# 12.3 Improving Sweep Linearity – Pulse and Digital Circuits

##### 12.3 IMPROVING SWEEP LINEARITY

A simple exponential sweep generator shown in Fig. 12.9(a) essentially produces a non-linear sweep voltage, unless τ >> TS or V >>Vs. In an exponential sweep generator, since the capacitor charges exponentially, the resultant sweep so generated is non-linear. 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 = vC), as the polarities of v and vC 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 vc are equal in magnitude and opposite in polarity, vi = 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 Ro = 0 At t = 0 the voltage across the capacitor is zero. Therefore,

viAvi = 0

vi(1 − A) = 0

vi = 0

vi = Avi = vo = 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 → ∞

At

t = ∞,vi = V

Hence,

vo = AV

We know that for an exponential sweep, es = Vs/V

where Vs is the sweep amplitude

V = total peak-to-peak excursion of the output swing.

In the case of Miller’s sweep, the total peak-to-peak excursion of the output swing, vo = AV.

Hence,

Substituting Eq. (12.32) in the equation:

where, Vs/V is the slope error of the exponential sweep.

Therefore,

Even if Ri 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 near-linear 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).

1. In the quiescent state (before the application of the trigger): The circuit conditions are adjusted such that when the input is not present Q1 is ON and in saturation. Therefore, the voltage at C1(collector of Q1)is VCE(sat) ≈ 0. Transistor Q2 is OFF since VBE2 ≈ 0. The voltage at C2 (collector of Q2)is VCC, vo = VCC. The voltage across the capacitor Cs is VCC.
2. When trigger is applied at t = 0. When the input signal goes negative, Q1 is OFF and the voltage at the collector of Q1 rises; Q2 is ON and the voltage at its collector is required to decrease abruptly to VCE(sat). Due to the capacitor, the voltage falls almost linearly. The capacitor Cs charges through RC1 and the small resistance RCS (saturation resistance) that exists between the collector and emittér terminals of Q2, which is driven into saturation as shown in Fig. 12.12(b). Hence, the output voltage decreases linearly from VCC to VCE(sat) in Ts and hence, is a negative-going ramp as shown in Fig. 12.12(d). Depending on the time constant employed, Ts may be less than or equal to Tg.
3. At the end of the trigger: Again at the end of the input pulse, at t = Tg, Q1 goes ON, Q2 goes OFF and the capacitor discharges through RC2 and the output again reaches VCC, as shown in Fig. 12.12(c). The waveforms are shown in Fig. 12.12(d).
4. Calculation of Ts:

From Fig. 12.12(b), the charging current of Cs:

FIGURE 12.12(a) A transistor Miller sweep generator

FIGURE 12.12(b) The circuit of Fig. 12.12(a) when Q1 is OFF and Q2 is ON

FIGURE 12.12(c) The discharge of Cs when Q1 is ON and Q2 is OFF

FIGURE 12.12(d) The waveforms of Miller’s sweep transistor

At

t = Ts, vo(t) = Vs

Therefore,

If

Vs = VCC, Ts = RC1Cs.

5. Calculation of Tr:

From Fig.12.12(c), the discharging current:

The change in voltage during Tr is once again Vs.

Therefore,

If Vs = VCC, then, Tr = R,C2Cs,

6. Calculation of the slope error: The slope error of the Miller’s sweep generator is given by the relation:

es = (Vs/VCC)[(1 + R/Ri) /|A|]

where, R is the resistance through which the capacitor charges when the switch is OFF. To calculate es, therefore, we have to calculate A and Ri of the common-emitter amplifier. The CE amplifier uses the h-parameter model as shown in Fig. 12.12(e):

We have from Fig. 12.12(e)

and

V2 = IoRC = −I2RC

Putting

V2 = −I2RC

in Eq. (12.38):

I2 = hfeI1 + hoe(I2RC)

Dividing by I1:

Therefore,

From Fig. 12.12(e), we also have the relation

Dividing by I1, we have:

FIGURE 12.12(e) A CE amplifier

A, the voltage gain is given by:

Ri and A can be calculated using Eqs. (12.41) and (12.42) and hence, the value of es.

##### EXAMPLE

Example 12.3: For the Miller’s sweep shown in Fig. 12.12(a), VCC = 25 V, RC2 = 5kΩ, RC1 = 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: hfe = 80, hie = 1kΩ, hoe = 1/40 kΩ and hre = 2.5 × 10−4.

Solution:

1. We have

Vs = VCC

Therefore,

2. Retrace time Tr = RC2 × Cs = 5 × 103 × 0.5 × 10−6 = 2.5 ms
3. Ri = hie + hreAIRC2 = 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 = vC, 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 Ri, i.e., vi = 0, Ri is replaced by a short circuit. As vi = 0, Avi = 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 Ro of the emitter follower is very small: vo ≈ 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 Ri:

Here, Ro is the output resistance of the emitter follower, which is small and Ri is its input resistance, which is large.

Therefore, Ro/Ri is negligible and A ≈ 1

Eq. (12.44) gives the peak-to-peak excursion of the output swing.

Therefore,

since A ≈ 1. If R = Ri, es(Bootstrap) = es

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, Ri >> 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 IB1 of Q1 gives rise to its collector current IC1(= hFE1IB1). IC1 becomes the base current IB2 for Q2 and there results a current IC2(= hFE2IB2 = hFE2 × hFE1IB1). Thus, we see that hFE = hFE2 × hFE1, 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: Ri = hie + (1 + hfe)RE.

For a Darlington pair, as hFE 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 C1 which is charged from the current through R1. The discharge transistor Q1, when ON, keeps V1 at VCE(sat) until a negative input pulse is applied. Q2 is an emitter follower with a low output resistance. Emitter resistance RE is connected to a negative supply (VEE) instead of referencing it to the ground to ensure that Q2 remains conducting even when its base voltage V1 is close to the ground. Capacitor C3, called the bootstrapping capacitance, has a much higher capacitance than C1. C3 is meant to maintain a constant voltage across R1 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, Q1 has sufficient base current. Therefore, Q1 is driven into saturation and the voltage V1 across the capacitor C1 is VCE(sat). V1 = VCE(sat) (point X), typically 0.2 V for Si, as shown in Fig. 12.16(b). Q2 is an emitter follower for which input is V1 and its output vo is:

vo = V1VBE2 = VCE(sat)VBE2 (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 VCE(sat) ≈ 0V, D1 is ON. The voltage across R1 is,

VR1 = VCCVD1VCE(sat)VCC

where, VD1 is the diode voltage when ON. Hence, the current I1 in R1 is VCC/R1 and is constant. The base current IB2 of Q2 is smaller than the collector current IC1 of Q1:

For Q1 to be in saturation:

IB1 > IB1(min)

Therefore,

For all practical purposes, when Q1 is ON, both V1 and Vo are zero.

(b) Sweep generation: At t = 0, as the trigger is applied, the voltage at the base of Q1 goes negative, Q1 is OFF, [see Fig. 12.16(c)]. The voltage at node K, VK = VCC + V1 and D1 is OFF and is an open circuit.

The voltage at node X is V1.

Therefore,

However, I1 = IC + IB2, where IC is the charging current of C1 and IB1 is the base current of Q2. As Q2 is an emitter follower with large input resistance, IB1 is small. As a result,

IC, the charging current of C1, is constant. Hence, the resultant sweep is linear. When D1 is OFF, the charging current I1 to C1 through R1 is supplied by C3 which is charged to VCC.

There is no current into the collector lead of Q1 and instead, this current flows through C1 charging it. As the voltage across the capacitor C1 varies as (I1/C)t, so does the output.

From Fig. 12.16(d), it is seen that the output vo varies linearly only when the duration of the gating signal (Tg) is small so that in this period vo does not reach VCC. However, if Tg is large, the output vo may reach VCC even before Tg. When vo = VCC, the voltage VCE2 of Q2 is practically zero (saturation). Q2 no longer behaves as an emitter follower. vo and V1 therefore remain at VCC. The current VCC/R1 now flows through C3, R1 and through the base-emitter diode of Q2; thereby changing the voltage across C3 by a small amount, ΔVC3.

(i) If Ts < Tg (i.e., vo reaches VCC before Tg) Then at

t = Ts

vo = Vs = VCC

From Eq. (12.46):

FIGURE 12.16(c) The circuit of Fig. 12.16(a) when Q1 is OFF and C1 charges

FIGURE 12.16(d) The output waveforms of the bootstrap sweep circuit

FIGURE 12.17 The calculation of I1

or

The amount of charge gained by C1 is QC1 = C1 vs

The amount of charge lost by C3 is QC3 = C3ΔVC3

Assume that during Ts the current through capacitors C1 and C3 are same.

Therefore, the amount of charge gained by C1 = the amount of charge lost by C3.

C3ΔVC3 = C1Vs

Therefore, at t = Ts, the voltage across C1 is Vs and the voltage across C3 is VC3 − (C1/C3)Vs and the output voltage is vo = AVs, as shown in Fig. 12.17.

IC = I1IB2

and

At

t = Ts

At,

t = Ts,

Therefore,

At t = 0, Vs = 0. Hence, from Eq. (12.48), IC = VC3/R1

Eq. (12.49) gives the expression for the slope error when the voltage VC3 changes during the sweep period.

(ii) On the other hand if Vs < VCC, the maximum sweep voltage is:

(c) Calculation of retrace time, Tr: At the end of the gate signal, at t = Tg, a current IB1 = (VCC/RB) again flows into the base terminal of Q1. Q1 once again tries to go into saturation. However, till such time VCE of Q1 is VCE(sat) (Q1 is in saturation), the collector current, iC1 remains constant at

As shown in Fig. 12.18, the current iR1 through R1 and the discharging current id of C1 now constitutes iC1, the collector current of Q1, neglecting the small base current IB2 of Q2,

Writing the KCL equation at node C:

iR1 remains approximately at VCC/R1(= I1) and the capacitor discharges with a constant current. The voltage across C1 falls, and consequently vo falls, since once again Q2 behaves as an emitter follower. The discharge current id from Eq. (12.51) is:

id = iC1iR1

FIGURE 12.18 The circuit to calculate id

Therefore V1 and vo fall linearly to the initial value. The voltage variation during the retrace time Tr 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 RB will have to be small to reduce the retrace time. However, if RB is too small, then the collector current of Q1 becomes large as:

This results in greater dissipation in Q1. During the period, T = Tg + Tr, though C3 is a large capacitor, it may still loose some charge. The circuit is said to have recovered completely only when the charge lost by C3 is regained. The minimum recovery time T1 for C3 can be found out as follows:

Charge lost by C3 in time T = VCC T/R1

The charging current of C3 = VEE/RE

Therefore, the charge recovered in time T1 = (VEE/RE)T1

Charge lost = charge regained

To reduce T1, VEE may be increased. However, this increases the quiescent current in Q2 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 C3 remains unaltered during the sweep duration, is given by the relation:

To calculate es, we have to calculate A and Ri. For a common collector amplifier we have:

since

hfc = −(1 + hfe) and hoc = hoe

since

hrc = 1 and hic = hie

From Eq. (12.56) dividing by Ri:

Therefore,

Using Eqs. (12.55), (12.56) and (12.57), it is possible to calculate es. 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, VCC = 15 V, VEE = −10 V, RB = 30 kΩ, R1 = 10 k Ω, RE = 5kΩ, C1 = 0.005 µF, C3 = 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 hFE = hfe = 50, hic = 1kΩ, 1/hoe = 40 kΩ, hrc = 1 Assume that the forward-biased 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 Vo and the collector current of Q1

Solution: Referring to the circuit in Fig. 12.16(a):

1. Sweep speed =

Vs(max) = VCC = 15 V = Sweep speed × Ts

i.e.,

15 = (3 × 105)Ts

Therefore, sweep time,

2. At the end of the input pulse, Q1 once again goes into saturation.

iC1 = hFEiB1 = 50 × 0.5 = 25 mA

The retrace time Tr is,

T = Tg + Tr = (60 + 3.18) = 63.18 μs

Recovery time T1 =

3. To find the slope error:

The current gain of the emitter follower is given by:

Therefore,

Input impedance of the emitter follower is given by:

Ri = hie + AI RE

Therefore,

A is the voltage gain of the emitter follower.

Ri = hie + AI RE = 1kΩ + 45.33 × 5kΩ = 227.67 kΩ

The slope error,

4. 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, hFE(min) = 100.

Solution:

Refer to the bootstrap circuit shown in Fig. 12.16(a).

RL = RE = 1

When Vo = 0,

When Vo = Vs(= 10V),

At Vo = 0,

At Vo = Vs,

ΔIB2 = 0.28 − 0.18 = 0.10 mA

I1 is much larger than IB2

Let

I1 = 100 × ΔIB2 = 100 × 0.10 mA = 10 mA

VR1 = VCCVD1VCE(sat) = 18 − 0.7 − 0.2 = 17.1V

For 1 per cent non-linearity due to discharge of C3:

ΔVC3 = 1 per cent of the initial VCC level

Initial

VC3 = VCC = 18 V

And C3 discharge current I1 is equal to 10 mA. Therefore,

The discharge time of C1 is 0.1ms which is 1/10 th of the charging time. For Q1 to discharge C1 in 1/10 th of the charging time,

IC1 = 10 × (C1charging current) = 10 × 10 mA = 100 mA

Choose RB = 20 kΩ, Q1 is to be biased OFF at the end of the input pulse.

|ΔVB| = vo − (pulse amplitude) = 0.7 − 5V = −4.3V

The charging current of C2is equal to the current through RB when Q1 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 VF = 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, VBB = 15 V, VYY = 30 V, R = 10 kΩ, C = 0.1 µF

We know that

VP = η VBB + VF = 0.6 × 15 + 0.7 = 9.7 V

From UJT characteristics, we have VV = 1.6 V

(a) Amplitude of the sweep = Vs = VPVV

= 9.7 − 1.6 = 8.1 V.

(b) Sweep speed error,

Displacement error,

FIGURE 12.20(c) The waveform of the sweep generator

(c) Sweep time,

Example 12.7: Using the characteristic of UJT shown in Fig. 12.21 (a), calculate the values of R, C, R1 and R2 of the relaxation oscillator shown in Fig. 12.21; (b) to generate a sweep with a frequency of 10 kHZ and amplitude of 10V, Tr is 0.5 % of T.

FIGURE 12.21(a) UJT characteristic; (b) UJT relaxation oscillator

Solution:

Given,

f = 10 kHz, Vs = 10 V, VBB = 15 V, VV = 1V, IP = 14 µA.

IV = 1 mA, η = 0.6

Therefore,

VP = Vs + VV = 10 + 1 = 11 V

R should lie between R(max) and R(min).

Let R = 150 kΩ.

Sweep time Ts =

Ts = 1.252RC

As Tr = 0.5% of T,

Tr = 0.5µs

Ts = TTr = 100 − 0.5 = 99.5μs

Therefore,

Return time Tr = R1C= 0.5µs

Therefore,

The value of R2 is usually higher than R1. So, choose R2 = 3kΩ.

Example 12.8: For the UJT relaxation oscillator shown in Fig.12.21(c), RBB = 3kΩ, R1 = 0.1 kΩ, η = 0.7, VV = 2V, IV = 10 mA, IP = 0.01 mA. (a) Calculate RB1 and RB2 under quiescent condition (i.e., when IE = 0). (b) Calculate the peak voltage, VP. (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 RB1 = 0.1 kΩ during the retrace time. (f) Calculate the voltage levels of VB1. (g)Plot the waveforms of the output voltage and VB1.

FIGURE 12.21(c) The UJT oscillator

Solution:

Given RBB = 3kΩ, η = 0.7.

1.

RB1 = 0.7 × 3 kΩ = 2.1 kΩ

We have

RBB = RB1 + RB2

FIGURE 12.21(d) The circuit to calculae Vp

Therefore,

RB2 = RBBRB1 = 3kΩ − 2.1 kΩ = 0.9 kΩ

2. The circuit that enables the calculation of VP is shown in Fig. 12.22(d). From Fig. 12.22(d):
3.

R(min) < R < R(max)

Choose

R = 100 kΩ

4.

Ts using the value of η is given as:

5.

T = Ts + Tr = 6.335 + 0.01735 = 6.352 ms

6. VB1 during charging of C is given as:

VB1 during discharge of C is given by:

FIGURE 12.21(e) The waveforms of the output and VB1

VF is the diode voltage when ON, Fig. 12.21(d).

7. Waveforms of the output and VB1 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 hFE(min) = 30, VCE(sat) = 0.3 V, VBE(sat) = 0.7 V, VBE(active) = VD = 0.6 V

FIGURE 12.22 A bootstrap sweep generator

Solution:

1. Current through R1 is:

Base Current of Q1 is:

Emitter current of Q2 = iE2 =

vo = VC1VBE2(active)

vo = VCE(sat) − VBE2(active) = 0.3 − 0.6 = −0.3V

Therefore,

Therefore,

iB2 = iE2iC2 = 0.97 × 10−3 − 0.938 × 10−3 = 0.032 mA

iC = I1iB2 = 1.14 × 10−3 − 0.032 × 10−3 = 1.108 mA

VCE2 = VCCvo = 18 + 0.3 = 18.3V

2. Sweep time Ts = R1C1 = 15 × 103 × 0.01 × 10−6 = 0.15 ms

Therefore,

Tg = 0.5 ms

Since

Ts < Tg, Vs = VCC = 18 V

Run time,

T = Tg + Tr = 0.5 + 0.075 = 0.575 ms

Example 12.10: Determine the values of R1, RE, RB, C1 and C3 of a bootstrap circuit shown in Fig. 12.23, if IC1 = 1.5 mA, IE2 = 1.75 mA, hFE(min) = 30, VCE(sat) = 0.3 V, VBE(sat) = 0.7 V and VBE(active) = VD1 = 0.6 V. Assume Ts = 12 ms.

Solution:

To calculate R1:
Applying KVL to collector loop of Q1, we have the collector voltage of Q1 when it is ON:

VC1 = VCC - VD1 - I1R1

To calculate RB:

Applying KVL to the base loop of Q1 we have VCCIB1RBVBE(sat) = 0

FIGURE 12.23 A bootstrap sweep generator

Choose,

Therefore,

To calculate C1:

Sweep Time Ts = R1C1

Therefore,

Given

Ts = 12 ms

The value of C3 should be larger than C1, therefore, choose C3 as 10 times of C1.

Therefore,

C3 = 9.43µF

To Calculate RE:

The output voltage vo = VCE(sat)VBE(active)

Therefore,

vo = 0.3 − 0.6 = −0.3

Applying KVL, to the input side of Q2 we have:

vo + VEEREiE2 = 0

Therefore,

##### SUMMARY
• Voltage sweep generators are based on the principle of electrostatic deflection.
• In an ideal sweep generator (saw-tooth generator), the sweep voltage increases linearly till Ts (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 fly-back time.
• Slope error es 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 ed 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 et 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 es is given by Vs/V where Vs is the sweep amplitude and V is the peak-to-peak excursion of the output swing. The larger the V, the smaller the deviation from linearity.
• Slope error es is also given as Ts where Ts 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: ed = (1/8)es = (1/4)et.
• A UJT is a negative resistance device and is used as a switch.
• The sweep duration of a UJT sweep generator is given by Ts = τ ln(1/(1 − η)) where, η is the intrinsic stand-off 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 Ts = τ ln (VBBVV)/(VBBVP) where VBB is supply voltage, VV and VP 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 es(Miller) = es × 1 + (R/Ri)/|A| and this is small as |A| is large.
• The slope error of a bootstrap sweep generator is es(Bootstrap)es(R/Ri) and is small only when Ri is large.
##### MULTIPLE CHOICE QUESTIONS
1. The application of a voltage time base generator is in a measuring instrument like:
1. CRO
2. Frequency counters
3. Digital voltmeter
4. Multimeter
2. The deflection of the electron beam in a CRO employing a voltage sweep generator is based on the principle of:
1. Electromagnetic deflection
2. Electrostatic deflection
3. Frequency multiplication
4. Counting
3. Errors es, ed and et define:
1. Current gain
2. Voltage gain
3. Deviation from linearity
4. Bandwidth
4. 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:
1. 5 %
2. 20 %
3. 200 %
4. 0.5 %
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:
1. 5 %
2. 100 %
3. 20 %
4. 10 %
6. For the output of a voltage sweep generator to be linear, the capacitor must be charged with a:
1. Constant current
2. Linearly varying current
3. Exponentially varying current
4. Zero current
7. A Miller sweep generator produces a:
1. Positive-going sweep
2. Negative-going sweep
3. Constant sweep voltage
4. Infinite output voltage
8. A bootstrap generator produces a:
1. Positive-going sweep voltage
2. Negative-going sweep voltage
3. Constant sweep voltage
4. Infinite sweep voltage
9. The slope error es and the displacement error ed are related as:
1. es = 2ed
2. 8 es = ed
3. es = 4 ed
4. es = 8 ed
10. The displacement error ed and the transmission error et are related as:
1. et = 4 ed
2. et = 2 ed
3. 4 et = ed
4. et = ed
11. The slope error es and the transmission error et are related as:
1. es = 2 et
2. es = 0.5et
3. es = et
4. es = 4et
12. The sweep duration of a UJT sweep generator is given as:
1. 0.69 τ
2. τ ln{(VBBVP)/(VBBVV)}
3. τ ln{(VBBVV)/(VBBVP)}
4. 1.4 τ
1. Define the three errors that specify deviation from linearity.
2. Draw the circuit of a UJT sweep generator and explain its operation.
3. Derive the expression for the frequency of oscillations of a UJT sweep generator in terms of η, the intrinsic stand-off ratio.
4. Derive the expression for the frequency of oscillations of a UJT sweep generator in terms of VV and VP.
5. Explain the basic principle employed in a Miller sweep to linearize the non-linear sweep.
6. Explain the basic principle employed in a bootstrap sweep to linearize a non-linear sweep.
1. Draw the circuit of a UJT sweep generator and explain its operation. Derive the expression for es, ed and et and obtain their inter-relationship.
2. 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.
3. 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.
4. 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
1. For the UJT relaxation oscillator shown in Fig. 12p.1, RBB = 4kΩ, R1 = 0.1 kΩ, η = 0.6, VV = 3V, IV = 10 mA, IP = 0.01 mA.
1. Calculate RB1 and RB2 under quiescent condition (i.e., IE = 0).
2. Calculate the peak voltage,VP.
3. Calculate the permissible value of R.
4. Calculate the frequency assuming that the retrace time is negligible. Also calculate f using the value of η.
5. Calculate the frequency, also considering the retrace time. Assume RB1 = 0.1 kΩ during the retrace time.
6. Calculate the voltage levels of VB1.
7. Plot the waveforms of the output voltage and VB1.

FIGURE 12p.1 The given UJT sweep circuit

2. The UJT relaxation oscillator is shown in Fig. 12p.2. Find:
1. The sweep amplitude.
2. The slope and displacement errors.
3. The duration of the sweep.
Given that VV = 3V, η = 0.6.

FIGURE 12p.2 The given UJT relaxation oscillator

3. 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 Tr = 1% of Ts, calculate VBB, VYY, R, R1, R2 and C. If the sweep duration is 300 µs, calculate the new value of C. The V-I 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

4. For the Miller’s sweep shown in Fig. 12.12(a), VCC = 24 V, RC2 = 2kΩ, RC1 = 10 kΩ and C = 1µF. The amplitude of the sweep is 18 V. (a) Calculate the sweep duration Ts; (b) the retrace time Tr; (c) frequency of the sweep generator and (d) the slope error. The transistor has the following parameters: hfe = 100, hie = 1kΩ, hoe = 1/20 kΩ and hre = 2.5 × 10−4.
5. The transistor used in the bootstrap circuit shown in Fig. 12p.4 has the following h-parameter values: hre = 2.5 × 10 − 4, hie = 1.1 kΩ, hfe = 60, 1/hoe = 40 kΩ. Assume VBE(sat) = VCE(sat) = 0. If the applied input gating voltage is a symmetrical square wave of the frequency 9.5 KHz, determine the time-base amplitude, the retrace time and the recovery time.

FIGURE 12p.4 The given bootstrap sweep circuit

6. 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 h-parameter values of transistor are, hfe = 90, 1/hoe = 35 kΩ, hie = 1kΩ and hrc = 1. Assume all forward-biased junction voltages are zero.

FIGURE 12p.5 A bootstrap sweep circuit

7. The transistor bootstrap has the following parameters: VCC = 20 V, VEE = −20 V, RB = 15 kΩ, R1 = 5kΩ, RE = 2.5 kΩ, C1 = 0.001 µF, C3 = 0.25 µF. The input gate has the amplitude of 1 V and a width of 50 µs. The transistor parameters are hfe = 60, hie = 2kΩ, 1/hoe = 10 kΩ, hre = 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 C3 and the recovery time and (d) the slope error. (ii) Plot the gate voltage, collector current iC1, and the output voltage vo.