Electronics

BIPOLAR JUNCTION TRANSISTOR

by

Prof. Fahmy El-Khouly

CHAPTER THREE

BIPOLAR JUNCTION TRANSISTORS

1-1 • TRANSISTOR CONSTRUCTION

constructed with three doped semiconductor regions separated by twop-n junctions

heavily doped emittermoderately doped collector lightly doped base

p-n-p

(b) p-n-p

Standard bipolar junction transistor (BJT) symbols

E

(a) n-p-n

The term bipolar refers to the use of both holes and electrons as carriers in the transistor structure

1- 2 • BASIC TRANSISTOR OPERATION

In order for the transistor to operate properly as an amplifier, the two p-n junctions must be correctly biased with external dc voltages

The operation of the p-n-p is the same as for the n-p-n except that the roles of the electrons and holes, the bias voltage polarities, and the current directions are all reverse

(a)n-p-n (b)p-n-p

FIGURE(1- 3 ) Forward-reverse bias of a bipolar transistor.

the base-emitter (BE) junction is forward-biased

and the base-collector (BC) junction is reverse-

biased.

FIGURE-(1-4)

illustration of

BJT action.

1-3 Transistor Currents

IE=IC+IBAs mentioned before, IB is very small compared to IE or IC.

The capital-letter subscripts indicate dc values.

1-4 • TRANSISTOR CHARACTERISTICS AND PARAMETERS

(a) n-p-n

When a transistor is

connected to dc bias

voltages, Vbb

forward-biases the

base-emitter junction,

and Vcc reverse-

biases the base-

collector junction.

the dc current gain of a transistor (βDC)

βDC =IC / IB

Typical values of βdc range from less than 20 to 200 or higher, βdc is usually designated as an equivalent hybrid parameter, hFE, on transistor data sheets

hFE = βDC

The ratio of the dc collector current Ic to the dc emitter current IE is the dc alpha

αdc =Ic / IE

Typically, values of αDC range from 0.95 to 0.99 or greater, but αDC is always less than 1

Relationship of βDC and αDC :--

IE / IC = IC / IC + IB / IC

1 / αDC = 1+ 1 /βDC

αDC = { βDC / ( 1 + βDC ) }

βDC = { αDC / (1 - αDC ) }

Current and Voltage Analysis

.IB: dc base current

IE: dc emitter current

IC: dc collector current

vBE: dc voltage at base

with respect to emitter

VCB. dc voltage at

collector with respect to

base

VCE - dc voltage at

collector with respect to

emitter

VBB : forward-biases the base-emitter junction VCC reverse-biases the base collector junctionVBE=0.7volts

VRB = VBB– VBE

IB * Rb = VBB– VBE

IB =( VBB– VBE ) / RB

VCE = VCC– IC RC

VCB = VCE– VBE

1– 6 Collector Characteristic Curves

FIGURE ( 1- 9 ) Collector characteristic curves

In linear region IC = βDC IB

1 - 7 Cutoff

Cutoff: Collector leakage current (ICEO) is extremely small and is usually neglected. Base-emitter and base-collector junctions are reverse-biased

1-8 Saturation

As IB increases due to increasing VBB, IC also

increases and VCE decreases due to the

increased voltage drop across RC- When the

transistor reaches saturation, IC can increase

no further regardless of further increase in IB.

Base-emitter and base-collector junctions are

forward-biased

1-9 DC Load Line

EXAMPLE

Determine whether or not the transistor in Figure

(1-15 ) is in saturation.

Assume VCE Sat=0.2V.

Solution First, determine Ic(sat)

IC(sat) = (VCC – VCE (sat.) ) / RC

= (10V-0.2V ) / Ik Ω

= 9.8 mANow, see if IB is large enough to produce IC(sat)

IB(sat) =(VBB – VBE) / RB

=(3 V - 0.7 V) /10 k Ω= 0.23 mA= 0.23 mA

IC = βdc ib = (50) (0.23 mA) = 11.5 mA

Therefore, the transistor is saturated

1-11 Maximum Transistor Ratings

*The transistor has limitations on its operation.

*maximum ratings are given for collector-to-base voltage, collector-to-emitter voltage, emitter-to-base voltage, collector current, and power dissipation.

*If VCE is maximum, Ic can be calculated as

*The product of VCE and Ic must not exceed the maximum power dissipation.

IC= (P D(Max)) / VCE

* If IC is maximum, VCE can be calculated by rearranging Equation

VCE =( P D(Max)) / Ic

Assume P D (max) is 500 mW, VCE max is 20 V, and Ic (max)

is 50 mA

EXAMPLE ( 6 )

The transistor in Figure (1-18 ) has the following

maximum ratings: PD (max) = 800 mW, VCE (max)

= 15 V, and IC (max) = 100 mA. Determine the

maximum value to which VCC can be adjusted

without exceeding a rating. Which rating would be

exceeded first?

ICVCEPD(Max)

100 mA

50 mA

33 mA

25 mA

5 V

10 V

15 V

20 V

500 mV

500 mV

500 mV

500 mV

Solution First, find IB so that you can determine IC.

IB = (VBB - VBE ) / RB = ( 5 –0.7 ) / 22k Ω = 195 uA

IC = βdc *IB = (100) * (195 uA) = 19.5 mAIC is much less than IC ( max) and will not change with VCC It is determined only by IB and βdc. The voltage drop across RC is

VRC = IC * RC = (19.5 mA) * ( 1 k Ω) = 19.5 V .

Now you can determine the value of VCC when

vCE = VCE (max) = 15 V.

VRC=VCC – VCE

VCC(max) = VCE(Max) + VRC = 15 V + 19.5 V

= 34.5 V

PD = VCE(max) * Ic = (15 V) * (19.5 mA) = 293 mW

Since PD (Max). is 800 mW, it is not exceeded when

VCC = 34.5 V. So, VCE (max) = 15 V is the limiting

rating in this case. If the base current is removed

causing the transistor to turn off, VCE(max) will be

exceeded first because the entire supply voltage,

VCC, will be dropped across the transistor.

1-18 • THE TRANSISTOR AS A SWITCH

(a) Cutoff— open switch

(b) Saturation — closed switch

1-19 Conditions in Cutoff

As mentioned before, a transistor is

in the cutoff region when the base-

emitter junction is not forward-biased.

Neglecting leakage current, all of the

currents are zero, and vce is equal to Vcc

VCE cutoff = VCC

1-20 Conditions in Saturation

when the base-emitter junction is forward-biased and there is enough base current to produce a maximum collector current, the transistor is saturated

IC sat= ( VCC – VCE sat ) /Rc

Since VCE sat is very small compared to VCC it can

usually be neglected.

The minimum value of base current needed to

produce saturation is

IB min = Ic sat / βdc

IB should be significantly greater than IB min to

keep the transistor well into saturation

EXAMPLE

For the transistor circuit in Figure (1-23), what is

vCE when Vin = 0 V?.

What minimum value of IB is required to saturate

this transistor if βdc is 200 Neglect vCE SAT .

Calculate the maximum value of RB when Vin = 5

V.

FIGURE )1-23)Vcc=10V,Rc =1k Ω

Solution

When Vin = 0 V, the transistor is in cutoff

(acts like an open switch) and

VCE=Vcc=10 V

Since Vce sat is neglected (assumed to be

0 V),

Vcc = 10 V

iCE sa t = vcc / Rc

= 10 / 1K Ω =10 mA

iB min = Ic sat / βdc =(10 mA/200) =50 uA

When the transistor is on, vBE= 0.7 V. The voltage

across RB is :-

VRB =VIN-VBE = 5V-0.7V=4.3V

Calculate the maximum value of RB needed to

allow a minimum IB of 50 uA by Ohm's law as

follows:

Rb max = ( VRB / IB(min) ) = ( 4.3 v/ 50 uA ) =

86 k Ω

A Simple Application of a Transistor Switch

transistor is in cutoff and, since there is no collector current, the LED does not emit light

When the square wave goes to its high level, the transistor saturates. This forward-biases the LED, and the resulting collector current through the LED causes it to emit light

CHAPTER FOUR

TRANSISTOR BIASING CIRCUITS

2-2 DC Bias

The dc operating point is often referred to as the Q-point

(quiescent point منطقة السكون).

If an amplifier is not biased with correct dc voltages on the input and

output, it can go into saturation or cutoff when an input signal is

applied.

(a) Linear operation: larger output has same shape as input except it is inverted

(C) Nonlinear operation: output voltage

limited (clipped) by saturation

dc operating point being too close to saturation.

The dc operating point (Q-point) being too close to cutoff.

(b) Nonlinear operation: output voltage limited (clipped) by

cutoff

2-3 Graphical Analysis

VCE = Vcc – Ic * Rc = 10 V - (20 mA)*(220 Ω) =

= 10 V - 4.4 V = 5.6 V .

VCE = 10 V - (30 mA)(220 Ω) = 10 V - 6.6 V = 3.4 V

VCE = 10 V (40 mA)* (220 Ω) = 10 V - 8.8 V = 1.2V

Kirchhoff's voltage law applied around the collector loop gives

Vcc – lc * Rc - VCE = 0

This results in a straight line equation for the load line of the form

y = m x + b as follows:

Ic = - (1 / Rc) VCE + Vcc / Rc

where –(l/Rc ) is the slope and Vcc / Rc is the y-axis intercept point.

2-6 Waveform Distortion

(a)Transistor driven into saturation

(B-)Transistor driven into cutoff

(c) Transistor driven into both saturation and cutoff

EXAMPLE 2-1

Determine the Q-point in Figure (2-7), and find the maximum

peak value of base current for linear operation. Assume βdc =

200.

IB = ( vBB - vBE )/ RB = (10V-0.7V) / 47K Ω =

198 uA

IC = βDC * IB=( 200)* (198 uA) = 39.6 mA

vCE. = VCC – IC * RC = 20 V - 13.07 V = 6.93 V

The Q-point is at IC = 39.6 mA and at VCE = 6.93 V. Since IC(cutoff) =0.

you need to know IC(sat) to determine how much variation in collector

current can occur and still maintain linear operation of the transistor.

IC sat = ( VCC / Rc) = (20V / 330 Ω) = 60.6mA

The dc load line is graphically illustrated in Figure (2-8),

showing that before saturation is reached, IC can increase an

amount ideally equal to

IC(sat) – ICQ = 60.6 mA - 39.6 mA =21 mA

However, IC can decrease by 39.6 mA before cutoff (IC = 0

V) is reached. Therefore, the limiting excursion is 21 mA

because the Q-point is closer to saturation than to cutoff.

The 21 mA is the maximum peak variation of the collector

current. Actually, it would be slightly less in practice

because VcE(sat) is not quite zero. The maximum peak

variation of the base current is determined as follows:

Ib peak = ( Ic max / βDC) = 21 mA / 200 = 105 uA

2-7 BASE BIAS

IB= (Vcc - VBE )/ RB

VCC – IC* RC- VCE=0

VCE = VCC -IC * RC

IC = βDC {(VCC – VBE ) / RB }

A more practical method is to use VCC as the single bias source, as shown

in Figure

2-10 • EMITTER BIAS

Although this method

of biasing requires

two separate dc

voltage sources, one

positive and the

other negative, it

does have an

important advantage

as you will learn.

VEE-IB*RB-VBE-IE*RE=0

Solving for VEE.

VEE = IB * RB + IE * RE + VBE

Since IC = IE and IC = βDC * IB

Substituting for IB,

(IE /βDC)*RB+IE*RE +VBE=VEE

Factoring out IE yields

IE {(RB /βDC) + RE} +VBE=VEE

Transposing VBE and then solving for IE

IE=( VEE – VBE) / {RE+(RB/ βDC)} ------- (2-4)

Since IC = IE.

IC = ( VEE - VBE ) / {RE+(RB/ βDC)} ---- (2-5)

VE=-VEE + IE*RE ----------------(2-6)

VB= VE +VBE-----------------------(2-7)

VC = VCC – ICRC ----------------- (2-8)

IC = IE yields

VCE = VCC – IC*RC - (-VEE + IE*RE)

= VCC + VEE – IC(RC + RE)

EXAMPLE (2-3)

Find IE, IC, and VCE in Figure (2-12) for βDC =

100 and VBE = 0.7 V. Draw the dc load line

showing the Q-point.

+10V

-10V

Solution

IE=( VEE – VBE) / {RE+(RB/ βDC)}

= (10V-0.7V)/ (4.7 k Ω +4.7 k Ω /100) = 1.80Ma

IC= IE = l.80 mA

VCE =VCC + VEE – IC*(RC + RE)

= 10 V + 10 V - 1.80 mA*(5.7 k Ω) = 9.74 V

IC sat ={ VCC - (-VEE)} / (RC+RE) =

={10 V - (-10 V)} / (5.7 k Ω)

=20 V / 5.7 k Ω= 3.51mA

The collector-to-emitter voltage at cutoff is

VCE (cutoff) = VCC - (-VEE) = 10 V - (-10 V) = 20 V

Δ IC(max)= IC sat – IC = 3.51 mA - 1.80 mA= 1.71 mA

2-13 VOLTAGE-DIVIDER BIAS

2-14 Input Resistance at the Base

R IN base = (V IN / I IN )

VIN=VBE + IE*RE

With the assumption that VBE « IE*RE. the equation reduces to

V IN = IE*RE

DC input resistance is RIN = VIN/ IIN.

Now, since

IE = IC = βDC *IBThe input current is the base current:

IIN=IBBy substitution,

RIN (base) = VIN / I IN = (βDC * IB * RE) / IBCanceling the IB terms gives

R IN (base) = βDC* RE ------------------(2-9)

2-15 Analysis of a Voltage-Divider Bias Circuit

If (βDC* RE ) ≥ 10 R2. then the formula simplifies to

VE = (VB)- VBE ------------------------- (2-11)

IE= VE / RE ---------------------- (2-12)

IC=IE ----------------------- (2-13)

VC=VCC -IC * RC ------ (2-14)

VCC – IC*RC –IE*RE- VCE= 0

Since IC = IE,

VCE = VCC – IC * RC – IE * RE

= VCC - IC (RC + RE) ------------ (2-15)

EXAMPLE (2-7)

Determine VCE and IC in Figure (2-22)

if βDC = 100,RC = 1k Ω ,RE = 560 Ω ,R1 =

10K Ω , VCC =10Vand R2 =5.6K Ω.

Solution

First, determine the dc input resistance at the

base to see if it can be neglected.

RIN (base) = βDC* RE = (100)*(560 Ω) = 56 k Ω

VB = { R 2/ ( R1+ R2)}* VCC

= {5.6k Ω / 15.6k Ω }* 10 V=3.59V

So, VE = VB – V BE =

= 3.59 V-0.7V= 2.89V

And IE = VE / RE = 2.89V / 56

= 5.16 mA

Therefore, IC = IE = 5.16 mA

VCE = VCC – IC (RC + RE)

= 10 V - 5.16 mA*(1.56 kΩ)

= 1.95 V

2-18 • COLLECTOR-FEEDBACK BIAS

IB = (VC - VBE ) / RB -----------------(2-16)

Lel's assume that IC » IB- The collector voltage is

VC = VCC - lC*RC

Also, IB = IC / βDC

Substituting for IB and VC in Equation (2-16),

IC / βDC = ( VCC - IC * RC- VBE ) / RB

{(IC*RB )/ βDC} +lC * RC = VCC - VBE.

Then you can solve for IC as follows:

IC { RC + (RB / βDC)} = VCC - VBE.

IC = {VCC - VBE} / { RC + (RB / βDC)} ----

Since the emitter is ground, VCE = VC

VCE = VCC - lC * RC ---------- ( 2-18)

EXAMPLE (2-10)

Calculate the Q-point values (IC and VCE) for the circuit in

Figure (2-28).RC =10 k ,RB = 100K ,VBE =0.7V, βDC = 100

+10 V

IC = (VCC- VBE ) / {RC +( RB / βDC) }

= (10 V - 0.7 V) / {lOk Ω +(100k Ω /100) }

= 845 uA

The collector-to-emitter voltage is

VCE = VCC – IC * RC = 10 V - (845 uA)*(10 k Ω) =

1.55 V