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ECM1106 - Electronics for Engineers Dr Mustafa M Aziz(2010)

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THE BIPOLAR JUNCTION TRANSISTOR (BJT)

1. Bias Stability

2. Transistor Switches

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1. Bias Stability

For amplifying signals, DC bias was used to fix ICand VCEfor constant IBon the linear or active

part of the characteristics. The position of the operating point depends on the current gain .

, however, changes with temperatureandfrom transistor to transistor. The consequence is ashift in the operating point on the characteristic curve outside the active region of the

characteristics leading to distortion, making the circuit not suitable for amplification.

Example: Determine the values of ICand VCEfor the following circuits:

1k5286k

typ=100

+15V

(a)

1k5286k

max

=200

+15V

(b)

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1.1. Voltage-Divider Bias Circuit

The circuit shown below provides DC bias that is almost independent of . The idea here is to

fix the emitter current IE(IC) through the emitter resistor REsuch that it is independent of .

In this case, IBis no longer fixed but will change to maintain the constant emitter (and collector)currents.

Resistors R1and R2form a voltage-divider to set the DC voltage level of the base terminal (to

ensure that the base-emitter junction is forward biased).

VCC

C1

R1

R2

RC

RE

vivo

C2

VCC

C1

R1

R2

RC

RE

vivo

C2

Aim: choose resistor values to set the operating point (i.e. VCEand IC) in the linear region of the

output characteristics.

1.1.1. Input Circuit

VCC

R1

R2 Ri

I1

I2

IB

VCC

R1

R2RE

I1

I2

IE

IB

Ri

VB

VCC

R1

R2 Ri

I1

I2

IB

VCC

R1

R2RE

I1

I2

IE

IB

Ri

VB

Resistance between base and ground: RiRE.

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Provided Riis much larger than R2such that:

RE10R2

then IB

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1.1.3. Self-Stabilising Network Action

IC increase

IE increase

VE=IERE increase

VBE=VB-VE decrease VB = constant

determined by

R1, R2 and VCC

IB decrease

VBE

IB

0

IC=IB decrease

IC increase

IE increase

VE=IERE increase

VBE=VB-VE decrease VB = constant

determined by

R1, R2 and VCC

IB decrease

VBE

IB

0 VBE

IB

0

IC=IB decrease

Any tendency to increase ICwill be met with a network reaction to decrease it.

1.1.4. Coupling Capacitors

The input capacitor C1serves two purposes:

1. Couples (superimposes) the AC input signal with the bias current.2. Blocks DC components of input signals (XC= ) that would affect the set DC bias

level.

The output capacitor C2is used to AC couple the output signal to a load and remove DC levels

produced by biasing stages.

The coupling capacitors should ideally provide zero reactance (XC=0, i.e. C) to AC signals.

Typical value for coupling capacitors are 1 10F.

1.2. Amplifier Design Considerations

RCis designed to drop about half of the supply voltage to provide a symmetrical voltage swing

of the output signal.

REcannot be unreasonably large because voltage across it limits range of V CE. VEis typically

set to be 1/4 to 1/0 of VCC.

The base resistor R2is chosen such that 10R2RE.

The value of R1is then determined from CC

21

2B VRR

R

V +=

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Example: The following circuit is used to amplify the signals coming from a microphone.

(a) Determine the values of R1, R2, REand RCfor the operating point indicated.

(b) If an input AC signal with amplitude 0.5V and 2kHz frequency is applied, sketch the input,

output and emitter voltage waveforms.

(c) Estimate the voltage gain of this circuit.

15V

10F

R1

R2

RC

RE

vivo

0.7V

10F

15V

10F

R1

R2

RC

RE

vivo

0.7V

10F

VCE (V)0 1 2 3 4 5 6 7 8 9 10

-3

0

3

6

9

12

15

18

21

24

27

30

IB=0A

20A

40A

60A

80A

100A

120A

140A

IC(mA)

VCE (V)0 1 2 3 4 5 6 7 8 9 10

-3

0

3

6

9

12

15

18

21

24

27

30

IB=0A

20A

40A

60A

80A

100A

120A

140A

IC(mA)

0 1 2 3 4 5 6 7 8 9 10-3

0

3

6

9

12

15

18

21

24

27

30

IB=0A

20A

40A

60A

80A

100A

120A

140A

IC(mA)

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1.3. Signal Voltage Gain

Increasing IBdue to input signal and hence ICincreases the voltage drop ICRC. But since VC=

VCC-ICRC, the output collector voltage decreases.

Likewise, decreasing IB due to input signal and hence IC, decreases voltage drop ICRC. Butsince VC= VCC-ICRC, the output collector voltage increases.

Therefore there is 180o phase shift between base (input) and collector (output) signals

inverting amplifier.

R1

R2

RC

RE

vivo

+VCC

vB

vC

R1

R2

RC

RE

vivo

+VCC

vB

vC

Signal voltage gain of amplifier,B

C

i

ov

vbase,atamplitudevoltagesignalAC

vcollector,atamplitudevoltagesignalAC

v

vA ==

Because base-emitter junction is forward biased, then vBvE, and hence:

EE

CC

B

C

vRi

Ri

v

v

A ==

But and since iCiEthen:

E

Cv

R

RA

Example: Compare the estimated signal voltage gain in the previous example with the calculated

one using the above equation.

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2. Transistor Switches

The second major application of transistors is electronic switching for computer and control

applications.

Typical transistor switching circuit is shown below, where it can be seen that the DC source isconnected to the collector (or output) side only (and not to the base).

VCC

RC

RB

vi

VCEorvo

IB

IC

VCC

RC

RB

vi

VCEorvo

IB

IC

When used as an electronic switch, the transistor switches between cutoffand saturation.

2.1. Cutoff

vi= 0V, IB= 0A and hence IC0mA.

The voltage drop across RCis ICRC= 0V resulting in:

VCEVCC

2.2. Saturation

Input signal vi forward biases B-E junction and

provides sufficient base current IB to increase IC to its

maximum (ICsat) such that most of the supply voltageVCCdrops across RC.

VCEis then very small (VCEsat0.1 - 0.3V) and can beneglected and replaced by short circuit. The collector

current (or value of RC to achieve saturation) is then

determined from:

C

CC

CsatR

VI =

RC

RB

0V

IC=0

+VCC

RC

+VCC

C

E

RC

RB

0V

IC=0

+VCC

RC

+VCC

C

E

RC

RB

vi

ICsat

+VCC

RC

+VCC

C

E

IB

ICsatRC

RB

vi

ICsat

+VCC

RC

+VCC

C

E

IB

ICsat

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To ensure saturation:

> CsatBI

I (typically set IB2ICsat/)

The level of IBis set through applying Kirchhoff's voltage law around the input circuit:

B

BEiB

R

VvI

=

Example: For the following circuit:

(a) What is VCEwhen vi= 0?

(b) What is the minimum value of IB that will saturate the

transistor?

(c) Calculate maximum RBwhen vi= 5V.

Example: For the following circuit:

(a) Determine ICsat.

(b) Determine IBat saturation.

(c) Plot the output voltage waveform across the collector.

820R

68k

+5V

=125

vi

5V

t0

820R

68k

+5V

=125

vi

5V

t0

1k

RB

vi

+10V

=200

1k

RB

vi

+10V

=200

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2.3. Inductive Load Drivers

Transistors are used to establish the necessary current to

energise (or drive) devices with coils such as relays and

motors.

When positive pulse is applied, the transistor is turned on

(saturation) to establish sufficient current to drive the

inductive load.

When the signal to the base is removed to turn off the

transistor (cutoff), the rapidly decreasing collector current

develops a large reverse voltage across the coil (VL =

LdI/dt).

This also induces a large positive voltage at the collector terminal which exceeds the breakdown

voltage of the transistor causing permanent damage to the transistor.

RB

+VCC

vi

t0

off

I

t0

Large voltage

spike, VL = LdI/dt

I

boomRB

+VCC

vi

t0

off

I

t0

Large voltage

spike, VL = LdI/dt

I

boom

Solution: add a diode in parallel with the inductive load diode forward biased during turn offto short out and dissipate the voltage spike and current.

RB

+VCC

vi

t0

Diode forward

biased during

turn off

ION

IOFF

RB

+VCC

vi

t0

Diode forward

biased during

turn off

ION

IOFF

RB

+VCC

vi

t0

on

I

RB

+VCC

vi

t0

on

I

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Example: A motor switch circuit is shown below. The motor is on when the control signal is 5V,

and off when the control signal is 0V.

(a) Determine the minimum IBrequired to turn on the motor (saturation).

(b) Determine the maximum value for RBto maintain saturation of the transistor.

(c) Sketch the input and output voltage waveforms for this circuit.(d) What protection you need to apply to prevent damage to the transistor during turn off?

RB

+5V

vi

t0

250mH

5R

=50

Motor

5V

RB

+5V

vi

t0

250mH

5R

=50

Motor

5V

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2.4. Logic Gates

Logic gates are the fundamental building blocks of digital integrated circuits and components.

They consist of transistors operating as switches in the saturation and cutoff regions.

For example, an AND gate implementation using transistors is shown.

10k

B

+5V

10k

A

3k3 C=A AND BA B C

0 0

0 1

1 0

1 1

Q1

Q210k

B

+5V

10k

A

3k3 C=A AND BA B C

0 0

0 1

1 0

1 1

Q1

Q2

If both inputs A and B have a turn-on voltage, then both transistors Q1 and Q2 are approximately

short circuits (saturation) providing a path from 5V supply to output terminal high or binary 1state.

If one or both transistors are off due to 0V at input terminal, an open circuit (cutoff) is placed in

series with the path from the 5V supply to output terminal 0V or binary 0 state.

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