555 TIMER

Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

Chapter-8 Monostable Multivibrators

1. A monostable multivibrator is used as a voltage-to-time converter. Find the time

period if R = 10 k ,C = 0.01 F, 5.0CC

BB

VV .

Solution:

Time period T is given by T = RC ln

BB

CC

VV

1

T =

5.0

11ln1001.01010 63 = 0.11 ms

3

1 1 9.09 kHz.0.11 10

fT

2. Design a collector-coupled monostable multivibrator using an n-p-n silicon transistor with hFE(min) = 40, VBE (cut off) 0 V and IB(sat) = 1.5IB(min). Given that: VCC = 10 V, IC(sat) = 5 mA, RC1 = RC2 = RC, VCE(sat) = 0.2 V and VBE(sat) = 0.7 V. If the pulse width required is 1 ms, calculate the value of C.

Fig.8p.2 The given circuit of the monostable multivibrator

Solution: (sat)

(sat)

10 0.2 9.8 V 1.96 k5 mA 5 mA

CC CEC

C

V VR

I

Dorling Kindersley India Pvt. Ltd 2010 1


2

(sat)2min

min

2 2min

3

5 mA 0.125 mA40

1.5 1.5 0.125 0.187 mA10 0.7 49.7 k

0.187 10

CC

B

CB

FE

B B

V VR

II

Ih

I I

R

T = 0.69RC 3 31 10 0.69 49.70 10

1 29.1 nF0.69 49.7

C

C

For the value of VBB to be fixed, consider Fig.2.1. If VBE(cut-off) = 0 V

Fig.2.1 Circuit to calculate VBB

(sat)2

1 2

1 2

2

12

2

22

( )

0.22

also( )

0

0.22

0.2 20.2

CE BB

BB

B BB

BB

BB BB

BB BB

BB

V VI

R RIf R R R

VIR

V VIR

VIRV V

R RV V

V

To find R1 = R2 = R



2 (sat)

(sat)1 2

2

1 5 mA 0.5 mA10 10

0.2 0.2 800 0.5 mA

C

CE BB

I I

V VR R

I

R1=R2=R = 400

3. A collector-coupled monostable multivibrator shown in Fig.8p.3 using Ge n-p-n transistors has the following parameters: VCC = 9 V, VBB = 9 V, RC = 2 k, R1 = 10 k, R2 = 20 k, R = 10 k, hFE = 4 'bbr = 0.2 k, C = 0.001 (sat) = 0.1 V and V0, F, CEV = 0.3

waveforms. (b) Find the pulse width.

V. (a) Calculate and plot the

Fig.8p.3 Circuit of the monostable

In the stable state: assume Q is OFF and Q is ON and in saturation, Fig.3.1. Solution:

1 2

Fig.3.1 Equivalent circuit in the stable state

N and in saturation or not, calculate IC2 and IB2 and verify whether IB2>>IB2min or not.

To verify whether Q2 is O



(sat)2

2

22min

min

2 2min

9 0.1 8.9 V 4.45 mA2 K 2 K

9 0.3 8.7 V= 0.87 mA10 K 10 K4.45 mA 0.11 mA

40

CC CEC

C

CCB

CB

FE

B B

V VI

RV V

IRI

Ih

I I

Hence Q2 is in saturation, as assumed. VC2 = 0.1 V, VB2 = 0.3 V. The voltage at the base of Q1 is

2 11 (sat)

1 2 1 2

( )

20 100.1 ( 9)10 20 10 20

0.066 3 2.934 V

B CE BBR R

V V VR R R

R

Hence Q1 is OFF. VC1=VCC=9 V VA = the voltage across the capacitor = VC1 VB2 = 9 0.3 = 8.7 V In the stable state Q1 is OFF, Q2 is ON. VC1 = 9 V, VC2 = 0.1 V VB1 = 2.934 V, VB2 = 0.3 V VA = 8.7 V In the quasi-stable state (on the application of a trigger): Q1 is ON and in saturation Q2 is OFF, Fig.3.2.

Fig.3.2 Circuit in the quasi-stable state

Is Q1 is really ON and in saturation? To verify this, calculate IC1, IB1 and verify whether IB1>>IB1min or not. If this is satisfied, Q1 is in saturation. IC1 = I1 +IR



(sat)1

1

1

1 1

1

11min

min

9 0.12 K

8.9 V2 K

4.45 mA

10 8.9 8.717.6 V

1.76 mA

4.45 1.766.21 mA

CC CE

C

R C

R

R

C R

C

CB

FE

V VI

R

I

AI R I R VI

II I I

I

II

h

6.21 mA=40

=0.155 mA

Now calculate IB1, using Fig.3.3.

Fig.3.3 Circuit to calculate IB1 at t = 0+



21

32

1 2 3

1

1 1min

9 0.32 108.7 V12 K0.725 mA

0.3 920 K

9.3 V=20 K0.465 mA

0.725 0.4650.26 mA

CC

C

BB

B

B

B B

V VI

R R

V VI

R

I I I

II I

Hence Q1, in the quasi-stable state, is in saturation. 1 1

2 2

2

2

0.1 V, 0.3 V

9 (0.725)(2)7.55V

9 (1.76)(10)9 17.6

8.6 V

C B

C CC C

B CC R

B

V VV V I R

V V I R

V

Alternately 2 1

(sat)

0.3 9 0.18.6 V

B C

CC CE

V V I RV V V

In the quasi-stable state, only VB2 changes as a function of time. At the beginning of the quasi-stable state VC1 = 0.1 V, VB1 = 0.3 V VC2 = 7.55 V, VB2 = 8.6 V and decays exponentially. At the end of the quasi-stable state:



Q1 goes into the OFF state and Q2 goes into the ON state, Fig.3.4.

Fig.3.4 Equivalent circuit at t = T+

(sat)'2

'

9 0.1 0.3 0.10.2 2

CC CEB

bb C

V V V VI

r R

'2

8.7 V=2.2 K

3.95 mABI

' '1 2

' '2 2 '

'2

9 (3.95)(2)1.1 V

(3.95)(0.2) 0.31.09 V

C CC B

B B bb

B

V V I R

V I r V

V

C

At the end of the quasi-stable state '1CV =1.1 V and jumps to VCC

VB1= 2.934 V '2 1.09 VBV

VC2 = 0.1 V The waveforms are now plotted in Fig.3.5.



Fig. 3.5 Waveforms

4. For a collector-coupled monostable multivibrator circuit shown in Fig. 8.1, R1 = R2 = R = 10 k , C = 0.01 F, R C = 1 k ,VCC = 10 V, hFE = 20. In the quasi-stable state, Q1 is in the active region with collector current of 2 mA. Find the time period and the value oV

f BB. Neglect junction voltages. IB(sat) = 1.5IB(min).

Solution:

The circuit, when Q1 is ON in the quasi-stable state, is given in Fig. 8p.4.

Fig. 4 Q1 ON in the Quasi-stable state.



I2= 1 1

10 0.909 mA.1 10

CC CC

C C

V V VR R R R

I3 =2 2

V mA.

10BB BB BBV V V

R R

The collector current of Q1, IC1= 2 mA. 1

1(min)2 0.1 mA.

20C

BFE

II

h From Fig 8p.4, IB1=I2 I3

0.1 mA =V

0.909 mA10

BB

809.01.0909.010

BBV VBB = 8.09 V.

3 31 1

3 6

10 2 10 1 10ln ln ln10

10 10 0.01 10 0.182 18.2 s.

CC C C CC C C

CC CC

V I R V V I RT

V V V

T

5. An emitter-coupled monostable multivibrator in Fig.8p.2 has the following parameters: VCC = 6 V, RC1 = RC2 = RE = 3 k, R = 50 k, V = 2.8 V and C = 0.01 F. npn silicon transistors with hFE = 50 and rbb' =100 are used. A trigger is applied at t = 0.

(a) Assume that Q1 is OFF and Q2 is ON at t = 0. Calculate the node voltages. Using your calculated values verify that Q1 is indeed OFF and Q2 is in saturation.

(b) Assume that Q1 is in the active region and Q2 is OFF at t = 0+. Calculate the node voltage and verify that Q1 is indeed in the active region and Q2 is OFF.

(c) Calculate the node voltages at t = T. (d) Calculate the node voltages at t = T+.

Fig.8p.2 Emitter-coupled monostable multivibrator



Solution: (a) At t = 0 (i.e. in the stable state) Q1 OFF, Q2 ON, saturation. The base and collector loops of Q2 are indicated in Fig. 5.1.

Fig. 5.1 Circuit to calculate the currents and voltages in the stable states Writing the KVL equations i.e. 5.3 V = 53IB2+3IC2 i.e. 5.8 V= 3IB2+6IC2 IB2 = 0.05 mA, IC2 = 1 mA

2

2 min(min)

2 2(min)

2

2 2 2

1 1 2

1 mA 0.02 mA50

Hence is in saturation( )

(0.05 1)31.05 3 3.15 V

2.8 3.150.35 V

CB

FE

B B

EN EN B C E

BE BN EN

II

hI I

QV V I I R

V V V

The base of Q1 is negative with respect to emitter by 0.35 V. As such Q1 is in the OFF state.

1

2

2 2 (sat)

2 2

6 V3.15 V

3.15 0.2 3.35 V

3.15 0.7 3.85 V

CN CC

EN EN

CN EN CE

BN EN

V VV VV V VV V V

Voltage across the capacitor terminals VA(t = 0) = VCN1 VBN2 = 6 V 3.85 V = 2.15 V. (b) Calculations at t = 0+ (i.e. at the beginning of the quasi-stable state)



At t = 0, a trigger is applied to drive Q1 into the active region and Q2 into the OFF state. The corresponding equivalent circuit is shown in Fig. 5.2.

Fig. 5.2 Equivalent circuit in the quasi-stable state

VEN = VEN1 = V VBE1 = 2.8 0.6 = 2.2 V

1 1

11 1 1 1

1

1

1

1 1

1

1

1

1

1

2.2 V 0.73 mA3 K

1(1 ) (1 )

500.73 mA ( )51

0.7 mA

0.7 mA50 3

3 2.15 V0.7 mA

3 50 2.15 V

B C

BC B C C

C F

C

C

C R

R

R A

R

R

I I

II I I IEI h

I

II I II I

I I VI

I II I

(1)

(2)

Solving for I1 and IR using Eqs. (1) and (2)



1

1 1 1

2

1

2

0.08 mA, 0.65 mA

6 (0.65)3 6 1.95 4.05 V6 V

2.2 V

6 (0.08)(50)2 V

R

CN CC C

CN

EN EN

BN CC R

I IV V I R

VV VV V I R

(c) Just prior to the completion of the quasi-stable state (t = T-) VBN2(T-) = VEN1 +

2 V

= 2.2 + 0.5 = 2.7 V 2 ( )( )

6 2.7 V 3.3 V=50 K 50 K

3.3 0.066 mA50 K

CC BNR

V V TI T

R

1 1

1 1 1

0.7 0.0660.634 mA

6 (0.634) 52.83 V

C R

CN CC C

I I I

V V I R

All other voltages except VBN2 remain unaltered. Voltage across the capacitor at t = T VA(T) = VCN1 VBN2 = 2.83 2.7 =0.13 V Is Q1 in the active region or not? Calculate VCB1 VCB1 = VCN1 VBN1 =4.05 2.8 =1.25 V Hence Q1 is in the active region. At the end of the quasi-stable state i.e. at t = T+ Q1 once again goes into the OFF state and Q2 goes into ON state and into saturation. As a result, there could be an overshoot at the base of Q2 and an overshoot at the collector of Q1. These overshoots are accounted for by taking the base spreading resistance into account. The resultant equivalent circuit is shown in Fig.5.3.




Fig.5.3 Equivalent circuit at t = T+

Writing the KVL equations of the two loops ' '

2 2' '

(sat) 2 2

(6.1 3 )

3 6CC A B C

CC CE B C

V V V I IV V I I

That is, ' '

2 2' '

2 2' '

2 2' ' '

2 2 2

' '2 2 (sat)

' ' '2 2 2 '

3.94 V 6.1 3

and 5.8V 3 6

0.226 mA, 0.854 mA

( ) (0.226 0.854)3(1.08)3 3.24 V

3.24 0.2 3.44 V

3.24 0.7 (0.226)(0.1)3.9

B C

B C

B C

EN B C E

CN EN CE

BN EN B bb

I II I

I IV I I R

V V V

V V V I r

'

1

6 V6 VCN CCV V

The waveforms can now be plotted

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