Electronics II Lecture 2(a): Bipolar Junction …Analysis of this transistor circuit to predict the dc voltages and currents requires use of Ohm’s law, Kirchhoff’s voltage law

Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458 • All rights reserved.

Electronic Devices and Circuit Theory, 10/eRobert L. Boylestad and Louis Nashelsky

Copyright @2013 by Dept. of Electrical Engineering, Electronics IICollege of Engineering, Maysan University

Electronics IILecture 2(a): Bipolar Junction Transistors

A/Lectr. Khalid ShakirDept. Of Electrical Engineering

College of EngineeringMaysan University



Transistor !• Transistor=Transfer+Resistor. When

Transistor operates in active region its inputresistance is high and output resistance islow. Transistor as a device which transfersits resistance from high to low and by thisproperty transistor amplifies/control inputcurrent signal.

• Transistor is a three terminal semiconductordevice used as amplifier, switch oroscillator.

• bipolar junction transistor (BJT ) is a typeof transistor that relies on the contact of twotypes of semiconductor junctions for itsoperation

2 Copyright @2013 by Dept. of Electrical Engineering,College of Engineering, Maysan University




Basic Transistor Operation



Look at this one circuit as two separatecircuits, the base-emitter(left side) circuitand the collector-emitter(right side)circuit.

Note that the emitter leg serves as aconductor for both circuits. The amountof current flow in the base-emitter circuitcontrols the amount of current that flowsin the collector circuit. Small changes inbase-emitter current yields a large changein collector-current.



Transistor Structure



The BJT (bipolar junction transistor) is constructed with three doped semiconductorregions separated by two pn junctions, as shown in Figure (a). The three regions arecalled emitter, base, and collector. Physical representations of the two types of BJTsare shown in Figure (b) and (c).

One type consists of two n regions separated by a p regions (npn), and other typeconsists of two p regions separated by an n region (pnp).



Transistor Currents



The directions of the currents in both npn and pnp transistors and their schematicsymbol are shown in Figure (a) and (b). Notice that the arrow on the emitter of thetransistor symbols points in the direction of conventional current. These diagramsshow that the emitter current (IE) is the sum of the collector current (IC) and thebase current (IB), expressed as follows:

IE = IC + IB



Transistor Characteristicsand Parameters



Figure shows the proper biasarrangement for npn transistorfor active operation as anamplifier. Notice that the base-emitter (BE) junction is forward-biased and the base-collector(BC) junction is reverse-biased.As previously discussed, base-emitter current changes yieldslarge changes in collector-emittercurrent. The factor of this changeis called beta().

= IC/IB

The ratio of the dc collector current (IC)to the dc emitter current (IE) is the alpha.

α = IC/IE






Example:Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA.

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IE = IC + IB = 3.65 mA + 50 μA = 3.70 mA

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The collector current is determinedby multiplying the base current bybeta.

Analysis of this transistor circuit to predictthe dc voltages and currents requires useof Ohm’s law, Kirchhoff’s voltage law andthe beta for the transistor.

Application of these laws begins with thebase circuit to determine the amount ofbase current.

Using Kichhoff’s voltage law, subtract the.7 VBE and the remaining voltage isdropped across RB.

Determining the current for the base withthis information is a matter of applying ofOhm’s law. VRB/RB = IB

.7 VBE will be used inmost analysis examples.

CB II






The base circuit VBB isdistributed across the base-emitter junction and RB in thebase circuit.

In the collector circuit wedetermine that VCC isdistributed proportionallyacross RC and thetransistor(VCE).



There are three key dc voltages and three key dc currents to be considered. Note thatthese measurements are important for troubleshooting.




IB: dc base current

IE: dc emitter current

IC: dc collector current

VBE: dc voltage across base-emitter junction

VCB: dc voltage across collector-base junction

VCE: dc voltage from collector to emitter






When the base-emitter junction is forward-biased,

VBE ≅ 0.7 V

VRB = IBRB : by Ohm’s law

IBRB = VBB – VBE : substituting for VRB

IB = (VBB – VBE) / RB : solving for IB

VCE = VCC – VRc : voltage at the collector with

VRc = ICRC respect to emitter

VCE = VCC – ICRCThe voltage across the reverse-biasedcollector-base junction

VCB = VCE – VBE where IC = βDCIB



When the base-emitter junction is forward-biased,

VBE ≅ 0.7 V

IB = (VBB – VBE) / RB

= (5 V – 0.7 V) / 10 kΩ = 430 μA

Example: Determine IB, IC, VBE, VCE, and VCB in the circuit of Figure. Thetransistor has a βDC = 150.




IC = βDCIB= (150)(430 μA)= 64.5 mA

IE = IC + IB= 64.5 mA + 430 μA= 64.9 mA

VCE = VCC – ICRC= 10 V – (64.5 mA)(100 Ω)= 3.55 V

VCB = VCE – VBE= 3.55 V – 0.7 V= 2.85 V

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Electronics II Lecture 2(a): Bipolar Junction …Analysis of this transistor circuit to predict the dc voltages and currents requires use of Ohm’s law, Kirchhoff’s voltage law