
Black
I
N
C
R
E
A
S
E

Carefully bend the leads of IC1, IC2,
IC3 and Q2 at right angles with pliers.
Install IC1, IC2 and Q2 in the
positions shown in Figure E. Fasten
in place using the parts shown in
Figure F. Spread the silicon grease
on the back of the transistor and ICs.
IC1 - LM-317 IC
IC2 - LM-337 IC
Q2 - 2N6124 Transistor
Install IC3 as shown in Figure Fa.
IC3 - LM-7805 IC
Mount the fuse holder to the top hole in the back of the chassis,
with the side lug up, as shown in Figure G. Fasten in place with
the fuse holder nut. After the holder is secure, insert the fuse and
screw the cap on.
Separate the wires of the line cord 3” from the end. Strip the
insulation off the end of all three wires to expose 1/4” of bare wire.
Insert 6” of the line cord into the bottom hole on the back
of the chassis, as shown in Figure G. Place the line cord
into the slot of the strain relief and squeeze the two
sections together with pliers. Then, insert the strain relief
into the hole.
One side of the transformer has only 2 black wires. Cut
these wires so that they are 4 1/2” in length (see Figure H).
Strip the insulation off the end of the wires to expose 1/4”
of bare wire.
Cut the red, blue and black wires on the other side of the transformer so that they are 3” in length. Strip the
insulation off the end of the wires to expose 1/4” of bare wire.
Cut the yellow wires flush with the transformer.
Install the transformer with the black wires side as shown in Figure J. Use an 8-32 x 3/8” screw, #8
lockwasher and an 8-32 nut on each side to fasten in place, as shown in Figure I.
-8-
Figure E
LM7805
LM-337
LM-317
2N6124
Mica
Figure F
Small 6-32 Nut
IC1, IC2,
Q2
Heat Sink
Insulator Washer
6-32 x 3/8”
Screw Silicon GreaseMica
Figure Fa
Small 6-32 Nut
IC3
LM7805
Heat Sink
Insulator Washer
6-32 x 3/8” Screw Insulator Washer
Figure G
Ribbed
Smooth
1/4”
Figure I
8-32 Nut
#8 Lockwasher
8-32 x 3/8” Screw
Figure H
4” Red
4” Red
4” Blue
4 1/2” Black
Side Lug
Pliers
Fuse Holder
Fuse
Cap
Washer
Nut
4” Blue
4” Black

-9-
Solder the black transformer wire (B), as shown in Figure J to lug 2 on the rocker switch.
Twist the black transformer wire (A) and the ribbed or flat line cord wire together. Solder the two wires to
lug 1 on the rocker switch, as shown in Figure J.
Slide the 3/4” diameter shrink tubing over the switch and shrink into place.
Solder the two red transformer wires to the holes marked T2 on the PC board.
Solder the black transformer wire to the hole marked CT1 on the PC board.
Solder the two blue transformer wires to the holes marked T1 on the PC board.
Slowly push the PC board stand-offs into the four holes on the bottom of the chassis. Push the PC board
down into place.
WIRING LINE CORD, FUSE, TRANSFORMER AND SWITCH
Figure J
PC Board
Chassis
Side Lug
End Lug
Smooth Line Cord
Flat or Ribbed Line Cord
12
3
(A) Black Red
Red
2” Tubing
Yellow
Yellow
Ribbed
Line Cord
Install the line cord ground lug to the chassis
using a 6-32 x 3/8” screw and a 6-32 large
nut in the location shown in Figure J.
Strip the insulation off of both ends of the
6” red 20 ga. wire to expose 1/4” of bare
wire. Solder one end of the wire to lug 3 on
the rocker switch, as shown in Figure J.
Slip the 2” section of shrink tubing over the
other end of the 6” 20 ga. red wire and the
smooth or round line cord wire. Solder the
line cord wire to the end lug on the fuse
holder, as shown in Figure J. Solder the 6”
20 ga. red wire to the side lug on the fuse
holder. Slide the shrink tubing over the fuse
holder, covering the two lugs. You may use
a hair dryer, heat gun (at lowest setting or
you will melt the tubing) or the heat
emitting from your soldering iron.
CAUTION: DO NOT touch any wires or
tubing with the iron.
Slip the 6” strip of red wire (leading from the
side lug of the fuse holder), the (A) and (B)
black transformer wire, and the ribbed line
cord wire through the 3/4” diameter piece of
shrink tubing (as shown in Figure J).
Blue
Black
Blue
(B) Black
6” 20 Ga.
Red
JK
T2 T2 T1
CT1
T1
W
VU
3/4” Shrink
Tubing
6-32 x 3/8” Screw
6-32 Large Nut

-10-
WIRE BINDING POSTS AND 317, 337
Solder the wires from the board to the binding posts, as shown in Figure K.
3” Blue wire from (G) on the PC board; To the Yellow post (–1.5 - 15V).
3” Orange wire from (H) on the PC board; To the Black post (common).
6” Blue wire from (D) on the PC board; To the Black
post (common).
6” Red wire from (C) on the PC board; To the
Red post (+5V 3A).
3” Red wire from (I) on the PC board; To the Red post (+1.5 - 15V).
Place the heat sink with ICs and
transistor in the position, as shown in
Figure K. Insert the wires from the PC
board, through the rectangular hole in
the chassis, to the ICs and solder into
place.
Tin the leads. Form the end of the wires
into a tight loop, for easy, tight
connection to leads, before you apply
solder.
Figure K
LM-317 LM-337
IHGDC
Black
Post
6” Red
6” Blue
3” Red
Red
Post
Black
Post
Yellow
Post
3” Orange
3” Blue
3 1/2” Red wire (W) from PC board; To
middle lead of LM-317.
3 1/2” Orange wire (S) from PC
board; To left lead of LM-317.
3 1/2” Blue wire (U) from PC board;To
right lead of LM-317.
3” Red wire (R) from PC board; To
middle lead of LM-337.
3” Blue wire (T) from PC board; To left
lead of LM-337.
3” Orange wire (V) from PC board; To
right lead of LM-337.
After wiring the ICs, be sure that none
of the leads touch each other and
cause a short.
Red
Post

-11-
WIRE 2N6124, 7805 & POTENTIOMETERS
Insert the wires from the PC board through the
rectangular hole in the chassis to the 2N6124
and LM-7805, solder into place, as shown in
Figure L.
5” Red wire (L) from the PC board; To
middle lead 0f 2N6124.
5” Orange wire (P) from the PC board; to
left lead of 2N6124.
5” Blue wire (N) from the PC board; to right
lead of 2N6124.
4 1/2” Red wire (O) from PC board; to
middle lead of LM-7805.
4 1/2” Blue wire (Q) from PC board; to left
lead of LM-7805.
4 1/2” Orange Wire (M) from PC board; to
right lead of LM-7805.
After wiring, be sure that the leads do not touch
each other and cause a short.
Solder the wires from the PC board to the
potentiometers, as shown in Figure L.
4” Red wire (E) from PC board; to middle
lug of the positive voltage pot.
4” Blue wire (F) from PC board; to right lug
on the positive voltage pot.
4” Orange wire (J) from PC board; to
middle lug on the negative voltage pot.
4” Red wire (K) from PC board; to right lug
on negative voltage pot.
4 1/2” Blue
5” Orange
4 1/2” Red
5” Blue
4 1/2” Orange
5” Red
ELENCO ELECTRONICS INC. XP-620
4” Blue
4” Red
EF
JK
4” Red
4” Orange
Negative VoltagePositive Voltage
Potentiometers
Figure L
QPO NML
2N6124 7805

-12-
FINAL ASSEMBLY
Fasten the heat sink to the chassis with two 6 x 3/8” black screws, as shown in Figure M.
Fit the cover onto the chassis. Fasten in place with two 6 x 3/8” truss head black screws on each side, as
shown in Figure M.
Peel off the protective paper from the bottom of the rubber feet and apply one to each corner on the bottom
of the chassis, as shown in Figure M.
TESTING THE XP-620 POWER SUPPLY
Testing the XP-620 Power Supply is very simple. Before applying power to the unit, be sure that all wiring and
soldering is firm. If so, obtain a digital voltmeter. Apply power to the XP-620 and measure the output voltages.
The variable voltage supplies should read between 2 and 16 volts. The 5V supply should read between 4.50
and 5.50 volts.
Next, short the output of each supply. They should turn off and recover, when the short is removed. If you have
a 10 ohm 10 watt resistor, place it across the output terminal (a 2 watt will work, but use it only a few seconds).
The output of the supply should not change more than .2 volts. Do the same on the 5V supply using a 2.5 ohm
12 watt resistor. Again, the output should not change more than .2 volts. In making this test, the voltmeter leads
should be clipped to the terminal directly and not to the load leads. This is to prevent errors in voltage drop due
to contact resistance of the load.
Should any of these tests fail, please refer to the troubleshooting guide.
Bottom View
Figure M
6 x 3/8” Screws
Feet
Feet
6 x 3/8” Screws Truss Head 6 x 3/8” Screws Truss Head

-13-
TROUBLESHOOTING GUIDE
No 2 - 15V Output Voltage
1) Check the AC voltage at anode of D1. It should read about 17VAC. If not, check the fuse, transformer,
ON/OFF switch or line cord.
2) Measure voltage at output of D1. It should read about 24VDC. If not, check D1, D3 and C5.
3) If 20V is OK, check IC1.
No Negative Voltage Output
1) Check the voltage at the output of D4. It should be –24VDC. Check D2, D4 and C6 and make sure that
they are not in backwards.
2) If DC is OK, then check IC2.
No 5V Output
1) Check the voltage across the transformer winding. It should read about 12 volts. If not, check the diode
bridge or C7.
2) Measure the DC voltage at the output of the diode bridge. It should read about 12 volts. If not, check the
diode bridge or C7.
3) If DC is OK, check IC3, Q1 and Q2.
Poor Regulation on any Supply
1) Check DC voltage at the input of the regulator. It should be greater than 18 for 2 - 15V output and 8V for
5V output.
2) Check AC ripple at the input of the regulator. It should be less than 5V for the variable supply and the 5V
supply.
3) If the ripple is greater, then check the diodes and its filter capacitor.
Fails to Shut Down on 5V Overload
1) Check transistor Q1 and resistors R3, R4 and R5.
Unable to Draw 3 Amps at 5 Volts
1) Check transistor Q2 and resistors R3 and R4.

-14-
CIRCUIT DESCRIPTION
Introduction
The Model XP-620 Power Supply features three solid-state DC power supplies. The first supply gives a negative
1.5 to 15 volts at 1 ampere. The third has a fixed 5V at 3 amperes. All DC supplies are fully regulated. A special
IC circuit keeps the output voltage within .2V when going from no load to 1 ampere. The output is fully protected
from short circuits. This supply is ideal for use in school labs, service shops or anywhere a precise DC voltage
is required.
The Positive 2-15V Power Supply
Figure 1 shows a simplified circuit diagram of the positive supply. It consists of a power transformer, a DC
rectifier stage and the regulator stage.
Transformer
The transformer T1 serves two purposes. First, it reduces the 120VAC input to 17VAC to 8VAC to allow the
proper voltage to enter the rectifier stages. Second, it isolates the power supply output from the 120VAC line.
This prevents the user from dangerous voltage shock should the user be standing in a grounded area.
AC to DC Converter
The AC to DC converter consists of diodes D1 and D3 and
capacitor C5. Transformer T1 has two secondary
windings
which are 180 degrees out of phase. The AC output at each
winding is shown in Figure 2A and 2B.
Diodes are semiconductor devices that allow current to flow
in one direction. The arrow in Figure 3 points to the
direction that the current will flow. Only when the
transformer voltage is positive will current flow through the
diodes. Figure 3 shows the simplest possible rectifier
circuit. This circuit is known as a half-wave rectifier. Here
the diode conducts only half of the time when the AC wave
is positive as shown in Figure 2C. Use of this circuit is
simple but inefficient. The big gap between cycles require
much more filtering to obtain a smooth DC voltage.
By addition of a second diode and transformer winding, we can fill in the gap between cycles as shown in
Figure 4. This circuit is called full-wave rectification. Each diode conducts when the voltage is positive. By
adding the two outputs, the voltage presented to capacitor C5 is more complete, thus easier to filter, as shown
in Figure 2E. When used in 60 cycles AC input power, the output of a full wave rectifier will be 120 cycles.
Capacitor C5 is used to store the current charges, thus smoothing the DC voltage. The larger the capacitor, the
more current is stored. In this design, 2200mF capacitors are used, which allows about 3 volts AC ripple when
one amp is drawn.
Figure 1
Figure 3
Figure 4
Figure 2
Simplified diagram of positive power supply
120VAC
Input 17VAC 20VDC 2 - 15V
Regulated
Output
Transformer
120V to 17V
AC to DC
Converter
Voltage
Regulator
Voltage Waveform for Supply
A) Transformer
Winding AB
B) Transformer
Winding BC
C) Output of
diode D1.
D) Output of
diode D2.
E) Total of diodes
D1 & D2.
20V
F) Output of capacitor C1
Ripple depends on load
current (expanded).
Half Wave Rectifier
Full Wave Rectifier

-15-
In practice, the current through the diodes is not as shown in Figure 2C. Because capacitor C5 has a charge
after the first cycle, the diode will not conduct until the positive AC voltage exceeds the positive charge in the
capacitor. Figure 5 shows a better picture of what the current flow looks like, assuming no loss in the diode.
It takes a few cycles for the voltage to build up on the capacitor. This depends on
the resistance of the winding and diode. After the initial start-up, there will be a
charge and discharge on the capacitor depending on the current drawn by the
output load. Remember current only flows through the diode when the anode is
more positive than the cathode. Thus, current will flow in short bursts as shown
in Figure 5C.
The DC load current may be one ampere, but the peak diode current may be three times that. Therefore, the
diode rating must be sufficient to handle the peak current. The 1N4001 has peak current rating of 10 amps.
Regulator Circuit
The regulator circuit in the Model XP-620 Power Supply consists of a LM-317 integrated circuit. This IC is
specially designed to perform the regulation function. Figure 6 shows a simplified circuit of how the LM-317 IC
works.
Transistors Q1 and Q2 form a circuit known as a differential amplifier. Transistor Q1 base is connected to a
stable 1.5V reference voltage. The base of Q2 is connected to the regulator output circuit through a voltage
divider network. The collector of transistor Q2 is connected to a current source. This basically is a PNP
transistor biased to draw about 1mA of current. Transistor Q2 sees the current source as a very high resistor
of about 1 meg ohms. Thus, the gain of transistor Q2 is extremely high.
Transistor Q5 is called the pass transistor. It controls the current reaching the
output. Transistors Q3 and Q4 are emitter followers. Their function is to raise
the impedance of the pass transistor. Note that transistors Q2, Q3, Q4 and Q5
and resistor R1 form a close loop. Also, note that the feedback to the base of
Q2 is negative, that is, when the base of Q2 goes positive, the output at emitter
Q5 goes negative. Now if the 2V output voltage goes down because of current
drain at the output, the base of Q2 will drop forcing the collector voltage of Q2
to go higher. This will bring the output voltage back to 2V. This is the basis of
all negative feedback regulators.
Another feature of the LM-317 regulator is to protect the IC against overload and output shorts. If the IC is
overloaded, the junction of an overload transistor will overheat. A transistor will sense this overheating and shut
down transistor Q5.
Figure 5
Figure 6
A) Transformer
Winding
B) Voltage C1
C) Current
through diodes
20V
Peak
20V
2V
Output
R1
R2
Divider
Q1
Q2
1.5V
Q3
Q4
Q5
Current
Source
Equalized
to 1 Meg.

-16-
The LM-317 IC is basically a 2V regulator. To be able to vary the output 2 - 15V,
we stack the IC on a DC voltage as shown in Figure 6A. When VR1 equals 0,
the output voltage is 2V as determined by the LM-317 IC. Note that the voltage
across R1 is always 2 volts. When R1 equals VR1, the voltage across VR1 will
equal the two volts across R1, therefore, the output voltage will be four volts.
When VR1 is 5 times R1, the output voltage is 12 volts. As you can see, varying
resistor VR1 will vary the voltage from 2V to 15V.
The Negative Voltage Regulator
The theory of the negative voltage regulator is the same as the previously discussed positive regulator. The
basic differences is that diodes D2 and D4 are reversed, producing a negative voltage across capacitor C6. The
LM-337 IC is designed to operate from a negative supply.
The 5 Volt Power Supply
In the previous discussion of the variable voltage regulators, the ICs can handle about 1A of current. In the
design of the 5V supply, we need 3A of current. To meet this current requirement we must add an external pass
transistor capable of delivering 3A.
Figure 7 shows a simplified 5V regulator with an external PNP pass
transistor. In this circuit, transistor Q1 is a power transistor capable of
delivering over 3A. Transistor Q2 is biased off until the LM-7805 IC
draws about .2A. When .2A is drawn by the LM-7805 IC, the voltage
drop across the 3 ohm resistor is .6V, enough to turn on transistor Q2.
Transistor Q2 takes over and delivers the current to the output. Note that
if the output voltage goes down, the LM-7805 regulator will draw more
current, forcing the output voltage back to 5V. Thus, the LM-7805
regulator controls the output voltage and keeps it at 5V.
Unfortunately, this circuit has no control of the output maximum current. If the output is shorted to ground
transistor Q2 will be overloaded and eventually be damaged. The LM-7805 IC will only draw the .2A it was
designed to handle and never heat up to turn itself off. Another transistor Q1 is added to limit maximum current.
Resistor R5 is added to sense the current in transistor Q2. When approximately 3A is drawn in transistor Q2,
the voltage drop in resistor R5 will turn on transistor Q1. This will force more current in the LM-7805 IC.
Eventually the LM-7805 IC will overheat turning itself off and thus limiting the circuit at about 3.2A.
The first .2A of current is drawn by the LM-7805 IC. The next 3A are drawn by transistor Q2. Thereafter, the
current is drawn by the LM-7805 IC until it overheats and turns itself off. This is a very effective circuit capable
of regulating the output voltage at a constant 5 volts and yet delivering over 3A of current.
Figure 6A
Figure 7
2 - 15V
R1
VR1
LM-317

-17-
1. AC voltage is supplied to the rectifier stages by the
A. step up transformer.
B. step down transformer.
C. 1 to 1 transformer.
D. AC to DC transformer.
2. The secondary windings of the transformer are . . .
A. 90Oout of phase.
B. 180Oout of phase.
C. 270Oout of phase.
D. 320Oout of phase.
3. Diodes allow current to flow . . .
A. when the anode is more negative than the cathode.
B. when the cathode is more positive than the anode.
C. in one direction.
D. when a negative or positive voltage is on the anode.
4. What circuit is more efficient for rectifying AC to DC?
A. Hartley oscillator.
B. Half-wave.
C. Schmitt trigger.
D. Full-wave.
5. The DC voltage is smoothed by using a . . .
A. half-wave rectification circuit.
B. small value capacitor with a high voltage value.
C. Large value capacitor.
D. 90Oout of phase rectification circuit.
6. An in efficient rectification circuit usually contains . . .
A. large gaps between cycles.
B. twice the AC voltage needed.
C. more diodes.
D. all of the above.
7. The maximum current that a diode can handle is determined by . . .
A. the transformer’s current rating.
B. the amount of AC ripple.
C. three times the diode rating.
D. peak current rating.
8. The LM317 will shut down when . . .
A. the output voltage is too high.
B. no current is being drawn.
C. the junction overheats.
D. the output voltage drops to 1.25V.
9. The LM317 regulator contains . . .
A. a pass transistor.
B. a constant current source.
C. a differential amplifier.
D. all of the above.
10. The LM317 is basically . . .
A. a 1.25V regulator.
B. a 6.25V regulator.
C. a 2.5V regulator.
D. a negative voltage regulator.
QUIZ
Answers: 1. B, 2. B, 3. C, 4. D, 5. C, 6. D, 7. D, 8. C, 9. A, 10. A

SCHEMATIC DIAGRAM
SPECIFICATIONS ON XP-620 POWER SUPPLY
Input Voltage 110-130VAC
Output Voltage 1) 1.5-15VDC @ 1A
(at 120V input) 2) 1.5-15VDC @ 1A
3) 5VDC @ 3A
Output Regulation 200mV each supply
Line Regulation 100mV each supply
Ripple Max 5mV rms
Current Protection 1A limit 2-15VDC
3A limit 5VDC
Short Protection 1A limit 2-15VDC
3A limit 5VDC
Output Impedance .025Weach supply
-18-

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