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08-BJT AC

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BJT – AC Analysis

DC and Small Signal Components

• In realistic and useful electronic circuit, the input can be decomposed into two separate components:  The DC component, V I  The small signal component, v constant e.g. V I =12V) i (t) • The DC component signal is not a function of time (as a • The small-signal component v i (t) is a function of time. This signal is an AC signal • This signal v i (t) is referred as the small-signal component because its magnitude is generally small for all time

t

DC and Small Signal Components

DC and Small Signal Components

• The purpose of DC analysis is to establish a Q-point (quiescent point) or DC operating point . The goal is to set the Q-point such that it does not go into saturation or cutoff when an ac signal is applied • If the Q-point is in active region , the transistor can operate as an amplifier . • The purpose of AC analysis is to obtain the gain (ratio of output voltage and input voltage) • To obtain – –

Current and Voltage Gain, A Input Impedance, Z i i and V i

Output Impedance, Z o

by adopting simple circuit model or transistor equivalent model.

AC Analysis – AC Equivalent Circuit

■ AC analysis;  Short all capacitors  Open circuit all DC Supplies and ground them

AC Analysis – AC Equivalent Circuit

• Transistor can be redrawn for small-signal AC analysis. • BJT is replaced with an equivalent circuit. • Transistor equivalent model will be introduced  The Hybrid Equivalent π Model

AC Analysis – Problem Solving Techniques 1. Analyze circuit with only dc sources to find Q point (DC analysis). 2. Replace the dc voltage sources with open-circuit and ground it 3. Replace the coupling and bypass capacitors with short-circuit 4. Replace transistor with its equivalent circuit, which is the small signal model or hybrid π model. 5. Analyze the small-signal equivalent circuit

AC Analysis

Equivalent Circuit for Common Emitter

• Note that for the capacitor has been shorted and the symbol used for the current is in small capital to indicate that AC signal is applied.

Hybrid π Model

βi b

g m r

  

I CQ V T

V T I CQ g m r

g m v be

   

i b

g m = transconductance V T = temperature equivalent voltage = 26mV at room temp

Common Emitter with Bypass Capacitor (C 3 ) Original Circuit After shorting the all the capacitors

Common Emitter with Bypass Capacitor (C 3 ) After replacing transistor with Hybrid π model, we get

Z ib = r π Z i = R 1 //R 2 //Z ib Z o = R C

Common Emitter with Bypass Capacitor (C 3 )

βi b

Z A vs A vs

   (

g m R C

( 

r

   )

R C R B because g m

)(

r

  

r

r

 

R B

)

or

Important Parameters (R E Bypass)

Z i Z o

 

r

R C A v A v A vs

v o v i v o

 (  

i b

)

R C

v o v i

v o v S

v i

(   )

R C

 (

i b

)

r

r

 

v o v i v x v S i A vs

v o v S Since Z i

 (   

r

 ;

r

v i

 )

R C Z i Z

i R B

(

v s

)

x Z i Z

i R B A vs

 ( 

r

   )

R C R B

Hybrid π Model (R E Un-bypass) (Example) • Determine V Th , R Th , I B , I C , and V CE • Draw an equivalent AC Circuit • Calculate r  , Z i , Z o , A vs and A i .

Z i Z ib

Hybrid π Model (R E Un-bypass) (Answer)

Z o

r

Z ib

  

g m r

Z i

R

1

A vs

v o v s g m

I C

,

V T

 26

mV V T

 ( 1 

R

2   )

R E Z ib r

   ( 1  

R C

 )

R E Z o

 (

Z i R C Z

i R S

)

Important Parameters (R E Un-bypass)

r

Z ib

  

g m r

Z i

R

1  ( 1 

R

2

Z ib g m

 )

R E

I V T C Z o

,

V T

 26

mV

R C A v

v o v i v o

  

i b R C A v

v o v i v i

A v

r

 

i b Z ib

  ( 1  

R C

 

i b

(

r

 )

R E

 ( 1   )

R E

)

A vs

v o v s

r

v i

   ( 1  

R C

 (

Z i

)

R E Z

i R S

(

Z i

)

v s Z

i R S

)

To be continued

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