1

EMT 112/4ANALOGUE ELECTRONICS 1

Power Amplifiers

SyllabusPower amplifier classification, class A, class B, class AB, amplifier distortion, class C and D, transistor power dissipation, thermal management.

2

Part III (cont’d)Power amplifier circuits

– Class AB

POWER AMPLIFIERS

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Class-ABPOWER AMP

Small biasing voltage to eliminate dead band

Qn & Qp are assumed matched transistors

4

Class-ABPOWER AMP

Various techniques are used in obtaining the bias voltage VBB in class AB power amplifier circuit.

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Class-AB – Diode BiasingPOWER AMP

The base-emitter of a BJT is basically a p-n junction and

hence exhibits similar characteristics as that of a diode –

voltage across a diode in a forward mode is almost

constant over a range of the diode current.

I D

V D

+V D-

I D

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Class-AB – Diode BiasingPOWER AMP

The above diode characteristic is utilised in biasing of the push-pull power amplifier

In the absence of input signal (vI = 0), most of IBias flows through D1 and D2

establishing a small bias voltage VBB for the base-emitter

junctions of Qn and Qp.

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Class-AB – Diode BiasingPOWER AMP

When vI is at its peak

+ve, iBn may be large.

IBias shall be sufficient to supply

both iBn and the

current through D1

and D2 in order to maintain the bias

voltage VBB.

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17/03/08

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3

A diode biasing class AB power amplifier is to meet the following specifications;

• RL = 8 ;

• output power to PL = 5 W;

• peak output voltage to be not more than 80% of VCC;

• minimum value of ID to be no less than 5 mA

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 (cont’d)

For both Qn and Qp;

75 A; 10 13 SQI

For D1 and D2;

A 103 14SDI

Determine;

a) IBias and VCC;b) The quiescent collector

currentc) iCn and iCp when the

output voltage is at its peak positive value

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution

LLo RPV rmsa)

V 32.685

rmspeak 2 oo VV

V 94.8

V 8.118.0

peak oCC

VV

Select VCC = 12 V

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Class-AB – Diode BiasingPOWER AMP

At the +ve peak of the output voltage;

L

oLEn R

Vii peak

maxpeak

1peak

peakEn

Bn

ii

EXAMPLE 3 – Solution (cont’d)

A 1.128

94.8

mA 14.776

12.1

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)

To maintain a minimum of 5 mA through the diodes;

DBn IiI Bias

Select IBias = 20 mA

514.7

mA 19.7

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)

b) Under quiescent

condition (vI = 0),

ID = 20 mA

(neglecting iBn);

SD

DTBB I

IVV ln2

14

3

103

1020ln026.02

V 416.1

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)

Assuming Qn and Qp are matched transistors;

V 708.02

BB

BEpBEn

V

VV

Hence;

TBB VVSQCQ eII 2/

mA 67CQI

026.0/708.01310 e

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)

c) At the peak +ve value of output voltage;

peakmax LEn ii

mA 14.7max Bni

mA 12.1

maxBias BnD iII

mA 5.314.720

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)

14

3

103

103.5ln026.02BBV

maxmax 1 EnCn ii

V 347.1

12.1751

75

A 105.1

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)

SQ

CnTBEn I

iVV

maxln

1310

105.1ln26

mV 781

BEnBBBEp VVV

V 566.0781.0347.1

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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)

TBEp VVSQCp eIi /

026.0/566.01310 e

mA 285.0

Hence, when the output voltage is at its peak positive value;

A 105.1Cni

and;mA 285.0Cpi

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Class-AB – Diode BiasingPOWER AMPEXERCISE 1

a) +5 V;

b) -5 V;

c) at its peak negative value.

For the same amplifier as in Example 3 above, find iCn and iCp when the output

voltage is;

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Class-AB – VBE Multiplier BiasingPOWER AMP

Thus, VBB can be set by selecting suitable values for R1 and R2.

2

1

R

VI BER

2

11

21

1 R

RV

RRIV

BE

RBB

Neglecting the base current of Q1;

The biasing circuit comprising Q1, R1, R2 and IBias, provides the biasing voltage VBB.

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Class-AB – VBE Multiplier BiasingPOWER AMP

A fraction of IBias flows through Q1, so that;

1

11 ln

S

CTBE I

IVV

Under quiescent condition, iCn and iCp are small and hence iBn and iBp are negligible.

As vI increases, iCn increases followed by an increase of iBn. IC1 decreases but VBE1 is almost constant.

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Class-AB – VBE Multiplier BiasingPOWER AMP

To facilitate adjustment of the ratio R1/R2 and hence, the value of VBB, a third resistor Rv is included in the circuit.

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Class-AB with Input Buffer TransistorsPOWER AMP

R1, R2 and the emitter-followers

Q1 and Q2 establish the required

quiescent bias.

R3 and R4 (usually of low values)

are incorporated to provide thermal stability.

The output voltage is approximately equal to the input voltage (emitter-follower)

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Class-AB with Input Buffer TransistorsPOWER AMP

When the input voltage vI increases, the base voltage of Q3 increases and the output

voltage vO increases. The

emitter current of Q3 increase to

supply the load current iO. The

base current of Q3 increases.

The increase in base voltage of Q3 reduces the voltage across, and the current through R1. This

means that iB1 and iE1 also

decrease.

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Class-AB with Input Buffer TransistorsPOWER AMP

Also when the input voltage vI increases, the voltage across R2

increases and iE2 and iE2

increase. The input current iI accounts for the reduction in iB1

and the increase in iB2 i.e.

12 BBI iii (Kirchhoff’s Current Law)

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Class-AB with Input Buffer TransistorsPOWER AMP

Neglecting and

we have;

2

2 1 R

VVvi

n

BEIB

,3Rv ,4Rv 3Bi 4Bi

and;

1

1 1 R

VvVi

p

EBIB

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Class-AB with Input Buffer TransistorsPOWER AMP

If;

RRRVVVV EBBE 21 , ,

and; pn

then;

RVvV

R

VVvi EBIBEII

11

RvI

1

2

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Class-AB with Input Buffer TransistorsPOWER AMP

Since the voltage gain is approximately unity, the output current is;

L

I

L

OO R

v

R

vi

The current gain is;

LI

Oi R

R

i

iA

2

1

which is quite substantial. A large current gain is desirable since the output stage must meet the power requirements.

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24/03/08

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Class-AB with Input Buffer TransistorsPOWER AMPEXAMPLE 4

(a) Determine the quiescent bias currents in all transistors;

(b) Calculate all the currents labeled in the figure and the current gain when vI = 10 V.

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Class-AB with Input Buffer TransistorsPOWER AMP

2121 EERR iiii

mA 2.72

6.015

EXAMPLE 4 – Solution

(a) For vI = 0 (quiescent currents);

Assuming all transistors are matched, the bias currents in Q3 and

Q4 are also approximately 7.2 mA

since the base-emitter voltages of Q1

and Q3 are equal and those of Q2 and

Q4 are equal.

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Class-AB with Input Buffer TransistorsPOWER AMP

L

I

L

OO R

v

R

vi

EXAMPLE 4 – Solution (cont’d)

(b) For vI = 10;

Because the voltage gain is approx. unity;

mA 100100

10

mA 1003 OE ii

mA 64.161

100

13

3

EB

ii

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Class-AB with Input Buffer TransistorsPOWER AMP

mA 2.2

2

106.015

EXAMPLE 4 – Solution (cont’d)

1

1 R

vVVi IBER

311 BRE iii

mA 56.064.12.2

μA 18.961

56.0

11

1

EB

ii

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Class-AB with Input Buffer TransistorsPOWER AMPEXAMPLE 4 – Solution (cont’d)

Since Q4 tends to turn off

when vI increases, iB4 is

negligible. Therefore;

222 R

Vvvii EBIRE

2

156.010

mA 2.12

μA 20061

mA 2.12

12

2

EB

ii

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Class-AB with Input Buffer TransistorsPOWER AMPEXAMPLE 4 – Solution (cont’d)

The input current;

12 BBI iii

μA 19118.9200

The current gain;

524191.0

100

I

Oi i

iA

37

L

i R

RA

2

1

Class-AB with Input Buffer TransistorsPOWER AMPEXAMPLE 4 – Solution (cont’d)

If the previous expression i.e.

we have;

610

1.02

2601

iA

is used,

The higher gain is due the fact that the base currents of Q3 and Q4 are neglected in deriving the expression.

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Class-AB with Input Buffer TransistorsPOWER AMPEXERCISES

Problems 8.36 through 8.38 in the text book