Download Fundamentals of Linear Electronics Integrated & Discrete

CHAPTER 4
The Bipolar
Transistor
OBJECTIVES
Describe and Analyze:
• Transistor architecture
• Transistor characteristics
• Transistors as switches
• Transistor biasing
• Transistor amplifiers
• Troubleshooting techniques
Introduction
The key characteristic of a bipolar transistor is that a
small amount of power in the base-emitter circuit
can control a larger amount of power in the
collector-emitter circuit.
Inside the Transistor
A small voltage applied base to emitter causes charge carriers to
flood into the base. Almost all of those carriers are then swept
into the collector. Some of them come out the base due to
electron-hole recombination at defects in the crystal structure.
Alpha () and Beta ()
Alpha is a key parameter of BJTs:  = IC / IE
Beta is a secondary parameter:  = IC / IB
With a little algebra you can get the relationship:
 =  / (1 - )
Using the values  = 0.99 and  = 0.98, you get
 = 99 and  = 49 respectively.
So you can see that small changes in alpha cause
large changes in beta. Two transistors with the
same part numbers can have two very different
values of beta.
Calculations with Beta
Suppose you measure the currents in a transistor
and find that IB = 0.2 mA and IC = 10 mA. Calculate
the value of IC for a base current of 1mA.
Find beta from data:
 = IC / IB = 10 / 0.2 = 50
Use calculated beta to find new IC:
IC =   IB = 50  1mA = 50 mA
Transistor Switches
Compared to mechanical switches, transistors used as
switches:
• Last much longer
• Can turn on and off much faster
Calculations for a
Transistor Switch
Suppose you want to use a transistor ( = 50) to turn
on an LED. The LED requires 40 mA for full
brightness. You will use a 12 Volt DC power supply.
What value should you use for a base resistor?
Base current needed: IB = IC /  = 40mA / 50 = 0.8 mA
For reliability, over-drive the base by a factor of 2:
Actual IB = 2  calculated IB = 2  0.8mA = 1.6 mA
Calculate base resistor: RB = 12V / 1.6mA = 7.5 k
Power Gain of a
Transistor Switch
Suppose a transistor ( = 100) is being used to turn
a 100 Watt load on and off. The load uses 50 Volts
DC. What is the power gain?
Find IC using I = P / V: IC = 100W / 50V = 2A
Find IB using beta: IB = IC /  = 2A / 100 = 20 mA
Find base-drive power assuming VBE = 0.7 Volts:
P = V  I = 0.7V  .02A = 14 mW
Power gain (AP) = PLOAD / PBASE:
AP = 100W / .014W = 7143
Transistor Data: Stress Limits
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Power Dissipation, Maximum: VCE  IC + VBE  IB
Collector Current, Maximum: IC(MAX)
Base Current, Maximum: IB(MAX)
VCE Maximum: VCEO
VCB Maximum: VCBO
VBE Reverse-Biased, Maximum: VEBO
Junction Temperature, Maximum: TJ(MAX)
Transistor Data: Signal
• DC Beta: hFE
hFE = IC / IB where currents IC and IB are DC
• AC Beta: hfe
hfe = iC / iB where currents iC and iB are signal currents
• Gain-Bandwidth Product: fT
fT is the frequency at which hfe = 1
Gain & Amplification
Voltage Gain: AV = VOUT / VIN
Current Gain: AI = IOUT / IIN
Power Gain: AP = POUT / PIN = AV  AI
Amplification requires power gain.
Classification of Amplifiers
Class A: Can be done with one device
Class B: Requires two devices, one for each half cycle
Class C: Requires resonant circuit to restore signal
shape to a sine-wave
Base-Bias
Simple one-resistor base-biasing is not practical due
to the large variations in beta between devices.
The Base-Biased Amplifier:
Input Impedance
Zin = rb ||   r’e where
r’e = 25mV / IE (approximately)
and rb = Rb for single resistor biasing
The Base-Biased Amplifier:
Output Impedance
Output Impedance: Zout = Rc
The Base-Biased Amplifier:
Voltage Gain
Voltage Gain: Av = rc / re where
Rc = Rc || RL and re = r’e
Characteristic Curves
• With Ib fixed, collector is a constant-current source.
• With fixed steps in Ib, the space between the Ic lines
shows change in beta with collector current.
Troubleshooting
• The base-emitter and base-collector junctions in a
transistor can be checked for opens and shorts by
measuring resistance with a DMM or VOM.
• You can usually distinguish the base-emitter from the
base-collector because the resistance will read lower
from base to collector. Actually, what your meter is
showing you is the voltage drop across the junction.
• The above test also lets you separate NPN
transistors from PNP transistors.