Chapter 5 Transistor Bias Circuits

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By: Dr Rosemizi Abd Rahim

Objectives

Discuss the concept of dc biasing of a transistor for linear operation

Analyze voltage-divider bias, base bias, and collector-feedback bias circuits.

Basic troubleshooting for transistor bias circuits

Lecture’s outline

• Objectives • Introduction• DC operating point• Voltage-divider bias• Other bias methods– Base bias– Emitter bias– Collector-feedback

bias

• Troubleshooting• Summary

IntroductionThe term biasing is used for application

of dc voltages to establish a fixed level of current and voltage.

Transistor must be properly biased with dc voltage to operate as a linear amplifier.

If amplifier is not biased with correct dc voltages on input and output, it can go into saturation or cutoff when the input signal applied.

There are several methods to establish DC operating point.

We will discuss some of the methods used for biasing transistors.

DC OPERATING POINT

The DC Operating Point• The goal of amplification in most cases is to

increase the amplitude of an ac signal without altering it.

• Improper biasing can cause distortion in the output signal.

The DC Operating PointThe purpose of biasing a circuit is to establish a proper stable dc operating point (Q-point). The dc operating point between saturation and cutoff is called the Q-point. The goal is to set the Q-point such that that it does not go into saturation or cutoff when an ac signal is applied.

• Q-point of a circuit: dc operating point of amplifier specified by voltage and current values (VCE and IC). These values are called the coordinates of Q-point.

• Refer to figure a, given IB = 200μA and βDC=100.

IC=βDCIB so IC=20mA and

• Figure b, VBB is increased to produce IB of 300μA and IC of 30mA.

• Figure c, VBB is increased to produce IB of 400μA and IC=40mA. So, VCE is:

•

VmAVRIVV CCCCCE 6.54.410)220)(20(10

VmAVRIVV CCCCCE 4.36.610)220)(30(10

VmAVRIVV CCCCCE 2.18.810)220)(40(10

DC Operating Point-DC load line• Recall that the collector characteristic curves graphically show the

relationship of collector current and VCE for different base currents.

• When IB increases, IC increases and VCE decreases or vice-versa. Each separate Q-point is connected through dc load line. At any point along line, values of IB, IC and VCE can be picked off the graph.

• Dc load line intersect VCE axis at 10V, where VCE=VCC. This is cutoff point because IB and IC zero. Dc load line also intersect IC axis at 45.5mA ideally. This is saturation point because IC is max and VCE=0.

DC Operating Point-Linear operation• Let’s look at the effect a

superimposed ac voltage has on the circuit. IB vary sinusoidally 100μA above and below Q-point of 300μA. IC vary up and down 10mA of its Q-point(30mA). VCE varies 2.2V above and below its Q-point of 3.4V.

• However, apply too much ac voltage to the base, the collector current might go into saturation or cutoff, resulting in a distorted or clipped waveform.

Variations in IC and VCE as a result of variation in IB.

Graphical load line illustration of transistor being driven intosaturation or cutoff

• When +ve IBQ peak is limited, transistor is in saturation.

• When –ve IBQ peak is limited, transistor is in cutoff.

Graphical load line for transistor in saturation and cutoff

Example 1• Determine Q-point in figure below and find the maximum

peak value of base current for linear operation. Assume βDC=200.

Solution • Q-point is defined by values of IC and VCE.

• Q-point is at IC=39.6mA and VCE=6.93V. Since IC(cutoff)=0, we need to know IC(sat) to determine variation in IC can occur and still in linear operation.

• Before saturation is reached, IC can increase an amount equal to: IC(sat) – ICQ = 60.6mA – 39.6mA = 21mA.

AkR

VVI

B

BEBBB 198

47

7.010

mAII BDCC 6.39)198(200

VmAVRIVV CCCCCE 93.6)330)(6.39(20

mAR

VI

C

CCsatC 6.60

330

20)(

Solution cont..• However, IC can decrease by 39.6mA before cutoff (IC=0) is

reached. Since the gap of Q-point with saturation point is less than gap between Q-point and cutoff, so 21mA is the max peak variation of IC.

• The max peak variation of IB is:

AmI

IDC

peakcpeakb

105

200

21)()(

VOLTAGE-DIVIDER BIAS

Voltage-Divider Bias• Voltage-divider bias is the

most widely used type of bias circuit. Only one power supply is needed and voltage-divider bias is more stable( independent) than other bias types. For this reason it will be the primary focus for study.

• dc bias voltage at base of transistor is developed by a resistive voltage-divider consists of R1 and R2.

• Vcc is dc collector supply voltage. 2 current path between point A and ground: one through R2 and the other through BE junction and RE.

Voltage divider bias

• If IB is much smaller than I2, bias circuit is viewed as voltage dividerof R1 and R2 as shown in Figure a.• If IB is not small enough to be neglected, dc input resistance RIN(base) must be considered. RIN(base) is in parallel with R2 asshown in figure b.

Input resistance at transistor base• VIN is between base and

ground and IIN is the current into base.

• By Ohm’s Law,

RIN(base) = VIN / IIN

• Apply KVL, VIN=VBE+IERE

• Assume VBE<<IERE, so VIN≈IERE

• Since IE≈IC=βDCIB,

VIN≈ βDCIBRE

• IN=IB, so

RIN(base)= βDCIBRE / IB

RIN(base) = DCRE

Analysis of Voltage-Divider Bias Circuit

Analysis of voltage divider bias circuit Total resistance from base to ground is:

A voltage divider is formed by R1 and resistance from base to ground in parallel with R2.

If DCRE >>R2, (at least ten times greater), then the formula simplifies to

EDC

baseIN

RR

RR

2

)(2

CCEDC

EDCB V

RRR

RRV

21

2

CCB VRR

RV

21

2

Analysis of Voltage-Divider Bias Circuit• Now, determine emitter voltage VE.

VE=VB – VBE

• Using Ohm’s Law, find emitter current IE.IE = VE / RE

• All the other circuit values IC ≈ IE

VC = VCC – ICRC

• To find VCE, apply KVL:VCC – ICRC – IERE – VCE =0

• Since IC ≈ IE,VCE ≈ VCC – IC (RC + RE)

Example 2• Determine VCE and IC in voltage-divider biased transistor

circuit below if βDC=100.

Solution 1. Determine dc input resistance at base to see if it can be

neglected.

2. RIN(base)=10R2, so neglect RIN(base). Then, find base voltage

3. So, emitter voltage

4. And emitter current

5. Thus, 6. And VCE is

kRR EDCbaseIN 56)560(100)(

VVk

kV

RR

RV CCB 59.310

6.15

6.5

21

2

VVVV BEBE 89.27.059.3

mAR

VI

E

EE 16.5

560

89.2

mAIC 16.5

VkmRRIVV ECCCCCE 95.1)56.1(16.510)(

Voltage-Divider Bias for PNP TransistorPnp transistor has opposite polarities from npn. To obtain pnp, required negative collector supply voltage or with a positive emitter supply voltage. The analysis of pnp is basically the same as npn.

Analysis of voltage bias for pnp transistor• Base voltage

• Emitter voltage

• By Ohm’s Law,

• And,

EEEDC

B VRRR

RV

21

1

BEBE VVV

E

EEEE R

VVI

CEEC

CCC

VVV

RIV

OTHER BIAS METHODS

BASE BIASEMITTER BIAS

COLLECTOR-FEEDBACK BIAS

Other bias methods - Base Bias• KVL apply on base circuit.

VCC – VRB – VBE = 0 or VCC – IBRB – VBE =0

• Solving for IB,

• Then, apply KVL around collector circuit. VCC – ICRC – VCE = 0

• We know that IC = βDCIB,

B

BECCB R

VVI

B

BECCDCC R

VVI

Base bias

• From the equation of IC, note that IC is dependent on DC. When DC vary, VCE also vary, thus changing Q-point of transistor.

• This type of circuit is beta-dependent and very unstable. Recall that DC changes with temperature and collector current. Base biasing circuits are mainly limited to switching applications.

Emitter Bias

Npn transistor with emitter bias

Emitter base•

DC

EB

II

Collector-Feedback BiasCollector-feedback bias is kept stable with negative feedback, although it is not as stable as voltage-divider or emitter. With increases of IC, VC decrease and causing decrease in voltage across RB, thus IB also decrease. With less IB ,IC go down as well.

Analysis of collector-feedback circuit• By Ohm’s Law,

• Collector voltage with assumption IC>>IB.

VC = VCC – ICRC

• And IB = IC / DC • So, collector current equation

• Since emitter is ground, VCE = VC.

VCE = VCC - ICRC

B

BECB R

VVI

DCBC

BECCC RR

VVI

/

TROUBLESHOOTING

TroubleshootingFigure below show a typical voltage divider circuit with correct voltage readings. Knowing these voltages is a requirement before logical troubleshooting can be applied. We will discuss some of the faults and symptoms.

All indicated faults

TroubleshootingFault 1: R1 Open

With no bias the transistor is in cutoff.

Base voltage goes down to 0 V.

Collector voltage goes up to10 V(VCC).

Emitter voltage goes down to 0 V.

Fault 2: Resistor RE Open

Transistor is in cutoff.

Base reading voltage will stay approximately the same.

Since IC=0, collector voltage goes up to 10 V(VCC).

Emitter voltage will be approximately the base voltage - 0.7 V.

TroubleshootingFault 3: Base lead internally open

Transistor is nonconducting (cutoff), IC=0A .

Base voltage stays approximately the same, 3.2V.

Collector voltage goes up to 10 V(VCC).

Emitter voltage goes down to 0 V because no emitter current through RE.

Fault 4: BE junction open

Transistor is in cutoff.

Base voltage stays approximately the same,3.2V.

Collector voltage goes up to 10 V(VCC)

Emitter voltage goes down to 0 V since no emitter current through RE.

TroubleshootingFault 5: BC junction open

Base voltage goes down to 1.11 V because of more base current flow through emitter.

Collector voltage goes up to 10 V(VCC).

Emitter voltage will drop to 0.41 V because of small current flow from forward-biased base-emitter junction.

TroubleshootingFault 6: RC open

Base voltage goes down to 1.11 V because of more current flow through the emitter.

Collector voltage will drop to 0.41 V because of current flow from forward-biased collector-base junction.

Emitter voltage will drop to 0.41 V because of small current flow from forward-biased base-emitter junction.

TroubleshootingFault 7: R2 open

Transistor pushed close to or into saturation.

Base voltage goes up slightly to 3.83V because of increased bias.

Emitter voltage goes up to 3.13V because of increased current.

Collector voltage goes down because of increased conduction of transistor.

SUMMARY

Summary

The purpose of biasing is to establish a stable operating point (Q-point).

The Q-point is the best point for operation of a transistor for a given collector current.

The dc load line helps to establish the Q-point for a given collector current.

The linear region of a transistor is the region of operation within saturation and cutoff.

Summary

Voltage-divider bias is most widely used because it is stable and uses only one voltage supply.

Base bias is very unstable because it is dependent. Emitter bias is stable but require two voltage supplies. Collector-back is relatively stable when compared to base bias, but not as stable as voltage-divider bias.