Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Chapter 12
Voltage Sweep Generators
1. For the UJT relaxation oscillator shown in Fig.12p.1, R BB = 4 kΩ, R 1 = 0.1 kΩ,  = 0.6,
V V = 3 V, I V = 10 mA, I P = 0.01 mA. (i) Calculate R B1 and R B2 under quiescent condition
(i.e., I E = 0). (ii) Calculate the peak voltage, V P. (iii) Calculate the permissible value of R.
(iv) Calculate the frequency assuming that the retrace time is negligible. Also calculate f
using the value of  . (v) Calculate the frequency, also considering the retrace time.
Assume R B1 = 0.1 kΩ during retrace time. (vi) Calculate the voltage levels of V B1 . (vii)
Plot the waveforms of the output voltage and V B1 .
Fig.12p.1 The given UJT sweep circuit
Solution:
Given R BB = 4 kΩ, = 0.6.
RB1
R
(i) 
= B1
RB1  RB 2 R BB
RB1
4 kΩ
R B1 = 0.6  4 kΩ = 2.4 kΩ
0.6 =
We have R BB = R B1 + R B2
 RB 2  RBB  RB1  4 kΩ  2.4 kΩ  1.6 kΩ
(ii) The circuit that enables the calculation of V P is shown in Fig. 1.1.
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Pulse and Digital Circuits
From Fig. 1.1,
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig. 1.1Circuit to calculate V P
V P = 0.7V + V BB 
( R B1  R1 )
(2.4  0.1)
= 0.7 + 20 
 12.89 V.
(4  0.1)
R BB  R1
(V BB  VV ) (20  3)
=
 1.7 kΩ
IV
10  103
(V  VP ) (20  12.89)

 711 kΩ
R max = BB
IP
0.01  103
R min < R< R max
Choose R = 100 kΩ.
(V  VV )
(20  3)
 100  103  100  109  ln
 8.7 ms.
(iv) T s = RC ln BB
(VBB  VP )
(20  12.89)
1
f=
 114.9 Hz.
8.7  103
T s using the value of  is given as:
1
1
T s = RC ln
 100  103  100  109 ln
 9.1 ms.
(1  )
1  0.6
1
f =
 109.8 Hz.
9.1  103
V
12.89
(v) T r = (R B1 +R 1 )C  ln P  (0.1  0.1)103  100  109  ln
 29.1  s.
VV
3
T = T s + T r = 9.1 + 0.0291 = 9.129 ms.
1
f=
 109.54 Hz.
9.129  103
(vi) V B1 during charging of C is given as:
R1
20  0.1
V B1 = V BB 

 0.487 V.
( R1  R BB ) 0.1  4
V B1 during discharge of C is given by:
0.1
(12.89  0.7)
V B1 = (V P – V F ) 

 6.09 V.
0.1  0.1
2
V F is the diode voltage when ON, Fig. 1.1.
(vii) Waveforms of the output and V B1 are shown in Fig. 1.2.
(iii) R min =
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig. 1.2 Waveforms of the output and V B1
2. For the UJT relaxation oscillator is shown in Fig.12p.2. Find (i) the sweep amplitude;
(ii) the slope and displacement errors; and (iii) the duration of the sweep. Given that V V =
3 V, η = 0.6.
Fig.12p.2 The given UJT relaxation oscillator
Solution:
The waveform of the sweep generator is shown in Fig. 2.1.
VBB  20 V, VYY  50 V, R  100 kΩ, C  0.1  F
Fig. 2.1 Waveform of the sweep generator
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
We know that V P   VBB  VF = 0.6  20  0.7  12.7 V
We have VV  3 V
(i) Amplitude of the sweep = Vs  VP  VV  12.7  3  9.7 V .
Vs
Vs
9.7


 0.206  20.6%
V VYY  VV 50  3
1
1
Displacement error, ed  es   20.6%  2.57%
8
8

(iii) Sweep time, T s  RC log e V YY  V V 
V  V 
 YY
P 
(ii) Sweep speed error, e s 
 50  3 
T s = 100  10 3  0.1  10 6  log e 

 50  12.7 
T s = 2.31 ms
3. (a) Design a UJT sweep circuit shown in Fig.12p.3(b) to generate a sweep of 15 V
amplitude and 3 ms duration, given that   0.6 . If the sweep error is 10 per cent and T r
= 1 per cent of T s , calculate V BB , VYY , R, R1 , R2 and C . If the sweep duration is 300  s ,
calculate the new value of C. The V–I characteristic of UJT is as shown in Fig. 12p.3(a)
Fig. 12p.3(b) The given UJT circuit
Fig. 12p.3 (a)The given UJT V– I characteristic
Solution:
From the characteristics, we have VV  3 V, IV  1  103 A, I P  0.3  103 A
Given V s = 15 V and e s = 0.1
VP  Vs  VV  15  3  18 V.
Peak-to-peak excursion of the output swing = Y VY – V V .
V
Vs
es  s 
V VYY  VV
0.1 
15
VYY  3
VYY  153 V.
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Given  = 0.6. We have Vp = V BB + V F .
V  VF 18  0.6
VBB  P

 29 V.
0.6

To calculate R and C
V  VV 
T s  RC log e  YY

 V YY  V P 
 153  3 
 RC log e 
 153  18 
Ts  0.105 RC
Ts
3  10 3

 28.57  10 3
0.105
0.105
V BB  V P
29  18
R max =

 36.66 k
IP
0.3  10 3
V V
29  3
Rmin  BB V 
 26 kΩ
IV
1  103
R should lie between 26 kΩ and 36 kΩ . Choose R = 33 kΩ
RC = 28.57  10 3
28.57  103
 0.865  F
C=
33  103
Given T r = 1 per cent of T s
T r = 0.01  3  10 3
 Tr  R1C
Tr 0.01  3  103
 R1 

 34.68  35 Ω
C
0.865  106
R2 is higher than R1 and must be of the order of several hundred ohms. Let
1
R2  10  R1  350 Ω . If T s is to be 300 s C should be
of the earlier value. By
10
reducing the value of C, the charging current will be reduced, but by reducing the value
of R the charging current is increased, which is undesirable. So the reduction of C is
advisable.
0.865  F
 Cnew 
 86.5 nF
10
 RC 
4. For the Miller’s sweep shown in Fig.12p.5, V CC = 24 V, R C2 = 2 k  , R C1 = 10 k  and
C = 1  F. The amplitude of the sweep is 18 V. (a) Calculate the sweep duration T s ; (b)
the retrace time T r ; (c) frequency of the sweep generator and (d) the slope error. The
1
and
transistor has the following parameters: h fe = 100, h ie = 1 k  , h oe =
20 kΩ
h re =2.5  10 4 .
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig.12p.5 Transistor Miller sweep
Solution:
(a)
Vs=
VCC
 Ts
RC1C
V
18
T s = s  RC1C =  10  10 3  1  10 6 =7.5 ms.
24
VCC
Vs
18
 RC 2 C =  2  10 3  1  10 6 =1.5 ms.
24
VCC
(c) T = T s + T r = 7.5 + 1.5 = 9 ms.
1000
f=
 111.11 Hz.
9
h fe
100
(d) A I =

 90.90
2
1  hoe RC 2
1
20
R i = h ie +h re A I R C2 = 1+(2.5  10 4 )(90.90)(2 )=(1 – 0.045)=0.955 k 
(b) Retrace time T r =
A=A I
RC 2
2
 190.36
 –90.90 
Ri
0.955
V
R
1
18
1
10

 (1 
) = 0.045
e s(Miller )= s   (1  C1 ) 
VCC A
Ri
24 190.66
0.955
e s = 4.5 per cent.
6. The transistor used in the bootstrap circuit shown in Fig.12p.6 has the following hparameter values. h re = 2.5  10–4, h ie = 1.1 k, h fe = 60, 1 / hoe = 40 kΩ. Assume V BE(sat)
= V CE(sat) = 0. If the applied input gating voltage is a symmetrical square wave of the
frequency 9.5 kHz, determine the time-base amplitude, retrace time and recovery time.
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig.12p.6 The given bootstrap sweep circuit
Solution:
V s = V CC = 12 V
At the end of the input pulse, Q 1 once again goes into saturation.
VCC
12

 0.4 mA
i B1 =
RB
30  103
i C1 = h fe i B1 = 60  0.4 = 24 mA
The retrace time T r is
Vs
12
C1
 0.06  106
VCC
12

 31.4 µs
Tr =
60
1
h fe
1


30  103 12  103
RB
R1
1
1

 0.105 ms
f 9.5  103
V
R
12 4
Recovery time T1  CC  E  T    0.105  10 3  35.08 µs.
VEE R1
12 12
T=
7. Find (a) the sweep amplitude and (b) the slope error for the bootstrap sweep generator
shown in Fig.12p.7, when a 2-kHz symmetrical square wave is applied as an input to it.
Plot to scale the input and output waveforms. The typical h-parameter values of transistor
are, h fe  90, 1 / hoe  35 kΩ, hie  1 kΩ and hrc  1 . Assume all forward-biased junction
voltages are zero.
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig.12p.7. A bootstrap sweep circuit
Solution:
Time period of input wave form T=
1
1
=
= 0.5 ms.
f 2  10 3
T
 0.25 ms
2
Maximum value of ramp voltage or sweep amplitude is equal to
VCC Tg
12  0.25  10 3
Vs 

= 10 V
R1C1 10  10 3  0.3  10 6
For an emitter follower, the current gain is
h fe  1
90  1
 79.68
AI 

1  hoe R E
5  10 3
1
35  10 3
Input impedance = Ri  hie  AI RE
The input is a symmetrical square wave, so the gate width Tg 
R i = 1  103  79.68  5  103  399.4 kΩ
h
1
 0.0025
We have (1  A)  ie =
Ri 399.4
Vs  R1
  (1  A
VCC  Ri
Sweep error = 0.29 per cent.
Sweep error e s 
 10  10

)  
 0.0025 = 0.0029

 12  399.4
7. The transistor bootstrap has the following parameters: V CC = 20 V, V EE = –20 V, R B =
15 kΩ, R 1 = 5 kΩ, R E = 2.5 kΩ, C 1 = 0.001 µF, C 3 = 0.25 µF. The input gate has
amplitude of 1 V and a width of 50 µs. The transistor parameters are h fe = 60, h ie = 2 kΩ,
1/h oe = 10 kΩ, h re = 10–4 and the forward-biased junction voltages are negligible. The
diode is ideal. (i) Evaluate (a) the sweep speed and the maximum amplitude of the sweep;
(b) the retrace time; (c) the peak voltage change across C 3 and the recovery time and (d)
the slope error. (ii) Plot the gate voltage, collector current i C1 , and the output voltage v o .
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig.12p.8 A practical bootstrap sweep generator
Solution:
(i) Referring to circuit in Fig.12p.8
V
I1
20
 CC 
 4  106 V/s
3
6
C1 R1C1 5  10  0.001  10
V s(max) = V CC = 20 V = sweep speed  T s
i.e. 20 = (4 106)T s
20

Sweep time, T s =
= 5 µs.
4  10 6
(b) At the end of the input pulse, Q 1 once again goes into saturation.
VCC
20
i B1 =

 1.33 mA
RB
15  103
i C1 = h fe i B1 = 60  1.33 mA = 79.8 mA
The retrace time T r is
Vs
20
C1
 1  109
VCC
20
= 0.26 µs
Tr =

60
1
h fe
1


3
5  103
RB
R1 15  10
T = T g +T r = (50+0.26) = 50.26 µs.
V
R
15 5
Recovery time T1  CC  E  T    63.18  106  47.385 µs
VEE R1
10 10
(c) To find the slope error:
The current gain of the emitter follower is given by:
(a) Sweep speed =
 AI 
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1  h fe
1  hoe RE

1  60
61

 48.8
1
1
1   2.5 .25
10
9
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Input impedance of the emitter follower is given by
R i = h ie + A I R E
h
 1 – A = ie
Ri
A is the voltage gain of the emitter follower.
R i = h ie + A I R E = 2 kΩ + 48.8  2.5 kΩ = 124 kΩ
h
2
1 – A  ie 
 0.016
Ri 124

R  V
(d) The slope error, e s = 1  A  1  s
Ri  VCC

5  20

e s = 0.016 
 0.056
124  20

e s = 5.6 per cent.
(ii) Using the above calculations, the waveforms can be sketched as shown in Fig. 8.1.
Fig. 8.1 Waveforms
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