12.3 IMPROVING SWEEP LINEARITY
A simple exponential sweep generator shown in Fig. 12.9(a) essentially produces a nonlinear sweep voltage, unless τ >> T_{S} or V >>V_{s}. In an exponential sweep generator, since the capacitor charges exponentially, the resultant sweep so generated is nonlinear. To get a linear sweep, the capacitor is required to charge with a constant current. Let us consider the methods of linearizing an exponential sweep. Introduce an auxiliary generator, v, as shown in Fig. 12.9(b). If v is always kept equal to the voltage across C(i.e.,v = v_{C}), as the polarities of v and v_{C} are opposite, the net voltage in the loop is V. Then i = V /R which is a constant. That means, the capacitor charging current is constant and perfect linearity is achieved. Let us identify three nodes X, Y and Z. In a circuit one terminal, is chosen as a reference terminal or the ground terminal. Ground in a circuit is nothing but an arbitrarily chosen reference terminal.
12.3.1 Miller Integrator Sweep Generators
Now let Z be the ground terminal. Then the circuit shown in Fig. 12.9(b) is redrawn as shown in Fig. 12.10(a). Again, redrawing the circuit shown in Fig. 12.10(a) with Z as the ground terminal results in the circuit shown in Fig. 12.10(b). Since v and v_{c} are equal in magnitude and opposite in polarity, v_{i} = 0. Hence, if the auxiliary generator is replaced by an amplifier with X and Z as the input terminals and Y and Z as output terminals, then the gain A of the amplifier should be infinity. Replacing the auxiliary generator by an amplifier with gain infinity, the circuit shown in Fig. 12.10(b) is redrawn as that shown in Fig. 12.10(c). The sweep generator shown in Fig. 12.10(c) is called the Miller integrator or Miller’s sweep.
Slope Error in Miller’s Sweep Generator. Consider the Miller’s sweep circuit shown in Fig. 12.11(a), in which the auxiliary generator is replaced by an amplifier with gain infinity. Thévenizing the circuit at the input, the Thévenin voltage source and its internal resistance as V^{′} and R^{′}.
FIGURE 12.9(a) The exponential sweep generator
The circuit of Fig. 12.11(a) can be redrawn as shown in Fig. 12.11(b).
FIGURE 12.9(b) The circuit to derive constant charging current
FIGURE 12.10(a) The circuit of Fig.12.9(b) with Z as ground
FIGURE 12.10(b) The sweep generator with Z as the ground terminal
FIGURE 12.10(c) Miller’s sweep generator
FIGURE 12.11(a) Miller’s sweep circuit
FIGURE 12.11(b) Miller’s circuit with Thévenin source and its internal resistance at t = 0
Let R_{o} = 0 At t = 0 the voltage across the capacitor is zero. Therefore,
v_{i} − Av_{i} = 0
v_{i}(1 − A) = 0
v_{i} = 0
v_{i} = Av_{i} = v_{o} = 0
As t → ∞, the capacitor is fully charged, hence, no current flows in it. Thus, to calculate the output voltage, the capacitor can be replaced by an open circuit. The resultant circuit is as shown in Fig. 12.11(c).
FIGURE 12.11(c) Miller’s circuit as t → ∞
t = ∞,v_{i} = V^{′}
Hence,
v_{o} = AV^{′}
We know that for an exponential sweep, e_{s} = V_{s}/V
where V_{s} is the sweep amplitude
V = total peaktopeak excursion of the output swing.
In the case of Miller’s sweep, the total peaktopeak excursion of the output swing, v_{o} = AV^{′}.
Hence,
Substituting Eq. (12.32) in the equation:
where, V_{s}/V is the slope error of the exponential sweep.
Therefore,
Even if R_{i} is small, as A is large, the slope error of a Miller’s sweep is very small. Hence, for all practical purposes this sweep generator produces a nearlinear sweep.
Transistor Miller Sweep Generator. Let us consider the working of the triggered transistor Miller’s sweep generator, as shown in Fig. 12.12(a).
 In the quiescent state (before the application of the trigger): The circuit conditions are adjusted such that when the input is not present Q_{1} is ON and in saturation. Therefore, the voltage at C_{1}(collector of Q_{1})is V_{CE(sat)} ≈ 0. Transistor Q_{2} is OFF since V_{BE2} ≈ 0. The voltage at C_{2} (collector of Q_{2})is V_{CC}, v_{o} = V_{CC}. The voltage across the capacitor C_{s} is V_{CC}.
 When trigger is applied at t = 0. When the input signal goes negative, Q_{1} is OFF and the voltage at the collector of Q_{1} rises; Q_{2} is ON and the voltage at its collector is required to decrease abruptly to V_{CE(sat)}. Due to the capacitor, the voltage falls almost linearly. The capacitor C_{s} charges through R_{C1} and the small resistance R_{CS} (saturation resistance) that exists between the collector and emittér terminals of Q_{2}, which is driven into saturation as shown in Fig. 12.12(b). Hence, the output voltage decreases linearly from V_{CC} to V_{CE(sat)} in T_{s} and hence, is a negativegoing ramp as shown in Fig. 12.12(d). Depending on the time constant employed, T_{s} may be less than or equal to T_{g}.
 At the end of the trigger: Again at the end of the input pulse, at t = T_{g}, Q_{1} goes ON, Q_{2} goes OFF and the capacitor discharges through R_{C2} and the output again reaches V_{CC}, as shown in Fig. 12.12(c). The waveforms are shown in Fig. 12.12(d).
 Calculation of T_{s}:
From Fig. 12.12(b), the charging current of C_{s}:
FIGURE 12.12(a) A transistor Miller sweep generator
FIGURE 12.12(b) The circuit of Fig. 12.12(a) when Q_{1} is OFF and Q_{2} is ON
FIGURE 12.12(c) The discharge of C_{s} when Q_{1} is ON and Q_{2} is OFF
FIGURE 12.12(d) The waveforms of Miller’s sweep transistor
At
t = T_{s}, v_{o}(t) = V_{s}
Therefore,
If
V_{s} = V_{CC}, T_{s} = R_{C1}C_{s}.
 Calculation of T_{r}:
From Fig.12.12(c), the discharging current:
The change in voltage during T_{r} is once again V_{s}.
Therefore,
If V_{s} = V_{CC}, then, T_{r} = R,_{C2}C_{s},
 Calculation of the slope error: The slope error of the Miller’s sweep generator is given by the relation:
e_{s} = (V_{s}/V_{CC})[(1 + R/R_{i}) /A]
where, R is the resistance through which the capacitor charges when the switch is OFF. To calculate e_{s}, therefore, we have to calculate A and R_{i} of the commonemitter amplifier. The CE amplifier uses the hparameter model as shown in Fig. 12.12(e):
We have from Fig. 12.12(e)
and
V_{2} = I_{o}R_{C} = −I_{2}R_{C}
Putting
V_{2} = −I_{2}R_{C}
in Eq. (12.38):
I_{2} = h_{fe}I_{1} + h_{o}e(−I_{2}R_{C})
Dividing by I_{1}:
Therefore,
From Fig. 12.12(e), we also have the relation
Dividing by I_{1}, we have:
FIGURE 12.12(e) A CE amplifier
A, the voltage gain is given by:
R_{i} and A can be calculated using Eqs. (12.41) and (12.42) and hence, the value of e_{s}.
EXAMPLE
Example 12.3: For the Miller’s sweep shown in Fig. 12.12(a), V_{CC} = 25 V, R_{C2} = 5kΩ, R_{C1} = 10 kΩ. The duration of the sweep is 5 ms. The sweep amplitude is 25 V. Calculate (a) the value of C; (b) the retrace time and (c) the slope error. The transistor has the following parameters: h_{fe} = 80, h_{ie} = 1kΩ, h_{oe} = 1/40 kΩ and h_{re} = 2.5 × 10^{−4}.
Solution:

We have
V_{s} = V_{CC}
Therefore,
 Retrace time T_{r} = R_{C2} × C_{s} = 5 × 10^{3} × 0.5 × 10^{−6} = 2.5 ms

R_{i} = h_{ie} + h_{re}A_{I}R_{C2} = 1 + (2.5 × 10^{−4})(−71.11)(5) = (1 − 0.0889) = 0.91 k Ω
12.3.2 Bootstrap Sweep Generators
Alternatively, in the circuit shown in Fig. 12.9(b) let Y be the ground terminal. The resultant circuit is shown in Fig. 12.13(a). Redrawing this circuit and replacing the auxiliary generator by an amplifier with X and Y as input terminals and Z and Y as the output terminals, the amplifier should have a gain of unity as v = v_{C}, as shown in Fig. 12.13(b).
FIGURE 12.13(a) The sweep generator, with Y as the ground terminal
FIGURE 12.13(b) The redrawn circuit of Fig. 12.13(a)
Replacing the generator by an amplifier, the circuit shown in Fig. 12.13(b) is redrawn as shown in Fig. 12.13(c). The sweep generator represented in Fig. 12.13(c) is called a bootstrap sweep generator because the increasing input at X is accompanied by an indentical rise in the output at Z, as the gain of the amplifier is unity.
Slope Error in Bootstrap Sweep Generators. Consider the bootstrap sweep generator shown in Fig. 12.14(a) in which the auxiliary generator is replaced by an amplifier with gain 1, which obviously is an emitter follower. If initially the capacitor is uncharged and if S is closed at t = 0, then the voltage across C and R_{i}, i.e., v_{i} = 0, R_{i} is replaced by a short circuit. As v_{i} = 0, Av_{i} = 0 and is also replaced by a short circuit. Hence, at t = 0, the circuit of Fig. 12.14(a) reduces to that in Fig.12.14(b).
From Fig.12.14 (b):
And as R_{o} of the emitter follower is very small: v_{o} ≈ 0. As t → ∞, C is fully charged and is open circuited and the resultant circuit is shown in Fig.12.14(c).
From Fig. 12.14(c)
FIGURE 12.13(c) A bootstrap sweep generator
FIGURE 12.14(a) A bootstrap sweep generator with the auxilary generator replaced by an amplifier
FIGURE 12.14(b) The circuit to calculate the output at t = 0
FIGURE 12.14(c) The circuit to calculate the output as t → ∞
Dividing by R_{i}:
Here, R_{o} is the output resistance of the emitter follower, which is small and R_{i} is its input resistance, which is large.
Therefore, R_{o}/R_{i} is negligible and A ≈ 1
Eq. (12.44) gives the peaktopeak excursion of the output swing.
Therefore,
since A ≈ 1. If R = R_{i}, e_{s}_{(Bootstrap)} = e_{s}
This means that the bootstrap circuit will not provide any improvement in linearity if the input resistance of the amplifier is small. For the output of the sweep generator to be linear, R_{i} >> R. As the emitter follower has a large input resistance, this requirement is normally satisfied. If not, a Darlington emitter follower may be used to derive a very large input resistance as shown in Fig. 12.15.
A Darlington pair is manufactured as a single transistor. In this transistor, the base current I_{B1} of Q_{1} gives rise to its collector current I_{C1}(= h_{FE1}I_{B1}). I_{C1} becomes the base current I_{B2} for Q_{2} and there results a current I_{C2}(= h_{FE2}I_{B2} = h_{FE2} × h_{FE1}I_{B1}). Thus, we see that h_{FE} = h_{FE2} × h_{FE1}, i.e., the current gain of the Darlington pair is the product of the current gains of the individual transistors. The input resistance of an emitter follower is given by: R_{i} = h_{ie} + (1 + h_{fe})R_{E}.
For a Darlington pair, as h_{FE} is very large, the input resistance of a Darlington emitter follower tends to be very large. However, the voltage gain A ≈ 1. A practical bootstrap ramp generator is shown in Fig. 12.16(a). The ramp is generated across capacitor C_{1} which is charged from the current through R_{1}. The discharge transistor Q_{1}, when ON, keeps V_{1} at V_{CE}(sat) until a negative input pulse is applied. Q_{2} is an emitter follower with a low output resistance. Emitter resistance R_{E} is connected to a negative supply (V_{EE}) instead of referencing it to the ground to ensure that Q_{2} remains conducting even when its base voltage V_{1} is close to the ground. Capacitor C_{3}, called the bootstrapping capacitance, has a much higher capacitance than C_{1}. C_{3} is meant to maintain a constant voltage across R_{1} and thus, maintain a constant charging current. We now discuss the operation of the circuit.
(a) Quiescent conditions: The voltages under quiescent conditions (before the application of a trigger) are calculated as illustrated below, using Fig. 12.16(b).
FIGURE 12.15 The Darlington emitter follower
When the input trigger signal is not present, Q_{1} has sufficient base current. Therefore, Q_{1} is driven into saturation and the voltage V_{1} across the capacitor C_{1} is V_{CE(sat)}. V_{1} = V_{CE(sat)} (point X), typically 0.2 V for Si, as shown in Fig. 12.16(b). Q_{2} is an emitter follower for which input is V_{1} and its output v_{o} is:
v_{o} = V_{1} − V_{BE2} = V_{CE(sat)} − V_{BE2} (point Y) ≈ 0.2 − 0.6 = −0.4 V for Si
FIGURE 12.16(a) A practical bootstrap sweep generator
FIGURE 12.16(b) The circuit to calculate voltages under quiescent condition
As the voltage now at node K is V_{CE(sat)} ≈ 0V, D_{1} is ON. The voltage across R_{1} is,
V_{R1} = V_{CC} − V_{D1} − V_{CE(sat)} ≈ V_{CC}
where, V_{D1} is the diode voltage when ON. Hence, the current I_{1} in R_{1} is V_{CC}/R_{1} and is constant. The base current I_{B2} of Q_{2} is smaller than the collector current I_{C1} of Q_{1}:
For Q_{1} to be in saturation:
I_{B1} > I_{B1(min)}
Therefore,
For all practical purposes, when Q_{1} is ON, both V_{1} and V_{o} are zero.
(b) Sweep generation: At t = 0, as the trigger is applied, the voltage at the base of Q_{1} goes negative, Q_{1} is OFF, [see Fig. 12.16(c)]. The voltage at node K, V_{K} = V_{CC} + V_{1} and D_{1} is OFF and is an open circuit.
The voltage at node X is V_{1}.
However, I_{1} = I_{C} + I_{B2}, where I_{C} is the charging current of C_{1} and I_{B1} is the base current of Q_{2}. As Q_{2} is an emitter follower with large input resistance, I_{B1} is small. As a result,
I_{C}, the charging current of C_{1}, is constant. Hence, the resultant sweep is linear. When D_{1} is OFF, the charging current I_{1} to C_{1} through R_{1} is supplied by C_{3} which is charged to V_{CC}.
There is no current into the collector lead of Q_{1} and instead, this current flows through C_{1} charging it. As the voltage across the capacitor C_{1} varies as (I_{1}/C)t, so does the output.
From Fig. 12.16(d), it is seen that the output v_{o} varies linearly only when the duration of the gating signal (T_{g}) is small so that in this period v_{o} does not reach V_{CC}. However, if T_{g} is large, the output v_{o} may reach V_{CC} even before T_{g}. When v_{o} = V_{CC}, the voltage V_{CE2} of Q_{2} is practically zero (saturation). Q_{2} no longer behaves as an emitter follower. v_{o} and V_{1} therefore remain at V_{CC}. The current V_{CC}/R_{1} now flows through C_{3}, R_{1} and through the baseemitter diode of Q_{2}; thereby changing the voltage across C_{3} by a small amount, ΔV_{C3}.
(i) If T_{s} < T_{g} (i.e., v_{o} reaches V_{CC} before T_{g}) Then at
t = T_{s}
v_{o} = V_{s} = V_{CC}
From Eq. (12.46):
FIGURE 12.16(c) The circuit of Fig. 12.16(a) when Q_{1} is OFF and C_{1} charges
FIGURE 12.16(d) The output waveforms of the bootstrap sweep circuit
FIGURE 12.17 The calculation of I_{1}
or
The amount of charge gained by C_{1} is Q_{C1} = C_{1} v_{s}
The amount of charge lost by C_{3} is Q_{C3} = C_{3}ΔV_{C3}
Assume that during T_{s} the current through capacitors C_{1} and C_{3} are same.
Therefore, the amount of charge gained by C_{1} = the amount of charge lost by C_{3}.
C_{3}ΔV_{C3} = C_{1}V_{s}
Therefore, at t = T_{s}, the voltage across C_{1} is V_{s} and the voltage across C_{3} is V_{C3} − (C_{1}/C_{3})V_{s} and the output voltage is v_{o} = AVs, as shown in Fig. 12.17.
I_{C} = I_{1} − I_{B2}
and
At
t = T_{s}
t = T_{s},
Therefore,
At t = 0, V_{s} = 0. Hence, from Eq. (12.48), I_{C} = V_{C3}/R_{1}
Eq. (12.49) gives the expression for the slope error when the voltage V_{C3} changes during the sweep period.
(ii) On the other hand if V_{s} < V_{CC}, the maximum sweep voltage is:
(c) Calculation of retrace time, T_{r}: At the end of the gate signal, at t = T_{g}, a current I_{B1} = (V_{CC}/R_{B}) again flows into the base terminal of Q_{1}. Q_{1} once again tries to go into saturation. However, till such time V_{CE} of Q_{1} is V_{CE}(sat) (Q_{1} is in saturation), the collector current, i_{C1} remains constant at
As shown in Fig. 12.18, the current i_{R1} through R_{1} and the discharging current i_{d} of C_{1} now constitutes i_{C1}, the collector current of Q_{1}, neglecting the small base current I_{B2} of Q_{2},
Writing the KCL equation at node C:
i_{R1} remains approximately at V_{CC}/R_{1}(= I_{1}) and the capacitor discharges with a constant current. The voltage across C_{1} falls, and consequently v_{o} falls, since once again Q_{2} behaves as an emitter follower. The discharge current i_{d} from Eq. (12.51) is:
i_{d} = i_{C1} − i_{R1}
FIGURE 12.18 The circuit to calculate i_{d}
Therefore V_{1} and v_{o} fall linearly to the initial value. The voltage variation during the retrace time T_{r} is:
Using Eq. (12.52), we have,
If the retrace time is large, it takes a longer time to initiate a new sweep cycle. From Eq. (12.53), it is seen that R_{B} will have to be small to reduce the retrace time. However, if R_{B} is too small, then the collector current of Q_{1} becomes large as:
This results in greater dissipation in Q_{1}. During the period, T = T_{g} + T_{r}, though C_{3} is a large capacitor, it may still loose some charge. The circuit is said to have recovered completely only when the charge lost by C_{3} is regained. The minimum recovery time T_{1} for C_{3} can be found out as follows:
Charge lost by C_{3} in time T = V_{CC} T/R_{1}
The charging current of C_{3} = V_{EE}/R_{E}
Therefore, the charge recovered in time T_{1} = (V_{EE}/R_{E})T_{1}
Charge lost = charge regained
To reduce T_{1}, V_{EE} may be increased. However, this increases the quiescent current in Q_{2} and hence, the dissipation in it.
(d) Calculation of the slope error of a bootstrap sweep circuit: The slope error of the bootstrap, assuming that the charge on C_{3} remains unaltered during the sweep duration, is given by the relation:
To calculate e_{s}, we have to calculate A and R_{i}. For a common collector amplifier we have:
since
h_{fc} = −(1 + h_{fe}) and h_{oc} = h_{oe}
since
h_{rc} = 1 and h_{ic} = h_{ie}
From Eq. (12.56) dividing by R_{i}:
Therefore,
Using Eqs. (12.55), (12.56) and (12.57), it is possible to calculate e_{s}. To understand the procedure to analyse and also design a bootstrap sweep let us consider the following examples.
EXAMPLE
Example 12.4: The transistor bootstrap circuit in Fig. 12.16(a) has the following parameters, V_{CC} = 15 V, V_{EE} = −10 V, R_{B} = 30 kΩ, R_{1} = 10 k Ω, R_{E} = 5kΩ, C_{1} = 0.005 µF, C_{3} = 1.0 µF. The input trigger is negative and has an amplitude of 2 V and a width of 60 µs. The transistor parameters are h_{FE} = h_{fe} = 50, h_{ic} = 1kΩ, 1/h_{oe} = 40 kΩ, h_{rc} = 1 Assume that the forwardbiased junction voltages are negligible. The diode is ideal. Evaluate (a) the sweep speed and the sweep duration; (b) retrace time and recovery time and (c) the slope error; (d) plot the gate voltage, the output voltage V_{o} and the collector current of Q_{1}
Solution: Referring to the circuit in Fig. 12.16(a):
 Sweep speed =
V_{s(max)} = V_{CC} = 15 V = Sweep speed × T_{s}
i.e.,
15 = (3 × 10^{5})T_{s}
Therefore, sweep time,
 At the end of the input pulse, Q_{1} once again goes into saturation.
i_{C1} = h_{FE}i_{B1} = 50 × 0.5 = 25 mA
The retrace time T_{r} is,
T = T_{g} + T_{r} = (60 + 3.18) = 63.18 μs
Recovery time T_{1} =
 To find the slope error:
The current gain of the emitter follower is given by:
Input impedance of the emitter follower is given by:
R_{i} = h_{ie} + A_{I} R_{E}
Therefore,
A is the voltage gain of the emitter follower.
R_{i} = h_{ie} + A_{I} R_{E} = 1kΩ + 45.33 × 5kΩ = 227.67 kΩ
The slope error,
 Using the above calculations, the waveforms can be sketched as shown in Fig. 12.19
FIGURE 12.19 The waveforms of the bootstrap circuit
EXAMPLE
Example 12.5: Design a transistor bootstrap sweep generator to provide an output amplitude of 10 V over a time period of 1 ms. The ramp is to be triggered by a negative going pulse with an amplitude of 5 V, a pulse width of 1 ms and a time interval between the pulses is 0.1ms. The load resistance is 1 kΩ and the ramp is to be linear within 1 per cent. The supply voltage is 18 V, h_{FE(min)} = 100.
Refer to the bootstrap circuit shown in Fig. 12.16(a).
R_{L} = R_{E} = 1kΩ
When V_{o} = 0,
When V_{o} = V_{s}(= 10V),
At V_{o} = 0,
At V_{o} = V_{s},
ΔI_{B2} = 0.28 − 0.18 = 0.10 mA
I_{1} is much larger than I_{B2}
Let
I_{1} = 100 × ΔI_{B2} = 100 × 0.10 mA = 10 mA
V_{R1} = V_{CC} − V_{D1} − V_{CE(sat)} = 18 − 0.7 − 0.2 = 17.1V
For 1 per cent nonlinearity due to discharge of C_{3}:
ΔV_{C3} = 1 per cent of the initial V_{CC} level
Initial
V_{C3} = V_{CC} = 18 V
And C_{3} discharge current I_{1} is equal to 10 mA. Therefore,
The discharge time of C_{1} is 0.1ms which is 1/10 th of the charging time. For Q_{1} to discharge C_{1} in 1/10 th of the charging time,
I_{C1} = 10 × (C_{1}charging current) = 10 × 10 mA = 100 mA
Choose R_{B} = 20 kΩ, Q_{1} is to be biased OFF at the end of the input pulse.
ΔV_{B} = v_{o} − (pulse amplitude) = 0.7 − 5V = −4.3V
The charging current of C_{2}is equal to the current through R_{B} when Q_{1} is OFF.
SOLVED PROBLEMS
Example 12.6: AUJT has characteristic as shown in Fig. 12.20(a) and the UJT relaxation oscillator is shown in Fig.12.20(b) Find the values of: (a) Sweep amplitude, (b) the slope and displacement errors, (c) the duration of the sweep. and (Assume η = 0.6 and V_{F} = 0.7 V for silicon.)
FIGURE 12.20(a) UJT characteristic; (b) UJT relaxation oscillator
Solution: The waveform of the sweep generator is shown in Fig. 12.20(c)
Given, V_{BB} = 15 V, V_{YY} = 30 V, R = 10 kΩ, C = 0.1 µF
We know that
V_{P} = η V_{BB} + V_{F} = 0.6 × 15 + 0.7 = 9.7 V
From UJT characteristics, we have V_{V} = 1.6 V
(a) Amplitude of the sweep = V_{s} = V_{P} − V_{V}
= 9.7 − 1.6 = 8.1 V.
(b) Sweep speed error,
Displacement error,
FIGURE 12.20(c) The waveform of the sweep generator
Example 12.7: Using the characteristic of UJT shown in Fig. 12.21 (a), calculate the values of R, C, R_{1} and R_{2} of the relaxation oscillator shown in Fig. 12.21; (b) to generate a sweep with a frequency of 10 kHZ and amplitude of 10V, T_{r} is 0.5 % of T.
FIGURE 12.21(a) UJT characteristic; (b) UJT relaxation oscillator
Solution:
Given,
f = 10 kHz, V_{s} = 10 V, V_{BB} = 15 V, V_{V} = 1V, I_{P} = 14 µA.
I_{V} = 1 mA, η = 0.6
Therefore,
V_{P} = V_{s} + V_{V} = 10 + 1 = 11 V
R should lie between R_{(max)} and R_{(min)}.
Let R = 150 kΩ.
Sweep time T_{s} =
T_{s} = 1.252RC
T_{r} = 0.5µs
∴ T_{s} = T − T_{r} = 100 − 0.5 = 99.5μs
Therefore,
Return time T_{r} = R_{1}C= 0.5µs
Therefore,
The value of R_{2} is usually higher than R_{1}. So, choose R_{2} = 3kΩ.
Example 12.8: For the UJT relaxation oscillator shown in Fig.12.21(c), R_{BB} = 3kΩ, R_{1} = 0.1 kΩ, η = 0.7, V_{V} = 2V, I_{V} = 10 mA, I_{P} = 0.01 mA. (a) Calculate R_{B1} and R_{B2} under quiescent condition (i.e., when I_{E} = 0). (b) Calculate the peak voltage, V_{P}. (c) Calculate the permissible value of R. (d) Calculate the frequency, assuming that the retrace time is negligible. Also calculate f using the value of η. (e) Calculate the frequency, considering the retrace time also. Assume R_{B1} = 0.1 kΩ during the retrace time. (f) Calculate the voltage levels of V_{B1}. (g)Plot the waveforms of the output voltage and V_{B1}.
FIGURE 12.21(c) The UJT oscillator
Solution:
Given R_{BB} = 3kΩ, η = 0.7.

R_{B1} = 0.7 × 3 kΩ = 2.1 kΩ
We have
R_{BB} = R_{B1} + R_{B2}
FIGURE 12.21(d) The circuit to calculae V_{p}
R_{B2} = R_{BB} − R_{B1} = 3kΩ − 2.1 kΩ = 0.9 kΩ
 The circuit that enables the calculation of V_{P} is shown in Fig. 12.22(d). From Fig. 12.22(d):

R_{(min)} < R < R_{(max)}
Choose
R = 100 kΩ

T_{s} using the value of η is given as:

T = T_{s} + T_{r} = 6.335 + 0.01735 = 6.352 ms
 V_{B1} during charging of C is given as:
V_{B1} during discharge of C is given by:
FIGURE 12.21(e) The waveforms of the output and V_{B1}
V_{F} is the diode voltage when ON, Fig. 12.21(d).
 Waveforms of the output and V_{B1} are shown in Fig.12.21(e).
Example 12.9: The bootstrap sweep circuit is shown in Fig.12.22. A square wave whose amplitude varies between 0 and −4 V and duration 0.5 ms is applied as a trigger. a) Calculate all the quiescent state currents and voltages. b) Determine the sweep amplitude, sweep time and sweep frequency. Assume h_{FE(min)} = 30, V_{CE(sat)} = 0.3 V, V_{BE(sat)} = 0.7 V, V_{BE}(active) = V_{D} = 0.6 V
FIGURE 12.22 A bootstrap sweep generator
 Current through R_{1} is:
Base Current of Q_{1} is:
Emitter current of Q_{2} = i_{E2} =
v_{o} = V_{C1} − V_{BE2(active)}
v_{o} = V_{CE(sat)} − V_{BE2(active)} = 0.3 − 0.6 = −0.3V
Therefore,
Therefore,
i_{B2} = i_{E2} − i_{C2} = 0.97 × 10^{−3} − 0.938 × 10^{−3} = 0.032 mA
i_{C} = I_{1} − i_{B2} = 1.14 × 10^{−3} − 0.032 × 10^{−3} = 1.108 mA
V_{CE2} = V_{CC} − v_{o} = 18 + 0.3 = 18.3V
 Sweep time T_{s} = R_{1}C_{1} = 15 × 10^{3} × 0.01 × 10^{−6} = 0.15 ms
Therefore,
T_{g} = 0.5 ms
Since
T_{s} < T_{g}, V_{s} = V_{CC} = 18 V
Run time,
T = T_{g} + T_{r} = 0.5 + 0.075 = 0.575 ms
Example 12.10: Determine the values of R_{1}, R_{E}, R_{B}, C_{1} and C_{3} of a bootstrap circuit shown in Fig. 12.23, if I_{C1} = 1.5 mA, I_{E2} = 1.75 mA, h_{FE(min)} = 30, V_{CE(sat)} = 0.3 V, V_{BE(sat)} = 0.7 V and V_{BE(active)} = V_{D1} = 0.6 V. Assume T_{s} = 12 ms.
Solution:
To calculate R_{1}:
Applying KVL to collector loop of Q_{1}, we have the collector voltage of Q_{1} when it is ON:
V_{C1} = V_{CC}  V_{D1}  I_{1}R_{1}
To calculate R_{B}:
Applying KVL to the base loop of Q_{1} we have V_{CC} − I_{B1}R_{B} − V_{BE(sat)} = 0
FIGURE 12.23 A bootstrap sweep generator
Choose,
Therefore,
To calculate C_{1}:
Sweep Time T_{s} = R_{1}C_{1}
Therefore,
Given
T_{s} = 12 ms
The value of C_{3} should be larger than C_{1}, therefore, choose C_{3} as 10 times of C_{1}.
Therefore,
C_{3} = 9.43µF
To Calculate R_{E}:
The output voltage v_{o} = V_{CE(sat)} − V_{BE(active)}
Therefore,
v_{o} = 0.3 − 0.6 = −0.3
Applying KVL, to the input side of Q_{2} we have:
v_{o} + V_{EE} − R_{E}i_{E2} = 0
Therefore,
SUMMARY
 Voltage sweep generators are based on the principle of electrostatic deflection.
 In an ideal sweep generator (sawtooth generator), the sweep voltage increases linearly till T_{s} (sweep duration) and abruptly falls to a small value.
 In a practical sweep generator, however, at the end of the sweep a small time may elapse before the sweep voltage reaches the initial value. This time interval is called the retrace time, restoration time or flyback time.
 Slope error e_{s} is defined as the ratio of the difference between the initial slope and the final slope to the initial slope and is expressed as a percentage.
 Displacement error e_{d} is defined as the ratio of the maximum difference between the actual sweep voltage and the linear sweep voltage to the sweep amplitude and is expressed as a percentage.
 Transmission error e_{t} is defined as the ratio of the difference in amplitude between a linear sweep voltage and the practical voltage to the amplitude of the linear sweep and is expressed as a percentage.
 Slope error e_{s} is given by V_{s}/V where V_{s} is the sweep amplitude and V is the peaktopeak excursion of the output swing. The larger the V, the smaller the deviation from linearity.
 Slope error e_{s} is also given as T_{s}/τ where T_{s} is the sweep duration and τ is the time constant of the sweep circuit. The larger the τ, the smaller the deviation from linearity.
 The interrelation between the three types of errors is given as: e_{d} = (1/8)e_{s} = (1/4)e_{t}.
 A UJT is a negative resistance device and is used as a switch.
 The sweep duration of a UJT sweep generator is given by T_{s} = τ ln(1/(1 − η)) where, η is the intrinsic standoff ratio and typically lies between 0.6 to 0.8.
 An alternate expression for the sweep duration of an UJT sweep generator is given as T_{s} = τ ln (V_{BB} − V_{V})/(V_{BB} − V_{P}) where V_{BB} is supply voltage, V_{V} and V_{P} are the valley and peak voltages of the UJT.
 The auxiliary generator in a Miller sweep generator is replaced by an amplifier with a gain of infinity.
 The auxiliary generator in a bootstrap sweep generator is replaced by an amplifier with unity gain.
 The slope error of a Miller sweep generator is given as e_{s(Miller)} = e_{s} × 1 + (R/R_{i})/A and this is small as A is large.
 The slope error of a bootstrap sweep generator is e_{s(Bootstrap)} ≈ e_{s}(R/R_{i}) and is small only when R_{i} is large.
MULTIPLE CHOICE QUESTIONS
 The application of a voltage time base generator is in a measuring instrument like:
 CRO
 Frequency counters
 Digital voltmeter
 Multimeter
 The deflection of the electron beam in a CRO employing a voltage sweep generator is based on the principle of:
 Electromagnetic deflection
 Electrostatic deflection
 Frequency multiplication
 Counting
 Errors e_{s}, e_{d} and e_{t} define:
 Current gain
 Voltage gain
 Deviation from linearity
 Bandwidth
 If in a simple exponential sweep generator, the supply voltage is 200 V and the required sweep amplitude is 10 V, and then the slope error is:
 5 %
 20 %
 200 %
 0.5 %
 In a simple exponential sweep generator, the time constant of the circuit is 1000 µs and the sweep duration is 100 µ s, then the slope error is:
 5 %
 100 %
 20 %
 10 %
 For the output of a voltage sweep generator to be linear, the capacitor must be charged with a:
 Constant current
 Linearly varying current
 Exponentially varying current
 Zero current
 A Miller sweep generator produces a:
 Positivegoing sweep
 Negativegoing sweep
 Constant sweep voltage
 Infinite output voltage
 A bootstrap generator produces a:
 Positivegoing sweep voltage
 Negativegoing sweep voltage
 Constant sweep voltage
 Infinite sweep voltage
 The slope error e_{s} and the displacement error e_{d} are related as:
 e_{s} = 2e_{d}
 8 e_{s} = e_{d}
 e_{s} = 4 e_{d}
 e_{s} = 8 e_{d}
 The displacement error e_{d} and the transmission error e_{t} are related as:
 e_{t} = 4 e_{d}
 e_{t} = 2 e_{d}
 4 e_{t} = e_{d}
 e_{t} = e_{d}
 The slope error e_{s} and the transmission error e_{t} are related as:
 e_{s} = 2 e_{t}
 e_{s} = 0.5e_{t}
 e_{s} = e_{t}
 e_{s} = 4e_{t}
 The sweep duration of a UJT sweep generator is given as:
 0.69 τ
 τ ln{(V_{BB} − V_{P})/(V_{BB} − V_{V})}
 τ ln{(V_{BB} − V_{V})/(V_{BB} − V_{P})}
 1.4 τ
SHORT ANSWER QUESTIONS
 Define the three errors that specify deviation from linearity.
 Draw the circuit of a UJT sweep generator and explain its operation.
 Derive the expression for the frequency of oscillations of a UJT sweep generator in terms of η, the intrinsic standoff ratio.
 Derive the expression for the frequency of oscillations of a UJT sweep generator in terms of V_{V} and V_{P}.
 Explain the basic principle employed in a Miller sweep to linearize the nonlinear sweep.
 Explain the basic principle employed in a bootstrap sweep to linearize a nonlinear sweep.
LONG ANSWER QUESTIONS
 Draw the circuit of a UJT sweep generator and explain its operation. Derive the expression for e_{s}, e_{d} and e_{t} and obtain their interrelationship.
 Explain the principle of working of a transistor constant current sweep generator with the help of a circuit diagram. Derive the expression for its slope error.
 Explain how a linearly varying sweep voltage is generated in a Miller sweep generator. Draw the circuit and explain its operation. Derive the expression for its slope error.
 Explain the basic principle of a bootstrap sweep generator. Draw the circuit and explain its operation. Derive the expression for its slope error.
UNSOLVED PROBLEMS
 For the UJT relaxation oscillator shown in Fig. 12p.1, R_{BB} = 4kΩ, R_{1} = 0.1 kΩ, η = 0.6, V_{V} = 3V, I_{V} = 10 mA, I_{P} = 0.01 mA.
 Calculate R_{B1} and R_{B2} under quiescent condition (i.e., I_{E} = 0).
 Calculate the peak voltage,V_{P}.
 Calculate the permissible value of R.
 Calculate the frequency assuming that the retrace time is negligible. Also calculate f using the value of η.
 Calculate the frequency, also considering the retrace time. Assume R_{B1} = 0.1 kΩ during the retrace time.
 Calculate the voltage levels of V_{B1}.
 Plot the waveforms of the output voltage and V_{B1}.
FIGURE 12p.1 The given UJT sweep circuit
 The UJT relaxation oscillator is shown in Fig. 12p.2. Find:
 The sweep amplitude.
 The slope and displacement errors.
 The duration of the sweep.
Given that V_{V} = 3_{V}, η = 0.6.FIGURE 12p.2 The given UJT relaxation oscillator
 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 η = 0.6. (b)If the sweep error is 10% and T_{r} = 1% of T_{s}, calculate V_{BB}, V_{YY}, R, R_{1}, R_{2} and C. If the sweep duration is 300 µs, calculate the new value of C. The VI characteristic of UJT is as shown in Fig. 12p.3(a).
FIGURE 12p.3(a) V–I characteristic of a given UJT
FIGURE 12p.3(b) UJT circuit
 For the Miller’s sweep shown in Fig. 12.12(a), V_{CC} = 24 V, R_{C2} = 2kΩ, R_{C1} = 10 kΩ and C = 1µF. 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 transistor has the following parameters: h_{fe} = 100, h_{ie} = 1kΩ, h_{oe} = 1/20 kΩ and h_{re} = 2.5 × 10^{−4}.
 The transistor used in the bootstrap circuit shown in Fig. 12p.4 has the following hparameter values: h_{re} = 2.5 × 10 − 4, h_{ie} = 1.1 kΩ, h_{fe} = 60, 1/h_{oe} = 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 timebase amplitude, the retrace time and the recovery time.
FIGURE 12p.4 The given bootstrap sweep circuit
 Find (a) the sweep amplitude and (b) the slope error for the bootstrap sweep generator shown in Fig.12p.5, when a 2 kHz symmetrical square wave is applied as an input to it. Plot to scale the input and output waveforms. The typical hparameter values of transistor are, h_{fe} = 90, 1/h_{oe} = 35 kΩ, h_{ie} = 1kΩ and h_{rc} = 1. Assume all forwardbiased junction voltages are zero.
FIGURE 12p.5 A bootstrap sweep circuit
 The transistor bootstrap has the following parameters: V_{CC} = 20 V, V_{EE} = −20 V, R_{B} = 15 kΩ, R_{1} = 5kΩ, R_{E} = 2.5 kΩ, C_{1} = 0.001 µF, C_{3} = 0.25 µF. The input gate has the amplitude of 1 V and a width of 50 µs. The transistor parameters are h_{fe} = 60, h_{ie} = 2kΩ, 1/h_{oe} = 10 kΩ, h_{re} = 10^{−4} and the forwardbiased 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}.