# 12.2 Exponential Sweep Generators – Pulse and Digital Circuits

##### 12.2 EXPONENTIAL SWEEP GENERATORS

In a simple voltage sweep generator, a capacitor (C) is allowed to charge to a voltage (V) through a resistance (R) with the time constant, RC, deciding the rate of charge of the condenser. However, as the capacitor charges exponentially, the resultant sweep voltage so generated may tend to be exponential or in other words, not necessarily linear. Thus, there is a need make arrangements to linearize an exponential sweep. In such arrangements, a constant current is used to charge the capacitor.

The three types of voltage sweep generators considered in this chapter include exponential sweep generators, Miller’s sweep generators, and bootstrap sweep generators.

A simple exponential sweep generator and its output are shown in Fig. 12.2(a) and (b), respectively. Initially, at t = 0, let the capacitor be uncharged. If now the switch S is open, then the capacitor tries to charge to the supply voltage V. At t = Ts (sweep duration), when the voltage across the capacitor is Vs, if the switch is suddenly closed, the voltage across the capacitor, ideally, is expected to abruptly fall to zero.

However, if the resistance offered by the switch is ideally not zero, there is a finite time delay before the signal reaches its initial value. This time delay is called the fly-back time, restoration time or retrace time (Tr), as shown in Fig.12.2(c).

Normally, Tr << Ts, so that TTs.

The voltage variation of the sweep voltage vs is given as:

vs = vf − (vfvi)et/τ

Here, vf = V and vi = 0. Therefore, vs = V − (V − 0)et/τ

We assume that after an interval Ts, when vs = Vs, the switch closes. Then the charge on the capacitor discharges with a negligible time constant and the voltage abruptly falls to zero at t = TS. From Eq. (12.4), we have:

FIGURE 12.2(a) A simple exponential sweep generator; and (b) output of the sweep generator

FIGURE 12.2(c) The waveform that depicts the sweep time and restoration time

The initial slope is:

The final slope is:

Therefore,

From Eq. (12.4), at t = TS, vs = Vs:

Hence,

Substituting Eq. (12.7) in Eq. (12.5) we get:

From Eq. (12.8), it is evident that es is small when V >>Vs, i.e., linearity improves only if the supply voltage (V) is large when compared to Vs, the sweep amplitude. Therefore, the disadvantage of a simple exponential sweep is that a linear sweep is generated only when the sweep amplitude is much smaller than the applied supply voltage, V. For example if Vs = 20 V, V = 100 V

And if Vs = 20 V, V = 1000 V

The above illustration explains that, for the same sweep amplitude, the smaller is the supply voltage the larger is the slope error. If the supply voltage is increased, the slope error decreases, which means linearity improves.

If t/τ << 1

We have from Eq.12.4:

Since vs = at t = Ts, for a linear sweep, then to the first approximation.

As this is a linear sweep:

Hence, for es to be small, τ >> Ts, i.e., the time constant employed in the circuit should be much larger than the sweep duration. If the actual sweep is non-linear, consider the first two terms given in Eq. (12.9):

Therefore, at: t = Ts:

This is a non-linear sweep. Hence, the transmission error et is:

Where is the amplitude of the linear sweep and Vs is the amphtude of the non-linear sweep.

From Eq. (12.11) we have:

If we relate es and et:

Displacement error, ed is:

From Eq. (12.12)

The deviation is maximum at t = (Ts/2)

Therefore,

At

t = TS    = Vs

Therefore,

Therefore,

From Eqs. (12.15) and (12.16), the interrelationship between the three types of errors is given as:

If we calculate one type of error, we can calculate the other types of errors also. If the capacitor is charged with a current I, the voltage across the capacitor C is (I/C)t. The rate of change of the voltage with time is called sweep speed. Therefore, sweep speed = I/C.

In the sweep generator described in Fig. 12.2(a), a switch S, which switches ON and OFF automatically at predetermined voltage levels, is used. The physical form of this switch could be a unijunction transistor (UJT).

#### 12.2.1 A Voltage Sweep Generator Using a UJT

A UJT and its dc circuit are shown in Figs. 12.3(a) and (b), respectively. A UJT consists of an n-type semiconductor bar with leads B1 and B2 drawn. The emitter is composed of a heavily doped p-type material. Let a bias voltage, VBB, be applied. RB1 is the resistance offered by the semiconductor bar from the emitter to base 1 and RB2 is the resistance offered by the semiconductor bar from the emitter to base 2.

From Fig. 12.3(b) we have:

where RBB = (RB1 + RB2) and η = (RB1/RBB) is the intrinsic stand-off ratio (lies between 0.6 and 0.8).

FIGURE 12.3(a) A unijunction transistor; and (b) its dc circuit

The p-type emitter and n-type semiconductor behave as a pn junction diode, D. As long as vi < ηVBB, D is OFF. When viηVBB, D is ON and a large number of charge carriers exist on the n-side, thus, reducing the resistance. The device conducts heavily and the switch S is said to be closed. The V–I characteristic of a UJT is shown in Fig. 12.3(c). From the characteristic shown in Fig. 12.3(c), it is seen that the UJT is ON when vi = VP, called the peak voltage, and is OFF when vi = VV, called the valley voltage. The device switches ON or OFF at predefined voltage levels. A practical UJT sweep generator is shown in Fig. 12.4(a) and its output is shown in Fig. 12.4(b). Initially, let the capacitor be uncharged. Once the power is ON, the voltage across the capacitor rises exponentially as:

FIGURE 12.3(c) The V–I characterstic of a UJT

At t = Ts, vs = ηVBB

Alternately,

vs = vf − (vfvi)et/τ

We have, from Fig. 12.4(b), vi = VV and vf = VBB

Therefore,

vs = VBB − (VBBVV) et/τ

FIGURE 12.4(a) A pratical UJT sweep generator; (b) its output

FIGURE 12.5(a) The resistance R1 is added to generate positive spikes

FIGURE 12.5(b) The circuit to calculate VB1 during the charging period

At t = Ts, vs = VP      VP = VBB − (VBBVV) eTs

Equations (12.20) and (12.21) give the sweep duration.

A small resistance (R1), as shown in Fig. 12.5(a), can be now included to derive a positive spike at B1, which can then be used to trigger some other circuit. The circuit, when the UJT is OFF (during the charging period), from which VB1 is calculated is shown in Fig. 12.5(b).

From the circuit shown in Fig. 12.5(b), we have:

FIGURE 12.5(c) The circuit to calculate VB1 when the UJT is ON

FIGURE 12.5(d) A UJT circuit to generate positive and negative spikes at B1 and B2

The circuit when the UJT is ON (during the discharge period), from which VB1 is calculated is shown in Fig. 12.5(c).

From the circuit shown in Fig. 12.5(c), VB1 is calculated as:

If negative spikes are also to be generated, a resistance R2 which is reasonably larger than R1 is connected to B2, as shown in Fig. 12.5(d). The resultant waveforms are plotted as shown in Fig. 12.5(e).

In the waveform shown in Fig. 12.5(e), the output vs is assumed to fall abruptly to VV at Ts. In practice, however, there will be a finite retrace time, as an additional resistance R1 is connected to B1. So, there is a need to calculate Tr to be able to calculate the time period (T) of the sweep (= Ts + Tr), as shown in Fig. 12.5(f). The output voltage for t > Ts is given as:

vs(t > Ts) = VPet/(RB1+R1)C

At t = Tr, vs = VV.

VV = VPeTr/(RB1+R1)C

Therefore,

From Eq. (12.21), we have:

Combining Eqs. (12.21) and (12.22), we now have the total time period, T = Ts + Tr.

The design of an exponential sweep generator using UJT as the switch is illustrated in Example 12.1.

FIGURE 12.5(e) The voltages at B1 and B2 and Vo

FIGURE 12.5(f) The practical exponential sweep

##### EXAMPLE

Example 12.1: Design a relaxation oscillator using a UJT, with VV = 3 V, η = 0.68 to 0.82, IP = 2µA, IV = 1 mA, VBB = 20 V, the output frequency is to be 5 kHz. Calculate the typical peak-to-peak output voltage.

Solution: The given UJT has the following parameters:

VV = 3V, IP = 2 µA, IV = 1 mA, η = 0.68 to 0.82

VP = VF + ηVBB

Therefore,

VP = 0.7 + (0.75)(20) = 15.7 V

Thus, R must be in the range 17k Ω to 2.15MΩ. If R is large, C must be very small. Therefore, choose R such that C is not very small.

Choosing R = 22 kΩ

Choose C = 6800 pF (a standard value)

Peak-to-peak sweep amplitude = 15.7 − 3 = 12.7 V.

#### 12.2.2 Generation of Linear Sweep Using the CB Configuration

The sweep voltage generated by an exponential sweep generator is non-linear as the current in the capacitor varies exponentially. However, to generate a linear sweep, the capacitor is required to charge with a constant current. As discussed in the earlier chapters, the collector current in the CE configuration may not remain constant with the variation in VCE. However, from the output characteristics of the CB configuration shown in Fig. 12.6, we see that for a constant value of IE, IC is independent of VCB and the curves are parallel to the VCB axis, except for a small range of values of VCB.

As can be seen from Fig. 12.6, IC remains practically constant with the variation in VCB, except, at smaller values of VCB. This suggests that if a capacitor is charged using the constant collector current of the CB configuration, the resultant sweep voltage must be a linear sweep voltage. Fig. 12.7(a) shows the circuit of a transistor constant current sweep generator using the CB configuration.

From Fig. 12.7(a), the current IE in the base loop is:

Let the switch S be open at t = 0. The collector current is constant and is given by the relation: IC = hFBIE Hence, C charges with the constant IC and the voltage across the capacitor varies linearly as a function of time. To determine the sweep voltage Vs, let us consider the small signal model of the transistor in the CB configuration as shown in Fig. 12.7(b).

Writing the KVL equation of the input loop:

Writing the KCL equation at the output node C:

FIGURE 12.6 The output characteristics of the CB configuration

FIGURE 12.7(a) The CB constant current sweep generator

FIGURE 12.7(b) The equivalent circuit of Fig. 12.17(a)

From Eq. (12.23):

Substituting this value of Ie in Eq.(12.24):

Therefore,

Let

Let

Applying Laplace transforms:

Put

s = 0

Therefore,

A =

Put

B = −

Substituting the values of A and B:

Taking inverse Laplace transform on both sides:

vs (t) = (1 − et/τ)

Putting the value of K:

Expanding et/τ as series and limiting only to the first two terms of the expansion:

FIGURE 12.8 A constant current sweep with a single supply

Eq. (12.27) reduces to:

At t = Ts, vs = Vs

Therefore,

Therefore,

where Ts is the sweep duration and Vs the sweep amplitude.

From Eqs. (12.25) and (12.30), we get:

The slope error

Therefore,

As the second term in the bracket tends to be small, es is small. The constant current sweep generator shown in Fig. 12.7(a) is modified such that a single source VYY is used to derive VEE and VCC sources, as shown in Fig.12.8.

In Fig. 12.8, D1 and D2 are Zener diodes. VEE is derived from the VYY source using these diodes such that VEE = (VZ1 + VZ2) where VZ1 and VZ2 are the breakdown voltages of D1 and D2. Similarly, the VCC source is also derived using the VYY source (the Zener current through R derives VCC). Hence, a single voltage source is used in the circuit shown in Fig. 12.8 when compared to the circuit shown in Fig. 12.7(a). RL is the load connected. Example 12.2 highlights the procedure to calculate the slope error of this circuit.

##### EXAMPLE

Example 12.2: For the circuit shown in Fig. 12.8, it is given that: VYY = 20 V, VZ1 = 6.8 V, VZ2 = 3.8 V, hrb = 3 × 10−4, hib = 20Ω, hob = 0.5µmhos, α = 0.98 and RE = 1kΩ. Find the slope error: (a) when RL = ∞ (b) when RL = 200 kΩ and (c) when RL = 50 kΩ.

Solution:

VEE = VZ1 + VZ2 = 6.8 + 3.8 = 10.6 V

VCC = VYYVEE = 20 − 10.6 = 9.4 V

Taking the junction voltages into account, the sweep voltage is:

Vs = VCCVBE = 9.4 − 0.6 = 8.8 V.

1. The slope error of a transistor constant current sweep is:

Vi = VEEVγ = 10.6 − 0.5 = 10.1 V and Vs = 8.8 V.

This error is small and hence, the sweep is linear.

2. If RL is connected as the load, then hob and RL are in parallel. The effective admittance is:

= hob +

If RL = 200 kΩ, then the effective admittance:

= 0.5 × 10−6 + 0.5 × 10−5 = 5.5 × 10−6mhos

In this case the slope error becomes large.

es ≈ 0.52 %

3. RL = 50 k Ω, then the effective admittance

= 0.5 × 10−6 + 0.2 × 10−4 = 20.5 × 10−6mhos

In this case, the slope error becomes large.

es ≈ 1.88%

The slope error progressively becomes large as RL decreases.