electronic troubleshooting et198b chapter 2

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:

• Ideal Diode, VP = 15V,ILoad= 20 mA, Freq=60Hz, C = 200µF• Find: Vrip , Vave

• 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

• See Page 21 Problems 2.1 – 2.8

v

x

xxx

C

ITVrip 67.1

10200

107.1610206

33

vvvVVV ripPave 2.1467.121152

1

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• Vpri = transformer primary

(line) voltage• Nsec = number of wire turns on the secondary winding

• Npri = number of wire turns on the primary winding

NOTE: The formula will work for Vrms , VP , or VP-P

• Usually specified by primary & secondary voltages in Vrms

pripri N

NVV sec

sec

Complete Halfwave Power Supply• All the discussed components plus:• Power cord, fuse, indicator lamp, and a switch

• Questions• Transformer turns ratio (Nsec / Npri)

• VP

Complete Halfwave Power Supply• Peak Reverse Voltage (PVR)• 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

Complete Halfwave Power Supply• Peak Reverse Voltage (PRV)• 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

Full-wave Rectifier• Circuit function • Peak Reverse Voltage

• 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

• 60Hz => Period of 1/60 = 16.67mS

• However T = 8.33 mS since it’s a Full-wave

• Vrip freq = 1/T = 120Hz

C

ITVrip

000.19996.0000250.0

0002499.0

250

33.830

F

mSmA

C

ITVrip

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

Troubleshooting Tree