Study Materials_EDC 01

BJT Biasing

1 Chapter Objectives

Understand construction and operation of BJT.

Sketch input and output characteristics of common-base, common-emitter and common-collector confi gurations.

Discuss the need for biasing in BJT circuits and draw DC load line and quiescent point.

Analyze fi xed bias, collector to base bias and voltage-divider bias circuits to determine Q-point and comment on stability.

Compare the diff erent biasing circuits.

Discuss bias compensation method.

1.1 INTRODUCTIONTransistor was invented by Walter H. Brattain and John Bardeen at Bell Laboratories in 1947. Transistor replaced vacuum tubes due to smaller size, light weight, less power consumption, lower operating voltages, etc. Transistors can perform the function of current amplifi cation and voltage amplifi cation as well as power amplifi cation. The amplifi cation in transistor is ob-tained by passing the weak signal from low-resistance region to high-resistance region. Hence, the device is named ‘transistor’ (transfer resistor).

1.2 BIPOLAR JUNCTION TRANSISTORS A bipolar junction transistor (BJT) is a three-terminal semiconductor device containing two pn junctions. When a p-type layer is placed between two n-type layers, an npn transistor is formed. Similarly, when an n-type layer is placed between two p-type layers, a pnp transistor is formed.

In each type of transistor, the middle region is called base of the transistor and other two regions are called emitter and collector. The physical size of the collector is greater than both emitter and base. The emitter is heavily doped while the base is lightly doped. The doping of the collector is in between that of emitter and base. The pn junction joining the base region and the emitter region is called the emitter-base junction. The pn junction joining the base region and the collector region is called collector-base junction. The term ‘bipolar’ refers to the use of both holes and electrons as charge carriers in the transistor structure. Figure 1.1 shows the transistor types and its symbols.

M01_XXXXXXXX_XX_C01.indd 1M01_XXXXXXXX_XX_C01.indd 1 19 07 12 3:23 PM19 07 12 3:23 PM

1.2 Electronic Circuits I

npn Transistor and Its Symbol

pnp Transistor and Its Symbol

C

B

E

CE n n

B

p

CE p p

B

n

C

B

E

Fig. 1.1 Transistor Types and Symbols

In the symbols shown in Fig. 1.1, the arrowhead in the emitter indicates the direction of conven-tional current which is opposite to the fl ow of electrons. In the npn transistor, conventional current fl ows out of emitter, while in the pnp transistor, the conventional current fl ows into the emitter.

1.3 BJT OPERATIONFor proper working of BJT, the emitter-base junction is forward biased by the voltage VEE and collector-base junction is reverse biased by the voltage VCC. The forward bias from the base to the emitter narrows the emitter-base depletion region and the reverse bias from the base to the collector widens the collector-base depletion region. Figure 1.2 shows the BJT operation with direction of conventional currents.

n np

VEE VCC

Fig. 1.2 BJT Operation

When the emitter-base junction is forward biased, the large number of majority carriers, i.e. electrons from the n-type emitter, will get pushed towards the base junction. If the forward-biased voltage is more than the cut-in voltage (0.7 V for the silicon transistor and 0.3 V for the germanium transistor), electrons will be diffused into the base junction. Since the base region is very thin and lightly doped, a very few of the electrons injected into the base recombine


1.4 Common-base Confi guration 1.3

with holes. This constitutes the base current IB. These diffused electrons are minority carriers in the base region. The minority carriers can easily cross the reverse-biased junction. Hence, most of the electrons diffuse to the reverse-biased collector-base junction and are swept across that junction under the infl uence of the electric fi eld established by VCC. This constitutes the collector current IC. Applying Kirchhoff’s current law to the transistor, IE = IB + IC which is graphically depicted in Fig. 1.3.

n np

VEE VCC

ICIE

IB

Fig. 1.3 Graphical Depiction of the Relationships Among the Emi er, Base and Collector Currents

BJT can be operated in three regions:

(i) Cut-off: In this region, both emitter-base and collector-base junctions are reverse biased.

(ii) Active: In this region, the emitter-base junction is forward biased and the collector-base junction is reverse biased.

(iii) Saturation: In this region, both emitter-base and collector-base junctions are forward biased.

1.4 COMMON-BASE CONFIGURATION Most of the circuits employing BJT are two-port networks. Since a two-port network has four terminals and BJT has three terminals, one of the terminals of BJT is made common to input and output circuits.

In common-base confi guration, input is applied between the emitter and the base and output is taken from the collector and the base. Thus, the base is common to both input and output circuits as shown in Fig. 1.4.

E C

B B

Input Output

Fig. 1.4 Common-base Confi guration

Figure 1.5 shows the experimental set-up to draw input and output characteristics in common-base confi guration.



V

mA mA

V

IE IC

VBE IBVCB VCCVEE

Fig. 1.5 Experimental Set-up to Draw Input and Output Characteristics

1.4.1 Input CharacteristicsThe input characteristics show the relation between input current IE and input voltage VBE for different values of output voltage VCB. For a given VCB, the input characteristic resembles the characteristic of forward-biased diode. Input current IE increases as input voltage VBE increases for a fi xed value of VCB. For a given value of VBE, IE increases with increase in VCB due to early effect.

As VCB increases, the width of the depletion layer in the base increases. Hence, the width of the base available for conduction decreases. The reduction in the width of the base due to increase in reverse bias is known as Early effect. Due to Early effect, the chance of recombi-nation of electrons with the holes in the base decreases. The base current decreases but more electrons can travel from emitter to collector terminals. Hence, collector current increases with increase in emitter current IE.

As reverse-biased voltage VCB further increases, at one stage the depletion region com-pletely occupies the base at which the collector-base junction breaks down. This phenom-enon is known as punch-through. Figure 1.6 shows the input characteristics of common-base confi guration.

Fig. 1.6 Input Characteristics

5

4

3

2

1

0.20 0.4 0.6 0.8

IE (mA)

VBE (V)

VCB = 20 V

VCB = 10 V

VCB = 0 V


1.4 Common-base Confi guration 1.5

1.4.2 Output CharacteristicsThe output characteristics show the relation between output current IC and output voltage VCB for different values of input current IE. There are three different regions in output characteristics as shown in Fig. 1.7.

(i) Cut-off region: In this region, both the junctions are reverse biased. When the emitter-base junction is reverse biased, the current due to majority carrier, i.e. IE, is zero. Since the collector-base junction is reverse biased, current due to minority car-riers fl ows from the collector to the base which is represented as ICBO. The current ICBO is so small (microamperes) in magnitude compared to the vertical scale of IC that it appears on the same horizontal line as IC = 0.

(ii) Active region: In this region, the emitter-base junction is forward biased and the col-lector-base junction is reverse biased. Once VCB reaches a value large enough to ensure a large portion of electrons enter the collector, collector current IC remains constant as shown by horizontal lines. As IE increases, IC increases.

(iii) Saturation region: In this region, both the junctions are forward biased. When VCB is negative, the collector-base junction is actually forward biased. Thus, graphs are drawn on the negative side of VCB. In this region, there is large change in collector cur-rent with small increase in voltage VCB.

Fig. 1.7 Output Characteristics

IE = 5 mA

IE = 4 mA

IE = 3 mA

IE = 2 mA

IE = 1 mA

IE = 0 mA

Cut-off region

Saturationregion

5

4

3

2

1

20 4 6 8

IC (mA)

VCB (V)

Active region

Current amplifi cation factor (α): It is defi ned as the ratio of change in collector current to the change in emitter current at constant collector-base voltage VCB.

αAC = II

C

E V constantCB

ΔΔ =

If only DC values are considered,

αDC = II

C

E



1.5 COMMON-EMITTER CONFIGURATION In common-emitter confi guration, input is applied between the base and the emitter and output is taken from the collector and the emitter. Thus, the emitter is common to input and output circuits as shown in Fig. 1.8.

Fig. 1.8 Common-emi er Confi guration

C

B

E E

OutputInput

Figure 1.9 shows the experimental set-up to draw input and output characteristics in common-emitter confi guration.


IE

V

IB

VBE

VCC

VBB

V VCE

IC

μA

mA

1.5.1 Input CharacteristicsThe input characteristics show the relation between input current IB and input voltage VBE for different values of output voltage VCE. It resembles the characteristics of the forward-biased diode. Input current IB increases as input voltage VBE increases for fi xed value of VCE. Figure 1.10 shows the input characteristics of common-emitter confi guration.

As reverse-biased voltage VCE increases, the depletion region in the collector base increases which decreases the width of the base available for conduction. Hence, IB decreases due to early effect and the graph shifts towards the X-axis.


1.5 Common-emi er Confi guration 1.7


VCE = 0 V

VCE = 10 V

VCE = 20 V50

40

30

20

10

0.20 0.4 0.6 0.8

IB (μA)

VBE (V)

1.5.2 Output CharacteristicsThe output characteristics show the relation between output current IC and output voltage VCE for different values of input current IB. The output characteristic has three different regions as shown in Fig. 1.11.


IB = 50 μA

IB = 40 μA

IB = 30 μA

IB = 20 μA

IB = 10 μA

IB = 0 μA

0

Cut-off region

Saturation region

IC (mA)

VCE (V)

5

4

3

2

1

2 4 6 8

Active region

(i) Cut-off region: In this region, both the junctions are reverse biased. When the emit-ter-base junction is reverse biased, the current due to majority carrier, i.e. IB, is zero. Since the collector-base junction is reverse biased, the current due to minority carriers fl ows from the collector to the emitter which is represented as ICEO.

(ii) Active region: In this region, the emitter-base junction is forward biased and the col-lector-base junction is reverse biased. As IB is maintained constant, current IC increases as reverse-biased voltage VCE increases.



(iii) Saturation region: In this region, both the junctions are forward biased. When VCE is reduced to a small value such as 0.2 V, the collector-base junction is actually forward biased (VCB = VCE – VBE = 0.2 – 0.7 = – 0.5 V). In this region, there is large change in collector current IC with small change in VCE.

Current amplifi cation factor (β ): It is defi ned as the change in collector current to the change in base current at constant collector-emitter voltage VCE.

βAC = II

C

B V constantCE

ΔΔ =


βDC = II

C

BRelation between α and βIE = IB + IC

Also,

βDC = II

C

B

= −I

I IC

E C

= −

II

II

1

C

E

C

E

= α

α−1DC

DC

α =⎛

⎝⎜⎞

⎠⎟II

C

EDC

If subscript DC is ignored,

β = αα−1

Collector current ICApplying Kirchhoff’s current law to the transistor,

IE = IB + IC (1.1)The collector current IC has two components:

(i) the current due to majority carriers, i.e. the fraction of emitter current which reaches the collector

(ii) the current due to minority carriers, i.e. leakage current which fl ows due to minority carriersIC = IC majority + IC minority

= αIE + ICO (1.2)


1.6 Common-collector Confi guration 1.9

For general purpose transistors, IC is measured in milliamperes and ICO is measured in microamperes or nanoamperes. ICO, like a reverse-biased diode, is temperature dependent and can be neglected in comparison with IC.

Substituting IE in Eq. (1.2), IC = α(IB + IC) + ICO

= α IB + α IC + ICO

IC(1 – α) = α IB + ICO

IC = αα−1

IB + α−

11

ICO

= β IB + (β + 1)ICO

1.6 COMMON-COLLECTOR CONFIGURATION In common-collector confi guration, input is applied between the base and the collector and output is taken from the emitter and the collector. Thus, the collector is common to input and output circuits as shown in Fig. 1.12.

Fig. 1.12 Common-collector Confi guration

E

B

C C

OutputInput

Figure 1.13 shows the experimental set-up to draw input and output characteristics in common-collector confi guration.


IC

V

IB

VBC

VEE

VBB

V VCE

IE

μA

mA

1.6.1 Input CharacteristicsThe input characteristics show the relation between input current IB and input voltage VBC for different values of output voltage VCE. The input voltage VBC is largely determined by the output voltage VCE.



Input current IB decreases to 0 as input voltage VCB increases slightly for fi xed value of VCE. For transistor,

VCE = VBE + VBC VBE = VCE – VBC

When VBC is increased keeping VCE constant, VBE decreases which decreases IB. Therefore, if the value of VBC is allowed to increase to a point where it is near to the value of VCE, the value of VBE approaches 0, and no base current will fl ow. Figure 1.14 shows the input characteristics of common-collector confi guration.


100

80

60

40

20

50 10 15

IB (μA)

VBC (V)

VCE = 5 V VCE = 10 V VCE = 15 V

1.6.2 Output Characteristics The output characteristics show the relation between output current IE and output voltage VCE for different values of input current IB. Since IC is approximately equal to IE, the common-collector output characteristics are identical to those of common-emitter output characteristics. Figure 1.15 shows the output characteristics of common-collector confi guration.


IB = 40 μA

IB = 30 μA

IB = 20 μA

IB = 10 μA

0

IE (mA)

VCE (V)

IB = 50 μA5

4

3

2

1

Cut-off region

2 4 6 8

Saturation region

Active region

IB = 0 μA


1.7 Load-line Analysis 1.11

Current amplifi cation factor (γ ): It is defi ned as the ratio of change in emitter current to the change in base current at constant collector-emitter voltage VCE.

γAC = II

E

B V constantCE

ΔΔ =


γDC = II

E

B

1.7 LOAD-LINE ANALYSIS The basic function of a transistor is to do amplifi cation. The weak signal is given to the tran-sistor and amplifi ed output is obtained from the collector. The process of raising the strength of weak signal without any change in its general shape is known as faithful amplifi cation. A transistor must be properly biased to operate as an amplifi er.

Figure 1.16 shows a basic common-emitter amplifi er. The capacitor CC1is a DC-blocking

capacitor and couples AC input signal to the base of the transistor. The capacitor CC2is used to

couple AC output of the amplifi er to load RL.

Fig. 1.16 CE Amplifi er

VS

RS

RB RC

CC1

Vo

+VCC

CC2

RL

DC analysis

For DC, f = 0,

XC = π f C

12

= ∞



The DC equivalent circuit is obtained by replacing all capacitors by open circuits as shown in Fig. 1.17.

Fig. 1.17 DC Equivalent Circuit

RB RC

VBE

VCE

+VCC

+

−−

+

IB IC

Load line

Applying Kirchhoff’s voltage law to the collector-emitter circuit,VCC – IC RC – VCE = 0

IC = – R1

C

VCE + VR

CC

C

This equation represents a DC load line with slope of –R1

C

and y-intercept of VR

CC

C

.

When IC = 0, i.e. the transistor is in the cut-off region,VCE = VCC

When VCE = 0, i.e. the transistor is in saturation region,

IC = VR

CC

C

Thus two end points are (VCC, 0) and ⎛

⎝⎜⎞

⎠⎟VR

0, CC

C

. A line passing through these points is called

DC load line as the slope of this line depends on the DC load RC.

Quiescent point

Applying Kirchhoff’s voltage law to the base-emitter circuit,VCC – IB RB – VBE = 0

IB = −V VR

CC BE

B


1.7 Load-line Analysis 1.13

This equation gives the value of base current. For this value of base current, output characteristic of the amplifier is plotted which intersects the DC load line at Q-point. Hence, Q-point indicates quiescent (inactive, still) value of collector-emitter voltage VCE and collector current IC. Figure 1.18 shows the DC load line and Q-point for common-emitter amplifier.

Fig. 1.18 Load Line and Q-point

VCEQ0 VCCVCE

VCCRC

ICQ

IC

IBQ

Q

Need for biasing

DC biasing is used to establish proper values of IC and VCE called the DC operating point or quiescent point or Q-point. The basic problem involved in the design of transistor circuits is establishing and maintaining the proper collector-to-emitter voltage and collector current in the circuit. This condition is known as transistor biasing. The biasing conditions must be maintained despite variations in temperature, variations in gain and leakage current and variation in supply voltages. For faithful amplifi cation, the following conditions must be satisfi ed:

(i) Proper zero signal collector current IC

(ii) Proper base-emitter voltage VBE

(iii) Proper collector-emitter voltage VCE

The value of IC and VCE is expressed in terms of operating point or quiescent point Q. For faithful amplifi cation, Q-point must be selected properly. The fulfi lment of the above condi-tions is known as transistor biasing.

While fi xing the Q-point it has to be seen that the output of the amplifi er is a proper sinusoidal waveform for sinusoidal input without distortion. If an amplifi er is not biased properly, it can go into saturation or cut-off when an input signal is applied. By fi xing the Q-point at different positions, we can observe the variation in collector current and collector-emitter voltage corresponding to a given variation of base current.

When the Q-point is located in the middle of the DC load line as shown in Fig. 1.19, sinusoidal waveform without distortion is obtained at the output.



Fig. 1.19 Q-point in the Active Region

IC

VCC

IBQ

ICQ

VCEQ

VCE

Q

0

When the Q-point is located near the saturation region as shown in Fig. 1.20, the collector current is clipped at the positive half cycle because the transistor is driven into saturation.

Fig. 1.20 Q-point Near Saturation Region

IC

VCC

IBQ

ICQ

VCEQ

VCE

Q

0

When the Q-point is located near the cut-off region as shown in Fig. 1.21, the collector cur-rent is clipped at the negative half cycle because the transistor is driven into cut-off.


1.8 Factors Aff ecting Stability of Q-point 1.15

Fig. 1.21 Q-point Near Cut-off Region

IC

VCC

IBQ

ICQ

VCEQ

VCE

Q

0

Hence, values of different resistances and voltages must be selected in such a way that the Q-point should be:

(i) in active region. (ii) on DC load line.

(iii) selected in middle of the DC load line to avoid clipping of signals.

1.8 FACTORS AFFECTING STABILITY OF Q -POINTThe collector current IC depends on reverse saturation current ICO, base-emitter voltage VBE and current gain β. These parameters are temperature dependent; i.e. as temperature changes, these parameters change. Hence, collector current IC changes. Due to this, the Q-point changes. Hence, the Q-point has to be stabilized against temperature variation.

(i) ICO: The collector current is given byIC = β IB + (β + 1)ICO

When a pn junction is reverse biased, there is a small amount of current due to fl ow of minority carriers across the junction. Since minority carriers are thermally generated, reverse saturation current ICO is extremely temperature dependent. The reverse satura-tion current ICO doubles for every 10oC rise in temperature. The fl ow of collector current produces heat at the collector junction. This increases the temperature, therefore reverse saturation current ICO increases. Hence, collector current IC again increases. This increase in IC increases the temperature of collector junction which increases ICO again. The effect is cumulative and at one stage IC is so large which damages the transistor. This process is known as thermal runaway and is shown in Fig. 1.22.



T°C

IC

ICOPD

Fig. 1.22 Thermal Runaway

(ii) VBE: The base-emitter voltage VBE decreases at the rate of 2.5 mV/oC; i.e. the device starts operating at lower voltages. Hence, base current IB changes. Since IC = β IB, collector current IC changes. Figure 1.23 shows the variation of VBE with temperature.

0 0.1

100°C 25°C −69°C

0.2 0.3 0.4 0.5

IC (mA)

VBE (mV)

Fig. 1.23 Variation of VBE with Temperature

(iii) β: The transistor parameter β is temperature and device dependent. β increases with the increase in temperature. The value of β is different even for transistors of the same type. If the transistor is replaced by another transistor even of the same type, the value of β is different. Hence, collector current IC changes. Figure 1.24 shows the variation of β with temperature.


1.8 Factors Aff ecting Stability of Q-point 1.17

0

100

β

50

100°C

25°C

−55°CIC (mA)

Fig. 1.24 Variation in β with Temperature

Table 1.1 shows typical parameters of silicon transistor at different temperatures. From this table, it is clear that as temperature changes, ICO, VBE and β change. Hence, collector current IC changes with the change in temperature.

Table 1.1 Typical Parameters of Silicon Transistor

t °C –65 25 175ICO, μA 1.95 × 10−3 1 33,000VBE, V 0.78 0.6 0.225β 25 55 100

There are two methods to stabilize variation in IC with these parameters:

(i) Thermal stabilization: In this method, resistive biasing circuits are used which allow IB to vary so as to keep IC relatively constant with variations in ICO, VBE and β. There are three confi gurations to bias the BJT:

• Fixed bias• Collector-to-base bias • Voltage-divider bias

This process of stabilizing Q-point is called thermal stabilization.(ii) Bias compensation: In this method, temperature-sensitive devices such as diodes,

transistors, thermistors, sensistors etc. are used which provide compensating voltage and current to stabilize variations in IC with VBE and ICO.

1.8.1 Stability FactorsThe rate of change of collector current with respect to collector leakage current ICO at constant VBE and β is called stability factor.



S = II

C

CO V and constantBE

ΔΔ β

=Δ − Δ

Δ − Δβ

I I

I IC C

CO CO V and constantBE

2 1

2 1

The rate of change of collector current with respect to VBE at constant ICO and β is called stability factor S ′.

S ′I

VC

BE I and constantCO

=ΔΔ β

=Δ − Δ

Δ − Δβ

I I

V VC C

BE BE I and constantCO

2 1

2 1

The rate of change of collector current with respect to b at constant ICO and VBE is called stability factor S ″.

S ″β

=ΔΔ

IC

I Vand constantCO BE

β β

=Δ − Δ

Δ − Δ

I IC C

I V2 1 and constantCO BE

2 1

The larger the value of stability factor, the more sensitive is the circuit to variations in that parameter.

The total change in collector current over a specifi ed temperature range is obtained by ex-pressing this change as the sum of individual changes due to three stability factors.

∆IC = S ∆ICO + S ′ ∆VBE + S ″ ∆β

1.9 FIXED-BIAS CIRCUITFigure 1.25 shows a fi xed-bias circuit. It is the simplest transistor DC bias confi guration.

Fig. 1.25 Fixed-bias Circuit

VS

RS

RB RC

CC1

Vo

+VCC

CC2

RL

DC analysis

For DC, f = 0,

XC = π f C

12

= ∞


1.9 Fixed-bias Circuit 1.19



RB RC

VBE

VCE

+VCC

+

−−

+

IB IC

Collector current ICApplying Kirchhoff’s voltage law to the base-emitter circuit,

VCC – IB RB – VBE = 0

IB = −V VR

CC BE

B

For the base-emitter circuit, the net voltage is VCC – VBE and the resistance is RB. When VCC and RB are selected for a circuit, IB is fi xed. Hence, the circuit is called fi xed-bias circuit.

IC = β IB

= β V V

RCC BE

B

−⎛

⎝⎜⎞

⎠⎟

The base current is controlled by the value of RB, and IC is related to IB by a constant β. But ΙC is not a function of resistor RC. Change in RC will not affect the value of IB or ΙC in the active region of the transistor. But change in RC will affect the value of VCE.

Collector-emitter voltage VCE

Applying Kirchhoff’s voltage law to the collector-emitter circuit,

VCC – IC RC – VCE = 0VCE = VCC – ICRC



Load-line analysis


VCC – IC RC – VCE = 0

IC = – R1

C

VCE + VR

CC

C

This equation represents a DC load line with slope of – R1

C


CC

C

.When IC = 0, i.e. transistor is in cut-off region,

VCE = VCC

When VCE = 0, i.e. transistor is in saturation region,

IC = VR

CC

C

Thus, two end points are (VCC, 0) and ⎛

⎝⎜⎞

⎠⎟VR

0, CC

C

. By joining these two end points, a DC load line is drawn.

From the base-emitter circuit,

IB = −V VR

CC BE

B

For this value of base current, we can establish the actual Q-point as shown in Fig. 1.27.


VCEQ0 VCCVCE

VCCRC

ICQ

IC

IBQ

Q

From the Fig. 1.27, it is clear that the saturation current for the circuit is IC sat = VR

CC

C. This is

the resulting current when a short circuit is applied between collector-emitter terminals.



Stability factors

Applying Kirchhoff’s voltage law to the base-emitter circuit,


IB = −V VR

CC BE

B

We know that

IC = β IB + (β + 1)ICO

= β −⎛

⎝⎜⎞

⎠⎟V V

RCC BE

B

+ (β + 1)ICO

= βV

RCC

B

– βV

RBE

B

+ (β + 1)ICO (1.3)

From Eq. (1.3), it is clear that collector current IC is function of ICO, VBE and β.

(a) Stability factor S: When ICO changes from ICO1 to ICO2

, IC changes from IC1 to IC2

. From Eq. (1.3), at t1°C,

IC1 =

βVR

CC

B

– βV

RBE

B

+ (β + 1)ICO1 (1.4)

At t2°C,

IC2 =

βVR

CC

B

– βV

RBE

B

+ (β + 1)ICO2 (1.5)

Subtracting Eq. (1.4) from Eq. (1.5),

IC2 – IC1

= (β + 1)(ICO2 – ICO1

)

−

−

I I

I IC C

CO CO

2 1

2 1

= β + 1

∴ S = ΔΔ

II

C

CO

= β + 1



(b) Stability factor S ′: When VBE changes from VBE1 to VBE2


. From Eq. (1.3), at t1°C,

IC1 =

βVR

CC

B

– βV

RBE

B

1 + (β + 1)ICO (1.6)

At t2°C,

IC2 =

βVR

CC

B

– βV

RBE

B

2 + (β + 1)ICO (1.7)


IC2 – IC1

= – βRB

(VBE2 – VBE1

)

I I

V VC C

BE BE

2 1

2 1

−

− = – β

RB

∴ S ′ = ΔΔ

IV

C

BE

= – βRB

(c) Stability factor S″: From Eq. (1.3),

IC = βV

RCC

B

– βV

RBE

B

+ β ICO [� (β + 1) ≈ β]

= β − +V V I R

R( )CC BE CO B

B

When β changes from β1 to β2, IC changes from β1 to β2. At t1°C,

IC1 =

β − +1 V V I RR

( )CC BE CO B

B

(1.8)

At t2°C,

IC2 =

β − +2 V V I RR

( )CC BE CO B

B

(1.9)

I

IC

C

2

1

= ββ

2

1



Subtracting one from both the sides,

I

IC

C

2

1

– 1 = ββ

2

1

– 1

−I I

IC C

C

2 1

1

= β β

β−2 1

1

β β

−

−2 1

I IC C2 1 = β1

IC1

∴ S ″ = β

ΔΔ

IC = β1

IC1

Example 1.1: For the circuit shown in Fig. 1.28, fi nd IC, VCE and S.

Fig. 1.28 Example 1.1

500 kΩ 3 kΩ

β = 100

+10 V

Solution:

(i) Applying Kirchhoff’s voltage law to the base-emitter circuit,


IB = −V VR

CC BE

B

= −×

10 0.7500 103



= 18.6 μAIC = β IB = 100 × 18.6 × 10–6 = 1.86 mA

(ii) Applying Kirchhoff’s voltage law to the collector-emitter circuit,VCC – IC RC – VCE = 0

VCE = VCC – IC RC

= 10 – 1.86 × 10–3 × 3 × 103

= 4.42 V(iii) S = β + 1 = 100 + 1 = 101

Example 1.2: For the circuit shown in Fig. 1.29, fi nd RB and RC with IC = 2 mA, VCE = 6 V, V CC = 12 V and β = 100.


RB RC

+12 V

Solution:

(i) IB = βIC = × −2 10

100

3

= 20 μA



RB = −V VI

CC BE

B

= 12 0.720 10 6

−× −

= 566 kΩ



(ii) Applying Kirchhoff’s voltage law to the collector-emitter circuit,


RC = −V VI

CC CE

C

= −× −

12 62 10 3

= 3 kΩ

Example 1.3: For a fi xed-bias circuit shown in Fig. 1.30, determine RB, IC, RC and VCE where VCC = 12 V, VC = 6 V, β = 80 and IB = 40 μA.


RB RC

VC

+12 V

Solution:



RB = −V VI

CC BE

B

= −× −

12 0.740 10 6

= 282.5 kΩ



(ii) IC = β IB = 80 × 40 × 10–6 = 3.2 mA(iii) Applying Kirchhoff’s voltage law to the collector-emitter circuit,

VCC – IC RC – VC = 0

RC = −V V

ICC C

C

= 12 63.2 10 3

−× −

= 1.875 kΩ(iv) VCE = VC = 6 V

Example 1.4: Using a CE amplifi er with fi xed bias, where IB = 20 μA, IE = 4 mA, VCE = 7.2 V and RC = 2.2 kΩ, determine IC, VCC, β and RB (Fig. 1.31).


RB 2.2 kΩ

+VCC

Solution:

(i) IE = IB + IC

IC = IE – IB = 4 × 10–3 – 20 × 10–6

= 3.98 mA(ii) Applying Kirchhoff’s voltage law to the collector-emitter circuit,

VCC – IC RC – VCE = 0VCC = IC RC + VCE

= 3.98 × 10–3 × 2.2 × 103 + 7.2= 15.96 V



(iii) β = II

C

B

= ××

−

−

3.98 1020 10

3

6 = 199

(iv) Applying Kirchhoff’s voltage law to the base-emitter circuit,VCC – IB RB – VBE = 0

RB = −V VI

CC BE

B

= −× −

15.96 0.720 10 6

= 762.8 kΩ

Example 1.5: For the fi xed-bias circuit, where α = 0.98, ICBO = 10 μA, RC = 4 kΩ, RB = 820 kΩ, VCC = 12 V, fi nd IC and VCE (Fig. 1.32).


820 kΩ 4 kΩ

+12 V

Solution:

(i) β = αα−1

= −0.98

1 0.98= 49

Applying Kirchhoff’s voltage law to the base-emitter circuit,VCC – IB RB – VBE = 0

IB = −V VR

CC BE

B

= −×

12 0.7820 103

= 13.78 μA IC = β IB + (β + 1)ICBO

= 49 × 13.78 × 10–6 + (49 + 1) × 10 × 10–6

= 1.17 mA




VCC – IC RC – VCE = 0VCE = VCC – IC RC

= 12 – 1.17 × 10–3 × 4 × 103

= 7.3 V

Example 1.6: Determine the percentage change in IC and VCE for the circuit shown in Fig. 1.33 when β changes from 90 to 135.


470 kΩ 2.7 kΩ

+16 V

Solution:

(a) For β = 90Applying Kirchhoff’s voltage law to the base-emitter circuit,


IB = −V VR

CC BE

B

= −×

16 0.7470 103

= 32.55 μAIC = β IB = 90 × 32.55 × 10–6 = 2.93 mA

Applying Kirchhoff’s voltage law to the collector-emitter circuit,VCC – IC RC – VCE = 0

VCE = VCC – IC RC = 16 – 2.93 × 10–3 × 2.7 × 103 = 8.09 V



(b) For β = 135IB = 32.55 μA

IC = β IB = 135 × 32.55 × 10–6 = 4.39 mA

VCE = VCC – IC RC

= 16 – 4.39 × 10–3 × 2.7 × 103

= 4.15 V

(i) % ∆IC = × − ××

− −

−

4.39 10 2.93 102.93 10

3 3

3 × 100

= 49.83%

(ii) % ∆VCE = −4.15 8.098.09

× 100

= −48.70%

When β increases by 50%, IC increases by 49.83% and VCE decreases by 48.70%.

Example 1.7: For the fi xed-bias confi guration of Fig. 1.33, determine (i) S, (ii) S′, (iii) S″ when β increases by 25% and (iv) the net change in IC if a change in operating conditions results in ICO increasing from 0.2 μA to 10 μA, VBE drops from 0.7 V to 0.5 V, and β increases by 25%.

Solution:β1 = 90β2 = 25% more than β1 = 112.5

IC1 = 2.93 mA (from Example 1.6)

(i) S = β1 + 1 = 90 + 1 = 91

(ii) S′ = –βRB

1 = –×

90470 103

= –1.91 × 10–4 Ω

(iii) S″ = β

IC

1

1 = × −2.93 1090

3

= 32.56 × 10–6 A

(iv) ∆IC = S ∆ICO + S′ ∆VBE + S″ ∆β = 91 (10 × 10–6 – 0.2 × 10–6) + (–1.91× 10–4)(0.5 – 0.7) + (32.56 × 10–6)(112.5 – 90)

= 91 × 9.8 × 10–6 + 1.91× 10–4 × 0.2 + 32.56 × 10–6 × 22.5 = 1.66 mA



1.10 MODIFIED FIXED-BIAS CIRCUITThe fi xed-bias circuit has higher thermal instability. The stability can be improved by using the emitter resistor RE which is connected between emitter and ground. Figure 1.34 shows a modifi ed fi xed-bias circuit. This circuit is also called an emitter feedback bias circuit.

Fig. 1.34 Modifi ed Fixed-bias Circuit

VS

RS

RE CE

RB RC

CC1

Vo

+VCC

CC2

RL

DC analysis

For DC, f = 0,

XC = π f C

12

= ∞



RB RC

RE

VBE

VCE

+VCC

+

−−

+

IB IC

IE


1.10 Modifi ed Fixed-bias Circuit 1.31


VCC – IB RB – VBE – IE RE = 0 VCC – IB RB – VBE – (β + 1)IBRE = 0

IB = β−

+ +V V

R R( 1)CC BE

B E

The emitter resistor, which is the part of the collector-emitter circuit, appears as (β + 1)RE in the base-emitter circuit. For the base-emitter circuit, the net voltage is VCC – VBE and the total resistance is the sum of RB and the refl ected resistance (β + 1)RE.

IC = β IB

= β V V

R R( 1)CC BE

B Eβ−

+ +⎡

⎣⎢

⎤

⎦⎥

Collector-emitter voltage VCEApplying Kirchhoff’s voltage law to the collector-emitter circuit,

VCC – IC RC – VCE – IE RE = 0VCC – IC RC – VCE – (IB + IC)RE = 0

VCE = VCC – IC RC – (IB + IC)RE

Load-line analysis

Applying Kirchhoff’s voltage law to the collector-emitter circuit,VCC – IC RC – VCE – IE RE = 0

Assuming IE ≈ IC,

VCC – IC(RC + RE) – VCE = 0

IC = – +R R1

C E

VCE + +

VR R

CC

C E

This equation represents a DC load line with slope of – +R R1

C E

and y-intercept of +

VR R

CC

C E


VCE = VCC


IC = +

VR R

CC

C E



Thus, two end points are (VCC, 0) and +

⎛

⎝⎜⎞

⎠⎟V

R R0, CC

C E

. By joining these two end points, a DC load line is drawn.

From the base-emitter circuit,

IB = β−

+ +V V

R R( 1)CC BE

B E



VCEQ VCCVCE

VCC

RC + RE

ICQ

IC

IBQQ

0

From the Fig. 1.36, it is clear that the saturation current for the circuit is IC sat = +

VR R

CC

C E

.

This is the resulting current when a short circuit is applied between collector-emitter terminals. The addition of emitter resistor reduces the collector saturation level below that obtained with a fi xed-bias confi guration using the same collector resistor.

Stability of Q-point

(i) Applying Kirchhoff’s voltage law to the base-emitter circuit,VCC – IB RB – VBE – IE RE = 0

VCC – IB RB – VBE – (IB + IC)RE = 0VCC – IB RB – VBE – IB RE – IC RE = 0

IB = − −

+V V I R

R RCC BE C E

B E

If reverse saturation current ICO increases, collector current IC increases. It will cause voltage drop across RE to increase which decreases base current IB. As IC depends on IB, decrease in IB reduces the original increase in IC. Hence, variation in IC with ICO is minimized and stability of Q-point is achieved.



(ii) From the base-emitter circuit,

IB = β−

+ +V V

R R( 1)B BE

B E

Generally,(β + 1)RE >> RB

β RE >> RB

IB = β−V VR

B BE

E

IC = β IB

= β V V

RB BE

Eβ−⎛

⎝⎜⎞

⎠⎟

= V V

RB BE

E

−

Hence, IC is independent of the value of β. Variation in IC with β is minimized and stability of Q-point is achieved.

Stability factors

Applying Kirchhoff’s voltage law to the base-emitter circuit,VCC – IB RB – VBE – IE RE = 0

VCC – IB RB – VBE – (IB + IC )RE = 0VCC – IB RB – VBE – IB RE – IC RE = 0

IB = V V I R

R RCC BE C E

B E

− −+

We know that,


= β V V I R

R RCC BE C E

B E

− −+

⎛

⎝⎜⎞

⎠⎟ + (β + 1)ICO

IC R

R R1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟=

VR R

CC

B E

β+

– V

R RBE

B E

β+

+ (β + 1)ICO (1.10)






. From Eq. (1.10), at t1°C,

IC1

RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B E

β+

– V

R RBE

B E

β+

+ (β + 1)ICO1 (1.11)

At t2°C,

IC2

RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B E

β+

– V

R RBE

B E

β+

+ (β + 1)ICO2 (1.12)


(IC2 – IC1

) R

R R1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟= (β + 1)(ICO2

– ICO1)

I I

I IC C

CO CO

2 1

2 1

−

− =

RR R

1

1 E

B E

ββ+

++

∴ S = II

C

CO

ΔΔ =

RR R

1

1 E

B E

ββ+

++

(b) Stability factor S′ : When VBE changes from VBE1 to VBE2


. From Eq. (1.10), at t1°C,

IC1

RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B E

β+

– V

R RBE

B E

1β

++ (β + 1)ICO (1.13)

At t2°C,

IC1 R

R R1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B E

β+

– V

R RBE

B E

2β

++ (β + 1)ICO (1.14)


(IC2 – IC1

) RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟= –

R RB E

β+

(VBE2 – VBE1

)

I I

V VC C

BE BE

2 1

2 1

−

− =

R RR

R R1

B E

E

B E

β

β

−+

++



= R R RB E E

ββ

−+ +

= R R( 1)B E

ββ

−+ +

∴ S′ = I

VC

BE

ΔΔ

= R R( 1)B E

ββ

−+ +


IC R

R R1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B E

β+

– V

R RBE

B E

β+

+ β ICO [� (β + 1) ≈ β]

IC R R

R R( 1)B E

B E

β+ ++

⎡

⎣⎢

⎤

⎦⎥ =

V V I R RR R

[ ( )]CC BE CO B E

B E

β − + ++

IC = V V I R R

R R[ ( )]

(CC BE CO B E

B E

ββ

− + ++ +1)

When β changes from β1 to β2, IC changes from IC1 to IC2

.At t1°C,

IC1 =

V V I R RR R

[ ( )]( 1)

CC BE CO B E

B E

1

1

ββ

− + ++

(1.15)

At t2°C,

IC2 =

V V I R RR R

[ ( )]( 1)

CC BE CO B E

B E

2

2

ββ

− + ++

(1.16)

I

IC

C

2

1

= R RR R

[ ( 1) ][ ( 1) ]

B E

B E

2 1

1 2

β ββ β

+ ++ +

Subtracting 1 from both the sides,

I

IC

C

2

1

– 1 = R RR R

[ ( 1) ][ ( 1) ]

B E

B E

2 1

1 2

β ββ β

+ ++ +

– 1

I I

IC C

C

2 1

1

− =

R R R RR R

[ ( 1) ] [ ( 1) ][ ( 1) ]

B E B E

B E

2 1 1 2

1 2

β β β ββ β

+ + − + ++ +



= R R R R R R

R R[ ( 1) ]B E E B E E

B E

2 1 2 2 1 1 2 1

1 2

β β β β β β β ββ β

+ + − − −+ +

= R R R R

R R( ) ( )

[ ( 1) ]B E B E

B E

2 1

1 2

β ββ β

+ − ++ +

= R R

R R( )( )

[ ( 1) ]B E

B E

2 1

1 2

β ββ β

− ++ +

I IC C

2 1

2 1

β β

−

− =

IC

1

1

β ×

R RR R( 1)

B E

B E2β+

+ +

∴ S″ = IC

βΔΔ

= IC

1

1

β ×

R RR R( 1)

B E

B E2β+

+ +



220 kΩ 1 kΩ

1 kΩ

+10 V

β = 100

Solution:

(i) Applying Kirchhoff’s voltage law to the base-emitter circuit,VCC – IB RB – VBE – IE RE = 0

VCC – IB RB – VBE – (β + 1)IB RE = 0

IB = V V

R R( 1)CC BE

B Eβ−

+ +

= 10 0.7220 10 (100 1)(1 10 )3 3

−× + + ×

= 28.9 μAIC = βIB = 100 × 28.9 × 10–6 = 2.89 mA



(ii) Applying Kirchhoff’s voltage law to the collector-emitter circuit,VCC – IC RC – VCE – IE RE = 0

VCC – IC RC – VCE – (IB + IC)RE = 0VCE = VCC – IC RC – (IB + IC)RE

= 10 – 2.89 × 10–3 × 1 × 103 – (28.9 × 10–6 + 2.89 × 10–3)(1 × 103)= 4.19 V

(iii) S = R

R R

1

1 E

B E

ββ+

++

= 100 1

1 100 1 10220 10 1 10

3

3 3

+

+ × ×× + ×

= 69.53

Example 1.9: In the circuit shown in Fig. 1.38, fi nd RC, VCE, RB, VB and RE.


RB RC

β = 80

+12 V

VC

VB

VE

RE

IC = 2 mA

VE = 2.4 V

VC = 7.6 V

Solution:

(i) Applying Kirchhoff’s voltage law to the collector circuit,VCC – IC RC – VC = 0

RC = V V

ICC C

C

−

= 12 7.62 10 3

−× −

= 2.2 kΩ



(ii) Applying Kirchhoff’s voltage law to the collector-emitter circuit,VCC – IC RC – VCE – VE = 0

VCE = VCC – IC RC – VE

= 12 – 2 × 10–3 × 2.2 × 103 – 2.4= 5.2 V

(iii) IB = IC

β= 2 10

80

3× −

= 25 μA

Applying Kirchhoff’s voltage law to the base-emitter circuit,VCC – IB RB – VBE – VE = 0

RB = V V V

ICC BE E

B

− −

= 12 0.7 2.425 10 6

− −× −

= 356 kΩ(iv) VB = VBE + VE = 0.7 + 2.4 = 3.1 V (v) IE = IB + IC = 25 × 10–6 + 2 × 10–3 = 2.025 mA

RE = VI

E

E

= 2.42.025 10 3× − = 1.185 kΩ

Example 1.10: In the circuit shown in Fig. 1.39, fi nd RC and RB such that VCE = 5 V, IC = 2 mA, VCC = 10 V, β = 100 and RE = 1 kΩ.


RB RC

1 kΩ

+10 V



Solution:

(i) IB = IC

β= 2 10

100

3× −

= 20 μA


VCC – IC RC – VCE – (IB + IC)RE = 0

RC = V V I I R

I( )CC CE B C E

C

− − +

= 10 5 (20 10 2 10 )(1 10 )2 10

6 3 3

3

− − × + × ××

− −

−

= 1.49 kΩ(ii) Applying Kirchhoff’s voltage law to the base-emitter circuit,

VCC – IB RB – VBE –IE RE = 0VCC – IB RB – VBE – (IB + IC)RE = 0

RB = V V I I R

I( )CC BE B C E

B

− − +

= 10 0.7 (20 10 2 10 )(1 10 )20 10

6 3 3

6

− − × + × ××

− −

−

= 364 kΩ



510 kΩ 2.4 kΩ

1.5 kΩ

+20 V



Solution:


VCC – IB RB – VBE – IE RE = 0VCC – IB RB – VBE – (β + 1)IB RE = 0

IB = V V

R R(CC BE

B Eβ−

+ +1)

= 20 0.7

510 10 (100 (1.5 10 )3 3

−× + +1) ×

= 29.18 μAIC = β IB = 100 × 29.18 × 10–6 = 2.92 mA


VCC – IC RC – VCE – (IB + IC)RE = 0VCE = VCC – IC RC – (IB + IC)RE

= 20 – 2.92 × 10–3 × 2.4 × 103 – (29.18 × 10–6 + 2.92 × 10–3)(1.5 × 103)= 8.57 V

(b) For β = 150

IB = V V

R R(CC BE

B Eβ−

+ +1)

= 20 0.7510 10 (150 1.5 103 3

−× + +1) ×

= 26.21 μAIC = β IB = 150 × 26.21× 10–6 = 3.93 mA


= 20 – 3.93 × 10–3 × 2.4 × 103 – (26.21 × 10–6 + 3.93 × 10–3)(1.5 × 103)= 4.63 V

(i) % ∆IC = 3.93 10 2.92 102.92 10

3 3

3

× − ××

− −

−× 100

= 34.59%

(ii) % ∆VCE = 4.63 8.578.57

− × 100

= −45.97%When β increases by 50%, IC increases by 34.59% and VCE decreases by 45.97%.



Example 1.12: For the circuit of the Fig. 1.40, determine (i) S, (ii) S’, (iii) S” when β increases by 25% and (iv) the net change in IC if a change in operating condition results in ICO increasing from 0.2 μA to 10 μA, VBE drops from 0.7 V to 0.5 V, and β increases by 25%.

Solution:

β1 = 100β2 = 25% more than β1 = 125

IC1 = 2.92 mA (From Example 1.11)

(i) S = R

R R

1

1 E

B E

ββ+

++

= 100 1

1 100 1.5 10510 10 1.5 10

3

3 3

+

+ × ×× + ×

= 78.1

(ii) S′ = R R( 1)B E

ββ

−+ +

= 100

510 10 (100 1)(1.5 10 )3 3−× + + ×

= –1.512 × 10–4 Ω

(iii) S ″ = IC

1

1

β ×

R RR R( 1)

B E

B E2β+

+ +

= 2.92 10

100

3× −

× 510 10 1.5 10

510 10 (125 1)(1.5 10 )

3 3

3 3

× + ×× + + ×

= 21.37 × 10–6 A(iv) ∆IC = S ∆ICO + S′ ∆VBE + S″ ∆β

= 78.1(10 × 10–6 – 0.2 × 10–6) + (–1.512 × 10–4)(0.5 – 0.7) + (21.37 × 10–6)(125 – 100) = 78.1 × 9.8 × 10–6 + 1.512 × 10–4 × 0.2 + 21.37 × 10–6 × 25 = 1.33 mA



1.11 COLLECTOR-TO-BASE BIAS CIRCUITThe stability can be improved if the resistor RB is returned to the collector terminal rather than to the battery terminal. Figure 1.41 shows a collector-to-base bias circuit. In this method, resistor RB is connected between the base and the collector. Hence, the circuit is called collector-to-base bias circuit. Although the Q-point is not totally independent of β, the sensitivity to changes in β or temperature variations is normally less than encountered for fi xed-bias or modifi ed fi xed-bias or emitter feedback bias confi gurations.

Fig. 1.41 Collector-to-base Bias Circuit

VS

RS

RB

RC

CC1

Vo

+VCC

CC2

RL

DC analysis

For DC, f = 0, XC =

f C1

2π = ∞



RB

VCE

RC

VBE

+VCC

−

+

+

−

IBIB + IC

IC


1.11 Collector-to-base Bias Circuit 1.43


VCC – (IB + IC)RC – IB RB – VBE = 0VCC – (β +1)IB RC – IB RB – VBE = 0

IB = V V

R R( )CC BE

B Cβ−

+ +1

The collector resistor, which is the part of the collector-emitter circuit, appears as (β +1)RC in the base-emitter circuit. For the base-emitter circuit, the net voltage is VCC – VBE and the total resistance is the sum of RB and the refl ected resistance (β +1)RC.

IC = β IB

= β β−

+ +1⎡

⎣⎢

⎤

⎦⎥

V VR R( )

CC BE

B C


VCC – (IB + IC)RC – VCE = 0VCE = VCC – (IB + IC)RC

Load-line analysis


VCC – (IB + IC)RC – VCE = 0

Assuming IB + IC ≈ IC,


IC = – R1

C

VCE + VR

CC

C

This equation represents a DC load line with slope of –R1

C


CC

C


VCE = VCC


IC = VR

CC

C



Thus two end points are (VCC, 0) and VR

0, CC

C

⎛

⎝⎜⎞

⎠⎟. By joining these two end points, a DC load

line is drawn.From the base-emitter circuit,

IB = V V

R R( )CC BE

B Cβ−

+ +1


Fig. 1.43 Load line and Q-point

VCEQ0 VCCVCE

VCCRC

ICQ

IC

IBQ

Q

From the Fig. 1.43, it is clear that the saturation current for the circuit is IC sat = VR

CC

C. This is

the resulting current when a short circuit is applied between collector-emitter terminals.



VCC – (IB + IC)RC – IB RB – VBE = 0VCC – IB RC – IC RC – IB RB – VBE = 0

IB = V V I R

R RCC BE C C

B C

− −+

If reverse saturation current ICO increases, collector current IC increases. It will cause voltage drop across RC to increase which decreases base current IB. As IC depends on IB, decrease in IB reduces the original increase in IC. Hence, variation in IC with ICO is mini-mized and stability of Q-point is achieved.


IB = V V

R R( )CC BE

B Cβ−

+ +1



Generally,(β + 1)RC >> RB

β RC >> RB

IB = V V

RB BE

Cβ−

IC = β IB

= β V V

RB BE

Cβ−⎛

⎝⎜⎞

⎠⎟

= V V

RB BE

C

−


Stability factors


VCC – (IB + IC)RC – IB RB – VBE = 0VCC – IB RC – IC RC – IB RB – VBE = 0

IB = V V I R

R RCC BE C C

B C

− −+

We know that,IC = β IB + (β + 1)ICO

= β V V I R

R RCC BE C C

B C

− −+

⎛

⎝⎜⎞

⎠⎟ + (β + 1)ICO

IC RR R

1 C

B C

β+

+⎛

⎝⎜⎞

⎠⎟= V

R RCC

B C

β+

– VR R

BE

B C

β+

+ (β + 1)ICO (1.17)




. From Eq. (1.17), at t1°C,

IC1

RR R

1 C

B C

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B C

β+

– V

R RBE

B C

β+

+ (β + 1)ICO1 (1.18)



At t2°C,

IC2

RR R

1 C

B C

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B C

β+

– V

R RBE

B C

β+

+ (β + 1)ICO2 (1.19)


(IC2 – IC1

)R

R R1 C

B C

β+

+⎛

⎝⎜⎞

⎠⎟= (β + 1)(ICO2

– ICO1)

I I

I IC C

CO CO

2 1

2 1

−

− =

RR R

1

1 C

B C

ββ+

++

∴ S = II

C

CO

ΔΔ

= R

R R

1

1 C

B C

ββ+

++

(b) Stability factor S′: When VBE changes from VBE1 to VBE2


. From Eq. (1.17), at t1°C,

IC1

RR R

1 C

B C

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B C

β+

– V

R RBE

B C

1β

++ (β + 1)ICO (1.20)

At t2°C,

IC2

RR R

1 C

B C

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B C

β+

– V

R RBE

B C

2β

++ (β + 1)ICO (1.21)


(IC2 – IC1

)R

R R1 C

B C

β+

+⎛

⎝⎜⎞

⎠⎟= –

R RB C

β+

(VBE2 – VBE1

)

I I

V VC C

BE BE

2 1

2 1

−

− =

R RR

R R1

B C

C

B C

β

β

−+

++

= R R RB C C

ββ

−+ +



= R R( 1)B C

ββ

−+ +

∴ S′ = I

VC

BE

ΔΔ

= R R( 1)B C

ββ

−+ +


IC R

R R1 C

B C

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

CC

B C

β+

– V

R RBE

B C

β+

+ β ICO [� (β + 1) ≈ β]

IC R R

R R( 1)B C

B C

β+ ++

⎡

⎣⎢

⎤

⎦⎥ =

V V I R RR R

[ ( )]CC BE CO B C

B C

β − + ++

IC = V V I R R

R R[ ( )]

(CC BE CO B C

B C

ββ

− + ++ +1)


.At t1°C,

IC1 =

V V I R RR R

[ ( )]( 1)

CC BE CO B C

B C

1

1

ββ

− + ++

(1.22)

At t2°C,

IC2 =

V V I R RR R

[ ( )]( 1)

CC BE CO B C

B C

2

2

ββ

− + ++

(1.23)

I

IC

C

2

1

= R RR R

[ ( 1) ][ ( 1) ]

B C

B C

2 1

1 2

β ββ β

+ ++ +


I

IC

C

2

1

– 1 = R RR R

[ ( 1) ][ ( 1) ]

B C

B C

2 1

1 2

β ββ β

+ ++ +

– 1

I I

IC C

C

2 1

1

− =

R R R RR R

[ ( 1) ] [ ( 1) ][ ( 1) ]

B C B C

B C

2 1 1 2

1 2

β β β ββ β

+ + − + ++ +

= R R R R R R

R R[ ( 1) ]B C C B C C

B C

2 1 2 2 1 1 2 1

1 2


+ + − − −+ +



= R R R R

R R( ) ( )

[ ( 1) ]B C B C

B C

2 1

1 2

β ββ β

+ − ++ +

= R R

R R( )( )

[ ( 1) ]B C

B C

2 1

1 2

β ββ β

− ++ +

I IC C

2 1

2 1

β β

−

−=

IC

1

1

β ×

R RR R( 1)

B C

B C2β+

+ +

∴ S″ = IC

βΔΔ

= IC

1

1

β ×

R RR R( 1)

B C

B C2β+

+ +

Example 1.13: For the circuit shown in Fig. 1.44, determine IC, VCE and stability factor S.


500 kΩ

β = 100

3 kΩ

+10 V

Solution:


VCC – (IB + IC)RC – IB RB – VBE = 0VCC – (β + 1)IB RC – IB RB – VBE = 0

IB = V V

R R( 1)CC BE

B Cβ−

+ +



= 10 0.7500 10 (100 1)(3 10 )3 3

−× + + ×

= 11.58 μA IC = β IB = 100 × 11.58 × 10–6 = 1.158 mA


VCE = VCC – (IB + IC)RC

= 10 – (11.58 × 10–6 + 1.158 × 10–3) (3 × 103) = 6.49 V

(iii) S = R

R R

1

1 C

B C

ββ+

++

= 100 1

1 100 3 10500 10 3 10

3

3 3

+

+ × ×× + ×

= 63.26

Example 1.14: For the circuit shown in Fig. 1.45, fi nd RC and RB.

RB

RC

+12 V

IC = 2 mAβ = 100

VCE = 6 V


Solution:

(i) IB = IC

β= 2 10

100

3× −

= 20 μA





VCC – (β + 1)IB RC – VCE = 0

RC = V V

I( 1)CC CE

Bβ−

+

= 12 6(100 1)(20 10 )6

−+ × −

= 2.97 kΩ(ii) Applying Kirchhoff’s voltage law to the base-emitter circuit,

VCC – (IB + IC)RC – IB RB – VBE = 0

RB = V V I I RI( )CC BE B C C

B

− − +

= 12 0.7 (20 10 2 10 )(2.97 10 )20 10

6 3 3

6

− − × + × ××

− −

−

= 265.4 kΩ



470 kΩ

9.1 kΩ

+22 V



Solution:

(a) For β = 90Applying Kirchhoff’s voltage law to the base–emitter circuit,

VCC – (IB + IC)RC – IB RB – VBE = 0

VCC – (β + 1)IB RC – IB RB – VBE = 0

IB = V V

R R( 1)CC BE

B Cβ−

+ +

= −

× + +1) × )22 0.7

470 10 (90 (9.1 103 3

= 16.41 μAIC = β IB = 90 × 16.41 × 10–6 = 1.48 mA




= 22 – (16.41 × 10–6 + 1.48 × 103) (9.1 × 103)

= 8.38 V

(b) For β = 135

IB = V V

R R(CC BE

B Cβ−

+ +1)

= −

× + +1) ×22 0.7

470 10 (135 (9.1 10 )3 3

= 12.47 μA

IC = β IB = 135 × 12.47 × 10–6 = 1.68 mA


= 22 – (12.47 × 10–6 + 1.68 × 10–3) (9.1 × 103)= 6.6 V

(i) % ∆IC = 1.68 10 1.48 101.48 10

3 3

3

× − ××

− −

− × 100

= 13.51%



(ii) % ∆VCE = 6.6 8.388.38− × 100


Example 1.16: For the collector-to-base bias confi guration of the Fig. 1.46, determine (i) S, (ii) S’, (iii) S” when β increases by 25% and (iv) the net change in IC if a change in operating condition results in ICO increasing from 0.2 μA to 10 μA, VBE drops from 0.7 V to 0.5 V, and β increases by 25%.



(i) S = R

R R

1

1 C

B C

ββ+

++

= 90 1

1 90 9.1 10470 10 9.1 10

3

3 3

+

+ × ×× + ×

= 33.59

(ii) S′ = R R( 1)B C

ββ

−+ +

= −× + + ×

90470 10 (90 1)(9.1 10 )3 3

= –6.93 × 10–5 Ω

(iii) S″ = IC

1

1

β ×

R RR R( 1)

B C

B C2β+

+ +

= 1.48 1090

3× −

× × + ×

× + + ×470 10 9.1 10

470 10 (112.5 1)(9.1 10 )

3 3

3 3


= 33.59(10 × 10–6 – 0.2 × 10–6) + (–6.93 × 10–5)(0.5 – 0.7) + (5.24 × 10–6)(112.5 – 90) = 33.59 × 9.8 × 10–6 + 6.93 × 10–5 × 0.2 + 5.24 × 10–6 × 22.5 = 0.45 mA


1.12 Modifi ed Collector-to-base Bias Circuit 1.53

1.12 MODIFIED COLLECTOR-TO-BASE BIAS CIRCUIT Figure 1.47 shows a modifi ed collector-to-emitter bias circuit. Emitter resistor RE is connected in the emitter terminal.

VS

RS

RE CE

RB

RC

CC1

Vo

+VCC

CC2

RL

Fig. 1.47 Modifi ed Collector-to-base Bias Circuit

DC analysis

For DC, f = 0,

XC = π f C

12

= ∞


RB

VCE

RC

RE

VBE

+VCC

−

+

+

−

IBIB + IC

IC

IE




Collector current IC


VCC – (IB + IC)RC – IB RB – VBE – IE RE = 0

VCC – (β +1)IB RC – IB RB – VBE – (β + 1)IB RE = 0

IB = V V

R R R( 1)( )CC BE

B C Eβ−

+ + +

The collector and emitter resistors, which are the part of the collector-emitter circuit, appear as (β + 1)(RC + RE) in the base-emitter circuit. For the base-emitter circuit, the net voltage is VCC – VBE and the total resistance is the sum of RB and the refl ected resistance (β + 1)(RC + RE).

IC = βIB

= β V V

R R R( 1)( )CC BE

B C Eβ−

+ + +⎡

⎣⎢

⎤

⎦⎥

Collector-emitter voltage VCECE


VCC – (IB + IC)RC – VCE – IE RE = 0

VCC – (IB + IC)RC – VCE – (IB + IC)RE = 0

VCE = VCC – (IB + IC)(RC + RE)

Load-line analysis


VCC – (IB + IC)RC – VCE – IE RE = 0


Assuming IB + IC ≈ IC,


IC = –+R R1

C E

VCE + +

VR R

CC

C E

This equation represents a DC load line with slope of – +R R1

C E

and y-intercept of +

VR R

CC

C E

.



When IC = 0, i.e. transistor is in cut-off region,

VCE = VCC


IC = +V

R RCC

C E

Thus, two end points are (VCC, 0) and VR R

0, CC

C E+⎛

⎝⎜⎞

⎠⎟. By joining these two end points, a DC

load line is drawn.From the base-emitter circuit,

IB = β

−+ + +

V VR R R( 1)( )

CC BE

B C E

For this value of the base current, we can establish the actual Q-point as shown in Fig. 1.49.

VCEQ0 VCCVCE

VCC

RC + RE

ICQ

IC

IBQQ


From the Fig. 1.49, it is clear that the saturation current for the circuit is IC sat = +V

R RCC

C E.

This is the resulting current when a short circuit is applied between collector-emitter terminals. The addition of the emitter resistor reduces the collector saturation level below that obtained with a collector-to-base bias confi guration using the same collector resistor.




VCC – (IB + IC)RC – IB RB – VBE – (IB + IC)RE = 0



VCC – IB RC – IC RC – IB RB – VBE – IB RE – IC RE = 0

IB = V V I R R

R R R( )CC BE C C E

B C E

− − ++ +

If reverse saturation current ICO increases, collector current IC increases. It will cause voltage drop across (RC + RE) to increase, which decreases base current IB. As IC depends on IB, decrease in IB reduces the original increase in IC. Hence, variation in IC with ICO is minimized and stability of Q-point is achieved.


IB = V V

R R R( 1)( )CC BE

B C Eβ−

+ + +

Generally,

(β + 1)(RC + RE) >> RB

β (RC + RE) >> RB

IB = β

−+

V VR R( )B BE

C E

IC = β IB

= β V V

R R( )B BE

C Eβ−

+⎡

⎣⎢

⎤

⎦⎥

= −+

V VR R

B BE

C E


Stability factors



VCC – IB RC – IC RC – IB RB – VBE – (IB + IC)RE = 0

VCC – IB RC – IC RC – IB RB – VBE – IB RE – IC RE = 0



IB = − − +

+ +V V I R R

R R R( )CC BE C C E

B C E

We know that


= β V V I R R

R R R( )CC BE C C E

B C E

− − ++ +

⎡

⎣⎢

⎤

⎦⎥ + (β + 1)ICO

IC R R

R R R1

( )C E

B C E

β+

++ +

⎡

⎣⎢

⎤

⎦⎥ =

VR R R

CC

B C E

β+ +

– V

R R RBE

B C E

β+ +

+ (β + 1)ICO (1.24)




. From Eq. (1.24), at t1°C,

IC1

R RR R R

1( )C E

B C E

β+

++ +

⎡

⎣⎢

⎤

⎦⎥ =

β+ +

VR R R

CC

B C E

– V

R R RBE

B C E

β+ +

+ (β + 1)ICO1 (1.25)

At t2°C,

IC2

R RR R R

1( )C E

B C E

β+

++ +

⎡

⎣⎢

⎤

⎦⎥ =

VR R R

CC

B C E

β+ +

– V

R R RBE

B C E

β+ +

+ (β + 1)ICO2 (1.26)


(IC2 – IC1

)R R

R R R1

( )C E

B C E

β+

++ +

⎡

⎣⎢

⎤

⎦⎥ = (β + 1)(ICO2

– ICO1)

I I

I IC C

CO CO

2 1

2 1

−

− = R R

R R R

1

1)C E

B C E

ββ

+

+( ++ +

∴ S = II

C

CO

ΔΔ

= R R

R R R

1

1)C E

B C E

ββ

+

+( ++ +





. From Eq. (1.24), at t1°C,

IC1

R RR R R

1( )C E

B C E

β+

++ +

⎡

⎣⎢

⎤

⎦⎥ =

VR R R

CC

B C E

β+ +

– V

R R RBE

B C E

1β

+ ++ (β + 1)ICO (1.27)

At t2°C,

IC2

R RR R R

1( )C E

B C E

β+

++ +

⎡

⎣⎢

⎤

⎦⎥ =

VR R R

CC

B C E

β+ + –

V

R R RBE

B C E

2β

+ ++ (β + 1)ICO (1.28)


(IC2 – IC1

) β+

++ +R R

R R R1

( )C E

B C E

= –R R RB C E

β+ +

(VBE2 – VBE1

)

I I

V VC C

BE BE

2 1

2 1

−

− =

R R RR R

R R R1

)B C E

C E

B C E

β

β

−+ +

+( ++ +

= R R R R RB C E C E

ββ β

−+ + + +

= R R R( 1)( )B C E

ββ

−+ + +

∴ S′ = I

VC

BE

ΔΔ

= R R R( 1)( )B C E

ββ

−+ + +


IC

R RR R R

1( )C E

B C E

β+

++ +

⎡

⎣⎢

⎤

⎦⎥ =

VR R R

CC

B C E

β+ +

– V

R R RBE

B C E

β+ +

+ β ICO [� (β + 1) ≈ β]

IC

R R RR R R( 1)( )B C E

B C E

β+ + ++ +

⎡

⎣⎢

⎤

⎦⎥ =

V V I R R RR R R

[ ( )]CC BE CO B C E

B C E

β − + + ++ +



IC = V V I R R R

R R R[ ( )]

( ( )CC BE CO B C E

B C E

ββ

− + + ++ +1) +


.At t1°C,

IC1 =

V V I R R RR R R

[ ( )]( 1)( )

CC BE CO B C E

B C E

1

1

ββ

− + + ++ + +

(1.29)

At t2°C,

IC2 =

V V I R R RR R R

[ ( )]( 1)( )

CC BE CO B C E

B C E

2

2

ββ

− + + ++ + +

(1.30)

I

IC

C

2

1

= R R RR R R

[ ( 1)( )][ ( 1)( )]

B C E

B C E

2 1

1 2

β ββ β

+ + ++ + +


I

IC

C

2

1

– 1 = R R RR R R

[ ( 1)( )][ ( 1)( )]

B C E

B C E

2 1

1 2

β ββ β

+ + ++ + +

– 1

I I

IC C

C

2 1

1

− =

R R R R R RR R R

[ ( 1)( )] [ ( 1)( )][ ( 1)( )]

B C E B C E

B C E

2 1 1 2

1 2

β β β ββ β

+ + + − + + ++ + +

= R R R R R R R R R R

R R R[ ( 1)( )]B C E C E B C E C E

B C E

2 1 2 1 2 2 2 1 1 2 1 2 1 1

1 2

β β β β β β β β β β β β β ββ β

+ + + + − − − − −+ + +

= R R R R R R

R R R( ) ( )

[ ( 1)( )]B C E B C E

B C E

2 1

1 2

β ββ β+ + − + +

+ + +

= R R R

R R R( )( )[ ( 1)( )]

B C E

B C E

2 1

1 2

β ββ β

− + ++ + +

I IC C

2 1

2 1

β β

−

− =

IC

1

1

β ×

R R RR R R( 1)( )

B C E

B C E2β+ +

+ + +

∴ S″ = IC

βΔΔ

= IC

1

1

β ×

R R RR R R( 1)( )

B C E

B C E2β+ +

+ + +




220 kΩ

β = 100

2.5 kΩ

1 kΩ

+10 V


Solution:



VCC – (β + 1)IB RC – IB RB – VBE – (β + 1)IB RE = 0

IB = V V

R R R( 1)( )CC BE

B C Eβ−

+ + +

= 10 0.7

220 10 (100 1)(2.5 10 1 10 )3 3 3

−× + + × + ×

= 16.21 μAIC = β IB = 100 × 16.21 × 10–6 = 1.621 mA


VCC – (IB + IC)RC – VCE – IE RE = 0 VCC – (IB + IC)RC – VCE – (IB + IC)RE = 0 VCE = VCC – (IB + IC)(RC + RE) = 10 – (16.21 × 10–6 + 1.621 × 10–3)(2.5 × 103+ 1 × 103) = 4.27 V



(iii) S = R R

R R R

1

1( )C E

B C E

ββ

+

++

+ +

= 100 1

1 100(2.5 10 1 10 )220 10 2.5 10 1 10

3 3

3 3 3

+

+ × + ×× + × + ×

= 39.36

Example 1.18: For the circuit shown in Fig. 1.51, determine IC, VC, VE and VCE.

690 kΩ

β = 100

6.2 kΩ

1.5 kΩ

+30 V


Solution:




IB = V V

R R R( 1)( )CC BE

B C Eβ−

+ + +

= 30 0.7

690 10 (100 1)(6.2 10 1.5 10 )3 3 3

−× + + × + ×

= 19.9 μAIC = β IB = 100 × 19.9 × 10–6 = 1.99 mA



(ii) VC = VCC – (IB + IC)RC

= 30 – (19.9 × 10–6 + 1.99 × 10–3)(6.2 × 103)= 17.5 V

(iii) VE = (IB + IC)RE

= (19.9 × 10–6 + 1.99 × 10–3)(1.5 × 103)= 3.02 V

(iv) VCE = VC – VE

= 17.5 – 3.02= 14.48 V


470 kΩ

9.1 kΩ

9.1 kΩ

+22 V

Fig. 1.52 Example 1.19Solution:




IB = V V

R R R( 1)( )CC BE

B C Eβ−

+ + +



= 22 0.7470 10 (90 (9.1 10 9.1 10 )3 3 3

−× + +1) × + ×

= 10.02 μA

IC = β IB = 90 × 10.02 × 10–6 = 0.9 mA




= 22 – (10.02 × 10–6 + 0.9 × 10–3)(9.1 × 103 + 9.1 × 103)

= 5.44 V

(b) For β = 135

IB = V V

R R R( ( )CC BE

B C Eβ−

+ +1) +

= 22 0.7

470 10 (135 (9.1 10 9.1 10 )3 3 3

−× + +1) × + ×

= 7.23 μA

IC = β IB = 135 × 7.23 × 10–6 = 0.98 mA


= 22 – (7.23 × 10–6 + 0.98 × 10–3)(9.1 × 103 + 9.1 × 103)

= 4.03 V

(i) % ∆IC = 0.98 10 0.9 100.9 10

3 3

3

× − ××

− −

− × 100

= 8.89%

(ii) % ∆VCE = 4.03 5.445.44

− × 100


Example 1.20: For the circuit of the Fig. 1.52, determine (i) S, (ii) S’, (iii) S” when β increases by 25% and (iv) the net change in IC if a change in operating condition results in ICO increasing from 0.2 μA to 10 μA, VBE drops from 0.7 V to 0.5 V, and β increases by 25%.





(i) S = R R

R R R

1

1)C E

B C E

ββ

+

+( ++ +

= 90 1

1 90(9.1 10 9.1 10 )470 10 9.1 10 9.1 10

3 3

3 3 3

+

+ × + ×× + × + ×

= 20.89

(ii) S′ = R R R( 1)( )B C E

ββ

−+ + +

= 90

470 10 (90 1)(9.1 10 9.1 10 )3 3 3−× + + × + ×

= – 4.23 × 10–5

(iii) S″ = IC

1

1

β × R R R

R R R( 1)( )B C E

B C E2β+ +

+ + +

= 0.9 1090

3× −

× 470 10 9.1 10 9.1 10470 10 (112.5 1)(9.1 10 9.1 10 )

3 3 3

3 3 3

× + × + ×× + + × + ×


= 20.89(10 × 10–6 – 0.2 × 10–6) + (−4.23 × 10–5)(0.5 – 0.7) + (1.93 × 10–6)(112.5 – 90) = 20.89 × 9.8 × 10–6 + 4.23 × 10–5 × 0.2 + 1.93 × 10–6 × 22.5

= 0.26 mA

1.13 VOLTAGE-DIVIDER BIAS CIRCUIT (SELF-BIAS CIRCUIT)In the previous confi gurations, the bias current IC and voltage VCE depend on the current gain β of the transistor. Figure 1.53 shows a voltage-divider bias circuit. Resistors R1 and R2 form a voltage-divider circuit. In this confi guration, the sensitivity to changes in β is quite small. If the circuit parameters are properly chosen, the bias current IC and voltage VCE are almost independent of β.


1.13 Voltage-divider Bias Circuit (Self-bias Circuit) 1.65

VS

RS

RE CE

R1 RC

CC1

Vo

+VCC

CC2

RLR2

Fig. 1.53 Voltage-divider Bias Circuit

1.13.1 Exact Analysis

DC Analysis

For DC, f = 0,

XC = π f C

12

= ∞


RE

R1 RC

+VCC

R2


The base circuit can be converted into Thevenin’s equivalent circuit as shown in Fig. 1.55.

VTh = VB = R

R R2

1 2+VCC

RTh = RB = R1 || R2 = R R

R R1 2

1 2+



As R1 and R2 divide the voltage VCC at the base, the circuit is called voltage-divider bias.

RB

VBE

RC

+VCC

IE

REVB

VCE

IB

IC

+−

−

+

Fig. 1.55 Thevenin’s Equivalent Circuit



VB – IB RB – VBE – IE RE = 0 VB – IB RB – VBE – (β + 1) IB RE = 0

IB = V V

R R( 1)B BE

B Eβ−

+ +

The emitter resistor, which is the part of the collector-emitter circuit, appears as (β +1)RE in the base-emitter circuit. For the base-emitter circuit, the net voltage is VB – VBE and the total resistance is the sum of RB and the refl ected resistance (β + 1)RE.

IC = β IB

= β β−

+ +⎡

⎣⎢

⎤

⎦⎥

V VR R( 1)

B BE

B E

Collector-emitter voltage VCECE






Load-line analysis


VCC – IC RC – VCE – IE RE = 0

Assuming IE ≈ IC,


IC = –R R

1

C E+VCE +

VR R

CC

C E+

This equation represents a DC load line with slope of –R R

1

C E+and y-intercept of

VR R

CC

C E+.

When IC = 0, i.e. transistor is in cut-off region,

VCE = VCC


IC = V

R RCC

C E+

Thus two end points are (VCC, 0) and V

R R0, CC

C E+⎛

⎝⎜⎞

⎠⎟. By joining these two end points, a DC

load line is drawn.From the base-emitter circuit,

IB = V VR R( 1)

B BE

B Eβ−

+ +For this value of base current, we can establish the actual Q-point as shown in Fig. 1.56.

VCEQ0 VCCVCE

VCC

RC + RE

ICQ

IC

IBQQ




From the Fig. 1.56, it is clear that the saturation current for the circuit is IC sat = V

R RCC

C E+.

This is the resulting current when a short circuit is applied between collector-emitter terminals.



VB – IB RB – VBE – IE RE = 0

VB – IB RB – VBE – (IB + IC)RE = 0

VB – IB RB – VBE – IB RE – IC RE = 0

IB = V V I R

R RB BE C E

B E

− −+

If reverse saturation current ICO increases, collector current IC increases. It will cause volt-age drop across RE to increase, which decreases base current IB. As IC depends on IB, decrease in IB reduces the original increase in IC. Hence, variation in IC with ICO is minimized and stabil-ity of Q-point is achieved.

Stability factors




VB – IB RB – VBE – IB RE – IC RE = 0

IB = V V I R

R RB BE C E

B E

− −+

We know that


= β V V I R

R RB BE C E

B E

− −+

⎛

⎝⎜⎞

⎠⎟ + (β + 1)ICO

IC RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟=

VR R

B

B E

β+

– V

R RBE

B E

β+

+ (β + 1)ICO (1.31)






. From Eq. (1.31), at t1°C,

IC1

RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

B

B E

β+

– V

R RBE

B E

β+

+ (β + 1)ICO1 (1.32)

At t2°C,

IC2 R

R R1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

B

B E

β+

– V

R RBE

B E

β+

+ (β + 1)ICO2 (1.33)


(IC2 – IC1

)R

R R1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟= (β + 1)(ICO2

– ICO1)

I I

I IC C

CO CO

2 1

2 1

−

− =

RR R

1

1 E

B E

ββ+

++

∴ S = II

C

CO

ΔΔ

= R

R R

1

1 E

B E

ββ+

++



. From Eq. (1.31), at t1°C,

IC1

RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

B

B E

β+

– V

R RBE

B E

1β

++ (β + 1)ICO (1.34)

At t2°C,

IC2

RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

B

B E

β+

– V

R RBE

B E

2β

++ (β + 1)ICO (1.35)


(IC2 – IC1

)R

R R1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟= –

R RB E

β+

(VBE2 – VBE1

)



I I

V VC C

BE BE

2 1

2 1

−

− =

R RR

R R1

B E

E

B E

β

β

−+

++

= R R RB E E

ββ

−+ +

= R R( 1)B E

ββ

−+ +

∴ S′ = I

VC

BE

ΔΔ

= R R( 1)B E

ββ

−+ +


IC

RR R

1 E

B E

β+

+⎛

⎝⎜⎞

⎠⎟ =

VR R

B

B E

β+

– V

R RBE

B E

β+

+ β ICO [� (β + 1) ≈ β]

IC

β+ ++

⎡

⎣⎢

⎤

⎦⎥

R RR R

( 1)B E

B E

= V V I R R

R R[ ( )]B BE CO B E

B E

β − + ++

IC = V V I R R

R R[ ( )]

(B BE CO B E

B E

ββ

− + ++ +1)


.At t1°C,

IC1 =

V V I R RR R

[ ( )](

B BE CO B E

B E

ββ

− + ++ +1)

1 (1.36)

At t2°C,

IC2 =

V V I R RR R

[ ( )](

B BE CO B E

B E

ββ

− + ++ +1)

2 (1.37)

I

IC

C

2

1

= R RR R

[ ( 1) ][ ( 1) ]

B E

B E

2 1

1 2

β ββ β

+ ++ +


I

IC

C

2

1

– 1 = R RR R

[ ( 1) ][ ( 1) ]

B E

B E

2 1

1 2

β ββ β

+ ++ +

– 1



I I

IC C

C

2 1

1

− =

R R R RR R

[ ( 1) ] [ ( 1) ][ ( 1) ]

B E B E

B E

2 1 1 2

1 2

β β β ββ β

+ + − + ++ +

= R R R R R R

R R[ ( 1) ]B E E B E E

B E

2 1 2 2 1 1 2 1

1 2


+ + − − −+ +

= R R R R

R R( ) ( )

[ ( 1) ]B E B E

B E

2 1

1 2

β ββ β

+ − ++ +

= R R

R R( )( )

[ ( 1) ]B E

B E

2 1

1 2

β ββ β

− ++ +

I IC C

2 1

2 1

β β

−

− =

IC

1

1

β ×

R RR R( 1)

B E

B E2β+

+ +

∴ S″ = IC

βΔΔ

= IC

1

1

β ×

R RR R( 1)

B E

B E2β+

+ +

1.13.2 Approximate Analysis


The input section of the voltage-divider bias circuit can be represented by the circuit shown in Fig. 1.57. The equivalent resistance between the base and the ground is (β + 1)RE. If the resistance (β + 1)RE is much larger than the resistance R2, the current IB will be much smaller than I2. If IB ≈ 0, then I1 = I2.

+VCC

R1

R2 ( β + 1) RE

IB

VB

I2

I1

Fig. 1.57 Input Section

The voltage across R2 can be found using the voltage-divider rule.



VB = R

R R2

1 2+VCC

Generally (β + 1)RE ≈ βRE. If βRE is at least ten times the value of R2, the approximate analy-sis can be used with a high degree of accuracy.

βRE ≥ 10R2

From the base-emitter circuit,VE = VB – VBE

IE = VR

E

E

= V V

RB BE

E

−

IC ≈ IE

Hence, IC is independent of the value of β.


VCC – IC RC –VCE – IE RE = 0VCE = VCC – IC RC – IE RE

= VCC – IC(RC + RE) (� IC ≈ IE)Hence, VCE is independent of the value of β.Thus, Q-point is independent of the value of β.

Example 1.21: For the circuit shown in Fig. 1.58, fi nd IC and VCE by approximate analysis. Compare the result with exact analysis.

+18 V

3.3 kΩ39 kΩ

8.2 kΩ 1 kΩ

β = 120




Solution:

If βRE ≥ 10R2, the approximate analysis can be used with a high degree of accuracy.

βRE = 120 × 1 × 103 = 120 kΩ

10R2 = 10 × 8.2 × 103 = 82 kΩ

Since βRE > 10R2, approximate analysis is used.

(i) VB = R

R R2

1 2+ VCC

= 8.2 10

39 10 8.2 10

3

3 3

×× + ×

× 18

= 3.13 V

VE = VB – VBE = 3.13 – 0.7 = 2.43 V

IE = VR

E

E

= 2.43

1 10 3× − = 2.43 mA

IC = 2.43 mA (� IC ≈ IE)


VCC – IC RC –VCE – IE RE = 0

VCE = VCC – IC RC – IE RE

= VCC – IC(RC + RE) (� IC ≈ IE)

= 18 – (2.43 × 10–3) (3.3 × 103 + 1 × 103)

= 7.55 V

Exact analysis

(i) Replacing the base circuit by its Thevenin’s equivalent (Fig. 1.59),

VB = 3.13 V

RB = R R

R R1 2

1 2+



= 39 10 8.2 1039 10 8.2 10

3 3

3 3

× × ×× + ×

= 6.78 kΩ

RB

VBE

RC

+VCC

IE

REVB

VCE

IB

IC

+−

−

+

Fig. 1.59 Thevenin's Equivalent Circuit



VB – IB RB – VBE – (β + 1)IB RE = 0

IB = V V

R R( 1)B BE

B Eβ−

+ +

= 3.13 0.7

6.78 10 (120 1)(1 10 )3 3

−× + + ×

= 19.02 μA

IC = β IB = 120 × 19.02 × 10–6 = 2.28 mA







= 18 – 2.28 × 10–3 × 3.3 × 103 – (19.02 × 10–6 + 2.28 × 10–3) (1 × 103)

= 8.18 V

Example 1.22: For the circuit shown in Fig. 1.60, fi nd IC, VCE and stability factor S.

+30 V

5 kΩ90 kΩ

10 kΩ 5 kΩ

β = 100


Solution:

(i) Replacing the base circuit by its Thevenin’s equivalent (Fig. 1.61),

VB = R

R R2

1 2+VCC

= 10 10

90 10 10 10

3

3 3

×× + ×

× 30

= 3V

RB = R R

R R1 2

1 2+

= 90 10 10 1090 10 10 10

3 3

3 3

× × ×× + ×

= 9 kΩ



RB

VBE

RC

+VCC

IE

REVB

VCE

IB

IC

+−

−

+




VB – IB RB – VBE – (β + 1)IB RE = 0

IB = V V

R R( 1)B BE

B Eβ−

+ +

= 3 0.7

9 10 (100 1)(5 10 )3 3

−× + + ×

= 4.47 μA

IC = β IB = 100(4.47 × 10–6) = 0.447 mA





= 30 – 0.447 × 10–3 × 5 × 103 – (4.47 × 10–6 + 0.447 × 10–3)(5 × 103)

= 25.5 V

(iii) S = R

R R

1

1 E

B E

ββ+

++



= 100 1

1 100 5 109 10 5 10

3

3 3

+

+ × ×× + ×

= 2.75

Example 1.23: Find RC, R1 and R2 of the circuit shown in Fig. 1.62, if IC = 1 mA, RE = 1 kΩ, VCC = 10 V, β = 100, S = 10 and VCE = 5 V.

+10 V

RCR1

R2 1 kΩ


Solution:

(i) IB = IC

β=

1 10100

3× −

= 0.01 mA




RC = V V I I R

I( )CC CE B C E

C

− − +

= 10 5 (0.01 10 1 10 )(1 10 )

1 10

3 3 3

3

− − × + × ××

− −

−

= 3.99 kΩ



(ii) S = RR R

1

1 E

B E

ββ+

++

10 =

R

100 1

1 100 1 101 10B

3

3

+

+ × ×+ ×

RB = 9.98 kΩ

Replacing the base circuit by its Thevenin’s equivalent (Fig. 1.63),

RB

VBE

RC

+VCC

IE

REVB

VCE

IB

IC

+−

−

+





VB = IB RB + VBE + (IB + IC)RE

= 0.01 × 10–3 × 9.98 × 103 + 0.7 + (0.01 × 10–3 + 1 × 10–3) (1 × 103)

= 1.71 V

V B = R

R R2

1 2+ VCC

Multiplying both the sides by R1,

R1VB = R R

R R1 2

1 2+ VCC = RB VCC



R1 = R V

VB CC

B

= 9.98 10 10

1.71

3× ×

= 58.36 kΩ

(iii) RB = R R

R R1 2

1 2+

9.98 × 103 = 58.36 × 103 × R2

58.36 × 103 + R2

R2 = 12.04 kΩ

Example 1.24: For the circuit shown in Fig. 1.64, fi nd RE, R1 and R2.

+16 V

R1

R2 RE

VCE = 8 V

VBE = 0.3 V

IC = 4 mA

S = 10β = 50

1.5 kΩ


Solution:

(i) IB = IC

β=

4 1050

3× −

= 0.08 mA



RE = V V I R

I ICC CE C C

B C

− −+



= 16 8 4 10 1.5 100.08 10 4 10

3 3

3 3

− − × × ×× + ×

−

− −

= 0.49 kΩ

(ii) S = R

R R

1

1 E

B E

ββ+

++

10 =

R

51

1 50 0.49 100.49 10B

3

3+ × ×+ ×

RB = 5.49 kΩ


RB

VBE

RC

+VCC

IE

REVB

VCE

IB

IC

+−

−

+





VB = IB RB + VBE + (IB + IC)RE

= 0.08 × 10–3 × 5.49 × 103 + 0.3 + (0.08 × 10–3 + 4 × 10–3) (0.49 × 10–3)

= 2.53 V

VB = R

R R2

1 2+ VCC



Multiplying both the sides by R1,

R 1 VB = R R

R R1 2

1 2+VCC

= RB VCC

R1 = R V

VB CC

B

= 5.49 10 16

2.53

3× ×

= 34.72 kΩ

(iii) RB = R R

R R1 2

1 2+

5.49 × 103 = 34.72 × 103 × R2

34.72 × 103 + R2 R2 = 4.74 kΩ


+18 V

1.2 kΩ

1 kΩ

33 kΩ

12 kΩ


Solution:


VB = R

R R2

1 2+ VCC



= 12 10

33 10 12 10

3

3 3

×× + ×

× 18

= 4.8 V

RB = R R

R R1 2

1 2+

= 33 10 12 1033 10 12 10

3 3

3 3

× × ×× + ×

= 8.8 kΩ

RB

VBE

RC

+VCC

IE

REVB

VCE

IB

IC

+−

−

+



VB – IB RB – VBE – IE RE = 0VB – IB RB – VBE – (β + 1)IB RE = 0

IB = V V

R R( 1)B BE

B Eβ−

+ +

= 4.8 0.7

8.8 10 (100 (1 10 )3 3

−× + +1) ×

= 37.34 μAIC = β IB = 100 × 37.34 × 10–6 = 3.73 mAIE = IB + IC = 37.34 × 10–6 + 3.73 × 10–3 = 3.77 mA




VCC – IC RC –VCE – IE RE = 0


= 18 – 3.73 × 10–3 × 1.2 × 103 − 3.77 × 10–3 × 1 × 103

= 9.75 V

(b) For β = 150

IB = V V

R R( 1)B BE

B Eβ−

+ +

= 4.8 0.7

8.8 10 (150 (1 10 )3 3

−× + +1) ×

= 25.66 μA

IC = β IB = 150 × 25.66 × 10–6 = 3.85 mA

IE = IB + IC = 25.66 × 10–6 + 3.85 × 10–3 = 3.88 mA


= 18 – 3.85 × 10–3 × 1.2 × 103 – 3.88 × 10–3 × 1 × 103

= 9.5 V

(i) % ∆IC = 3.85 10 3.73 103.73 10

3 3

3

× − ××

− −

− × 100

= 3.22%

(ii) % ∆VCE = 9.5 9.759.75− × 100

= −2.56%

When β increases by 50%, IC increases by 3.22% and VCE decreases by 2.56%.

Example 1.26: For the circuit of the Fig. 1.66, determine (i) S, (ii) S′, (iii) S″ when β2 is 25% more than β1 and (iv) the net change in IC if a change in operating condition results in ICO in-creasing from 0.2 μA to 10 μA, VBE drops from 0.7 V to 0.5 V, and β increases by 25%.



Solution:

β1 = 100

β2 = 125


(i) S = R

R R

1

1 E

B E

ββ+

++

= 100 1

1 100 1 108.8 10 1 10

3

3 3

+

+ × ×× + ×

= 9.01

(ii) S′ = R R( 1)B E

ββ

−+ +

= 100

8.8 10 (100 1)(1 10 )3 3−× + + ×

= –9.11 × 10–4

(iii) S″ = IC

1

1

β ×

R RR R( 1)

B E

B E2β+

+ +

= 3.73 10

100

3× −

× 8.8 10 1 10

8.8 10 (125 1)(1 10 )

3 3

3 3

× + ×× + + ×

= 2.71 × 10–6 A

(iv) ∆IC = S ∆ICO + S′ ∆VBE + S″ ∆β

= 9.01(10 × 10–6 – 0.2 × 10–6) + (–9.11 × 10–4)(0.5 – 0.7) + (2.71 × 10–6) (125 – 100)

= 9.01 × 9.8 × 10–6 + 9.11 × 10–4 × 0.2 + 2.71 × 10–6 × 25

= 0.33 mA


1.13 Voltage-divider Bias Circuit (Self-b ias Circuit) 1.85

SUMMARY OF BJT BIASING CIRCUITS

Confi guration Circuit Equation Load line and Q-PointFixed bias

RB RC

+VCC

IB = −V VR

CC BE

B

IC = β −⎛

⎝⎜⎞

⎠⎟V V

RCC BE

B

VCE = VCC – ICRC

S = β + 1

S' = − βRB

S" = 1

1

β

IC

VCEQ0 VCC VCE

VCCRC

ICQ

IC

IBQQ

Modifi ed fi xed bias

RB RC

+VCC

RE

IB = ( 1)β

−+ +

V VR R

CC BE

B E

IC = β ( 1)β

−+ +

⎡

⎣⎢

⎤

⎦⎥

V VR R

CC BE

B E


S = 1

1

ββ+

++R

R RE

B E

S' = ( 1)

ββ

−+ +R RB E

S" = 1

1

β

IC ( 1)2β

++ +R R

R RB E

B E

VCEQ VCC VCE

VCC

RC + RE

ICQ

IC

IBQQ

0

(Continued )

M01_XXXXXXXX_XX_C01.indd 85M01_XXXXXXXX_XX_C01.indd 85 7/20/12 3:40 PM7/20/12 3:40 PM


Confi guration Circuit Equation Load line and Q-PointCollector-to-base bias

RB

RC

+VCC

IB = ( 1)β

−+ +

V VR R

CC BE

B C

IC = β ( 1)β

−+ +

⎡

⎣⎢

⎤

⎦⎥

V VR R

CC BE

B C

S = 1

1

ββ+

++R

R RC

B C

S' = ( 1)

ββ

−+ +R RB C

S" = 1

1

β

IC

( 1)2β+

+ +R R

R RB C

B C

VCEQ0 VCC VCE

VCCRC

ICQ

IC

IBQQ

Modifi ed collector-to-base bias

RB

RC

+VCC

RE

IB = ( 1)( )β

−+ + +

V VR R R

CC BE

B C E

IC = β ( 1)( )β

−+ + +

⎡

⎣⎢

⎤

⎦⎥

V VR R R

CC BE

B C E


S = 1

1)

ββ

+

+( ++ +R R

R R RC E

B C E

S' = ( 1)( )

ββ

−+ + +R R RB C E

S" = 1

1

β

IC

( 1)( )2β+ +

+ + +R R R

R R RB C E

B C E

VCEQ0 VCC VCE

VCC

RC + RE

ICQ

IC

IBQQ

SUMMARY OF BJT BIASING CIRCUITS (Continued)

(Continued )


1.14 Miscellaneous Bias Circuits 1.87

Confi guration Circuit Equation Load line and Q-PointVoltage-divider bias

R1 RC

+VCC

RE

R2

IB = ( 1)β−

+ +V V

R RB BE

B E

IC = β ( 1)β−

+ +⎡

⎣⎢

⎤

⎦⎥

V VR R

B BE

B E

S = 1

1

ββ+

++R

R RE

B E

S' = ( 1)

ββ

−+ +R RB E

S" = 1

1

β

IC

( 1)2β+

+ +R R

R RB E

B E

VCEQ0 VCC VCE

VCC

RC + RE

ICQ

IC

IBQQ

1.14 MISCELLANEOUS BIAS CIRCUITSThere are many bias circuits that do not match the basic models discussed so far. These circuits are discussed in this section. For each circuit, fi rst base current has to be calculated. Then col-lector current and output voltage can be determined to locate Q-point.

Example 1.27: In the circuit shown in Fig. 1.68, fi nd RB and VCE.+18 V

−4 V

4 kΩ

200 ΩRB

IC = 2 mAβ = 50




Solution:DC equivalent circuit (Fig. 1.69)

VBE

RC

+VCC

−VEE

IE

RERB

VCE

IB

IC

+−

−

+


(i) IB = IC

β=

2 1050

3× −

= 40 μA

IE = IB + IC = 40 × 10–6 + 2 × 10–3 = 2.04 mA


–IB RB – VBE – IE RE + VEE = 0

RB = V V I R

IEE BE E E

B

− −

= 4 0.7 2.04 10 200

40 10

3

6

− − × ××

−

−

= 72.3 kΩ


VCC – IC RC – VCE – IE RE + VEE = 0 VCE = VCC + VEE – IC RC – IE RE

= 18 + 4 – 2 × 10–3 × 4 × 103 – 2.04 × 10–3 × 200 = 13.59 V



Example 1.28: Determine IB and VE for the circuit shown in Fig. 1.70.

−6 V

β = 120

1.2 kΩ

330 kΩ

+6 V

VE


DC equivalent circuit (Fig. 1.71)

−VEE

RE

+VCC

VE

+

−VBE

RB

IE


(i) Applying Kirchhoff’s voltage law to the base-emitter circuit, VCC – IB RB – VBE – IE RE + VEE = 0

VCC – IB RB – VBE – (β + 1)IB IE + VEE = 0

IB = V V VR R( 1)

CC EE BE

B Eβ+ −+ +

= 6 6 0.7

330 10 (120 1)(1.2 10 )3 3

+ −× + + ×

= 23.72 μA(ii) IE = (β + 1)IB

= (120 + 1) (23.72 × 10–6) = 2.87 mA



Applying Kirchhoff’s voltage law to the emitter circuit,

VE – IE RE + VEE = 0

VE = IE RE – VEE

= 2.87 × 10–3 × 1.2 × 103 – 6

= –2.547 V

Example 1.29: Determine IE, VCE and VC in the circuit shown in Fig. 1.72.+10 V

−8 V

2.2 kΩ

1.8 kΩ



VBE

RC

+VCC

−VEE

IE

RE

VCE

VCIB

IC

+−

−

+





– VBE – IE RE + VEE = 0

IE = V V

REE BE

E

−

= 8 0.72.2 103

−×

= 3.318 mA

IE ≈ IC = 3.318 mA


VCC – IC RC – V CE – IE RE + VEE = 0

VCE = VCC + VEE – (RC + RE)IC (� IE ≈ IC)

= 10 + 8 – (1.8 × 103 + 2.2 × 103) (3.318 × 10–3)

= 4.73 V

(iii) Applying Kirchhoff’s voltage law to the collector-emitter circuit,

VC – VCE – IE RE + VEE = 0

VC = VCE + IE RE – VEE

= 4.73 + 3.318 × 10–3 × 2.2 × 103 – 8

= 4.03 V

Example 1.30: Determine IE and VCE for the circuit shown in Fig. 1.74.

−20 V

β = 90

2 kΩ240 kΩ




Solution:


+

+−

−

−VEE

VCE

VBE

RB RE

IE

IB



–IB RB – VBE – IE RE + VEE = 0

–IB RB – VBE – (β + 1)IB RE + VEE = 0

IB = V V

R R( 1)EE BE

B Eβ−

+ +

= 20 0.7

240 10 (90 (2 10 )3 3

−× + +1) ×

= 45.73 μA

IE = (β + 1)IB

= (90 + 1) (45.73 × 10–6)

= 4.16 mA


–VCE – IE RE + VEE = 0

VCE = VEE – IE RE

= 20 – 4.16 × 10–3 × 2 × 103

= 11.68 V



Example 1.31: Find RC and R1 for the circuit shown in Fig. 1.76.+16 V

R1

30 kΩ 1 kΩ

VBE = 0.2 V

IE = 2 mA

VCE = 6 V α = 0.985

RC


DC equivalent circuit (Fig. 1.77)(i) V2 = VBE + IE RE

= 0.2 + 2 × 10–3 × 1 × 103

= 2.2 V

R1

RC

RE

VCE

+VCC

+

−

V1

+

−

V2

+

−

I1 IC

IB

IC + I1

IE

R2

I2




I2 = VR

2

2

= 2.2

30 103×= 73.3 μA

β = 1

αα−

= 0.985

1 0.985−= 65.66

IB = I

1E

β +=

2 1065.66 1

3×+

−

= 30 μA

IC = IE – IB = 2 × 10–3 – 30 × 10–6 = 1.97 mA

I1 = I2 + IB = 73.3 × 10–6 + 30 × 10–6 = 103.3 μA


VCC – (IC + I1)RC – VCE – IE RE = 0

RC = V V I R

I ICC CE E E

C 1

− −+

= 16 6 2 10 1 101.97 10 103.3 10

3 3

3 6

− − × × ×× + ×

−

− −

= 3.859 kΩ


VCC – (IC + I1)RC – R1I1 – V2 = 0

R1 = V V I I R

I( )CC C C2 1

1

− − +

= 16 2.2 (1.97 10 103.3 10 )(3.859 10 )103.3 10

3 6 3

6

− − × + × ××

− −

−

= 56.15 kΩ



Example 1.32: Determine IB, IC, VE and VCE for the circuit shown in Fig. 1.78.+18 V

−18 V

+

+−

−VBEVE

VCE

VC

6.8 kΩ

9.1 kΩ

510 kΩ

510 kΩ

β = 130


Solution:


(a) (b)

+VCC+VCC

−VEE

RC

IC

IE

RE

VC

VBE

VB

VB

IB RBVCE

VE

+

+−

−

−VEE

R1

R2

I


Applying Kirchhoff’s voltage law to the circuit, VCC – I(R1 + R2) + VEE = 0

I = V V

R RCC EE

1 2

++



= 18 18

510 10 510 103 3

+× + ×

= 35.29 μA

Applying Kirchhoff’s voltage law to the lower part of the circuit of Fig. 1.79(a),

VB – I R2 + VEE = 0

VB = –VEE + I R2

= –18 + 35.29 × 10–6 × 510 × 103

= –2.1 V

RB = R R

R R1 2

1 2+

= 510 10 510 10510 10 510 10

3 3

3 3

× × ×× + ×

= 255 kΩ

(i) Applying Kirchhoff’s voltage law to the base-emitter circuit of Fig. 1.79(b),

–VB – IB RB – VBE – IE RE + VEE = 0

–VB – IB RB – VBE – (β + 1)IB RE + VEE = 0

IB = V V VR R( 1)

EE B BE

B Eβ− −

+ +

= 18 2.1 0.7

255 10 (130 (6.8 10 )3 3

− −× + +1) ×

= 13.27 μA(ii) IC = β IB = 130 × 13.27 × 10–6 = 1.73 mA

(iii) IE = IB + IC = 13.27 × 10–6 + 1.73 × 10–3 = 1.74 mA

Applying Kirchhoff’s voltage law to the emitter circuit,

VE – IE RE + VEE = 0

VE = IE RE – VEE

= 1.74 × 10–3 × 6.8 × 103 – 18

= –6.17 V



(iv) Applying Kirchhoff’s voltage law to the collector-emitter circuit of Fig. 1.79(b),

VCC – IC RC – VCE – IE RE + VEE = 0

VCE = VCC + VEE – IC RC – IE RE

= 18 + 18 – 1.73 × 10–3 × 9.1 × 103 – 1.74 × 10–3 × 6.8 × 103

= 8.43 V

Example 1.33: Determine the currents IE and IB and the voltages VCB and VCE for the circuit shown in Fig. 1.80.

4.7 kΩ

β = 75

5 V

3.6 kΩ

8 V


Solution:


VCC

RC

VCB

VCE

VBE

VEE

RE +

+−− +

−

IE IC


(i) Applying Kirchhoff’s voltage law to the emitter-base circuit,

–VEE + IE RE + VBE = 0

IE = V V

REE BE

E

−



= 5 0.74.7 103

−×

= 0.91 mA

(ii) IB = I

1E

β +

= 0.91 1075 1

3×+

−

= 11.97 μA

(iii) IC = β IB = 75 × 11.97 × 10−6 = 0.9 mAApplying Kirchhoff’s voltage law to the collector-base circuit,

VCC – IC RC – VCB = 0

VCB = VCC – IC RC

= 8 – 0.9 × 10–3 × 3.6 × 103

= 4.76 V

(iv) Applying Kirchhoff’s voltage law to the entire outside perimeter of the circuit,

–VEE + IE RE + VCE + IC RC – VCC = 0

VCE = VEE + VCC – IE RE – IC RC

= 5 + 8 – 0.91 × 10–3 × 4.7 × 103 – 0.9 × 10–3 × 3.6 × 103

= 5.48 V

Example 1.34: Find RE and RC for the circuit shown in Fig. 1.82.

12 V

RC

4 V

REα = 0.99

IE = 1.1 mA

VCE = 7 V




Solution:


VCC

RC

VCBVBE

VEE

RE +

+−

−

IE ICVCE− +


(i) Applying Kirchhoff’s voltage law to the emitter-base circuit,

–VEE + IE RE + VBE = 0

RE = V V

IEE BE

E

−

= 4 0.71.1 10 3

−× −

= 3 Ω

(ii) Applying Kirchhoff’s voltage law around the transistor terminal,

VCE = VCB + VBE

VCB = VCE – VBE = 7 – 0.7 = 6.3 V

IC = α IE = 0.99 × 1.1 × 10–3 = 1.089 mA

Applying Kirchhoff’s voltage law to the collector-base circuit,

VCC – IC RC – VCB = 0

RC = V V

ICC CB

C

−

= 12 6.31.089 10 3

−× −

= 5.234 kΩ



1.15 BIAS COMPENSATIONThe biasing circuits provide stability of Q-point against variations in ICO, VBE and β. The collector-to-base bias circuit and voltage-divider bias circuit use negative feedback to stabilize Q-point which reduces the amplifi cation of the signal. If this reduction in gain is intolerable, compensation techniques are used. In this method, temperature-sensitive devices such as diodes, transistors, thermistors and sensistors etc. are used which provide compensating volt-age and current to stabilize variations in IC with VBE and ICO.

1.15.1 Diode Compensation for VBEA circuit utilizing the voltage-divider stabilization technique and diode compensation is shown in Fig. 1.84.

R1 RC

R2

RE

+VCC

V

R

D

RC

VB

RE

RB

VBE

+VCC

VVD

R

D

+

−

IB

IC

+

−

Fig. 1.84 Diode Compensation for VBE

The diode is forward biased by the voltage V and current-limiting resistor R. We know that VBE decreases by 2.5 mV/oC; i.e. the device starts operating at lower voltage which changes IB and IC and hence, Q-point. If the diode is of the same type and material as the transistor, the voltage across the diode will have the same temperature coefficient (–2.5 mV/oC) as the base-emitter voltage VBE. The variation in VBE and voltage across the diode will be equal and opposite. Hence, they cancel out and IB and IC become indepen-dent of VBE.


VB – IB RB – VBE – IE RE + VD = 0 VB = IB RB + IE RE (� VBE = VD = 0.7 V)


1.15 Bias Compensation 1.101

≈ IE RE (� IE >> IB)

IE ≈ IC = VR

B

E = constant

1.15.2 Diode Compensation for ICOFor germanium transistors, changes in ICO with temperature play the important role in collector current stability. A circuit utilizing the voltage-divider stabilization technique and diode compensation for ICO is shown in Fig. 1.85. In this circuit, the diode is kept in reverse-biased condition. The diode D is connected in place of resistance R2. If the diode and the transistor are of the same type and material, the reverse saturation current IO of the diode will increase with temperature at the same rate as the transistor saturation current ICO.

IB

D

+VCC

+

−

RCR1

I

Io

IC

VBE

Fig. 1.85 Diode Compensation for ICO

From Fig. 1.85,

I = V V

RCC BE

1

−

For germanium transistor VBE = 0.2 V which is very small and can be neglected.

I ≈ VRCC

1= constant

IB = I – IO



We know that


Substituting the value of IB,

IC = β (I – IO) + (β + 1)ICO

= β (I – IO) + β ICO [� (β +1) ≈ β]

= β [(I – (IO + ICO)]

As temperature increases, IO and ICO will change by the same amount.

IC = β IO (� IO = ICO)

As I is constant, IC remains essentially constant. Thus, changes in ICO with temperature are compensated by the diode.

1.15.3 Thermistor and Sensistor Compensation In this method of transistor compensation, temperature-sensitive resistive elements are used instead of diodes or transistors. A circuit utilizing the voltage-divider stabilization technique and thermistor is shown in Fig. 1.86 to minimize the increase in collector current due to chang-es in ICO, VBE and β.

R1 RC

RT

RE

+VCC

R2

Fig. 1.86 Thermistor Compensation

Thermistor RT has a negative temperature coeffi cient. It is a temperature-sensitive resistor whose resistance decreases exponentially as temperature increases. When temperature increases,


1.15 Bias Compensation 1.103

ICO and IC increase. But with rise in temperature, resistance of the thermistor decreases. Hence, increased collector current fl ows through thermistor into RE. Hence, voltage drop across RE increases. This voltage drop provides negative feedback such that IB decreases and hence, IC decreases.

An alternative compensation technique using thermistor is to place RT in place of R2 as shown in Fig. 1.87. When temperature increases, ICO and IC increase. But with rise in temperature, resistance of the thermistor decreases. The voltage drop across RT decreases, which decreases the forward-biasing base voltage. Hence, IB and IC decrease.

+VCC

R1

RT RE

RC

Fig. 1.87 Thermistor Compensation

Instead of a thermistor, it is possible to use a temperature-sensitive resistor with a positive temperature coeffi cient, i.e. sensistor. Its resistance increases exponentially as temperature increases. It is connected in place of R1 in voltage-divider bias circuit as shown in Fig. 1.88.

+VCC

R2 RE

RCRT

Fig. 1.88 Sensistor Compensation



When temperature increases, ICO and IC increase. But with rise in temperature, resistance of the sensistor increases, which decreases current fl owing through it. Hence, current through R2 decreases, which reduces the voltage drop across it. This decreases the forward-biasing base voltage. Hence, IB and IC decrease.


Study Materials_EDC 01

Documents

Transcript of Study Materials_EDC 01