Class B Output Stage. Class B Operation Multisim Simulations Class B Power Efficiency. ESE319 Introduction to Microelectronics

(1)

ESE319 Introduction to Microelectronics

Class B Output Stage

●

Class B Operation

●

Multisim Simulations

(2)

ESE319 Introduction to Microelectronics

Class B Amplifier Operation

V _B_{=0V ⇒ v}_I_{≥0.7 V}

NOTE: 1. when v_I < 0.7V, i_C = 0

2. a 2nd_{class B BJT is needed to}

(3)

ESE319 Introduction to Microelectronics

Quick Review

Power output to load: Power from dc sources:

Power Conversion Efficiency: P_{L av}₌1 2 V _{o− peak}2 R_L Power dissipation P_{L av}_max₌ 1 2 V _CC2 R_L = 1 2 V CC I _max=1 4 =0.25

Let max V_0-peak = V_CC PD av=2VCC I PD avmax=2VCC I

=

1

4 V

_o2_{− peak}

I R

_L

V

_CC P_Disp_=2V_CC I₋1 2 V_o2_{− peak} R_L P_Disp_max_=2V _CC I

Class-A Amp Power Inventory

f(V_o-peak) max f(V_o-peak)

(4)

ESE319 Introduction to Microelectronics

The Class B Output Stage

I. For v_I > V_BEN = 0.7 V, Q_N = ON, Q_p = OFF. a.

b. Q_N saturates if v_O > V_CC - V_CENsat.

c. Q_P is cut off for v_I > -0.7 V.

The output voltage v_O follows the input v_I- V_BEN

up to v_O = V_CC – V_CENsatwhere Q_N saturates.

II. For v_I < -V_EBP = -0.7 V, Q_P = ON, Q_N = OFF.

For v_I > 0.7 V V_BEN=vI−vO v_O_=v_I_{−0.7 V} For v_I < -0.7 V V_EBP=vO−vI v_O=vI0.7 V

0. For -0.7 V ≤ v_I ≤ 0.7 V, Q_N and Q_P cutoff. => v O = 0 V i_EN i_EP i_L_=i_EN_≈i_CN i_CN i_CP

(5)

ESE319 Introduction to Microelectronics

The Class B Output Stage

I. For v_I > V_BEN = 0.7 V, Q_N = ON, Q_p = OFF.

II. For v_I < -V_EBP = -0.7 V, Q_P = ON, Q_N = OFF.

a.

b.

Q_P saturates if v_O < - V_CC + V_ECPsat

c. Q_N is cut off for v_I < 0.7 V.

The output voltage v_O follows the input v_I+ V_EBP

down to v_O = - V_CC + V_ECPsatwhere Q_Psaturates.

For v_I > 0.7 V V_BEN=vI−vO v_O_=v_I_{−0.7 V} For v_I < -0.7 V V_EBP=vO−vI v_O=vI0.7 V

0. For -0.7 V ≤ v_I ≤ 0.7 V, Q_N and Q_P cutoff. => v O = 0 V i_EN i_EP _i L=−iEP≈−iCP i_CN i_CP

(6)

ESE319 Introduction to Microelectronics

Class B Amplifier VTC Plot

0.7 V_v_i_V _CC_−V _CENsat_0.7V v_o_=0V −0.7V vi0.7V v_o_=v_i_0.7V For: −VCC−V ECPsat−0.7V vi−0.7V For: For: 0.7 V -0.7 V Q_N saturated Q_P saturated Class A VTC v_o_=v_i_−0.7V

(7)

ESE319 Introduction to Microelectronics

Class B Crossover Distortion

Crossover distortion in audio

power amps produces unpleasant

sounds.

(8)

ESE319 Introduction to Microelectronics

(9)

ESE319 Introduction to Microelectronics

(10)

ESE319 Introduction to Microelectronics

Multisim Simulation Class B Crossover Distortion

The “dead band” from -V_EBP = -0.7 V < v_i < V

BEN = 0.7 V causes crossover

(11)

ESE319 Introduction to Microelectronics

Class B Power Analysis

The average power to the load resistor

R_L: P_{L av}=V o−rmsIo−rms= V_o2_−rms R_L = 1 2 V_o2_{− peak} R_L P_V CCav= 1 T

∫

₀ T 2 V_CCi_CN dt₌ 1 T

∫

₀ T 2 V_CCi_Ldt₌ 1 T

∫

₀ T 2 V _CC Vo− peak R_L sin 2_ T tdt Simplifying Assumptions:

1. Ignore the dead-zone in VTC.

2. Let V_CEN-sat= V_ECP-sat = 0 V => maximum v_O swing is _−V

CC≤vo≤VCC

The average power delivered by the positive +V_CC power

supply over the positive half-cycle 0 ≤ t ≤ T/2 (i_C = 0 from T/2 ≤ t ≤ T ):

v_O=V_o_{− peak}sint

same as Class A

(12)

ESE319 Introduction to Microelectronics

Class B Power Analysis - cont.

∫

sin_{ t dt=−}1

 cost 

Recalling from calculus

P_V CCav=− 1 2_ V_o_{− peak} R_L VCC{cos 2 T  T 2 −cos 2 T 0} P_V_CC_av₌ 1 T V CC V _{o− peak} R_L − T 2_ cos 2_ T t t = 0 t = T/2 P_V CCav=− 1 2_ V _{o− peak} R_L VCC{cos−cos0} P_V _av₌ 1  V _o_{− peak} R VCC -1 1

From previous slide

P_V CCav= 1 T

∫

₀ T/2 V_CC Vo− peak R_L sin 2_ T tdt= 1 T VCC V_o_{− peak} R_L

∫

₀ T/2 sin_ 2 T tdt

(13)

ESE319 Introduction to Microelectronics

Class B Power Analysis - cont.

P_{D av}_=P_V CCavP−VCCav= 2  V _o_{− peak} R_L V CC

Power delivered by the negative –V_CC power source is equal to that for +V_CC:

The power conversion efficiency of the class B amplifier (accurate for |V_o-peak| > 0.7 V) is: = PL av P_{D av}= 1 2 V _o2_{− peak} R_L 2  V_o_{− peak} R_L V CC = 4 V_{o− peak} V_CC P_−V CCav=PVCCav= 1  V _{o− peak} R_L V CC Hence:

(14)

ESE319 Introduction to Microelectronics

Class B Power Analysis - cont.

= 4 V_o_{− peak} V_CC ⇒max=  4 =0.785

Maximum Class B power conversion efficiency:

Class B power dissipated in the transistors: P_Disp_=P_{D av}_−P_{L av}₌ 2  V_o_{− peak} R_L VCC− 1 2 V _o2_{− peak} R_L Since we assume _−V _CC_≤v_o_≤V_CC or 78.5 %

Recall for Class A: max=0.25 or 25 %

Power Dissipation:

(15)

ESE319 Introduction to Microelectronics

Class B Power Analysis - cont.

P_Disp_=P_{D av}_−P_{L av}₌ 2  V_{o− peak} R_L VCC− 1 2 V _{o− peak}2 R_L To estimate maximum P

Disp: set derivative of PDisp w.r.t. Vo-peak to zero, i.e.

P_Disp_max₌ 4 2 V_CC2 R_L − 1 2 4 2 V_CC2 R_L = 2 2 V _CC2 R_L d P_Disp d V _o_{− peak} = 2  V_CC R_L − V _o_{− peak} R_L =0 Vo− peak= 2  VCCV CC =>

½ P_Disp(max)is dissipated in Q_N and ½ P_Disp(max) is dissipated in Q_P.

V _o_{− peak}= 2

 VCC

with

What value of V

_o-peak

results in the maximum P

_Disp

?

(16)

ESE319 Introduction to Microelectronics

Class B Power Analysis Summary

Power output to load:

P_{D av}_max₌ 2



V_CC2 R_L

Power from sources:

Power Conversion Efficiency: =  4 V_{o− peak} V_CC P_{L av}₌1 2 V _o2_{− peak} R_L Transistor dissipation Q_N + Q_P: P_Disp₌ 2  V_o_{− peak} R_L VCC− 1 2 V _{o− peak}2 R_L PDispmax= 2 2 V_CC2 R_L P_{L avmax}₌1 2 V_CC2 R_L max=  4 ≈0.785 P_{D av}₌ 2  V _{o− peak} R_L V CC V _{o− peak}₌ 2  VCC

Max dissipation at: max−Disp=0.5

max V_0-peak = V_CC

(17)

ESE319 Introduction to Microelectronics

Power Dissipation Function Plot

Not accurate for small V .

V_o_{− peak} Let V_CC = 12 V and R_L=100 = 7.63 V 0.7 V P_Disp(max) = 0.29 W 0.20 W P_Disp_−B= 2  V _o_{− peak} R_L VCC− 1 2 V_o2_{− peak} R_L P_Disp P_Disp_max₌2 VCC 2 2R_L=0.29 W

(18)

ESE319 Introduction to Microelectronics

Power Conversion Efficiency vs.

Output Voltage Amplitude

Let V_CC = 12 V 0 2 4 6 8 10 12 V Volts Class B=  4 V _o_{− peak} V_CC 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 Efficiency max−B=0.785 P_Disp_max _V o− peak= 2  VCC @ Class A= 1 4 V _o_{− peak} V _CC max−A=0.25

(19)

ESE319 Introduction to Microelectronics

Class B Audio Amplifier Example

It is required to design a Class B output stage to deliver an average power of 20 W to an

8 speaker load.

To avoid saturation of the transistors and non-linear distortion, the power supply V_CC is to be

5V greater than peak output voltage.

1. Determine the V_CC required.

2. Determine peak current drawn from each power supply.

3. Determine the total power drawn from the power supplies.

4. Determine the power conversion efficiency. 5. Determine the max power each transistor

(20)

ESE319 Introduction to Microelectronics

Class B Audio Amplifier Example - cont.

SOLUTION:

1. Determine the V_CC required.

P_{L av}₌1 2 V _o2_{− peak} R_L =20 W ⇒V o− peak=



2 PL−av RL=



2∗20∗8=17.9 V P_{L av}_{=20 W} and R_L_=8 V _CC_=V_o_{− peak}_{5V =17.9V 5V =23 V}

2. Determine peak current drawn from each power supply.

I_{o− peak}₌V o− peak R_L =

17.9 V

(21)

ESE319 Introduction to Microelectronics

Class B Audio Amplifier Example - cont.

3. Determine the total power drawn from the power supplies.

SOLUTION cont.:

P_{D av}₌ 2  V_o_{− peak} R_L V CC= 2  17.9 V 8_ 23V =32.8W

4. Determine the power conversion efficiency.

= 4 V _{o− peak} V_CC =  4 17.9V 23V =0.61 or 61 %

5. Determine the max power each transistor must dissipate safely. P_DispN _max_=P_DispP_max₌1

2 PDispmax= 1 2 V _CC2 R_L = 1 2 23 V 2 8_ =6.7W

(22)

ESE319 Introduction to Microelectronics

Multisim Simulation – Max Power Dissipation

P_V CCav= 1  V_o_{− peak} R_L VCC P_V CCav PL−av P_−V CCav= 1  V_o_{− peak} R_L VCC = 0.265 W = 0.265 W P_{L av}₌1 2 V_o2_{− peak} R_L = 0.240 W  = PL av = 0.2198 =0.45 v_O=vI−0.7 V v_O=vI ± 0.7 V  = PL av = 0.240 =0.45 V_i-peak = 7.63 V P_−V CCav

(23)

ESE319 Introduction to Microelectronics

Multisim Simulation – Max Power

Conversion Efficiency

P_V CCav PL av sim= P_{L av} P_V _avP−V av = 0.6772 0.42310.4239=0.78 V i-peak = 12.7 V P_−V CCav v_O=vI± 0.7 V

(24)

ESE319 Introduction to Microelectronics

Anticipating Class AB – Eliminating

Crossover Distortion

If we bias the bases of the emitter follower pair, we can

set both transistors on the verge of conduction – or have

them conduct a small bias current with zero signal input:

V

_BB

2 ≈0.7 V.

Class AB is a compromise

between Class and Class B

(25)

ESE319 Introduction to Microelectronics

Summary

Class B advantages:

1. Much higher power conversion efficiency than class A for large signal amplitudes.

2. Zero power dissipation with zero input.

Class B disadvantage:

Higher distortion than class A, especially at low input amplitudes.

Class AB operation is a compromise mode:

1. Delivering better linearity than class B, with reduced efficiency.

2. Delivering better efficiency than class A, with reduced linearity.