Ch. 8 Introduction to Amplifier

Ch. 8

Introduction to Amplifier

Amplification

Process increasing the power (or voltage or current) of the AC signal (input/output)

General Amplifier Model

Fundamental properties of Amplifier:

Gain (A), Input impedance (Z_in), Output impedance (Z_out)

Gain: the ratio of an output value to its corresponding input signal

ac output voltage Voltage gain ~

ac input voltage ac output current Current gain ~

ac input current ac output power Power gain ~

ac input power

out v

in

out i

in

out p

in

A v v A i

i A p

p













Example

1. Find the votlage gain from the above diagram.

250 mV 400 V 625

out v

in

A V

V  

 

2. When the input signal changes to V_in = 240 V, the output voltage is given by 625 240 V 150 mV

out v in

V  A V    

Input (Z

) and Output Impedance (Z

)

When a source is connected to the amplifier, the amplifier plays a role of a load to the source.

The value of the load corresponds to the input impedance (Z_in).

The input voltage is given by

in

in S

S in

v v Z

R Z

 

Input (Z

) and Output Impedance (Z

)

When the amplifier is connected to a load resistance, there is some measurable value of an internal source impedance, which is the output impedance (Z_out).

The load voltage is given by

where

L

L out out v in

out L

v v R v A v

Z R

 



Example

1. Calculate the value of V_in

2. Calculate the value of V_L

1.5 kΩ

2 mV 1.875 mV <

1.6 kΩ

in

in S

S in

S

v v Z

R Z

v

 

 

1.2 k

300 mV 240 mV <

1.5 k

L

L out

out L

out

v v R

Z R

v

 

  



Combined Effects of the Input and Output Circuits

The combination of the input and output circuits can cause a significant reduction in the effective voltage gain.

( )

(15 mV) 980 14.7 mV 1000

(340)(14.7 mV) 5 V 1.2 k

(5 V) 4.14 V

1.45 k 4.14 V

276 ~ 340

15 mV

in

in S

S in

out v in

L

L out

o

L v eff

ut L

S

v

v v Z

R Z

v A

A v Ef

v R

fe v

ctive Voltage Gain

v Z R

v A

  



 

 

 



  



  



v

v

Ideal Voltage Amplifier

( )

1

1

1 1 1

1 1

in

in S

S in

out v in

L

L out

out L

L v eff

S out L S

L v

out L S

in L

v S v

S out

S in ou

L

out

t L

L i

n n

S

i

v v

v v v Z

R Z

v A v v v R

Z R

A R

v Z R v

A R

Z R v

Z R

A v A

R Z

R Z Z R v

Z R

 



 

  



 



   

   

The ideal voltage amplifier requires 1. infinite gain

2. infinite input impedance

3. zero output impedance 0

v

in in S

out L out

A

Z v v

Z v v

 

   

  

Three Types of BJT Amplifiers

1. Common-emitter amplifier

2. Common-collector amplifier

3. Common-base amplifier input: Base

output: Collector

input: Base output: Emitter

input: Emitter output: Collector Out-of-phase V_out Optimized power gain

Decibel (dB)

dB power gain :

A logarithmic unit to express the ratio of output power to input power

( )

10 log 10 log

p dB p

in

A A P

  P

Example

A given amplifier has values of P_in = 100 mW and P_out = 10 mW, 0.1 W, 1 W, 10 W, 100 W.

Calculate the dB power gain of the amplifier.

( )

( )

( )

( )

( )

10 mW

10 log 10 dB

100 mW 0.1 W

10 log 0 dB 0.1 W

10 log 1 W 10 dB 0.1 W

10 log 10 W 20 dB 0.1 W

100 W

10 log 30 dB 0.1 W

p dB

p dB

p dB

p dB

p dB

A A A A A

  

 

 

 

 

Decibel (dB)

How to convert a given dB power gain value to the standard numeric form of A_p

( )

( )

1 10

( )

10 log log 10

10

Ap dB

p dB

p dB p p

A  A  A 

A 

Example

1. A given amplifier has a power gain of A_p(dB) = 3 dB. What is the ratio of output power to input power? Calculate the amplifier output power for P_in = 50 mW.

( ) 3

( )

1 10 10 0.3

log 10 10 10 2

10 50 mW

100 mW

Ap dB

p dB out out

p

in

out p in

A P P

A P

P A P

       

 

2. A given amplifier has values of P_in = 100 mW and A_p(dB) = -3 dB. Calculate the amplifier output power.

( )

( )

1 10 0.3 1

log 10 10

10 2 100 mW

50 mW

Ap dB

p dB out out

p

in

out p in

A P P

A P

P A P

 

     

 

Decibel (dB)

Positive dB means a gain, while negative dB means a loss.

log 2 0.3

In multistage amplifiers, the total gain of the circuit is the sum of the individual amplifier dB gain ~ dB gains are additive

,1 ,2

,2 ,1

,1 ,2

,2 ,1 ,2 ,1 ,2

( ), ( ),1 ( ),2

,1 ,1 ,2 ,1 ,2

20 dB 10 log , 6 dB 10 log ,

10 log 10 log 10 log 10 log 26 dB

out out

in out

in in

out out out out out

p dB total p dB p dB

in in in in in

P P

P P

P P

P P P P P

A A A

P P P P P

  

     

         

     

Milli-decibel (dBm)

Decibel notation for the reference input power of P_in = 1 mW

10 log

1 mW

out dBm

P  P

Example

When the output rating of an amplifier is given as P_dBm = 50 dBm, calculate the output power.

10 5

10 log 50 dBm 1 mW 10 1 mW 10 100 W 1 mW

PdBm

out

dBm out

P  P   P     

dB Voltage Gain

The dB voltage gain is defined as

The standard voltage gain is given by

( ) 20 log 20 log ^out

v dB v

in

A A v

v

 

   

 

( )

( )

1 20

log 10

20

Av dB

v dB v

A  ^ A 

Example

An amplifier has values of V_in = 10 mV and V_out = 1 V. Calculate the dB voltage gain.

( )

20 log 20 log 20 log 1 40 dB 0.01

out

v dB v

in

A A v

v

   

       cf. P = V²/R