Bistable Multivibrator design.pdf

Pulse and Digital Circuits

Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

Chapter 9 Bistable Multivibrators

1. Design a fixed-bias bistable multivibrator using Ge transistors having h FE (min) (min) = 50, V CC CC = 10 V and V BB = 10 V, V CE(sat) CE(sat) = 0.1 V, V BE(sat) BE(sat) = 0.3 V, I C(sat) C(sat) = 5 mA and assume I assume I B(sat) B(sat) = 1.5 I B(min) B(min) . Solution:

RC



VCC

 V CE (sat)



10  0.1 V

I C 2

 1.98

R2



k Ω V



5 mA

9.9 V 5 mA

 ( V BB )

I 2

Choose I 2

1



10

I C 2

0.5 mA



 R2 

I B 2min

0.3  10 0 .5 I C 2







h FE min

10.3 V

=20.6 k Ω 0.5 mA 5 mA =0.1 mA 50

If Q 2 is in saturation I B 2



1.5 1. 5 I B 2 m in

= 0.15 mA I 1



I

I

2  B 2

= 0. 0.5 mA mA +0 +0 . 1 5 m A RC



R1



R1 ( RC



VCC

 V 

I 1 





0 . 6 5 mA mA

10



0 .3

0 .6 5 m A



9 .7 V 0 .6 5 m A



14.92 k Ω

R1 )  RC

1 4 . 9 2  1 . 9 8  1 2 . 9 4

kΩ .

2. For a fixed-bias bistable multivibrator shown in Fig. 9p.2 using n–p–n Ge n–p–n Ge transistor V CC V, R C = 1 k Ω, R 1 = 10 k Ω, R 2 = 20 k Ω, h FE(min) = 40, V BB = 10 V. Calculate: CC = 10 V, R (a) Stable-state currents and voltages assuming Q 1 is OFF and Q 2 is ON and in saturation. Verify whether Q 1 is OFF and Q 2 is ON or not. (b) the maximum load current.

© Dorling Kindersley India Pvt. Ltd 2010

1



Fig. 9p.2 The fixed-bias bistable multivibrator Solution: Assume V CE CE (sat) (sat) = 0.1 V, V BE (sat) (sat) = 0.3 V

Calculate V B1 B1 to verify whether Q 1 is OFF or not. R2 R1 0.1  20 (10)10 V B1  VCE (sat)  ( V BB )   R1  R2 R1  R2 10  20 10  20 0.066  3.333  3.267 V Hence Q 1 is OFF VC1  V CC  10 V 

To verify whether Q 2 is in saturation or not: Calculate I Calculate I B2 B2 , I C C2 To calculate I calculate I B2 B2 . Consider the cross-coupling circuit shown in Fig.2.1.

Fig. 2.1 Circuit to calculate the base current of Q 2


2


I 1



I 2



VCC  V 



10  0.3

RC  R1

I B 2

V

 V BB



1  10 

0.3  10

R2 


20

9.7 V 11 kΩ



10.3 20



=0.88 mA 0.515 mA

0.88  0.51

0.37 mA To calculate I C2 Consider the cross-coupling network shown in Fig. 2.2. 

Fig. 2.2 Circuit to calculate the collector current of Q 2


3


I 3 

I 4



VCC


 V CE (sat)

RC 10  0.1



R1

9.9 mA



1 K VCE (sat) 

 V BB

R2

10.1

=0.336 mA 30 K I C 2  I 3  I 4 



9.9  0.34  9.56 mA

I B 2min I B 2





I C 2 h FE min



9.56 mA 40

=0.24 mA

I B 2min

Hence Q 2 is verified to be in saturation. VC 2  0.1 V, V B 2  0.3 V . V C1 = V CC – I 1 R C = 10 – (0.88)1 = 9.12 V Hence the stable-state currents and voltages are as follows: V C1 = 9.12 V, V B1 = –3.267 V V C2 = 0.1 V I B2 = 0.37 mA, I C2 = 9.56 mA To find the maximum load current or minimum load resistance, consider Fig.2.3.

Fig. 2.3 Circuit to calculate maximum load current I L is maximum (I L(max) ) when I B2 is I B2(min)


4



I B2(min) = 0.2 mA I 2 = 0.51 mA I 1(min) = I 2 + I B2(min) =0.51+0.24 =0.75 mA V C1 (min) = I 1(min) R 1 +V  0.75  10  0.3

σ



7.8 V

I 

VCC

 V C 1(min)



12  7.8

RC

1



4.2 mA

I Lmax = I – I 1(min) = 4.2 mA – 0.75 mA =3.45 mA 7.8 V  2.26 kΩ. R L (min) = 3.45 mA 3. Design a self-bias bistable multivibrator shown in Fig.9p.2 with a supply voltage of – 12 V. A p-n-p silicon transistors with h FE (min) = 50, V CE (sat) = –0.3 V, V BE (sat) = –0.7 V and I C 2 = –4 mA are used.

Fig. 9p.2 Self-bias bistable multivibrator Solution:

1 Assume V EN = V CC 3 I C2 = –4 mA



1 3

 12  4


V

5


I B2(min) =

4

mA

50


 0.08

mA

Choose I B2 = 1.5 I B2(min) = –0.12 mA ( I C 2 + I B2 ) = –4 – 0.12 = –6.12 mA R E



RC





V EN 2 I B 2

4.12

V

I C 2



VCC

 VCE (sat)  V EN 2



mA

0.97 k Ω

I C

12  0.3  4

Let V BN 2 R2

4



I 2

4



mA 1

7.7

V

4

mA 1

 1.925

k 

 4 mA  0.4 mA I C 2  10 10  VEN 2  V   4  0.7  4.7 V 

V BN 2

4.7

V

 11.75 k Ω I 2 0.4 mA Choose R 2 = 12 k  Find I 2 for this R 2 V 4.7 V  0.392 mA I 2  BN 2  R2 12.0 K 

RC  R1 





VCC I 2

12 

 V BN 2 

I B 2

4.7

7.3



0.392  0.12

0.512

V mA

 14.26

k Ω

( RC  R1 ) = 14.26 k Ω R 1 = ( RC



R1 )  RC

 14.26  1.925  12.33 kΩ

Choose R 1 =12 k  Note: Choose the nearest standard values. 4. A self-bias bistable multivibrator uses Si transistors having h FE (min) = 50. V CC = 18 V, R 1 = R 2 , I C( sat) = 5 mA. Fix the component values R E , R C , R 1 and R 2. Solution: 1 1 Assume V EN = V CC   18  6 V 3 3 and I C( sat) = 5 mA I B2(min) =

5 mA

 0.1 mA 50 Choose I B2 = 1.5 I B2(min) =0.15 mA


6



( I C 2 + I B2 ) = 5 + 0.15 = 5.15 mA V EN 2 6V   1.16 k Ω R E  I C 2  I B 2 5.15 mA RC 



VCC

 VC E (sat)  V EN 2

I C

18  0.3  6

V BN 2

RC  R1



RC  R1



R1



R1



V CC  V BN 2 I 2  I B 2

2.34  R1 0.15 R12

V CC  V BN 2 VBN 2  I B 2 R2 



R1 (V CC  V BN 2 ) VBN 2  R1 I B 2

R1 (18  6.7)

 4.25 R1  15.67 

 14

( 4.25) 2

11.3 R1



6.7  0.15 R1

4.25  R2

11.7 V

 2.34 k Ω 5 mA  V   6  0.7  6.7 V

5 mA  VEN 2



6.7  0.15 R1 0

 4  0.15  15.67

2  0.15 k Ω .

5. For a Schmitt trigger in Fig. 9p.4 using n–p–n silicon transistors having h FE (min) = 40, the following are the circuit parameters: V CC = 15 V, R S = 0, R C 1 = 4 k Ω, R C 2 = 1 k Ω, R 1 = 3 k Ω, R 2 = 10 k Ω and R E = 6 k Ω. Calculate V 1 and V 2 .


7



Fig. 9p.4 The Schmitt trigger circuit Solution: From the given data, if Q 2 is in the active region, typically, V BE 2 = 0.6 V and let h FE = 40. To calculate V 1 : R E (1+h FE ) = 6(1+40) =246 k Ω R '  R2 / /( RC 1  R1 )  10 k / /(4 k+3 k)  4.11 k

V'

 V CC 

V EN 2



(V '

V EN 2 

R2 ( RC 1



R1

 V BE 2 )



R

'





R1  R2

R t



 8.82

V

h FE )

246



0.769

 R 2 )

 R1  R 2

4 (3 4

 R E (1 

3  10

R C 1 ( R1 R C 1

4  3  10

 8.08 V 4.11  246  V   8.08  0.5  8.58 V 1

To calculate V 2 : R2 10



10

R E (1  h FE )

(8.82  0.6) 

V 1  V EN 2

R t

R2 )

 15 





3

10) 

10



3 . 0 5 k Ω


8


 Rt 

R E"




0.769  3.05  2.35 kΩ

(1 

1 h FE

) RE



41  6

I C 1 V 2



(V '

 V  2 )

 Rt 

( RC 1



" RE

6.15 kΩ

R2

'

 Vt  V  V CC 



40 

R1



R2 )

(8.82  0.5)



2.35  6.15

 15 

10 4  3  10

 8.82

V

0.978 mA

"

 V BE 1  I C 1 RE

V 2 

0.6 V  (0.978 mA)(6.15 k Ω)

0.6 V  6.01 V  6.61 V Hence for the given Schmitt trigger V 1 = 8.58 V V 2 = 6.61 V 

6. The self-bias transistor bistable multivibrator shown in Fig. 9p.3 uses n–p–n Si transistors. Given that V CC = 15 V, V CE(sat) = 0.2 V, V = 0.7 V, R C = 3 k  ,R 1 = 20 k  ,R 2 = 10 k  , R E = 500 Ω. Find: σ

(i) Stable-state currents and voltages and the h FE needed to keep the ON device in saturation. (ii) f (max) , if C 1 = 100 pF. (iii) The maximum value of I CBO that will still ensure one device is OFF and the other is ON. (iv)The maximum temperature up to which the multivibrator can work normally if I CBO at 25°C = 20 µA.

Fig. 9p.3 The given self-bias bistable multivibrator


9



Solution: (i) To calculate I B2 , consider the base circuit of Q 2, Fig. 6.1.

Fig.6.1. Circuit to calculate V thb and R thb of Q 2 . From Fig. 6.1, V thb

 V CC 

R2 RC  R1  R2

=

15  10



150

3  20  10

33

= 4.54

V Rthb

 R2

( RC  R1 ) =

10  (3  20) 3  20  10



230 33

=6.96k 

(ii) To calculate I C2 , consider the collector circuit of Q 2 , Fig. 6.2. .

Fig. 6.2. Circuit to calculate V thc and R thc of Q 2

V thc

 V CC 

Rthc

R1  R2 RC  R1  R2  RC


=

(R1  R2 ) =

15  ( 20  10) 3  20  10

3  30 33



90 33



450 33

=13.6 V

=2.72 k 

10



Now let us draw the base and collector circuits of Q 2 , Fig. 6.3.

Fig. 6.3. Circuit to calculate I B2 and I C2 Writing the KVL equations of the input and output loops 4.54 – 0.7= (6.96+0.5) I B2 + 0.5 I C2 13.6 – 0.2 = 0.5 I B2 + (2.72+0.5) I C2 Eqs. (1) and (2) are simplified as 3.84 = 7.46 I B2 + 0.5 I C2 13.4 = 0.5 I B2 + 3.22 I C2 Solving Eqs. (3) and (4) for I B2 and I C2 we get I B2 = 0.263 mA I C2 = 3.75 mA 3.75 h FE = = 14.25 0.263 The h FE that keeps the ON device in saturation is 14.25.

(1) (2) (3) (4)

V EN 2 = ( I B2 + I C2 ) R E = (0.263+3.75)0.5 = 2 V V CN 2 = V EN 2 + V CE (sat) = 2 + 0.2 = 2.2 V V BN 2 = V EN 2 + V  = 2 + 0.7 = 2.7 V. V BN 1 = V CN 2 

R2

2.2  10

22

=0.733 V R1  R2 20  10 30 V BE 1 =V BN 1 – V EN 2 = 0.733 – 2 = –1.26 V 



Hence Q 1 is OFF  V CN 1 should be V CC . But actually it is less than V CC . I 1 =

V CC  V BN 2 RC  R1

=

15  2.7 3  20



0.534 mA.

V CN 1 = V CC – I 1 R C = 15 – (0.534)(3) =13.4 V. f max



R1



R2



2 R1 R2 C 1

(20  10)103 2  20  103  10  103  100  1012



750 kHz

(iii) V BE 1 was calculated as –1.26 V. This voltage exists at the base of Q 1 to keep Q 1 OFF. Till such time the voltage at B 1 of Q 1 is 0 V, let us assume that Q 1 is OFF, Fig. 6.4.To calculate R B and hence I CBO R B , short V EN (though I E1 = 0, there exists a


11



voltage V EN at the first emitter) and V CE(sat) sources. From Fig. 6.4, it is seen that R B is the parallel combination of R 2 and ( R 1 + R E ).

Fig. 6.4. Circuit to calculate I CBO R B R B



R2 ( R1  RE )



10  20.5



10  20.5

6.72 kΩ

Until I CBo(max) R B = V BE 1, Q 1 will be OFF.  I CBo(max) =

1.26 V



6.72 k Ω

0.187 mA  187  A

(iv) I CBo at 25°C = 20 µA I CBo(max)



187

I CB 0



20

9.35

n

9.35 = 2 n=

log 9.35



log 2

0.3

T

10 T 2

 25

10 T 2

0.97









3.23

n

3.23

25  32.3  57.3C

7. (a) Design a Schmitt trigger shown in Fig. 9p.4 with UTP of 8 V and LTP of 4 V. Si transistors with h FE = 40 and I C = 5 mA are used. The supply voltage is 18 V. The ON transistor is in the active region for which V BE = 0.6 V, V CE = 2.0 V. (b) Calculate R e1 for eliminating hysteresis.


12



Fig. 9p.4 The Schmitt trigger circuit Solution: Till UTP is reached Q 1 is OFF and Q 2 is ON and in active region. Just at V 1 (UTP) Q 1 goes ON and Q 2 goes OFF. Just prior to this, Q 2 is ON and Q 1 is OFF, Fig. 7.1.

Fig. 7.1 Circuit when Q 1 is OFF and Q 2 is ON V 1 = UTP = V BN 2 = 8 V I E = I C2 = 5 mA V EN = V EN2 = V BN2 – V BE2 V EN2 = 8 – 0.6 = 7.4 V. V 7.4 R E  EN 2  I E 5 mA


13



= 1.48 k Ω Choose R E = 1.5 k  If Q 2 is in the active region and V CE = 2 V I C2 R C 2 = V CC – V CE – V EN 2 18  2.0  7.4

 RC 2 



5

8.6 V 5 mA

 1.72

k Ω

Choose R C 2 = 1.75 k  I 2



R2



1

I C 2 10 V BN 2





8V



I 2

I B 2min

0.5 mA

0.5 mA

I C 2



=16 k Ω

5 mA

h FE



40

0.125 mA

I B 2

 1.5  I B 2 min  1.5  0.125 

I B 2



I 2



0.1875 mA+0.5 mA  0.6875 mA

( RC 1  R1 )  R1

0.1875 mA

 14.55

VCC

 V BN 2

( I B 2



18  8



I 2 )



0.6875

10 0.6875

 14.55

kΩ

k Ω  RC 1

At LTP = 4 V, consider the Fig. 7.2.

Fig. 7.2 Circuit at LTP V BN 2 = V BN 1 = 4 V = LTP = V 2 Let I 1 be the current in R 2 I 1



I C1


V BN 2



R2 

I E 1

4V



16 k Ω 

V2

 V BE 1

R E

0.25 mA 

4  0.6 1.5 k Ω

14



I C 1 = 2.27 mA Writing the KVL equation of the outer loop consisting of R C1 , R 2 and R 1 , V CC = ( I C 1 + I 1 ) R C 1 + I 1 ( R 1 + R 2 ) = ( I C1 + I 1 ) R C 1 + I 1 (14.55 k  – R C 1 + R 2 ) V CC = I C 1 R C 1 + I 1 (14.55 k  + R 2 ) V  I (14.55  R2 ) RC 1  CC 1 I C 1 18  0.25(14.55  16)

10.36

 4.56 k Ω 2.27 2.27 R C 1 = 4.56 k  R 1 = ( R C1 – R 1 ) – R C 1 =14.55 – 4.56 = 9.99 k  



Choose R 1 = 10 k  and R C1 = 4.5 k  . The designed Schmitt trigger circuit is shown in Fig. 7.3 with component values.

Fig. 7.3 Designed Schmitt trigger (b) To eliminate hysteresis R e1 is added in series with the emitter of Q 1, Fig. 7.4, such that V 1 – V 2 = V H = ( I C1 + I B1 ) R e1 4V  1.76 k  R e1 = 2.27 mA


15



Fig. 7.4 R e1 connected to eliminate hysteresis 8. (i) Design a Schmitt trigger in Fig.9p.5 with UTP of 8 V and LTP of 4 V. Si transistors with h FE = 40 and I C = 4 mA are used. The supply voltage is 12 V. The ON transistor is in saturation for which V BE = 0.7 V, V CE (sat) = 0.2 V. (ii) Calculate R e1 for eliminating hysteresis. (iii) Find R e2 to eliminate hysteresis.

Fig. 9p.5 The given Schmitt trigger circuit Solution: (i) Till UTP is reached, Q 1 is OFF and Q 2 is ON and in saturation region. Just at V 1 (UTP) Q 1 goes ON and Q 2 goes OFF. Just prior to this, Q 2 is ON and Q 1 is OFF, Fig. 8.1.


16



Fig. 8.1 Circuit when Q 1 is OFF and Q 2 is ON V 1 = UTP = V BN 2 = 8 V I E = I C2 = 4 mA V EN = V EN 2 = V BN 2 – V BE 2 V EN 2 = 8 – 0.7 = 7.3 V. V 7.3  1.825 k Ω R E  EN 2  I E 4 mA If Q 2 is in the saturation region and V CE = 0.2 V I C 2 R C 2 = V CC – V CE – V EN 2 12  0.2  7.3 4.5 V  RC 2    1.125 k Ω 4 4 mA 1 I 2  I C 2  0.4 mA 10 V 8V R2  BN 2  =20 k Ω I 2 0.4 mA I B 2min I B 2 I B 2



I C 2

4 mA

h FE



40

0.1 mA

 1.5  I B 2 min  1.5  0.1 



I 2

R1 )





0.15 mA

0.15 mA  0.4 mA  0.55 mA

VCC

 V BN 2

( RC 1



R1

7.27 k Ω  RC 1





( I B 2



I 2 )



12  8 0.55



4 0.55

 7.27

kΩ

At LTP = 4 V, consider Fig. 8.2.


17



Fig. 8.2 Circuit at LTP V BN 2 = V BN 1 = 4V = LTP = V 2 Let I 1 be the current in R 2 I 1



V BN 2 R2

I C1



RC 1



I E 1

=

4V 20 k Ω V2

=0.2 mA

 V BE 1

4  0.7

 1.8 mA R E 1.825 k Ω Writing the KVL equation of the outer loop consisting of R C1 , R 2 and R 1 , V CC = ( I C1 + I 1 ) R C 1 + I 1 ( R 1 + R 2 ) V CC = ( I C1 + I 1 ) R C 1 + I 1 (7.27 – R C 1 + R 2 ) V CC = I C 1 R C 1 + I 1 (7.27+ R 2 ) V  I 1 (7.27  R2 ) RC 1  CC I C 1 



12  0.2(7.27  20)

10.36

 3.6 k Ω 1.8 2.27 R 1 =( R C1 – R 1 ) – R C 1 R 1 = 7.27 – 3.6 = 3.67 k  (ii) To eliminate hysteresis R e1 is added in series with the emitter of Q 1 , Fig. 8.3, such that V 1 – V 2 = V H = ( I C1 + I B1 ) R e1 4V =2.22 k Ω R e1 = 1.8 mA




18



Fig. 8.3 R e1 connected to eliminate hysteresis (iii) To eliminate hysteresis R e2 is added in series with the emitter of Q 2, Fig. 8.4, such that

Fig. 8.4 R e2 connected to eliminate hysteresis V '

VCC R2



RC 1



R1



12  20



R2

'

3.6  3.67  20

R = R 2 //( R C1 + R 1 ) =

20  7.27



27.27

 8.8

V

5.33 k Ω

We know, V 2



(V '

 V BE 2 ) 

4  (8.8  0.7) 


(1  h FE ) R E R '(1  h FE )( Re 2  R E ) 41  1.825

5.33  (41)( Re 2

 1.825)

 V 

 0.5

19



3.5 

606 80.1  41 Re 2

143.5 R e2 =325.65 R e2 =2.26 k  .


20

Bistable Multivibrator design.pdf

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