Download Electronic Troubleshooting

Power Supply Basics
In-depth Treatment
• Most electronic device need:
- DC (Direct-Current) power
- Batteries are one source:
- They can become very costly
- Due to replacement costs and labor
- Better Source – DC Power Supply
- Converts Available AC (Alternating Current)
power to DC
- Requires a device that permits flow in
one direction (Rectifiers)
- A Diode
Solid State Diode
• When the P material is
sufficiently positive with
respect to the N material
current flows
• With Reverse polarity
Current doesn’t flow
Diode Packages
• Part numbers usually start as 1N as in 1N4001
• Many types of cases are possible
• Range from glass to stud mountable metal cases
(for heat dissipation)
Characteristic Curve
Characteristic Curve
• In the flat region
• When VF is between Breakdown and 0.7 V (forward bias
voltage) for silicon based diodes almost no current flows
• When VF is between Breakdown and 0.3 V (forward bias
voltage) for germanium based diodes almost no current
flows
• Region to the right
• When VF is higher than the forward bias voltage current
increases rapidly for a slight voltage increase
Characteristic Curve
• Region to the left
• When VR is less than the Breakdown voltage current
increases rapidly for a slight voltage decrease and the
diode is destroyed
• Except for tunnel and Zener diodes which are designed to operate in that
region
• Diodes designed as rectifiers are destroyed when operated in this region
• Rectifying diodes come with a range of Breakdown voltages
• 25V to thousands of volts
• Always use ones with Breakdown voltages greater than any reverse
voltage
• Calculated based upon instantaneous voltages not RMS values (peak-topeak voltages)
Measuring Diode Characteristics
Measuring Diode Characteristics
• For the drawings on the previous page
• E = 6VDC and R=10Ω
• Find VF and VR
1.
2.
An ideal diode
For a real diode with a breakdown voltage (VB) of 50VDC
Half-wave Rectifier
• Sample circuit in Fig A
• Voltages taken at points “X”
and “Y” with respect to
Ground
• O-scope not a meter
Half-wave Rectifier
• Input Voltage 24 VP-P at “X”
• Output Voltage at “Y”
•
Only the Positive ½ cycle of the input
reaches the output
•
Is the Diode a real one or ideal?
•
How can you tell?
• Calculation Notes:
• Usually ignore the diode VF if the voltages
involved are 10 times larger than VF
• Except on Tests and quizzes in this course
Filter Capacitors
• Used to smooth the pulses
from a rectifier Circuit
• On the first pulse seen by the
capacitor
• It’s voltage closely matches the
rising input voltage
• After the input voltage reaches
its peak value the current to the
load resistor is supplied from the
capacitor
• The output voltage or VY slowly
decreases
• The rate of decrease is
dependent upon the value of R
times C - the Time Constant (τ)
Filter Capacitors
• Used to smooth the pulses from a rectifier Circuit
• On the first pulse seen by the capacitor
• The output voltage or VY slowly decreases
• This RC time constant (τ) should be long compared to the time
between pulses
t
• The output voltage

decreases per the equation
O
Y
C
P
after reaching its peek voltage. t= elapsed time, VP = max VO
» The voltage decreases until the next pulse is higher than the
output voltage
V V V V e
• On the second and following pulses
• The process repeats per the previous sketch
• The Cap is charged during the positive going transition of the pulse
• The cap supplies the current at other times
Filter Capacitors
• Used to smooth the pulses from a rectifier Circuit
• On the second and following pulses
• Key aspects of the resulting output voltage
• It still has pulses, but they are smaller than before the Cap
• The pulses alternate plus and minus around an average output
voltage (Vave)
• The peak to peak value of the alterations is known as the ripple
voltage (Vrip)
• Calculating Vrip , where:
• I = Load current in Amps
• T = time between recharging pulses
• C = filter capacitance in farads
IT
Vrip 
C
• Calculating VAVE , where
1
1
VAVE  VP  Sec  VF  Vrip  VP Cap  Vrip
2
2
Calculating Vrip & VAVE
Effects of Load Current & Filter Capacitance
• Anything that makes the RC time constant
smaller increases the ripple voltage (Vrip)
• A decreased RL causes increased current and more Vrip
• A change to a lower value of capacitance causes a smaller
τ and more Vrip
Effects of Load Current & Filter Capacitance
• Anything that makes the RC time constant smaller
increases the ripple voltage (Vrip)
• Decreased RC time constant (τ) can be caused by:
• Improper
replacement parts
Installed during a
repair
• Change in the value
of the filter capacitor
with age – especially
with electrolytic Caps
Effects of Load Current & Filter Capacitance
• Example Problem:
• Ideal Diode, VP = 15V,
ILoad= 20 mA, Freq=60Hz, C = 200µF
• Find: Vrip , Vave



IT
20 x103 x 16.7 x103
Vrip 

 1.67v
6
C
200 x10
Vave  VP  1 Vrip  15v  1 1.67v   14.2v
2
2
• Practice. Repeat with:
• Ideal Diode, VP(source) = 30V, ILoad= 25 mA, 60Hz source, C =
250µF
• Repeat both for a silicon diode that has VF = 0.7v
Transformers
• In power supplies
• Used to step-up/down VP
to a desired value based
upon the required DC
output voltage
• Construction
• The core is usually
laminated sheets of iron
or steel
• Minimizes losses due to
eddy currents
• At least two windings of
insulated wires (Fig B)
Transformers
• Electrical characteristics
• Voltages on the secondary and Primary Windings
are related by the following equation:
• Vsec = transformer secondary
output voltage
sec
• Vpri = transformer primary
sec
pri
(line) voltage
pri
• Nsec = number of wire turns on the secondary winding
• Npri = number of wire turns on the primary winding
V
V
N
N
NOTE: The formula will work for Vrms , VP , or VP-P
• Usually specified by primary & secondary voltages
in Vrms
Complete Halfwave Power Supply
• All the discussed components plus:
• Power cord, fuse, indicator lamp, and a switch
• Questions
• Transformer turns ratio (Nsec / Npri)
• VP
Reverse Voltage
• It is equal to the sum of the voltage on the Cap
and the VP from the next ½ cycle of the input
waveform
• In other words the A simple rule for VB rating on a
Diode is that it must exceed
the VP-P of the transformer's
secondary voltage
Reverse Voltage
• Given 12 Vrms on the secondary
• 16.97VP ~ 17.0VP or Approx 34VP-P
• Thus on the positive ½ cycle the Cap is charged to
16.3V
• On the negative ½ cycle the anode side of the diode
reaches – 17V
• About 33.3 V ~ 34V across
the diode
Full-wave Rectifier
Full-wave Rectifier
• Characteristics
• For given components
it will have twice the pulse
rate on the output.
• In a power supply it will have ½ the ripple voltage
• Recharges the capacitor twice every input cycle
• For a given output voltage the total secondary voltage will be
double that of a ½ wave rectifier
• Like a ½ wave - reversing the diode connections will result in
negative pulses on the output
• Circuit function
• On half the input cycle
• Point A is driven positive with respect to point B and the center
tap of the transformer which is at ground
Full-wave Rectifier
• Circuit function
• On half the input cycle
• Causes diode D1 to conduct and D2 is back biased and not
conducting
• Current flows through RL and D1
• On the next ½ cycle of the input
• Point B is driven positive with respect to point A and the
center tap of the transformer which is at ground
• Causes diode D2 to conduct and D1 is back biased and not
conducting
• Current flows through RL and D2
• On both ½ cycles
• Current through RL flows in the same direction
Full-wave Rectifier
• Circuit function
• Filter Capacitors
• Need to smooth out the pulsating output to a DC voltage
• For the same load , output voltage , and acceptable ripple
voltage – the capacitor can be ½ the farad value
• Same equation for Vrip and Vave work but the the period is ½
since there are two pulses per input cycle
Rectifier Assemblies
• Full-wave Rectifier diode assemblies are available
• Come packaged for
either positive or negative
outputs
Full-wave Rectifier
• Example Problems
• With the given circuit:
• Vpri = 120VRMS @ 60 Hz, Vsec =20V
• Find: VP, and PRV
• For C= 250 μF, ILoad = 30 mA
• Find: Vrip and the freq of the ripple voltage
Full-wave Rectifier
• Example Problems
• Find: VP, and PRV
• VP=VC-Peek = 10V x 1.414 – 0.7V = 14.14- 0.7V = 13.44V
• PRV ~ 13.44V – (-14.14V) = 27.58V ~ 2x VC-Peek ~ 2xVDC
• Find: Vrip and the freq of the ripple voltage
IT
Vrip 
C
However T = 8.33 mS since it’s a Full-wave
• 60Hz => Period of 1/60 = 16.67mS
•
Vrip 
IT
30mA  8.33mS
0.0002499


 0.9996  1.000
C
250 F
0.000250
• Vrip freq = 1/T = 120Hz
Full-wave Rectifier
• Example Problems
• Find: The freq of the ripple voltage
• 60Hz source and a ½ Wave Rectifier
• 60Hz source and a Full Wave Rectifier
• 400 Hz source and a Full Wave Rectifier
• 800 Hz source and a ½ Wave Rectifier
Panasonic Tape Recorder
Bridge Rectifier
Bridge Rectifier
• Characteristics
• Doesn’t need a center tapped transformer
• Uses four diodes connected in a bridge circuit
• Circuit function
• When top of transformer is positive with respect to
the bottom
• D2 and D3 are forward biased and conduct
• When the polarity is reversed
• D1 and D4 are forward biased and conduct
• Current flows through RL in the same direction
regardless of polarity
• PRV equals approximately the voltage across RL
Complete Bridge Power Supply
• Also needs a filter capacitor
• VO without a load equals VP-Sec after accounting for the forward
bias voltages of two diodes that are between the load and the
transformer leads.
• The VF for both diodes are usually ignored for VO of 15V or higher
except for quizzes and tests in this course
Complete Bridge Power Supply
• Problems
• Example
• Vsec = 20 Vrms , IL =100mA
C=400µF
• Find: VO , Vrip , and PRV
VO  VP  2VF  1.41x20vrms  1.4v  28.2v  1.4v  26.8v


IT 0.1Ax 8.35 x10 3
Vrip 

 2.0875V , Aprox  2V
6
C
400 x10
PRV  VP  VO  VF  27.5v
Complete Bridge Power Supply
Walk through
assuming 6VP on the
transformer
secondary
Bridge Rectifier Assemblies
Voltage Doubler Circuits
• Full-wave Voltage Doubler
• Produces VO that is twice as large as VP on the secondary
terminals of the transformer. Freq of Vrip is twice input Freq.
• Circuit Operation
• When top of transformer is positive, D1 conducts and charges C1
to VP with the polarity shown
• When Bottom of transformer is positive, D2 conducts and
charges C2 to VP with the polarity shown
• The voltage across the resistor is approx 2 x VP
Voltage Doubler Circuits
• Half-wave Voltage Doubler
• Produces VO that is twice as large as VP on the secondary
terminals of the transformer. Freq of Vrip is equals input Freq
• Circuit Operation
• When bottom of transformer is positive, D1 conducts and
charges C1 to VP with the polarity shown
• When Top of transformer is positive, D2 conducts and charges C2
to VP + VC1 (approx 2 VP) with the polarity shown
• The voltage across the resistor is approx 2 x VP
Power Supply Component Failures
• Overview
• Each component’s typical failures and testing methods
are described
• Diodes
• Failure characteristics
• Either shorted or open
• They don’t get weaker
• Simple meter tests
• Exact resistances aren’t important
• Should be high in
one direction and low in the other
direction
• Reverse or leakage current should
be negligible compared to
forward current
Power Supply Component Failures
• Diodes
• Replacement Considerations
• Forward current Max rating
• PRV or Peak Inverse Voltage (PIV) rating
• Replacement parts should meet or exceed the original part
specification
• Electrolytic Capacitors
• Common source of Power Supply failures
• Failure modes
• Open
• Leaky
• Dry out
Power Supply Component Failures
• Electrolytic Capacitors
• Open Capacitors
• Don’t filter ripple or store a charge
• Leaky capacitors
• Appear to have a low resistance shunt
• Normal resistance should be a few hundred thousand ohms as
measured by a meter
• Note: the meter should initially spike to a very low resistance
and then climb to a high reading
• Best if viewed on an analog meter - See the next slide
• As the problem worsens the shunt resister appears to be smaller
and smaller
• Some can become so leaky they appear as a short
• Can blow fuses, damage diodes and transformers
Testing Electrolytic Capacitors
Power Supply Component Failures
• Electrolytic Capacitors
• Capacitors that are drying out
• Lose capacitance
• Which causes poor filtering and excessive ripple voltages
• Simple Test
• Bridge suspected filtering Cap with a known good one of
approximately the same rated value
• If the ripple problems decrease – replace the capacitor
• Replacement Considerations
• Replacement working voltage of should be > the original
• Working voltage – WVDC
• Usually designed for a few volts above the DC output voltage
Power Supply Component Failures
• Transformers
• Don’t usually fail in power supplies
• Failure modes
• A winding may open if excessive current is drawn
• An internal short due to insulation breakdown
• Massive failures are possible
• Also shorted turns – causes changes in output voltages
• Open windings
• Easily found with an ohmmeter
• Test each winding
• Likely location is the connection of the actual winding to the
external wires
• Can be repaired by carefully opening the winding and splicing
Power Supply Component Failures
• Transformers
• Shorted turn on a winding
• Testing
• Disconnect the secondary
from the circuit.
• Without a secondary short
the bulb will be very dim
due to the reflected
impedance
• With a short in the secondary the bulb will be bright.
Locating Problems in Power Supplies
• Power Supplies and system failures
• Power Supply Problems are some of the most common
causes for system failures
• Should be one of the first things checked
• Check Test Point 1 for the proper output voltage and
acceptable level of ripple voltage
• If correct – the power supply is good – Else, test the next most
likely system component
• Else use troubleshooting tree/flowchart
Troubleshooting
Tree
Troubleshooting Tree
• Sample problem walk through
1. Primary Winding is open –follow the flowchart
•
•
•
•
•
•
•
•
Measurement at TP 1 is 0 V
Go to right at first decision and a quick check of the fuse yields
a “NO” – go down
Measure transformer secondary at points 2 and 3 => 0V
Go down
Measure trans former primary voltage a points 4 and 5 => OK
Go Left
Remove power source and measure primary winding with an
ohmmeter
Open winding is found.
2. No Flowchart
•
•
Start at the output and work back towards the input
Make voltage and waveform checks to isolate the problem