Power Amplifier Characteristics

Chapter 13 Output Stages and Power Amplifiers

¾ 13.1 General Considerations

¾ 13.2 Emitter Follower as Power Amplifier

¾ 13.3 Push-Pull Stage

¾ 13.4 Improved Push-Pull Stage

¾ 13.5 Large-Signal Considerations

¾ 13.6 Short Circuit Protection

¾ 13.7 Heat Dissipation

¾ 13.8 Efficiency

¾ 13.9 Power Amplifier Classes

Why Power Amplifiers?

¾ Drive a load with high power.

¾ Cellular phones need 1W of power at the antenna.

¾ Audio systems deliver tens to hundreds of watts of power.

¾ Ordinary Voltage/Current amplifiers are not suitable for such applications

Chapter Outline

Power Amplifier Characteristics

¾ Experiences low load resistance.

¾ Delivers large current levels.

¾ Requires large voltage swings.

¾ Draws a large amount of power from supply.

¾ Dissipates a large amount of power, therefore gets “hot”.

Power Amplifier Performance Metrics

¾ Distortion - Nonlinearity

¾ Power Efficiency

¾ Voltage Rating – Transistor Breakdown voltage

Power Delivered to Load Power Drawn from Supply η =

Emitter Follower Large-Signal Behavior I

¾ As V_in increases V_out also follows and Q₁ provides more current.

(

)

For 8 ,

1/ 0.8 , thus, 32.5 mA

v L L m

L

m C

A R R g

R

g I

= +

= Ω

= Ω =

¾ If V_in rises from 0.8 V to 4.8V, V_out=4.0V, I_L=500mA

C m

T

g I

= V

∵

≈ 0 V 0.8 V

≈

Emitter Follower Large-Signal Behavior II

¾ For V_in=0.8V, V_out=0V, I_L=0mA, I_E1=32.5mA

¾ For V_in≈0.7V, V_out≈-0.1V, I_L=12.5mA, I_C1≈I_E1=20mA

¾ When all current (32.5mA) flows to the load, V_out=-0.26V

¾ However, as V_in decreases, V_out also decreases, shutting off Q₁ and resulting in a constant V_out.

=32.5mA =32.5mA

Example 13.1: Emitter Follower

For 0.5

211 mV by a few iterations

in out

V V

V

=

⇒ ≈ −

1 1 1

1 1

1

,

ln( / ) ln 1

out

in BE out C

L

BE T C S

out

in T out

L S

V V V V I I

R

V V I I

V V V I V

R I

− = + =

=

⎡ ⎛ ⎞ ⎤

− ⎢ ⎜ + ⎟ ⎥ =

⎝ ⎠

⎣ ⎦

¾ V_in=0.5V, V_out=?

By KVL

By KCL

Example 13.1: Emitter Follower

( )

1 1 1

ln

in T C L

S

V V I I I R

= I + −

¾ For what V_in, Q₁ carries only 1% of I₁?

1 1

For 0.01 390 mV

C in

I I

V

≈ ⋅

⇒ ≈

(

)

Linearity of an Emitter Follower

¾ As V_in decreases the output waveform will be clipped, introducing nonlinearity in I/O characteristics.

Turn-off

Push-Pull Stage

¾ As V_in increases, Q₁ is on and pushes current into R_L.

¾ As V_in decreases, Q₂ is on and pulls current out of R_L.

I₁

×

Example 13.2: I/O Characteristics for Large V

¾ For positive V_in, Q₁ shifts the output down and for negative V_in, Q₂ shifts the output up.

V

= V

-V

for large +V

V

= V

+|V

| for large -V

V_BE1 V_BE2

Overall I/O Characteristics of Push-Pull Stage

¾ However, for small V_in, there is a “dead zone” (both Q₁ and Q₂ are off) in the I/O characteristic, resulting in gross nonlinearity.

At least V_BE ≈ 600 ~ 700 mV before BJT turns on

Example 13.3: Small-Signal Gain of Push-Pull Stage

¾ The push-pull stage exhibits a gain that tends to unity when either Q₁ or Q₂ is on.

¾ When V_in is very small, the gain drops to zero.

Gain= ^out

in

V V

∂

∂

Example 13.4: Sinusoidal Response of Push-Pull Stage

¾ For large V_in, the output follows the input with a fixed DC offset, however as V_in becomes small the output drops to zero and

causes “Crossover Distortion.”

For V_in well above 600 mV, either Q₁ or Q₂ serves as an emitter follower.

Within the dead zone, V_out ≈ 0 V

Improved Push-Pull Stage

¾ With a battery of V_B inserted between the bases of Q₁ and Q₂, the dead zone is eliminated.

V

= V

+ |V

|

If Q₁ is to remain on, then V₁= V_out+V_BE1 If Q₂ is to remain on, then V₂= V_out-|V_BE2|

B 1 2

BE1 BE 2

V V V

V V

= −

= +

out B BE 1 in

V =V −V +V

B BE1 BE 2

V =V + V

∵

out BE 2 in

V V V

⇒ = +

B BE 1 BE 2

V V V 0

− + − =

Implementation of V

¾ Since V_B=V_BE1+|V_BE2|, a natural choice would be two diodes in series.

¾ I₁ in figure (b) is used to bias the diodes and Q₁. The diodes carry no

current, exhibiting a zero voltage drop

solution

Example 13.6: Current Flow I

I

1 1 2

in B B

I = I − I + I

1 2

Usually unless 0 flows even when = 0.

B B out

in out

I I V

I V

≠ =

1 2 1 2

if , β = β I

= I

When V_out =0 V, I_L

through R_L is 0. Then I_C1=|I_C2|

Example 13.8: Current Flow II

1 2

0 when 0 if

in out

I = V = I = I

1

→

D BE

V ≈ V V

≈ V

1 in D1

V =V +V

out 1 BE1 in D1 BE1

V = −V V =V +V −V

B1 B 2

I ≈ I

Addition of CE Stage

¾ A CE stage (predriver) is added to provide voltage gain from the input to the bases of Q and Q .

Used extensively in high-power output stages

Bias Point Analysis

¾ For bias point analysis for V_out=0, the circuit can be simplified to the one on the right, which resembles a current mirror.

V_A=0 V_out=0

, 1

1 3

, 1 S Q

C C

S D

I I I

= I ⋅

if

A D1 BE1

A BE1 D1

A BE1 D1

V V V 0

V V V

V 0 V V

− − + =

= −

= =

Current mirror

D 1 T

V / V C 3 S ,D1

I = I e

C 3

D1 T

S ,D1

V V ln I

= I

A

Emitter size difference between Q₁ and D₁

Summary

¾ Push-Pull stage could be a possible solution to solve the nonlinearity in emitter follower, but there is still nonlinearity due to dead zone near small V_in

¾ Need to understand large signal behavior of emitter follower.

¾ As V_in decreases the output waveform will be clipped, introducing nonlinearity in I/O characteristics.

¾ Two diodes in series are used to implement the battery of V_B.

¾ With a battery of V_B inserted between the bases of Q₁ and Q₂, the dead zone is eliminated.

¾ A CE stage (predriver) is added to provide voltage gain from the input to the bases of Q₁ and Q₂.

Small-Signal Analysis

1 2

N out

v −v = v_π = v_π

(

)( )

1 2

out N out

m m N out

L

v v v

g g v v

R r

r

= − + + −

1 2

1

out L

N L

m m

v R

v R

g g

=

+ +

1 2 1 2

if 1 (+ g_m + g_m )(r_π r_π ) >>1 Assuming

2r_D is small

Emitter follower having a transconductance of g_m1+g_m2

out N out

V

in in N

v v v

A = v = v v

Small-Signal Analysis

( )( )

4 1 2 1 2

v m m m L

A = − g r

r

g + g R

I_X

+ -

V_X

(

) 0

X X out

V I r

r

v

− + + =

(

)( )

out

X m m X out L

v

I g g V v R

=

⎡ + + − ⎤

⎣ ⎦

(

)(

)

X

/

N m m L

V I

R g g r

r

R r

r

=

= + +

( )( )

4 1 2 1 2 1 2

1 2

1

out L

m m m L

in L

m m

v R

g g g r r R r r

v R

g g

π π π π

⎡ ⎤

= − ⎣ + + ⎦

+ +

(

)

if 1 R

g

+ g

>>

Example 13.9: Output Resistance Analysis

3 4

1 2 1 2 1 2

||

1

( )( || )

O O

out

m m m m

r r

R ≈ g g + g g r

r

+ +

¾ If β is low, the second term of the output resistance will rise, which will be problematic when driving a small resistance.

Assuming 2r_D is small

Example 13.9: Output Resistance Analysis

I_X

1 2

v

= v

3 4

o o

r r

+ -

1 2

1 2 3 4

X

o o

r r

v V

r r r r

π π

π

π π

= − +

(

)

1 2 3 4 1 2 3 4

X

X m m X

o o o o

r r

I V g g V

r r r r r r r r

π π

π π π π

= + +

+ +

(

)(

)

1

o o

X

X m m

r r r r

V

I g g r r

π π

π π

= +

+ +

(

)(

)

if 1 g + g r

r

>> 1

||

( )( || )

O O

out

r r

R ≈ +

+ +

Example13.10: Biasing

1 2

Predriver (CE stage): 5

OutputStage: 0.8 for 8 2 100,

V

V L

npn pnp C C

A

A R

I I

β β

=

= = Ω

= = ≈

¾ Compute the required bias current.

( ) ( )

( ) ( )

( )

1 1

1 2 1 2

1 2

1 2

4 1 2 1 2

1

4 3 4

2 4 Thus, 6.5 mA

|| 400 || 200 133

|| 1 4

133 195 μA

m m m m

C C

m m m L

m C C

g g g g

I I

r r

g r r g g R

g I I

π π

π π

− −

−

+ = Ω => ≈ ≈ Ω

≈ ≈

= Ω Ω = Ω

⎡ ⎤

⋅ ⋅ + ⎣ + ⋅ ⎦ =

=> = Ω => ≈ ≈

m

r^π g

=

β

∵

1 2

0.8 8 8 1

m m

g g

=

+ +

C m

T

g I

= V

∵

Problem of Base Current

¾ 195 µA of base current in Q₁ can only support 19.5 mA of

collector current, insufficient for high current operation (500 mA for 4 V on 8 Ω).

Modification of the PNP Emitter Follower

¾ Instead of having a single PNP as the emitter-follower, it is now combined with an NPN (Q₂), providing a lower output

resistance.

(

)

1

out

1

m

R ^≈ β + g

Gain

(

)

3

1 1 1

out L

in L

m

v R

v R

g r

β

=

+

+ +

( )

2 2 2 2 3

2 2 2 3

C B C

m m in out

I I I

g v_π g v v

β β

β

= = −

= − −

I_C3 I_B2

I_C2

( ) ( )

3 2 3

3

out in out

m in out m in out

L

v v v

g v v g v v

r

R

β ⁻

− − + − − = −

By KCL

v

I_X +

-

v

( )

2 2 2 3

m m in out

g v_π = −

β

( ) (

)

2 3

3

Comparing with the standard EF,

1 1

1 1

out

1

m m

r g g

r

β β

= ≈

+ + +

Output Resistance

By KCL

3 2 3

3

0

X

X m X m X

I V g V g V

r

β

− + + + =

Example 13.11: Input Resistance

( )

3

2 3

3 2 3

1

1 1 ( 1)

L

in in in

L

m

in L

i v v R

r R

g

r R r

π

π

β β β

⎛ ⎞

⎜ ⎟

⎜ ⎟

= −

⎜ + ⎟

⎜ + ⎟

⎝ ⎠

= + +

i

3 in out

v v

r

= −

Additional Bias Current

¾ I₁ is added to the base of Q₂ to provide an additional bias current to Q₃ so the capacitance at the base of Q₂ can be

charged/discharged quickly. Additional pole at the base of Q₂. carries a small current

2 in

0

out EB

Min V

V V

≈

≈

Example 13.12: Minimum V

2

3 2

in BE

out EB BE

Min V V

V V V

≈

≈ +

Requirement: no saturation

HiFi Design

¾ As V_out becomes more positive, g_m rises and A_v comes closer to unity, resulting in nonlinearity.

¾ Using negative feedback, linearity is improved, providing higher fidelity.

C m

T

g I

= V

∵

1 2

1

out L

N L

m m

v R

v R

g g

=

+ +

∵

1 2

1

out in

V R

V ≈ + R

∵

Short-Circuit Protection

¾ Q_s and r are used to “steal” some base current away from Q₁ when the output is accidentally shorted to ground, preventing short-circuit damage.

r: small resistance

Q_s reduces the base drive of Q₁

if the current exceeds a certain level

0.7 V V

→

Disadvantages:

- r raises the output impedance

- V_r under normal operating condition reduces the maximum output swing

Emitter Follower Power Rating

( )

0

0 1

2

1 1 1

1

sin

1 sin

2 if

T

av C CE

T P

CC P

L

P P P

CC CC

L L

P I V dt

T

V t

I V V t dt

T R

V V V

I V I V I

R R

ω ω

= ⋅

⎛ ⎞

= ⎜ + ⎟⋅ −

⎝ ⎠

⎛ ⎞

= ⋅ − = ⎜⎝ − ⎟⎠ =

∫

∫

¾ Maximum power dissipated across Q₁ occurs in the absence of a signal, i.e. with V_p=0: P_av,max=I₁V_CC

( )

sin

ω t = − 1 cos 2 ω t / 2

Example 13.13: Power Dissipation

( )

1 0 1

1

1

sin

I p EE

EE

P I V t V dt

T

I V

ω

= −

= − ⋅

∫

¾ Avg Power Dissipated in the Current Source I₁

Push-Pull Stage Power Rating

( )

/2

, 0

/2 0

2

1 1 sin

sin

4 4

T

av NPN C CE

T P

CC P

L

CC P P P CC P

L L L

P I V dt

T

V t

V V t dt

T R

V V V V V V

R R R

ω ω

π π

= ⋅

= − ⋅

⋅ ⎛ ⎞

= − = ⎜ − ⎟

⎝ ⎠

∫

∫

¾ No power for half of the period.

( )

sin

ω t = − 1 cos 2 ω t / 2

/2

0^T

cos 2 ω tdt = 0

∫

For V_p≈0 or V_p≈4V_CC/π, the power dissipated in Q₁ approaches zero

2 2

2 2

cos 2 sin cos cos 1 sin

t t t

t t

ω ω ω

ω ω

= −

= −

Push-Pull Stage Power Rating

( )

/2

, 0

/2 0

2

1

sin

1 sin

4 4

T

av NPN C CE

T P

CC P

L

CC P P P CC P

L L L

P I V dt

T

V t

V V t dt

T R

V V V V V V

R R R

ω ω

π π

= ⋅

= − ⋅

⋅ ⎛ ⎞

= − = ⎜ ⎝ − ⎟ ⎠

∫

∫

¾ No power for half of the period.

¾ Maximum power occurs between V_p=0 and 4V_cc/π.

2

,max ^CC2 ^P

when 2

av P

L

V V V

P V

R π

π

= ⋅ = ⋅

Push-Pull Stage Power Rating

¾ Maximum power occurs between V_p=0 and 4V_CC/π.

, ,

when

av PNP av NPN CC EE

P = P V = − V

2

,max ^CC2 ^P

when 2

av P

L

V V V

P V

R π

π

= ⋅ = ⋅

( )

, /2

/2

2

1

sin

1 sin

4 4

T

av PNP T C CE

T P

P EE

T L

EE P P P EE P

L L L

P I V dt

T

V t

V t V dt

T R

V V V V V V

R R R

ω ω

π π

= ⋅

⎛ ⎞

= − ⋅ − ⎜ ⎟

⎝ ⎠

− ⋅ ⎛ − ⎞

= − = ⎜ ⎝ − ⎟ ⎠

∫

∫

Heat Sink

¾ Heat sink, provides large surface area to dissipate heat from the chip.

Thermal Runaway Mitigation

1 2

1 2

, 1 , 2

1 2 , 1 , 2

1 2

1 2

, 1 , 2

1 2 , 1 , 2

1 2 1 2

, 1 , 2 , 1 , 2

ln ln

ln

ln ln

ln

With the same ,

D D

D D T T

S D S D

D D T

S D S D

C C

BE BE T T

S Q S Q

C C T

S Q S Q T C C D D

S D S D S Q S Q

I I

V V V V

I I

I I

V I I

I I

V V V V

I I

V I I

I I V I I I I

I I I I

+ = +

=

+ = +

=

=

¾ Using diode biasing prevents thermal runaway since the

currents in Q₁ and Q₂ will track those of D₁ and D₂ as long as their I_s’s track with temperature.

/

, ^{D T}

V V

D S D

I ≈ I e

/

, ^{BE T}

V V

C S Q

I ≈ I e

Efficiency

Power Delivered to Load

Power Drawn From Supply Voltage

out out ckt

P

P P

η = =

+

¾ Efficiency is defined as the average power delivered to the load divided by the power drawn from the supply

Emitter Follower

( )

2

2 1 1

1

2

2 2

if and 4

P L

EF

P L CC P EE

P P

EE CC

CC L

V R

V R I V V I V

V V

I V V

V R

η =

+ − − ⋅

= = − =

¾ Maximum efficiency for EF is 25% if V_P approaches V_CC. Load

Tr.

Current source

Example 13.15: Efficiency of EF

( )

( )

1

2

2 1 1

2 2

/2

With = and 2

2 2

2

2 2

Thus,

1 15

P CC

P CC

EE CC

L L

P L

EF

P L CC P EE

P L

CC CC

P L CC P CC

L L

EF V V

V

I V V V

R R

V R

V R I V V I V

V R

V V

V R V V V

R R

η

η

= − =

= + − − ⋅

=

+ − + ⋅

=

¾ EF designed for full swing operates with half swing.

Efficiency

Push-Pull Stage

2

2

2 2

2 4

4

P L PP

CC

P P P

L L

P CC

V R

V V V V

R R

V V η

π π

= + ⎛ ⎜ ⎝ − ⎞ ⎟ ⎠

=

¾ Maximum efficiency for PP is 78.5% for V_P≈V_CC.

Example 13.16: Efficiency incl. Predriver

( )

1

2

2

2 ( )

1 4

1 1 1 1 4

P L

P L

P CC P

CC EE

L L

P

CC CC

P CC

I V R

V R

V V V

V V

R R

V

V V

V V

β

η

π β

π β

π β

=

=

+ −

= +

=

+

Power Amplifier Classes

Class A: High linearity, low efficiency

Class B: High efficiency, low linearity

Class AB: Compromise between

Class A and B

Summary

¾ Using negative feedback, linearity is improved, providing higher fidelity

¾ Small signal analysis push-pull stage: gain, R_in, and R_out.

¾ Composite transistor: lower R_out and higher R_in

¾ Efficiency is defined as the average power delivered to the load divided by the power drawn from the supply:

Emitter Follower (25%), Push-Pull (78.5%)

¾ Power rating of devices.

¾ Amplifier classes: Class A, Class B, and Class AB