15.5 SYNCHRONIZATION OF A SWEEP CIRCUIT WITH SYMMETRICAL SIGNALS
So far, we have considered the synchronization of relaxation circuits with external sync signals that are essentially pulses only. However, we can also synchronize a relaxation circuit such as a sweep generator with sync signals that could as well be gradually varying signals like sinusoidal signals. Let us consider the output of a UJT sweep generator that is to be synchronized with a slowly varying sinusoidal signal, as shown in Fig. 15.13(a).
Let it be assumed that the breakdown voltage of the UJT varies sinusoidally in the presence of the sync signal. Here, Vpo is the quiescent breakdown voltage of the UJT and Vp is the breakdown voltage in the presence of the sync signal. It is possible that synchronization can be effected with T = To. If this happens, then the period of the sweep is not altered by the sync signal and the sweep amplitude is also unaffected. The sweep cycle, as a result, terminates at Vpo, which means that the sweep terminates on its own at points labeled ‘o’ in Fig. 15.13(a).
Earlier, when synchronization was achieved using a pulse train as sync signals, it was observed that for synchronization to take place it was imperative that the spacing between pulses (Tp) should be less than or equal to the natural time period (To) of the relaxation circuit. It was also observed that a pulse could prematurely and reliably terminate a sweep cycle (reduce the sweep duration) but will not be able to extend the sweep duration. However, when it is a case of synchronization with symmetric signals, synchronization is always possible whether T (Tp in the case of a pulse train) is less than or equal to To or T is greater than To.
FIGURE 15.13(a) The synchronization of a sweep generator with sinusoidal sync signal
It is seen from Fig. 15.13(b) that if the sweep voltage meets the VP curve at a point above Vpo for T > To, say X, then the duration of the sweep is lengthened ( > To). On the other hand, if the sweep voltage meets the VP curve at a point below Vpo for T < To, the duration of the sweep is shortened ( < To).
FIGURE 15.13(b) The timing relationship between the sweep voltage and the time-varying breakdown voltage when (i) T = To, (ii) T > To, then > To and (iii) T < To, then < To
Let us summarize this with the help of Fig. 15.13(c):
- When T = To, the sweep is terminated at ‘o’ on the Vpo line, leaving the period and the amplitude of the sweep unaltered.
- When T > To, if the sweep terminates at a point say, X that lies between ‘o’ and positive maximum, at A, then the sweep is lengthened and its duration is greater than To. This lengthening is maximum when the sweep terminates at A.
- When T < To, if the sweep terminates at a point, say Y, that lies between ‘o’ and the negative maximum at B, then the sweep is shortened and its duration is smaller than To. This shortening is maximum when the sweep terminates at B. To calculate the range of synchronization let us consider an example.
FIGURE 15.13(c) The synchronization when T ≠ To
Example 15.1: A UJT sweep operates with a valley voltage of 4 V and peak voltage of 16 V. A sinusoidal synchronizing voltage of 3 V peak is applied as a sync signal. η = 0.5. If the natural frequency of the sweep is 1 kHz, over what range of sync signal frequency will the sweep remain in 1:1 synchronization with the sync signal.
The quiescent breakdown voltage, Vpo = 16 V.
Peak-to-peak amplitude of the synchronizing signal = 3 V
In the absence of the sync signal, the peak-to-peak swing of the sweep = Vpo − VV = 16 − 4 = 12 V
In the presence of the sync signal, the sweep amplitude must therefore lie in the range
(12 − 1.5) = 10.5 V and (12 + 1.5) = 13.5 V.
Time period of the sync signal,
Amplitude of the natural sweep signal = 16 − 4 = 12 V.
And this sweep amplitude is generated in 1ms. Therefore, the time required to generate a sweep of, 10.5 V is
and the corresponding frequency is,
The time required to generate a sweep of 13.5 V is
and the corresponding frequency is
It is seen from the above calculations that the sweep generator remains synchronized as the frequency of the sync signal varies from 889 c/s to 1143 c/s.
Let us now consider the operation of a sweep circuit as a frequency divider using sinusoidal signal as sync signal, as shown in Fig. 15.14(a). The solid lines represent the sweep with a time period To and the sync signal with a time period T. The natural sweep terminates on Vpo line. In the presence of the sinusoidal sync signal, if the sweep meets the sync signal above the Vpo line at X, the new time period of the sweep is Ts. Thus, the sweep period To changes to Ts as a result of the sync signal. The sync signal completes three cycles during the period Ts, resulting in the division by a factor 3 as Ts = 3T (counting ratio of 3).
If now the amplitude of the sync signal is increased (dashed line), keeping the time period the same as T, as shown in Fig. 15.14(b), the sweep meets the sync signal at Y between O and B. The duration of the sweep is shortened (dashed line) resulting in a 2:1 synchronization (a counting ratio of 2). If the amplitude of the sync signals further increases, it could result in a counting ratio of 1. Hence, this circuit can operate as a counter.
Increasing the amplitude of the sync signal, in principle, can cause 1:1 synchronization. The sweep is terminated prematurely when it meets the sync signal below the Vpo line. Beyond this point once again the sweep voltage increases and this time it will terminate on the sync signal above the Vpo line. Therefore, the actual sweep waveform consists of the alternate sweeps of short and long durations. The suggestion therefore is that if this sweep is used to cause deflection of the electron beam along the X-axis in a CRO, it is preferable to use a sync signal of smaller amplitude, as shown in Fig. 15.14(c).
FIGURE 15.14(a) 3:1 synchronization of the sweep with symmetric sync signal
FIGURE 15.14(b) 2:1 Synchronization of the sweep with a symmetric sync signal
FIGURE 15.14(c) The excessive amplitude of the sync signal in a sweep resulting in sweeps of long and short durations
Example 15.2: The UJT relaxation oscillator shown in Fig. 15.15(a) is to be used as 3:1 divider for pulses which occur at 2.5 kHz rate. The available supply voltage is 30 V. The pulses are applied at the base B2. Draw the waveforms and calculate the pulse amplitude. Given R = 2 kΩ, C = 0.5 μF, η = 0.5 and Vγ = 0.6 V.
FIGURE 15.15(a) The UJT relaxation circuit with a sync signal
Solution: From the given data:
Pulse frequency f = 2.5 kHz
Pulse time period Tp = = 0.4 ms
For 3:1 synchronization, the frequency of the UJT oscillator should be 1/3rd of the pulse frequency.
Output time period To = 3Tp = 3 × 0.4 ms = 1.2 ms
Let VA be the peak-to-peak pulse input and it is negative.
vC = vf (1 − e−t/τ)
At t = 1.2 ms, vC = VPf
Vpf = VP − (−VA) = VP + VA VP = ηVBB + Vγ
t = 1.2 ms, vo = VP + VA
vf (1 − e−t/τ) = (ηVBB + Vγ) + VA
30(1 − e−1.2×10−3/2×103×0.5×10−6) = (0.5 × 30 + 0.6) + VA
VA = 30(1 − 0.3) − 15.6 = 21 − 15.6 = 5.4 V
The waveforms are drawn as shown in Fig. 15.15(b).
FIGURE 15.15(b) The waveforms
Example 15.3: A free running relaxation oscillator shown in Fig. 15.16(a) has sweep amplitude of 100 V and a period of 1 ms. Synchronizing pulses are applied to the device such that the breakdown voltage is lowered by 50 V at each pulse. The synchronizing pulse frequency is 4 kHz. What is the amplitude and the frequency of the synchronized oscillator waveform?
FIGURE 15.16(a) The UJT oscillator; (b) the waveforms
To = 1ms (fo = 1 kHz) and Vs = 100 V
TP = × 10−3 = 0.25 ms since fp = 4 kHz Tp < To
The unsynchronized sweep amplitude reaches 100 V in 1ms. So it reaches 50 V in (50 × 1)/100 = 0.5 ms.
In 0.5 ms, two sync pulses occur and consequently, 2:1 synchronization is achieved. The amplitude of the sweep is 50 V. The waveforms are shown in Fig. 15.16(b).
- If, there are a number of waveform generators in a pulse or a digital system and all of them arrive at some reference point in their cycle at the same time, then these waveform generators are said to be running in synchronism.
- If two waveform generators operate at two different frequencies — one completes only one cycle and during the same time the other generator completes precisely a fixed integral number of cycles, then these two generators are said to be running in synchronism but with the frequency division.
- A relaxation circuit is one in which a capacitor charges during a finite time interval and abruptly discharges when the voltage across the capacitor reaches a predetermined level.
- Sweep generators, blocking oscillators and multivibrators are some examples of relaxation circuits.
- Synchronization of a relaxation circuit can be achieved by either the external pulse train or the symmetrical signals like sinusoidal signals as sync signals.
- To establish synchronization using pulse train as sync signals, it becomes necessary that the spacing between the sync pulses should be less than or equal to the sweep duration of the relaxation circuit.
- Synchronization cannot be achieved if the spacing between the sync pulses is larger than the sweep duration or if the amplitude of the sync signals is very small.
- If the magnitude of the sync signals is appreciably large, synchronization is always possible.
- A monostable multivibrator can be used as a frequency divider.
- When a synchronizing input pulse is applied to a frequency divider, there will be a finite time delay before the output is available. This time delay is called phase delay.
- If the phase delay is subjected to periodic variations then it is called the phase jitter.
- When symmetric signals like sinusoidal signals are used as sync signals, synchronization is possible even if the time period T (Tp in the case of pulse train) is more than the natural time period of the relaxation circuit To.
- When T > To, the duration of the sweep cycle is lengthened () and when T < To, the duration of the sweep cycle is shortened ().
MULTIPLE CHOICE QUESTIONS
- Sweep generators, multivibrators and blocking oscillators are classified as:
- Relaxation circuits
- Multistage amplifiers
- Sampling gates
- Clipping and clamping circuits
- When synchronization with the frequency division is implemented in a pulse or a digital system, this circuit can be called as a:
- Scale-of-two circuit
- Counting circuits
- Schmitt trigger
- Sweep generator
- Timing interval in a relaxation circuit is established by the:
- Charging and discharging of a capacitor
- Charging of a capacitor
- Discharging of a capacitor
- None of the above
- A relaxation circuit can be synchronized by a train of sync pulses. For synchronization to take place:
- It is possible to synchronize the output of a UJT relaxation circuit with symmetric sync signals when:
- T > To
- T < To
- T ≤ To and T ≥ To
- None of the above
- The time interval between the instant the input pulse is applied to a frequency divider to the instant it appears at the output is called:
- Phase delay
- Resolution time
- Propagation delay
- Storage time
- The phase delay that varies due to the cumulative effect of the variations in the device characteristics, supply voltage and noise is termed as:
- Noise margin
- Resolution time
- Propagation delay
- Phase jitter
- When synchronizing the output of a UJT relaxation circuit with symmetric signals, if T > To, the duration of the sweep is:
- Remains unchanged
- None of the above
- When synchronizing the output of a UJT relaxation circuit with symmetric signals, if T < To, the duration of the sweep is:
- Remains unchanged
- None of the above
SHORT ANSWER QUESTIONS
- Explain the principle of frequency synchronization and division.
- It is said that synchronization with pulses is not possible if the spacing between the pulses (Tp) is larger than the time period (To) of the UJT oscillator. Is synchronization possible with symmetric signals if T > To. Explain.
- What is meant by jitter in a frequency divider?
- What are the factors responsible for the instability in the natural time period of a frequency divider?
- Is it possible to achieve frequency synchronization with division using symmetric signals as sync signals?
- What is the effect of amplitude of the symmetric sync signals on frequency synchronization of a relaxation oscillator, with division?
LONG ANSWER QUESTIONS
- With the help of a neat circuit diagram and waveforms, explain the method to achieve frequency synchronization using pulse train as sync signals.
- What is meant by synchronization with frequency division? Explain, with suitable waveforms, the procedure to obtain 3:1 and 5:1 synchronization.
- Explain suitable methods to synchronize both the time periods in an astable multivibrator. It is required that the time period T2 be synchronized with a division of 2:1 and the time period T1 be synchronized with a division of 3:1.
- Show, with the help of suitable waveforms, how a monostable multivibrator can be used as a frequency divider.
- What is phase jitter? Explain the method to eliminate it.
- Explain how synchronization and synchronization with division is possible with symmetric signals as sync signals.
- A UJT sweep operates with a valley voltage of 3 V and a peak voltage of 15 V. A sinusoidal synchronizing voltage peak of 6 V is applied as a sync signal. If the natural frequency of the sweep is 2 kHz over what range of sync signal frequency will the sweep remain in 1:1 synchronization with the sync signal.
- The UJT relaxation oscillator shown in Fig. 15p.2 is to be used as 2:1 divider for pulses which occur at a 2 kHz rate. The available supply voltage is 30 V. The pulses are applied at the base B2. Calculate the pulse amplitude and plot the waveforms. Given R = 2 kΩ, C = 0.5 μF, η = 0.5 and Vγ = 0.6 V.
FIGURE 15p.2 The given UJT relaxation circuit with sync signal
- A free running relaxation oscillator shown in Fig. 15p.3 has a sweep amplitude of 100 V and a period of 0.5 ms. Synchronizing pulses are applied to the device such that the breakdown voltage is lowered by 50 V at each pulse. The synchronizing pulse frequency is 4 kHz. What is the amplitude and frequency of the synchronized oscillator waveform.
FIGURE 15p.3 The given UJT circuit