Lecture 8: Lecture 8:

BIPOLAR JUNCTION BIPOLAR JUNCTION TRANSISTORSTRANSISTORS

Semester II2010/2011

Code: EEE2213

BJT STRUCTUREBasic structure of the bipolar junction transistor (BJT) determines its operating characteristics.

Constructed with 3 doped semiconductor regions called emitter, base, and collector, which separated by two pn junctions.

2 types of BJT;

(1)npn: Two n regions separated by a p region

(2)pnp: Two p regions separated by an n region.

BIPOLAR: refers to the use of both holes & electrons as current carriers in the transistor structure.

Base-emitter junction: the pn junction joining the base region & the emitter region.

Base-collector junction: the pn junction joining the base region & the collector region.

A wire lead connects to each of the 3 regions. These leads labeled as;

E: emitter

B: base

C: collector

BASE REGION: lightly doped, & very thin

EMITTER REGION: heavily doped

COLLECTOR REGION: moderately doped

Standard BJT Symbols

BASIC BJT OPERATIONFor a BJT to operate properly as an amplifier, the two pn junctions must be correctly biased with external dc voltages.

Figure: shows a bias arrangement for npn BJTs for operation as an amplifier.

In both cases, BE junction is forward-biased & the BC junction is reverse-biased. called forward-reverse bias.

Look at this one circuit as two separate circuits, the base-emitter(left side) circuit and the collector-emitter(right side) circuit. Note that the emitter leg serves as a conductor for both circuits. The amount of current flow in the base-emitter circuit controls the amount of current that flows in the collector circuit. Small changes in base-emitter current yields a large change in collector-current.

The heavily doped n-type emitter region has a very high density of conduction-band (free) electrons.

These free electrons easily diffuse through the forward-based BE junction into the lightly doped & very thin p-type base region (indicated by wide arrow).

The base has a low density of holes, which are the majority carriers (represented by the white circles).

A small percentage of the total number of free electrons injected into the base region recombine with holes & move as valence electrons through the base region & into the emitter region as hole current (indicated by red arrows).

BJT operation showing electron flow.

When the electrons that have recombined with holes as valence electrons leave the crystalline structure of the base, they become free electrons in the metallic base lead & produce the external base current.

Most of the free electrons that have entered the base do not recombine with holes because the base is very thin.

As the free electrons move toward the reverse-biased BC junction, they are swept across into the collector region by the attraction of the positive collector supply voltage.

The free electrons move through the collector region, into the external circuit, & then return into the emitter region along with the base current.

The emitter current is slightly greater than the collector current because of the small base current that splits off from the total current injected into the base region from the emitter.

Transistor CurrentsThe directions of the currents in both npn and pnp transistors and their schematic symbol are shown in Figure (a) and (b). Arrow on the emitter of the transistor symbols points in the direction of conventional current. These diagrams show that the emitter current (IE) is the sum of the collector current (IC) and the base current (IB), expressed as follows:

IE = IC + IB

BJT CHARACTERISTICS & PARAMETERS

Figure shows the proper bias arrangement for npn transistor for active operation as an amplifier. Notice that the base-emitter (BE) junction is forward-biased by VBB and the base-collector (BC) junction is reverse-biased by VCC. The dc current gain of a transistor is the ratio of the dc collector current (IC) to the dc base current (IB), and called dc beta (DC).

DC = IC/IB

The ratio of the dc collector current (IC)to the dc emitter current (IE) is the dc alpha. α DC = IC/IEα DC = IC/IE

Ex 4-1 Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA.

Ex 4-1 Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA.

7350

65.3

A

mA

I

I

B

CDC

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

986.070.3

65.3

mA

mA

I

I

E

CDC

The collector current is determined by multiplying the base current by beta.

Thus, IC= βDC * IB

Analysis of this transistor circuit to predict the dc voltages and currents requires use of Ohm’s law, Kirchhoff’s voltage law and the beta for the transistor;

Application of these laws begins with the base circuit to determine the amount of base current. Using Kichhoff’s voltage law, subtract the VBE =0.7 V, and the remaining voltage is dropped across RB .

Thus, VRB = VBB - VBE

.

Determining the current for the base with this information is a matter of applying of Ohm’s law. VRB

/RB = IB

What we ultimately determine by use of Kirchhoff’s voltage law for series circuits is that, in the base circuit, VBB is distributed across the base-emitter junction and RB in the base circuit. In the collector circuit we determine that VCC is distributed proportionally across RC and the transistor(VCE).

BJT Circuit AnalysisThere are three key dc voltages and three key dc currents to be considered. Note that these 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 respect to the grounded emitter

VRc = ICRC

VCE = VCC – ICRC : voltage at the

collector with

respect to the emitter

The voltage across the reverse-biased collector-base junction

VCB = VCE – VBE where IC = βDCIB

Ex 4-2 Determine IB, IC, IE, VBE, VCE, and VCB in the circuit of Figure. The

transistor has a βDC = 150.

Ex 4-2 Determine IB, IC, IE, VBE, VCE, and VCB in the circuit of Figure. The

transistor has a βDC = 150.

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

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

Since the collector is at a higher voltage than the base, the collector-base junction is reverse-biased.

Gives a graphical illustration of the relationship of collector current and VCE with specified amounts of base current. With greater increases of VCC , VCE continues to increase until it reaches breakdown, but the current remains about the same in the linear region from 0.7V to the breakdown voltage.

Collector Characteristic Curves

Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 μA to 25 μA in 5 μA increment. Assume βDC = 100 and that VCE does not exceed breakdown.

Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 μA to 25 μA in 5 μA increment. Assume βDC = 100 and that VCE does not exceed breakdown.

IC = βDC IB

IB IC

5 μA 0.5 mA10 μA 1.0 mA15 μA 1.5 mA20 μA 2.0 mA25 μA 2.5 mA

CutoffWith no IB , the transistor is in the cutoff region and just as the name implies there is practically no current flow in the collector part of the circuit. With the transistor in a cutoff state, the full VCC can be measured across the collector and emitter(VCE).

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

SaturationOnce VCE reaches its maximum value, the transistor is said to be in

saturation.

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.

DC Load LineThe dc load line graphically illustrates IC(sat) and cutoff for a transistor.

DC load line on a family of collector characteristic curves illustrating the cutoff and saturation conditions.

Active region of the transistor’s operation.

Ex 4-4 Determine whether or not the transistors in Figure is in

saturation. Assume VCE(sat) = 0.2 V.

mAk

VV

R

VVI

C

satCECCsatC

8.90.1

2.010

)()(

Ex 4-4 Determine whether or not the transistors in Figure is in

saturation. Assume VCE(sat) = 0.2 V.

First, determine IC(sat)

mAmAII

mAk

V

k

VV

R

VVI

BDCC

B

BEBBB

5.11)23.0)(50(

23.010

3.2

10

7.03

Now, see if IB is large enough to produce IC(sat).

Thus, IC greater than IC(sat). Therefore, the transistor is saturated.

Maximum Transistor RatingsA transistor has limitations on its operation. The product of VCE

and IC cannot be maximum at the same time. If VCE is maximum, IC can be calculated as

CE

DC V

PI (max)

Ex 4-5 A certain transistor is to be operated with VCE = 6 V. If its maximum power rating is 250 mW, what is the most collector current that it can handle?

mAV

mW

V

PI

CE

DC 7.41

6

250(max)

Ex 4-6 The transistor in Figure 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?

First, find IB so that you can determine IC.

The voltage drop across RC is.

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

VCE(max) will be exceeded first because the entire supply voltage, VCC will be dropped across the transistor.

VRc = ICRC = (19.5 mA)(1.0 kΩ) = 19.5 VVRc = VCC – VCE when VCE = VCE(max) = 15 VVCC(max) = VCE(max) + VRc = 15 V + 19.5 V = 34.5 V

mAAII

Ak

VV

R

VVI

BDCC

B

BEBBB

5.19)195)(100(

19522

7.05

Ex 4-6 The transistor in Figure 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?

Derating PD(max)

P D(max) is usually specified at 25°C.

At higher temperatures, P D(max) is less.

Datasheets often give derating factors for determining P D(max) at any temperature above 25°C.

Ex 4-7 A certain transistor has a P D(max) of 1 mW at 25°C. The derating factor is 5 mW/ °C. What is the P D(max) at a temperature of 70°C?

Transistor DatasheetRefer Figure 4-20 (a partial datasheet for the 2N3904 npn transistor).

The maximum collector-emitter voltage (VCEO) is 40V.

The CEO subscript indicates that the voltage is measured from collector to emitter with the base open. VCEO= VCE(MAX)

The maximum collector current is 200 mA.

* Other characteristics can be referred from the datasheet.

A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?

A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?

PD(max) = 800 mWVCE(max) = 15 VIC(max) = 100 mA.

VCC(max) = VCE(max) + VRc = 40 V + 19.5 V = 59.5 V

PD = VCE(max)IC = (40V)(19.5mA) = 780 mW

However at the max value of VCE, the power dissipation is

Power Dissipation exceeds the maximum of 645 mW specified on the datasheet.

THE BJT AS AN AMPLIFIERAmplification of a relatively small ac voltage can be had by placing the ac signal source in the base circuit.

Recall that small changes in the base current circuit causes large changes in collector current circuit.

The ac emitter current : Ie ≈ Ic = Vb/r’e

The ac collector voltage : Vc = IcRc

Since Ic ≈ Ie, the ac collector voltage : Vc ≈ IeRc

The ratio of Vc to Vb is the ac voltage gain : Av = Vc/Vb

Substituting IeRc for Vc and Ier’e for Vb : Av = Vc/Vb ≈ IcRc/Ier’e

The Ie terms cancel : Av ≈ Rc/r’e

Ex 4-9 Determine the voltage gain and the ac output voltage in Figure if r’e = 50 Ω.

The voltage gain : Av ≈ Rc/r’e = 1.0 kΩ/50 Ω = 20The ac output voltage : AvVb = (20)(100 mV) = 2 V

THE BJT AS A SWITCHA transistor when used as a switch is simply being biased so that it is in cutoff (switched off) or saturation (switched on). Remember that the VCE in cutoff is VCC and 0V in saturation.

Conditions in Cutoff & Saturation

C

satCECC

satC R

VVI )(

)(

DC

satC

B

II

)(

(min)

A transistor is in the cutoff region when the base-emitter junction is not forward-biased. All of the current are zero, and VCE is equal to VCC

VCE(cutoff) = VCC

When the base-emitter junction is forward-biased and there is enough base current to produce a maximum collector current, the transistor is saturated.The formula for collector saturation current is

The minimum value of base current needed to produce saturation is

Ex 4-10 (a) For the transistor circuit in Figure, what is VCE when VIN = 0 V?

(b) What minimum value of IB is required to saturate this transistor if βDC is 200? Neglect VCE(sat). (c) Calculate the maximum value of RB when VIN = 5 V.

Ex 4-10 (a) For the transistor circuit in Figure, what is VCE when VIN = 0 V?

(b) What minimum value of IB is required to saturate this transistor if βDC is 200? Neglect VCE(sat). (c) Calculate the maximum value of RB when VIN = 5 V.

AmAI

I

mAk

V

R

VI

DC

satCB

C

CCsatC

50200

10

100.1

10

)((min)

)(

kA

V

I

VR

B

RB

B 8650

3.4

(min)(max)

(a) When VIN = 0 VVCE = VCC = 10 V

(b) Since VCE(sat) is neglected,

(c) When the transistor is on, VBE ≈ 0.7 V.VRB = VIN – VBE ≈ 5 V – 0.7 V = 4.3 V

Calculate the maximum value of RB

Transistor ConstructionTransistor Construction

There are two types of transistors: • pnp • npn

The terminals are labeled: • E - Emitter• B - Base• C - Collector

pnppnp

npnnpn

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Transistor OperationTransistor OperationWith the external sources, VEE and VCC, connected as shown:

• The emitter-base junction is forward biased• The base-collector junction is reverse biased

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Currents in a TransistorCurrents in a Transistor

The collector current is comprised of two currents:

BICIEI

minorityCOI

majorityCICI

Emitter current is the sum of the collector and base currents:

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Common-Base ConfigurationCommon-Base Configuration

The base is common to both input (emitter–base) and output (collector–base) of the transistor.

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Common-Base AmplifierCommon-Base Amplifier

Input CharacteristicsInput Characteristics

This curve shows the relationship between of input current (IE) to input voltage (VBE) for three output voltage (VCB) levels.

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This graph demonstrates the output current (IC) to an output voltage (VCB) for various levels of input current (IE).

Common-Base AmplifierCommon-Base Amplifier

Output CharacteristicsOutput Characteristics

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Operating RegionsOperating Regions

• Active – Operating range of the amplifier.

• Cutoff – The amplifier is basically off. There is voltage, but little current.

• Saturation – The amplifier is full on. There is current, but little voltage.

4848

EI

CI

Silicon)(for V 0.7BEV

ApproximationsApproximations

Emitter and collector currents:

Base-emitter voltage:

4949

Ideally: = 1 In reality: is between 0.9 and 0.998

Alpha (Alpha ())

Alpha () is the ratio of IC to IE :

EI

CIα dc

Alpha () in the AC modeAC mode:

EI

CIα

Δ

Δac

5050

Transistor AmplificationTransistor Amplification

Voltage Gain:

V 50kΩ 5ma 10

mA 10

10mA20Ω

200mV

))((RL

IL

V

iI

LI

EI

CI

iR

iViIEI

Currents and Voltages:

5151

250200mV

50V

iV

LVvA

Common–Emitter ConfigurationCommon–Emitter Configuration

The emitter is common to both input (base-emitter) and output (collector-emitter).

The input is on the base and the output is on the collector.

5252

Common-Emitter CharacteristicsCommon-Emitter Characteristics

Collector Characteristics Base Characteristics

5353

Common-Emitter Amplifier CurrentsCommon-Emitter Amplifier Currents

Ideal CurrentsIdeal Currents

IE = IC + IB IC = IE

Actual CurrentsActual Currents

IC = IE + ICBO

When IB = 0 A the transistor is in cutoff, but there is some minority current flowing called ICEO.

μA 0

BICBO

CEO α

II

1

where ICBO = minority collector current

5454

ICBO is usually so small that it can be ignored, except in high power transistors and in high temperature environments.

Beta (Beta ())

In DC mode:

In AC mode:

represents the amplification factor of a transistor. ( is sometimes referred to as hfe, a term used in transistor modeling calculations)

B

C

I

Iβ dc

constantac

CEV

B

C

I

I

5555

Determining from a Graph

Beta (Beta ())

108

A 25

mA 2.7β 7.5VDC CE

100

μA 10

mA 1

μA) 20 μA (30

mA) 2.2mA (3.2β

7.5V

AC

CE

5656

Relationship between amplification factors and

1β

βα

1α

αβ

Beta (Beta ())

Relationship Between Currents

BC βII BE 1)I(βI

5757

Common–Collector ConfigurationCommon–Collector Configuration

The input is on the base and the output is on the emitter.

5858

Common–Collector ConfigurationCommon–Collector Configuration

The characteristics are similar to those of the common-emitter configuration, except the vertical axis is IE.

5959

VCE is at maximum and IC is at minimum (ICmax= ICEO) in the cutoff region.

IC is at maximum and VCE is at minimum (VCE max = VCEsat = VCEO) in the saturation region.

The transistor operates in the active region between saturation and cutoff.

Operating Limits for Each ConfigurationOperating Limits for Each Configuration

6060

Power DissipationPower Dissipation

Common-collector:

CCBCmax IVP

CCECmax IVP

ECECmax IVP

Common-base:

Common-emitter:

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Transistor Specification SheetTransistor Specification Sheet

6262

Transistor Specification SheetTransistor Specification Sheet

6363

Transistor TestingTransistor Testing• Curve TracerCurve Tracer

Provides a graph of the characteristic curves.

• DMMDMMSome DMMs measure DC or hFE.

• OhmmeterOhmmeter

6464

Transistor Terminal Transistor Terminal IdentificationIdentification

6565