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:  = I C / I E Beta is a secondary parameter:  = I C / I B 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 I B = 0.2 mA and I C = 10 mA. Calculate the value of I C for a base current of 1mA. Find beta from data:  = I C / I B = 10 / 0.2 = 50 Use calculated beta to find new I C : I C =   I B = 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: I B = I C /  = 40mA / 50 = 0.8 mA For reliability, over-drive the base by a factor of 2: Actual I B = 2  calculated I B = 2  0.8mA = 1.6 mA Calculate base resistor: R B = 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 I C using I = P / V: I C = 100W / 50V = 2A Find I B using beta: I B = I C /  = 2A / 100 = 20 mA Find base-drive power assuming V BE = 0.7 Volts: P = V  I = 0.7V .02A = 14 mW Power gain (A P ) = P LOAD / P BASE: A P = 100W /.014W = 7143

Transistor Data: Stress Limits Power Dissipation, Maximum: V CE  I C + V BE  I B Collector Current, Maximum: I C(MAX) Base Current, Maximum: I B(MAX) V CE Maximum: V CEO V CB Maximum: V CBO V BE Reverse-Biased, Maximum: V EBO Junction Temperature, Maximum: T J(MAX)

Transistor Data: Signal DC Beta: h FE h FE = I C / I B where currents I C and I B are DC AC Beta: h fe h fe = i C / i B where currents i C and i B are signal currents Gain-Bandwidth Product: f T f T is the frequency at which h fe = 1

Gain & Amplification Voltage Gain: A V = V OUT / V IN Current Gain: A I = I OUT / I IN Power Gain: A P = P OUT / P IN = A V  A I 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 / I E (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 || R L 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.