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EE

Solved Problems on BJT

Sedra/Smith 5

th

th

ed.

By Turki Almadhi,

EE Dept., KSU,

Riyadh, Saudi Arabia

25/07/

3 mA; 1 1 1.

ln( ) 0.747 V

0.747 0 0.747 V 0.9901 3 2.9703 mA 1 10 2.9703 2 4.0594 V (which verifies active mode)

BE T

E SE S S S V V (^) E E SE BE T SE BE B E E E C E E C B E C

I I I I I

I I e V V I I V V V V V I I I

V

I I I

    

    

                 

   3  2.9703  29.7  10 ^3 mA=29.7 A

1 1

1 1

1 1 1 1 1 26 1

5, that means the pnp transistor is operating in the active mode. Given that 10 A 10 10 0.625 A 1 16 15 10 9.375 A; 16

BE BE T T

EC E B

C B V V V V C S S C S

V

I I

I I

I I e I I e I e

   (^)  

15 15 (^115) 2 2 2

59.4 10 A

59.4 10 A 59.4 10 29.

The power BJT has an emitter-base junction area 29.3 times larger than the small signal BJT.

BE T

EBJ V EBJ (^) V S C

A

A

I I e

 (^)     

 ^   

3

3

Assuming the transistor is in active mode: 0.8 0.8^ ( 3)^ 2.2 1 mA 2.2 2. (^1) 19.61 10 mA 1 51 50 33.78 10 0.980 mA 0 3 2.2 0.980 0.844 V. 0 (0.844) 0.844 0.4 the C

E E

B E

C B B C BC

V I

I I

I I V V V

 





    ^    

                     BJ is reverse -biased the transistor is in active mode as assumed!

(i) Note: the negative value of V indicates that the base current is going (into) the base which is the right direction for an npn BJT. 0 ( 1.5) (^) 0.15 mA 10 (current is in mA because the resistance is

B

I B

in k .) 0.7 1.5 0.7 2.2 V ( 9) 2.2 9 6.8 (^) 0.68 mA 10 10 10 0.68 0.15 0.53 mA 9 0.53 10 3.7 V 1.5 3.7 5.2 V 0.4 V, which means the transistor is operating in the active mode.

E B E E C E B C BC

V V

I V

I I I

V V



  ^  ^   

(ii) 0 0 0.7 V 0.7 V

C C B E B B E C

I I

I I

I V V V

  

(i) R 100 k Assuming the transistor is in active mode: ( 1)

101 5 (0.7^ )^ 4.3^100 100 1 101 2.16 V 2.16 2.16 mA 1 0.7 2.16 0.7 2.86 V 5 1 5 (^100 ) 2.86 V 101 0

B

B E E E E E

E E B E C C E BC B C

I I

V V V V

V I

V V

V I I

V V V

 ^     

   0.4 V the BJT is in active mode as assumed.

(ii) R 10 k Assuming the transistor is in active mode: ( 1)

101 5 (0.7^ )^ 4.3^10 10 1 101 3.91 V 3.91 3.91 mA 1 0.7 3.91 0.7 4.

B

B E E E E E

E E B E

I I

V V V V

V I

V V

 ^     

( )

V

5 1 5 (^100 ) 1.13 V

3.48 V 0.4 V! the BJT is saturated.

Restarting, and considering V 0.2 V :

C C E BC B C

CE sat E C B E C B

E E E

E E E

V I I

V V V

I I I

V V V

V V V

V V V

 ^  

 ^ ^  ^ 

     5.23 2.49 V.

0.2 2.69 V; 0.7 3.19 V.

E E C E B E

V V

V V V V

2.325 V 2.325 1.9375 mA;

(^100 ) 1.918 mA 101 0.7 2.325 0.7 3.025 V 9 2.2 ; 9 2.2 1.918 4.78 V 3.025 4.78 -1.755 V 0.4 V the BJT is in active mode as assumed.

E E

C E B E C C BC B C

m C

V I

I I

V V

V I

V V V

g I

3

1 2

1.918 (^) 73.8 mS; 26 (^100 10) 1.355 k

9.64 1.355 1.18 k 1.19 (^) 0. 1.19 10 ( ) 77.3 77.

T T T B C m in th be in sig in sig o m be C L be o be V o^ o^ be sig be sig

V

r V^ V I I g R R R r R r v R v R R v g v R R v v v A v^ v^ v v v v

  

^ 

    6 ( 77.3) 8.2 V/V

o o L (^) 77.3 o in o 45.6 A/A be i in i L be

v i R i R v v i R i R v

Small signal equivalent circuit