14.3 Astable Blocking Oscillators – Pulse and Digital Circuits

14.3 ASTABLE BLOCKING OSCILLATORS

Now let us consider astable operation. In an astable operation, the device switches state automatically. However, in astable blocking oscillators, a trigger is used to initiate regeneration after which the circuit oscillates on its own. Here, two types of astable blocking oscillators are considered: a diode-controlled astable blocking oscillator and an RC-controlled astable blocking oscillator. In a diode-controlled astable blocking oscillator, a diode circuit is used to trigger regeneration and the operation then switches to astable mode whereas in an RC-controlled astable blocking oscillator, an RC circuit is used for the astable operation.

14.3.1 Diode-controlled Astable Blocking Oscillators

A diode-controlled astable blocking oscillator is shown in Fig. 14.4(a). If a negative pulse is applied at the collector of Q, the base current starts to flow and regeneration takes place as discussed earlier (Section 14.2.2). The waveforms indicated in Fig. 14.3(e) and Fig. 14.3(f), for 0 ≤ ttp are generated. When regeneration takes place, the diode combination (D1 and D2) is OFF and the circuit is essentially a monostable circuit. The amplitude and the duration of the pulse have the values previously calculated for the monostable circuit (for VBB << VCC). At the end of the pulse, the magnetizing current is im.

FIGURE 14.4(a) A diode-controlled astable blocking oscillator

Or

Therefore,

From Eq. (14.27 b), tp is calculated as:

With RL → ∞

Substituting the Eq. (14.31) in Eq. (14.29):

which shows that the peak magnetizing current is independent of the inductance. At t = tp, the transistor is cut-off; the diode combination conducts and the inductor current now flows through the winding capacitance of the transformer, as shown in Fig. 14.5(a).

If Rf = 0, Vγ appears directly across L and C. Hence, the collector voltage rises above VCC by Vγ. Since, for t > tp:

Then, shifting the time origin to the end of the pulse, we obtain:

The magnetizing current im decreases linearly with time after the pulse ends. The diode current also falls to zero at a time tf.

FIGURE 14.5(a) When Q1 is OFF the diode combination is ON, transistor is cut-off; (b) the ringing circuit

FIGURE 14.6 The waveforms of a diode-controlled astable blocking oscillator

At tf, im = 0, from Eq. (14.33):

And from Eq. (14.32):

Therefore,

After im is reduced to zero, the diode becomes an open circuit. This results in an under-damped ringing circuit shown in Fig. 14.5(b). Since the shunt capacitor is initially charged to Vr, a sinusoidal oscillation of amplitude Vr and time period 2π begins (dashed curve). After one quarter of a cycle, that is, at the time ta = 2π/4 = 1.57, VC falls below VCC. There results a negative swing at the collector and because of the polarity inversion, this appears as a positive swing at the base. When VBE exceeds the cut-in voltage of the transistor Q, regeneration takes place again. Q again turns ON and the cycle repeats itself. Hence, without the need for any further external trigger, a new cycle starts. Thus, at the end of the pulse generated by the monostable circuit, the diode combination conducts and the capacitor charges to Vγ. Also, the LC (C is the distributed capacitance of the inductance, represented as a lumped capacitance) circuit oscillates when D is OFF, leading to the regeneration of the pulse and hence, sustained oscillations. As the diode combination causes the circuit to function in the astable mode, this astable multivibrator is called a diode-controlled circuit. The time period of oscillations is T = tp + tf + ta. The collector and base voltages are nearly rectangular. The waveforms are plotted as shown in Fig. 14.6.

To understand the procedure to make the calculations and plot the waveforms let us consider Example 14.3.

EXAMPLE

Example 14.3: The astable blocking oscillator (diode-controlled) shown in Fig. 14.7 has the following parameters: L = 5.2 mH, C = 90 pF, VCC = 10 V, R = 500 Ω, Vr = 0.6 V, n = 1 and VBB = 0.5 V. Calculate (a) the period and the duty cycle of the free oscillations; (b) the peak voltages and currents and (c) the current in the magnetizing inductance at the end of one cycle.

FIGURE 14.7 The diode-controlled astable blocking oscillator

Solution: (a) Since VBB << VCC

T = tp + tf + ta = 10.4 + 87 + 1.07 = 98.47μs

   The duty cycle is defined as the ratio of tp to T.

(b)

   The collector voltage extends from:

 

VCCV = 10 − 5 = 5 V to VCC + Vγ = 10 + 0.6 = 10.6 V

   The base voltage extends from:

 

nV tonVγ (i.e. 5 V to − 0.6 V)

   The emitter current is constant and is given as:

   The base current is maximum at t = 0 and is given as:

   The collector current is maximum at:

   Therefore,

   The peak magnetizing current is:

(c) At T = tp + tf, the capacitor is charged to a voltage Vr and the magnetizing current is zero. The LC circuit then rings and in one quarter of a cycle (in time ta) the capacitive energy is transformed to magnetic energy.

   Since

   Therefore,

   The current im goes below zero by the amount Im.

   If we want a lower duty cycle (i.e., a pulse train rather than a typical square-wave output), we must increase tf relative to tp.

   Hence, the duty cycle will be at minimum if in place of the diode combination, a single p-n Ge diode for which Vγ ≈ 0.1 V is used. Then with VCC = 10 V, and n = 1,

   Thus, tp/tf of 0.02 is possible.

   If the duty cycle has to be large, in place of a diode, use a Zener diode with VZ = 3.8 V. Then the duty cycle is:

14.3.2 RC-controlled Astable Blocking Oscillators

The circuit shown in Fig. 14.8(a) is an RC-controlled astable blocking oscillator. Here, the RC network, comprising R1 and C1, is provided in the emitter circuit. VBB now is a positive voltage.

The operation of the circuit can be explained as follows. Initially let Q be ON. Then the capacitor C1 tries to charge to VCC through the small dc resistance of the transformer winding in the collector lead and R, which is a relatively small resistance. The moment the voltage V1 on C1 is greater than (VBBVγ), as the base-emitter diode of Q gets reverse-biased, Q goes into the OFF state. Hence, the charge on C1 discharges with a time constant R1 C1. Once again, when V1 reaches (VBBVγ), there is a base current; as a result, there is a collector current and the regenerative action takes place and Q once again switches into the ON state. This process is automatically repeated, resulting in an astable operation. As R1 and C1 control the timing operation, this astable circuit is called an RC-controlled astable blocking oscillator. The collector and base waveforms are similar to that of the monostable multivibrator with emitter timing as shown in Fig. 14.3(e).

FIGURE 14.8(a) A RC−controlled astable blocking oscillator

FIGURE 14.8(b) The waveforms of an RC-controlled astable blocking oscillator

During the pulse duration tp, Q is ON and the capacitor is recharged and attains a voltage V1, which is larger than the voltage it had at the beginning of the pulse. Q now is OFF for a time period tf, during which period C1 discharges to the voltage (VBB - Vγ), at which instant Q again goes ON and the cycle repeats itself. The waveforms are shown in Fig. 14.8(b).

(i) Calculation of tp:

Similar to the relation obtained earlier in the Eq. (14.30) we have for the circuit shown in Fig. 14.8(a):

We can calculate tp, using Eq. (14.35). If tp/RC1 << 1, expanding etp/RC1 as series and retaining only the 1st two terms:

Substituting the 1st two terms of this expansion in Eq. (14.35):

For a very large C1, << 1. Hence, Eq. (14.36) reduces to:

This gives the pulse width of the monostable blocking oscillator.

(ii) Calculation of tf:

We have vC(t) = vf − (vfvi)et/τ From Fig. 14.8(b), we have vi = V1 and vf = 0. Thus,

 

vC(t) = V1 et/R1C1

At

t = tf, vC(tf) = VBBVγ

                      VBBVγ = V1 etf/R1C1

FIGURE 14.8(c) The circuit to calculate V

To calculate tf, we need to know V1, the maximum voltage on C when Q is ON. The amplitude of the trigger V is calculated using the circuit in Fig. 14.8(c):

From Fig. 14.8(c):

VCC = V + nV + VBB

Therefore,

V (n + 1) = VCCVBB

The amplitude of the trigger is:

To get V1 during charging, from the waveforms shown in Fig. 14.8(b):

vC(t) = vf(vfvi)et/τ

vf = nV + VBB and vi = V BBVγ

vC(t) = nV + VBB − (nV + VBBVBB + Vγ)et/RC1

At

t = tp, vC(t) = V1

V1 = nV + VBB − (nV + Vγ)etp/RC1

Substituting Eq. (14.39)

since Vγ is small. This equation is simplified as:

To calculate the value of V1, we first calculate tp, and then use that value of tp in Eq. (14.40) to calculate V1. Having calculated V1, substitute this value of V1 in Eq. (14.38) to calculate tf. The time period of the oscillation, T = (tp + tf) and the frequency of oscillations is the reciprocal T.

EXAMPLE

Example 14.4: For the RC-controlled astable blocking oscillator shown in Fig. 14.8(a) following are the parameters: VCC = 15 V, n = n1 = 1, VBB = 2 V, R = 100 Ω, R1 = 1000 Ω, RL = 1000 Ω, C1 = 0.01 µF, L = 5 mH.
(a) Calculate the amplitude of the trigger; (b) tp, the pulse duration; (c) V1, the maximum voltage to which C1 charges; (d) tf, the discharge period of C; (e) T, the time period of the astable multivibrator and its frequency of oscillations and (f) the duty cycle.

Solution:

(a) We have from Eq. (14.39):

(b) From Eq. (14.37):

(c) From Eq. (14.40):

(n + 1)V1 = nVCC + VBBn(VCCVBB)etp/RC1

2V1 = 17 − 13 × 0 = 17,   V1 = 8.5 V

(d) tf is given by Eq. (14.38) as:

(e) T = tp + tf = 45 + 17.35 = 62.35 µs

(f) Duty cycle =

14.3.3 Effect of Core Saturation on Pulse Width

So far the pulse width was calculated based on the assumption that magnetizing core is not saturated. Let

V = Voltage across the primary

n = Number of turns on the primary

A = Area of cross section of the core

φ = Magnetic flux

FIGURE 14.9(a) The variation of flux density with time

FIGURE 14.9(b) The variation of the magnetizing current with time

= Rate of change of flux linkages

From Faraday’s law of electromagnetic induction, the voltage induced in the primary winding is proportional to the rate of change of flux linkages.

If there are n turns in the primary then;

If B is the flux density then

If there is a change in φ, B also changes incrementally. Therefore, from Eq. (14.42) we get:

Putting Eq. (14.43) in Eq. (14.41) we get:

Figure 14.9(a) shows the variation of flux density as a function of time.

From Fig. 14.9(a), dt = tp and dB = Bm.

Therefore, Eq. (14.44) is written as:

We know that:

Therefore,

If we assume that the core losses are zero and the magnetizing core gets saturated, the magnetizing current im becomes large. For 0 ≤ tt1, im and iC vary linearly as a function of time. During t1 to tp from Eq. (14.46), im and iC increase rapidly, as shown in Figs. 14.9(b) and (c).

FIGURE 14.9(c) The variation of the collector current with time

FIGURE 14.9(d) The variation of the base current with time

At t = tp, iC is equal to hFE iB. As a result, the device quickly switches into the OFF state and the pulse gets terminated, as shown in Fig. 14.9(d). From Eqs. (14.13) and (14.44), we have:

Therefore,

Let the flux density be increased from 0 to the maximum value (Bm) in the interval tp then:

Therefore,

It can be observed from Eq. (14.47) that the pulse width (tp) is independent of the transistor parameter hFE. which is temperature dependent. It depends only on VCC and the parameters of the core. Thus, the pulse can be stable.

The initial value of the flux density depends on the magnetic properties of the core. The hysteresis curve of the core is shown in Fig. 14.10. Let us now try to calculate the pulse width using the entire hysteresis curve. At t = 0, there is no current in the winding, the core flux density B = −Bm. When the blocking oscillator is triggered, as H increases, B also increases from – Bm to + Bm till H = HC. At H = HC the core gets saturated and the pulse ends. tp can be thus calculated by substituting B = 2Bm in Eq. (14.47).

Therefore,

As the core gets saturated, B the flux density remains at Bm. Subsequently, the rate of change of the flux density is zero, i.e., dB/dt = 0. So no voltages are induced in the transformer and regeneration is not possible. To trigger the oscillator after the pulse ends, it should be ensured that the flux density is reset to – Bm. This resetting to – Bm can be accomplished by passing a current through an auxiliary winding with Ra connected in series, as shown in Fig. 14.11.

When the transistor is turned OFF, the current VCC/Ra in the auxiliary winding supplies the necessary opposite ampere turns, so that H becomes more negative than – HC and hence, B = −Bm and the oscillator comes back to the normal condition.

FIGURE 14.10 The hysteresis curve

FIGURE 14.11 The resetting of the flux density

14.3.4 Applications of Blocking Oscillators

  1. As a transformer is used as a coupling element for regeneration, the monostable blocking oscillator can be used to derive both positive and negative pulses by using a centre-tapped auxiliary winding. The pulses so derived may, if need be, used to trigger other circuits.
  2. When used in the astable mode, square waves with varying duty cycles can be derived using a blocking oscillator. In fact, the duty cycle can be very small. The smaller duty cycle means sharp pulses.
  3. Unlike conventional multivibrators, blocking oscillators use a transformer. Thus, they can generate pulses with large peak power, i.e., pulses may have a large amplitude and the associated current can also be large. However, in cases where the duty cycle is small, the average power is small.
  4. As blocking oscillators can generate pulses with small mark duration and varying repetitive frequencies, these circuits can be used in pulse synchronization where there is a need to synchronize the outputs of a number of generators (see Chapter 15 for a complete discussion).
  5. Blocking oscillators find applications in frequency dividers and counters, discussed in detail in Chapter 15.
SOLVED PROBLEMS

Example 14.5: Calculate the pulse amplitude and the width for the triggered blocking oscillator shown in Fig. 14.12(a). Si transistor uses hFE = 50, VBE(sat) = 0.7 V, VCE(sat) = 0.2 V, magnetizing inductance of the collector winding L is 5 mH, VBB = −2 V, VCC = 10 V, R = 1.5 KΩ.

Solution:

From the circuit in Fig. 14.12(b),

iC = 2iB + im

iE = iC + iB = 2iB + im + iB = 3iB + im

Writing for the voltage at node E:

10 − V − 0.2 = −2 + 2V − 0.7

3V = 10 + 2 + 0.7 − 0.2

FIGURE 14.12(a) The triggered blocking oscillator

FIGURE 14.12(b) The equivalent circuit of Fig. 14.12(a) when core saturates

iC = 2iB + im = 0.0024− −554.6t + 832t = 0.0024 + 277.4t

At t = tp

iC = hFEiB

0.0024 + 277.4 tp = 50(0.0012 − 277.3 tp)

0.0024 + 277.4 tp = 0.06 − 13865 tp

14142.4 tp = 0.0576

Example 14.6: For the RC-controlled astable blocking oscillator shown in Fig. 14.8(a), VCC = 18 V, n = n1 = 1, VBB = 1.5 V, R = 100 Ω, R1 = 1000 Ω, RL = ∞, C1 = 0.01 µF, L = 3 mH. (a) Calculate the amplitude of the trigger; (b) tp, the pulse duration; (c) V1, the maximum voltage to which C1 charges; (d) tf, the discharge period of C; (e) T, the time period of the astable multivibrator and its frequency of oscillations and (f) the duty cycle.

Solution:

(a) We have from Eq. (14.39)

(b) From Eq. (14.37)

(c) From Eq. (14.40)

(n + 1)V1 = nVCC + VBBn(VCCVBB)etp/RC1

2V1 = 19.5 − 16.5 × 0    2V1 = 19.5    V1 = 9.75 V

(d) tf is given by Eq. (14.38) as

(e) T = tp + tf = 30 + 22.77 = 52.77 µs

(f) Duty cycle =

SUMMARY
  1. Blocking oscillators are regenerative circuits that use a pulse transformer and can be operated either in an astable or a monostable mode.
  2. The main advantage of a blocking oscillator is that it can generate pulses of peak power.
  3. In a monostable blocking oscillator using base timing, the disadvantage is that the pulse duration is dependent on hFE of the transistor. As hFEvaries with temperature, its pulse width can not be stable.
  4. A monostable blocking oscillator with emitter timing generates a pulse whose width is independent of hFE. Hence, the duration of the pulse is stable.
  5. In a diode-controlled astable blocking oscillator, the diode is responsible for regeneration and it is possible to vary the duty cycle by proper choice of the diode.
  6. An RC-controlled astable blocking oscillator can give lower duty cycles.
  7. Saturation of the magnetic circuit of the transformer influences the pulse width of a blocking oscillator.
MULTIPLE CHOICE QUESTIONS
  1. The type of feedback that is employed in a blocking oscillator is:
    1. Regenerative feedback
    2. Degenerative feedback
    3. No feedback
    4. None of the above
  2. The main advantage of a blocking oscillator is that it can generate a pulse of:
    1. High peak power
    2. Low output voltage
    3. Low peak power
    4. None of the above
  3. In a blocking oscillator, on the application of a trigger, if the circuit generates a single pulse, it is called:
    1. Astable blocking oscillator
    2. Monostable blocking oscillator
    3. Bistable blocking oscillator
    4. Free-running blocking oscillator
  4. In a blocking oscillator, on the application of a trigger, if the circuit generates a square wave, it is called:
    1. Monostable blocking oscillator
    2. Free-running blocking oscillator
    3. Flip-flop
    4. Counter
  5. In a monostable blocking oscillator with base timing, the main disadvantage is that:
    1. T varies with the variation of hFE
    2. T remains unchanged with the variation of hFE
    3. T is inversely proportional to hFE
    4. None of the above
  6. The improvement in a monostable blocking oscillator with emitter timing when compared to the circuit that employs base timing is:
    1. T is independent of hFE
    2. T is proportional to hFE
    3. T is inversely proportional to hFE
    4. T depends only on transistor parameters
  7. The duty cycle in an astable blocking oscillator employing diode control can be varied by the choice of:
    1. Diode forward voltage
    2. Diode reverse voltage
    3. Turns ratio of the pulse transformer
    4. Magnetic circuit
  8. An astable blocking oscillator can be used as a:
    1. Frequency divider and counter
    2. Time-base generator
    3. Sampling gate
    4. Logic gate
SHORT ANSWER QUESTIONS
  1. Explain the principle of working of a blocking oscillator.
  2. Name the two types of monostable blocking oscillators and discuss their relative performance.
  3. Explain the principle of operation of an astable blocking oscillator with a diode control.
  4. Compare the performance of a diode-controlled astable blocking oscillator with an RC-controlled astable blocking oscillator.
LONG ANSWER QUESTIONS
  1. Draw the circuit of a monostable blocking oscillator with base timing and explain its operation. Calculate its pulse width. Draw the waveforms.
  2. Explain with the help of a neat circuit diagram the working of a monostable blocking oscillator with emitter timing and obtain the expression for its pulse width. Plot the waveforms.
  3. Draw the circuit of an astable blocking oscillator with a diode control and explain its operation. Plot the waveforms. Derive the expression for its frequency.
  4. With the help of a neat circuit diagram and waveforms explain the working of an astable blocking oscillator with RC control. Obtain the expression for its frequency of oscillations.
UNSOLVED PROBLEMS
  1. For the monostable blocking oscillator in Fig. 14.1(a), following are the parameters: VCC = 15 V, R = 0.5 kΩ, L = 3 mH, hFE = 20 and n = 1.
    (a) Calculate and plot the base current and collector current waveforms and calculate the pulse width. (b) If due to temperature variation, hFE increases to 30, then calculate the pulse width.
  2. For the blocking oscillator with emitter timing shown in Fig. 14.3(a), it is given that n = 1, n1 = 1, L = 3 mH, R = 0.5 kΩ, hFE = 20 and VCC = 15 V.
    Calculate (a) the amplitude of the trigger; (b) the value of RL that allows pulse formation; (c) pulse width with hFE = 20 and also calculate tp using the approximate relation; (d) the base, collector and emitter currents and (e) plot the current waveforms.
  3. For the diode-controlled astable blocking oscillator shown in Fig. 14.4(a), L = 3 mH, C = 100 pF, VCC = 15 V, R = 500 Ω, Vγ = 4.4 V, n = 1 and VBB = 0.7 V, RL = ∞.
    (a) Calculate the amplitude of the trigger pulse; (b) time period of the oscillations and the frequency, and (c) the duty cycle.
  4. For the RC−controlled astable blocking oscillator shown in Fig. 14.8(a), following are the parameters: VCC = 20 V, n = n1 = 1, VBB = 3 V, R = 100 Ω, R1 = 1000 Ω, RL = 1000 Ω, C1 = 0.01 µF, L = 2 mH.
    1. Calculate the amplitude of the trigger; (b) tp, the pulse duration; (c) V1, the maximum voltage to which C1 charges; (d) tf, the discharge period of C; (e) T, the time period of the astable multivibrator and its frequency of oscillations and finally (f) the duty cycle.