L16.hegde.unlocked quiz

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Lecture 16 ESC 201A : Introduction to Electronics Transistor Circuits, and Transistor Amplifiers rhegde Dept of EE IIT Kanpur

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Page 1: L16.hegde.unlocked quiz

   

Lecture  16  ESC  201A  :  Introduction  to  

Electronics    

Transistor  Circuits,    and  Transistor  Amplifiers  

rhegde

Dept of EE IIT Kanpur

Page 2: L16.hegde.unlocked quiz

Show  this  circuit  is  an  inverter  (I/p  switches  

between  0  and  +5V)  :  HW

+5  V

+  5  V

5  KΩ

100  KΩOutput

+Vcc

Vo

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Example-3

100Fβ =

we assume that transistor is in forward active mode and carry out analysis

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0.7 21.5BBB

B

VI AR

µ−

= =2.15C F BI I mAβ= =

5 1.45CE C CV I R V= − × = −Since VCE < 0.2V, our assumption is incorrect and transistor is actually in saturation mode.

IC IB

+ VBE -

+ VCE -

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In saturation mode: C F BI Iβ≠

The transistor model in saturation is

B

E

CICIB

0.7V 0.2V

5 0.2 1.63CI mAK−

= =

Base current is same as before:

IC IB

4.745.216.1

===A

mAII

B

Cforced µ

β

AIB µ5.21=

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Example-4

100Fβ =

Find IC and VCE

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Example-4

100Fβ =

Find IC and VCE

0.7 0BB B B E EV I R I R− + + + =

IE = IB + ICIE = (β +1)IB

0.7 14.29(1 )BB

BB E

VI AR R

µβ

−= =

+ +

IC = β IB

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1.429C F BI I mAβ= =

( 1) 1.443E F BI I mAβ= + =

0CC C C CE E EV I R V I R− + + + =2.129CEV V=

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16V

3.6kΩ

470kΩ

510Ω

+ VCE -

IC

β=120

Example-5 Find IC and VCE

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16V

3.6kΩ

470kΩ

510Ω

+ VCE -

IC

β=120

IC+IB

IC+IB

IB

mAmAII

mAkk

I

IkIkI

BC

B

BBB

9.10158.0120

0158.05101214706.3121

7.0160510)121(7.04706.3)121(16

=×==

=×++×

−=

=++++−

β

BBBC IIII 121)1( =+=+ β

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16V

3.6kΩ

470kΩ

510Ω

+ VCE -

IC

β=120

IC+IB

IC+IB

IB mAImAI

C

B

9.10158.0

=

=

VVVkI

CE

CEB

13.87.04700158.007.0470

=+×=

=−+

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Forward Active Mode Base Emitter (BE) junction is forward biased and Base Collector (BC) junction is reverse biased

0.7BEV V≅FB

C

II

β=

Current Gain

0; 0; 0B C EI I I≅ ≅ ≅ Transistor acts like an open circuit

Cut off Mode Both the junctions are reverse biased

Saturation Mode Both the junctions are forward biased

0.7BEV V≅ 0.5BCV V≅ 0.2BE BCV V V− ≅

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Let  us  analyze  this  circuit  •  Choose  RC  =  1  kΩ  •  Vi  <  Vγ

o  Transistor  in  cut  off  o  IB=0;  IC=0  o  V0=VCC  

 

V CC 5  V

R C R B

47  k Ω V i

V 0

I B

I C

V BE +

-­‐‑ V CE + -­‐‑

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Vo  vs  Vi    

0

1

2

3

4

5

6

0 0.2 0.4 0.6 0.8

V o(V)

Vi  (V)

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Vo  as  Vi  increases  …  •  RC  =  1  kΩ  •  Vi  >  Vγ  

o  KVL  in  the  BE  loop:   V CC 5  V

R C R B

47  k Ω V i

V 0

I B

I C

V BE +

-­‐‑ V CE + -­‐‑

BEBBi VRIV +=

B

BEiB R

VVI −=

Load line?

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Vi  >  Vγ  

•  KVL  in  the  CE  loop:  

•  As  Vi  increases  …  

CECCCC VRIV +=

V CC 5  V

R C R B

47  k Ω V i

V 0

I B

I C

V BE +

-­‐‑ V CE + -­‐‑

B

BEiB R

VVI −=

CCCCCE RIVVV −==0

BC II β=100=β

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Vo  as  Vi  increases  …  

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                                                                       How  low  can  it  go?  V0 =VCE =VC −VE = −VBC +VBEVBC =VBE −VCEVBE = 0.7 V CC

5  V R C

R B 47  k Ω

V i

V 0

I B

I C

V BE +

-­‐‑ V CE + -­‐‑

VVCE 2.0≅When  

VVBC 5.0=Both  CB  and  BE  juncMons  are  forward  biased   and   the   transistor   enters   into  saturaMon  

Add  and  Subtract   VB

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Display  this  on  the  DSO  

Please Verify

βforced < βF

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Vi  >  Vγ  

•  KVL  in  the  CE  loop:  

 

•  As  Vi  increases  …..  

CECCCC VRIV +=

V CC 5  V

R C R B

47  k Ω V i

V 0

I B

I C

V BE +

-­‐‑ V CE + -­‐‑

B

BEiB R

VVI −=

CCCCCE RIVVV −==0

BC II β=100=β

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What  happens  if  RC  increases?  

Slope  represents  gain  in  active  region

What is the transition point if Rc=15kΩ?

V CC 5  V

R C R B

47  k Ω V i

V 0

I B

I C

V BE +

-­‐‑ V CE + -­‐‑

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Transistor Circuit Analysis

Circuit VO+vo

VIN(dc)

vin

VIN(dc)

VOdc Circuit small signal Circuit

vin

vo

dc analysis Small signal analysis

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Small Signal Model or ac Model

cb

e

+vπ-

gmvπ

T T

BQ CQ

V VrI Iπ β= =

CQm

T

Ig

V=

The small signal model is valid only when

26 300TkTv V mV at Kqπ << = =

This model is valid for both npn and pnp transistors

IB1

IB2

IB4

IB3

r0 not completely flat vπ

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Complete Analysis: dc +ac

1. Dc analysis Capacitor is like an open circuit under dc

vin

v0

V0

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1. Dc analysis 0.7CC

BQB

VIR−

=

CQ BQI Iβ= ×

( )O CC CQ CV dc V I R= − ×

mAIBQ 0215.0=

mAICQ 15.2=

VV 85.20 = Can you plot the load line?

β=100

V0

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DC voltage sources are shorted and dc current sources are open circuited

Analysis is done at frequencies for which impedance due to capacitor is small so that capacitor can be considered as short.

Small signal Analysis

vin

v0

vin

v0 v0

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b

e

+vπ-

gmvπ

cNext, the transistor is replaced by its small signal model

Small signal Analysis

vin

v0

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Small signal Analysis

rπ+vπ-

gmvπb c

e

vin

vout

RB RC200K 1K

vin

v0

v0

vin

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rπ+vπ-

gmvπb c

e

vin

vout

RB RC200K 1K

rπ+vπ-

e

voutbvin RB gmvπ RC

c

inv vπ =

o m Cv g v Rπ= − ×Simplify

om C

in

v g Rv=−

vin

v0

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rπ+vπ-

e

voutbvin RB gmvπ RC

c

om C

in

v g Rv=−

861000086.00 −=×−=−= Cmin

Rgvv

SmVmA

VI

gT

CQm 086.0

2515.2

===

vin

v0

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Common Emitter (CE) Amplifier

R1

R2 CE

VCC

RC

RE

CB

VS

VSrπ

+vπ−

RB

b

e

c

gmvπ

RC

vo

Emitter is the common terminal between input and output ports !

vin

vin

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Common Emitter (CE) Amplifier

R1

R2 CE

VCC

RC

RE

CB

VS

56 kΩ

8.2 kΩ 1.5 kΩ

6.8 kΩ

22 V

βF=90

vin

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dc  Analysis  

RC

VCC

RERB

VB

RC

RE

VCCVCC

R1

R2

Ω=+

×== k

kkkkRRRB 15.72.8562.856

21

2.  Apply  Thevenin’s    theorem  

VRR

RVV CCB 81.22.642.822

21

2 ==+

×=

1.5 kΩ

6.8 kΩ

7.15 kΩ

2.81 V

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RC

VCC

RERB

VB 1.5 kΩ

6.8 kΩ

7.15 kΩ

2.81 V

kIkI EQBQ 5.17.015.781.2 ++=

)11(;F

CQEQF

CQBQ II

II

ββ+==

mAI

kIkI

CQ

CQCQ

32.1

5.19011

9015.711.2

=

⎟⎠

⎞⎜⎝

⎛ ++=

SmVmA

VI

gT

CQm 0528.0

2532.1

===

Page 35: L16.hegde.unlocked quiz

CE

RC

RE

CB

VSRB

VSRB

RC

VSRB rπ

gmvπ+vπ−

RC

vo

35968000528.00 −=×−=−== Cmin

v RgvvA

6.8 kΩ

vin

vin

vin