Electronic Troubleshooting ET198B Chapter 2 • 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
Electronic Troubleshooting ET198B Chapter 2. 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 - PowerPoint PPT Presentation
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Electronic TroubleshootingET198B Chapter 2
• 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 (semiconductor or tube)
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-to-peak voltages)
Measuring Diode Characteristics
Measuring Diode Characteristics• For the drawings on the previous page
• E = 6VDC and R=10Ω• Find VF and VR
1. An ideal diode2. 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• The output voltagedecreases per the equation after reaching its peek voltage. t= elapsed time, VP = max VO
» The voltage decreases until the next pulse is higher than the output voltage
• 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
t
PCYO eVVVV
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
• Calculating VAVE , where
C
ITVrip
ripCapPripFSecPAVE VVVVVV2
1
2
1
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 partsInstalled 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:
Full-wave Rectifier• Example Problems• Find: The freq of the ripple voltage
• 50Hz source and a ½ Wave Rectifier
• 50Hz source and a Full Wave Rectifier
• 400 Hz source and a Full Wave Rectifier
• 800 Hz source and a ½ Wave Rectifier
• See problems 2.9 & 2.10 on page 23
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 Full-wave 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 Full-wave Power SupplyWalk through
assuming 6VP on the transformer secondary
Key Points•Frequency of ripple voltage will be twice the frequency of the primary•Different PRV from both ½ and full wave
• PRV always depend on circuit
• PVR = VOut + VF
•No transformer center tap in the circuit.
Complete Full-wave Power Supply• Problems
• Example• Vsec = 20 Vrms , IL =100mA C=400µF • Find: VO , Vrip , and PRV
• Additional problems• Problems 2-11 and 2-12 on pages 23 and 24
vvvvvxVVV rmsFPO 8.264.12.284.12041.1
VAproxV
x
xxA
C
ITVrip 2,0875.2
10400
1035.81.06
3
vVVVPRV FOP 5.27
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
• 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
Power Supply Component Failures
Testing Electrolytic Capacitors
• 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 shortthe bulb will be very dim due to the reflected impedance• With a short in the secondary the bulb will be bright.
Power Supply Component Failures
• 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
Locating Problems in Power Supplies
Troubleshooting Tree
• Sample problem walk through1. 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