Download Electronic Troubleshooting

Electronic Troubleshooting
Chapter 3 Bipolar Transistors
• Most devices still require some individual
(discrete) transistors
• Used to customize operations
• Interface to external devices
• Understanding their operation is critical to
understanding the operation of digital or analog ICs
Construction & Symbols
• They use two PN junctions
• Two versions, but both have narrow center regions (aka
Base)
• NPN transistors
• Base is P material (holes ) are the primary charge carriers and the
collector and emitter regions are made fro n materials
Construction & Symbols
• They use two PN junctions
• Two versions ( continued )
• NPN transistors
• Usually used with positive power supplies
• The arrow on the symbol indicates the
direction of conventional current flow
• PNP transistors
• Usually used with negative power supplies
• The arrow on the symbol indicates the
direction of conventional current flow
• Construction
• Most are made from silicon, but some are
made from germanium
NPN Transistor Current Flow
NPN Current Flow
• With the base emitter
PN junction forward biased
• Current flows
• Emitter – Collector flow is much larger than Emitter-Base current
• The Emitter-Base current controls the amount of EmitterCollector current.
• Key Equations:
β = hFE = IC/IB
IE = IC + IB
IC = β IB
β values range from 20 - 200
if β is large, IE ~ IC
NPN Input Characteristics
NPN Current Flow
• Sample problem
• IB = 10µA, β = 99
• Find: IC and IE
I C    I B  99 10A  990A
I E  I C  I B  990A  10A
• Thus with a large β, IC ~ IE w/error = 1%
• In class Exercises
• If a transistor has β = 150 and IB = 6mA, what is IC.
• What value of base current will make IE = 2mA if β = 80
Typical Transistor Packages
Acts like a Variable Resistor
• Walk-through
• Set R = to the lamp’s resistance and
assume the lamp lights dimly
• Reducing R brightens the lamp & increases I
• Increasing R dims the lamp & decreases I
• Replace R with a transistor
• Lamp brightness is controlled by IB
which controls IC
• Decreasing VBB to 0v will decrease
IC to approx 0A
• Small leakage current will
still exist and it varies with
temperature
Equivalent Circuits
• Transistor cutoff and saturated equivalent
circuits
• In cutoff IC = 0A and all of VCC is dropped across the transistor
• In Saturation only an insignificant amount of VCC is dropped
across the transistor.
Biasing the Transistor
• Biased Transistors
• When used as an amplifier
the transistor is operated
between cutoff and saturation
• Output signals shouldn’t go to saturation
or cutoff
• Base-biased Circuit
• If RB is adjusted for IC = 1mA
VCE  VCC  I C RC  12  1mAx5000  7v
• Near ½ VCC
Biasing the Transistor
• Base-biased Circuit
• Apply small input signal through a Cap
• Causes changes in IB around the level
established by RB the and VCC.
• The changes in ΔIC = Δ IB x β
• The changes around the DC
level set by RB and VCC will equal
Δ IC x RC
• Finding IB is critical
• For base biased circuits
like on this and the next slide
with lower supply voltages
VBE is critical
Biasing the Transistor
• Base-biased Circuit
• Finding IB is critical
• Assume RB = 1MΩ,
β = 80, RC = 3KΩ
• Find: IB , IC , VCE
IB 
VCC  VBE 12v  0.7v

 11.3A
RB
1M
I B  12 A 
VCC
RB
I C    I B  80 11.3A  0.904mA  0.904mA
I C    I B  80 12A  0.96mA
VCE  VCC  RC I C  12v  0.904mA  3K  9.29V  9.3V
VCE  VCC  RC I C  12v  0.96m  3K  9.12V  9.1V
Biasing the Transistor
• Base-biased Circuit
• Distorted output
• The quiescent value of VCE is critical to
having non distorted outputs
• Positive clipping indicates that the
quiescent level of IC is too low
• Negative clipping indicates that
the quiescent level of IC is too high
• Such circuits are unstable to
variations due to temperature and age
Biasing the Transistor
• Base-biased Circuit
• In Class Exercises
• Problems
• 3-5
» Given: VCC = 15V, RB = 750 K ohms
β = 60, RC = 5000 ohms
» Find: IB, IC, VCE
• 3-6
» Given: VCC = 10V, RB = 470 K ohms, β = 90,
RC = 3300 ohms
» Find: IB, IC, VCE
• 3-7
» Given: VCC = 15V, RB = ??? K ohms
» β = 60, RC = 5000 ohms
» Find: Maximum value of RB that will cause transistor to go
into saturation
Biasing the Transistor
• Voltage Divider & Emitter
Resister Bias
• Provides a more stable
quiescent value for VB
• Small changes in one component
don’t get multiplied by β
V  VBE
I E  IC  B
RE
if VBE negligible =>>
IC 
VB
RE
• If IE increases in value due to heat and VE increases, VB should
be more stable and VB shouldn’t change much since it depends
on a ratio. Then VBE decreases and then decreases IC , thus
the resulting change is less than expected from the initial
change
Biasing the Transistor
• Voltage Divider & Emitter
Resister Bias
• Provides a more stable
quiescent value for VB
• Since β isn’t required to determine IC
VB  VCC
R2
R1  R2
IC 
VCC R2
R1  R2 RE
• Thus even if β changes the gain of the transistor amplifier
circuit is stable
Biasing the Transistor
• Voltage Divider & Emitter
Resister Bias
• Example Problem
• Given: VCC=20v, R1=30KΩ,
R2=10KΩ, RE= 2KΩ, RC=3KΩ
• Find: IC and VC
VCC R2
20V 10 K
IC 

 2.5mA
R1  R2 RE 30 K  10 K2 K
VC  VCC  RC I C  20v  2.5mA  3K  12.5V
• In Class Problems: 3-9 to 3-12 on page 60
• Input
AC Signal Amplifier
• Characteristics
• AC signal coupled through a cap
• ib depends upon the resistance of the
BE junction and β
• re = resistance of the BE
junction = 25mV/IE
• rin = β(rE+re)= resistance seen by the
input signal
• rin = βre when rE= 0 ohms
• Higher quiescent IE yields
lower re and rin , and higher
quiescent IB and larger Δib
• Sample Problem
• Given: IC = 1mA and β =100, RE= 0 ohms
• Find: rin
AC Signal Amplifier
• Input
• Sample Problem
re 
25mV
 25
1mA
rin    re  100  25  2500
• Circuit Analysis
vin  5mVP
• Biasing Values (DC values)
IB 
VCC
 10 A
RB
I C    I B  1mA
VC  VCC  I C RC  7v
• AC values – See Top of slide plus
ib 
5mV
 2A
2500
ic  peak  ib  peak    2A 100  200A
AC Signal Amplifier
• Circuit Analysis
• AC values
vo  peak  ic  peak RC  200A  5,000  1v
Av 
vo
1v

 200
vin 5mv
AC Signal Amplifier
• Circuit Analysis
• Add a load on the output
• Under AC analysis VCC
appears to the output AC
signal as a ground
• Thus RC and RL form
a current divider
• rL = RC || RL
• Thus
and
vo  ic  rL
vo
ic  rL
rL
Av 


vin ib    re re
Assumes rE = 0 ohms
AC Signal Amplifier
• Circuit Analysis
• Voltage Divider and RE
Variation
• Add a AC By-Pass Cap across RE
• The AC signal will see a short to
ground
• The DC bias voltages will see
the Cap as an open
• An open bypass Cap will result in
significantly less gain – rE goes
from 0 ohms to RE is usually
much larger than re . See
equation to the right.
• Other variations RE1 and RE2.
• Walk through analysis – re
varies with temperature and IE
rL
rL
Av 

re  RE RE
AC Signal Amplifier
• Typical Problems
• Transistor Problems
• Failures
• They don’t become weaker
• They open or short
• Open Transistor
• No IC
» DC voltage at the collector will = VCC
• Shorted Transistor
• Will have 0 volts across the transistor
• Other Component problems
• Resistors
• Values can change with age and heat
• Yields abnormal bias voltages
• Coupling Capacitors
• Can become leaky, short or open
• Leads to circuit malfunctions, i.e., bias voltage not blocked
AC Signal Amplifier
• Typical Problems Multistage amplifiers
• Coupling Capacitor
• An open would stop the AC output
of the first stage from reaching the
second stage
• A short would allow the DC biasing
voltages on the first stage to send
the second stage into saturation
• Manufacturing/Repair
• Solder Bridges
• As the one shown indicates
a shorted transistor – Cured
by removing it
• Break/Opens in foil tracks
• Will appear as open components
Identifying the Basic Amplifier
• Characteristics
• Best practice when done on actual equipment
• Sample Commercial circuits used
• Part of Sony Tape recorder Model CF-320
• Part of Panasonic Tape recorder Model RF-7400
• Sample Circuit Analysis
• Sony Tape recorder Model CF-320
• See circuit on the next slide & page 53 of the textbook
• Driver stage using Q4
• Single stage audio amplifier
• Voltage divider in the base
» R35, R36, R37
» C29 from junction of R35 and R36 to ground serves to
smooth power supply variations
» C28 servers to filter out high frequencies and to prevent
oscillations
Sony Model CF-320
Identifying the Basic Amplifier
• Sample Circuit Analysis
• Sony Tape recorder Model CF-320
• Driver stage using Q4
• Emitter Circuit
» R38 key to determining the VE
» R43 provides feedback from the secondary of the output
transformer. Has minimal effect on DC biasing levels
• Collector Circuit
» Doesn’t have a RC, but uses the primary of transformer T2
» Drives a push-pull power amplifier that consists of Q8 and Q9
» R39 and C30 are connected across the primary of T2 to filter out
high frequency noise
• Panasonic Tape recorder Model RF-7400
• Circuit is on the next slide and page 54 of the textbook
• Analysis that focuses upon another amplifier circuit starts with the
third from the bottom paragraph on page 53
Panasonic Model RF-7400
Finding Single Stage Amp Problems
• Characteristics
• Usually uses a Voltage Divider
with RE configuration
• Troubleshooting
• Check Power supply voltages
• Check VC
•
If normal –Transistor and all biasing
resisters are normal
•
Causes of low VC are:
•
•
•
•
Larger RC value
Larger IC value causes a lower VC
Larger VB value causes larger IC and lower VC
Cause of a high VB
•
•
VC  VCC  I C RC
Lower R1 value
Larger R2 value
Finding Single Stage Amp Problems
• Troubleshooting
•
Key points
•
•
•
Understand the operation of the circuit
Take some measurements to determine current operation
The above steps will lead to the problem.
•
The flow chart on the next slide demonstrates the
measurement and thought process for troubleshooting a
single stage amplifier
• Flowchart
•
Walk through all the branches
• Homework
•
Problems 3-17 through 3-22
Troubleshooting Flowchart
The values in the
flowchart are per the
above circuit.