7.5 Emitter-coupled Astable Multivibrators – Pulse and Digital Circuits

7.5 EMITTER-COUPLED ASTABLE MULTIVIBRATORS

The main drawback of a collector-coupled astable multivibrator is that the circuit may not necessarily oscillate when VCC is switched ON. It is possible that both Q1 and Q2 may be ON because of the inherent symmetry in the circuit. For oscillations to build up, one device may need to be switched into the OFF state by connecting its base to the emitter for a short duration. This disadvantage can be overcome in an emitter-coupled astable multivibrator as shown in Fig. 7.16(a). The circuit operates in such a fashion that Q1 switches between cut-off and saturation, and Q2 switches between the cut-off and active regions.

When Q1 is OFF, VCN1 = VCC1 = VBN2. If Q2 (when ON) is to be in the active region, VBN2 should be less than VCN2. For this, VCC1 < VCC2 and to avoid driving Q1 into saturation, VBB < VCC1. Therefore, VBB < VCC1 < VCC2. The voltages VBB and VCC1 are derived from VCC2 source using potential divider networks comprising R1, R2 and R3, R4, respectively. From Fig. 7.16(a) we have:

FIGURE 7.16(a) An emitter-coupled astable multivibrator

R1 and R2 are chosen such that they do not load the biasing circuit comprising R3 and R4.

To understand the operation of the circuit, assume that initially Q1 is ON and Q2 is OFF. The voltage VEN1 = VBBVσ and the capacitor C charges from this voltage source, as shown in Fig. 7.16(b).

From Fig. 7.16(b), we have VEN2 = VEN1VA. As the capacitor voltage rises exponentially, VEN2 falls exponentially to zero, since VEN1 is constant. When VEN2 = VBN2Vγ, Q2 begins to conduct. As a result, VEN2 rises. As E1 and E2 are connected through a condenser, VEN1 also rises by the same amount. As VEN1 rises further, Q1 comes out of saturation. Therefore, VCN1 rises and so does VBN2. This causes a further increase in the current in Q2. Hence, both VEN1 and VEN2 rise further. Q1 then goes into the OFF state and Q2 into the active region. This process is repeated. To plot these waveforms, the voltages are calculated as illustrated here:

When Q1 is ON and Q2 is OFF at t = T1−:

FIGURE 7.16(b) Charging of C from VEN1 source

VCN1(T1−) = VBN2(T1−) = VEN1 + VCE(sat) = VBBVσ + VCE(sat)

As the capacitor C charges, the voltage VEN2 falls exponentially to zero.

Q2 begins to conduct when VEN2 falls to VCN1(T1−).

At t = T1+, Q2 conducts and Q1 goes into the OFF state because of the regenerative action.

VCC1, as IB2RC1 and VBE2 are small when compared to VCC1.

As Q2 switches into the ON state; VEN2 rises by VD. As E2 and E1 are connected through C, VEN1 also rises by VD, where,

VD = VEN1(T1+) − VEN1(T1−) = VEN2(T1+) − VEN2(T1−)

Therefore,

Substituting Eqs. (7.32), (7.33) and (7.36) in Eq. (7.37):

V1VCC1, if the junction voltages are small and IB2 is small.

Therefore,

VEN1(T1+) = VEN2(T1+) = V1VCC1

At t = T1+, Q2 is ON and the current in

And the current in

At this time Q1 is OFF. Hence the sum of these two emitter currents must be supplied by Q2.

Hence IE2IC2 = VCC1/Re and IB2 = IC2/hFE, where Re is the parallel combination of Re1 and Re2.

In the above circuit, Re1 = Re2 and Re = Re1//Re2.

FIGURE 7.16(c) Waveforms of the emitter voltages

V1 = VCC1IB2RC1VBE2VCE(sat) + Vγ     VD = V1VEN1(T1+)

Also,

VCN2(T1+) = VCC2IC2RC2

The variations of VEN1 and VEN2 are plotted as shown in Fig. 7.16(c).

The time periods T2 and T1 are calculated as illustrated:

T2 is the time period for which Q1 is OFF,

VEN1(t) = vf − (vfvi)et/Re1C

vf = 0, vi = V1

VEN1(t) = V1et/Re1C

At t = T1, VEN1(T1) = VBBVσ

VBBVσ = V1eT1/Re1C

From which, if Vσ is small,

However, neglecting junction voltages, small currents from Eq. (7.38), we have:

V1 = VCC1

If this is a symmetric astable multivibrator, T1 = T2 = T/2. The operation of this circuit and plotting of the waveforms is better understood by considering an example.

EXAMPLE

Example 7.2: Consider the circuit in Fig. 7.16 (a) with VCC2 = 18 V, R1 = 250 kΩ, R3 = 200 kΩ, R2 = R4 = 1 MΩ, Re1 = Re2 = 3 kΩ, RC1 = 0.5 kΩ, RC2 = 0.2 kΩ, C = 0.1 μF and hFE = 30. Si devices are used. Calculate the voltages in the circuit and plot the waveforms to the scale. Also obtain the frequency of oscillations.

Solution:

Given VCC2 = 18 V, R3 = 250 kΩ and R4 = 1MΩ

When Q1 is ON and Q2 is OFF at t = T1

VCN2(T1−) = VCC2 = 18 V   VEN1(T1−) = VBBVσ = 12 − 0.7 = 11.3 V

VCN1(T1−) = VBN2(T1−) = VBBVσ + VCE(sat) = 11.3 + 0.3 = 11.6 V

VEN2 exponentially falls to zero.

VEN2(T1−) = VBN2(T1−) − Vγ = 11.6 − 0.5 = 11.1 V

Q2 begins to conduct when VEN2 falls to 11.1 V.

At t = T1+, Q2 conducts and Q1 goes into the OFF state because of the regenerative action. Re is the parallel combination of Re1 and Re2. In the above circuit,

IC2RC2 = (10 mA)(0.2 kΩ) = 2 V

We have Vγ = 0.5V, VBE2 = 0.6 V, Vσ = 0.7 V, VCE(sat) = 0.3 V

V1 = VCC1IB2RC1VBE2VCE(sat) + Vγ = 15 V − 0.165 V − 0.6 V − 0.3 V + 0.5 V = 14.435 V

VCN1(T1+) = VCC1IB2RC1 = 15 − 0.1655 = 14.8345 V

VD = V1VEN1(T1+) = 14.435 − 11.3 = 3.135 V

VCN2(T1+) = VCC2IC2RC2 = 18 − 2 = 16 V

The waveforms are shown in Fig. 7.17.

FIGURE 7.17 The waveforms of an emitter-coupled astable multivibrator

We have,

Re2 = 3 kΩ, C = 0.1 μF

7.5.1 Advantages of Emitter-coupled Astable Multivibrators

The advantages of an emitter-coupled astable multivibrator over a collector-coupled astable multivibrator are:

1. The emitter-coupled astable multivibrator starts oscillating the moment power is switched ON. There is no need to induce oscillations.
2. As the second collector is not involved in the regeneration loop, loading is avoided at this collector.
3. The first base again is not involved in the regeneration loop.
4. The pulses at the collectors will have sharp rising edges.
5. The circuit has only one timing capacitance C and thus, it is convenient to adjust the frequency.
6. Q2 in the quasi-stable state is not driven into saturation but is held in the active region; thereby the storage time is reduced, which enables faster switching.
7. IC1 and IC2 when Q1 and Q2 are ON are stable because of the feedback provided in an emitter-coupled circuit.

7.5.2 Disadvantages of Emitter-coupled Astable Multivibrators

The disadvantages of an emitter-coupled astable multivibrator over a collector-coupled astable multivibrator are:

1. The circuit will not function unless the operating conditions are properly adjusted.
2. Since a single tuning condenser is used, T1 and T2 are nearly the same. Unsymmetric operation with variable duty cycle is difficult to implement.
3. This circuit is more complex than a collector-coupled astable multivibrator.
4. When the ON device is held in the active region, the dissipation in the device is large.
SOLVED PROBLEMS

Example 7.3: For the astable multivibrator shown in Fig. 7.1:

1. Determine the time period and the frequency of oscillations if R1 = R2 = R = 10 kΩ, C1 = C2 = 0.01 μF.
2. Determine the time period and the frequency of oscillations if R1 = 1 kΩ, R2 = 10 kΩ, C1 = 0.01 μF, C2 = 1 μF.

Solution:

1. Given R1 = R2 = R = 10 kΩ, C1 = C2 = 0.01 μF.

This is a symmetric astable multivibrator.

2. Given R1 = 1 kΩ, R2 = 10 kΩ, C1 = 0.01 μF, C2 = 0.1 μF

This is an un-symmetric astable multivibrator:

T = 0.69(R1C1 + R2C2) = 0.69(1 × 103 × 0.01 × 10−6 + 10 × 103 × 0.1 × 10−6) = 0.69 ms

Example 7.4: For the astable multivibrator shown in Fig. 7.1:

1. Find the value of C to provide symmetrical oscillations if R = 10 kΩ and f = 10 kHz.
2. Determine the values of capacitors to provide a train of pulses 0.1 ms wide and at a frequency of 1 kHz, if R1 = R2 = 1 kΩ
3. Find the minimum value of RC in a symmetric astable if R = 10 kΩ and hFE = 50.

Solution:

1. Given R = 10 kΩ and f = 10 kHz, for a symmetrical astable multivibrator.
2. Given T1 = duration of the pulse = 0.1 ms, f = 1 kHz, for the un-symmetric astable multivibrator:
3. We have R = hFERC

Example 7.5: For the multivibrator shown in Fig. 7.13(a), VCC = 20 V, VBB = 10 V, R1 = R2 = R = 10 kΩ, C1 = C2 = C = 0.01 μF. Find the time period and the frequency.

Solution:

Example 7.6: In an astable multivibrator if R2 = 60 kΩ, R1 = 40 kΩ, C1 = C2 = 2.9 nF, find its frequency and duty cycle.

Solution:

T2 = 0.69 R1C1 = 0.69 × 40 × 103 × 2.9 × 10−9 = 80 μs = 0.08 ms

T1 = 0.69R2C2 = 0.69 × 60 × 103 × 2.9 × 10−9 = 120 μs = 0.12 ms

T = T1 + T2 = 0.08 + 0.12 = 0.2 ms

Example 7.7: A symmetrical collector-coupled astable multivibrator has the following parameters:

VCC = 10 V, RC = 1 kΩ, R = 10 kΩ, and C = 0.01 μF. Silicon transistors with hFE = 50 and rbb = 0.2 kΩ are used. Plot the waveforms and calculate the overshoot. Also plot the waveforms if the circuit uses p-n-p transistors.

Solution:

Given VCC = 10V, RC = 1 kΩ, R = 10 kΩ, C = 0.01 μF, rbb = 0.2 kΩ and hFE = 50.

Assume Vσ = 0.7V and VCE(sat) = 0.2V.

Let Q2 be ON and in saturation, then Q1 is OFF. After a time interval T1, Q1 is ON and Q2 is OFF. Suddenly when a device switches from the OFF state into the ON state, there is an overshoot at its base. Similarly, when the device switches its state from ON to OFF there is an overshoot at its collector. To account for these overshoots the base spreading resistance is taken into account. Let Q2 switch from OFF to ON and Q1 from ON to OFF, then the equivalent circuit is shown in Fig. 7.18(a).

When , then . Hence, R is omitted in Fig. 7.18(b).

FIGURE 7.18(a) The circuit of the astable multivibrator when Q2 switches from OFF to ON

FIGURE 7.18(b) The circuit when R is much larger than RC

Let δ and δ be the overshoots at the collector of Q1 and the base of Q2.

Vσ = 0.7 V Vγ = 0.5 V and VCE(sat) = 0.2 V

But, δ = δ, since the collector Q1 and the base of Q2 are connected through a condenser C and identical changes are required to take place at these two nodes.

The waveforms are shown in Fig. 7.18(c). For the astable multivibrator circuit using p–n–p transistors, the waveforms are shown in Fig. 7.19.

FIGURE 7.18(c) The waveforms of the circuit in Fig. 7.21(a)

FIGURE 7.19 The Waveforms of the circuit using p–n–p transistors

Example 7.8: A symmetric astable multivibrator with vertical edges (see in Fig. 7.20) has the following parameters, VCC = 15 V, RC = 3 kΩ, R1 = R2 = R = 30 kΩ, R3 = 2 kΩ and silicon transistors with hFE = 50 are used. C1 = C2 = C = 0.01 μF. Verify whether the ON device is in saturation or not. Find its frequency.

FIGURE 7.20 An astable multivibrator with vertical edges

Assume Q1 is OFF and Q2 is ON and in saturation. If Q2 is ON and in saturation

VC2 = VCE(sat) = 0.2V, VB2 = Vσ = 0.7 V then D2 is ON. The collector load is R3||RC.

Hence, Q2 is in saturation.

To find, f:

For a symmetric astable multivibrator:

SUMMARY
1. An astable multivibrator is essentially a square-wave generator.
2. An astable multivibrator is used as a clock in digital circuits.
3. If the time periods T1 and T2 are not equal, the astable multivibrator is called an un-symmetric astable multivibrator. If T1 = T2 = T/2, then it is called a symmetric astable multivibrator.
4. The ratio of T1 to T is called the duty cycle and is expressed as a percentage.
5. In a symmetric astable multivibrator, the frequency is given by f = 0.7/RC cycles/s.
6. There could be rounding-off of the rising edge of a pulse in an astable multivibrator because of a small recharging current associated with the condenser.
7. An astable multivibrator can be used as a voltage-controlled oscillator (VCO) or a voltage-to-frequency converter (VFC).
8. An astable multivibrator can also be used a frequency modulator.
9. An emitter-coupled astable multivibrator starts oscillating the moment power is switched ON.
MULTIPLE CHOICE QUESTIONS
1. An astable multivibrator is a:
1. Square-wave generator
2. Sweep generator
3. Ramp generator
4. None of the above
2. For the time period of an astable multivibrator to be stable, in the quasi-stable state, the ON device is required to be driven into:
1. Saturation
2. The active region
3. The cut-off region
4. The reverse-biased condition
3. The duty cycle of a symmetric multivibrator is:
1. 50%
2. 75%
3. 25%
4. 100%
4. The frequency of oscillations of a symmetric astable multivibrator is given as:
1. 0.7τ c/s
2. 2.2τ c/s
3. None of the above
5. To ensure that the ON device is in saturation, the condition that should be satisfied in a symmetric astable multivibrator is given by:
1. RhFERC
2. RhFERC
3. R should be infinity
4. R should be zero
6. The ratio of trec/(T/2) of a symmetric astable is given by:
1. 3.2/hFE
2. hFE/3.2
3. 3.2 hFE
4. hFE
7. An astable multivibrator can be used as a:
1. Voltage-controlled oscillator
2. Gating circuit
3. Ramp generator
4. Exponential generator
8. An astable multivibrator can also be used as a:
1. Frequency modulator
2. Gating circuit
3. Ramp generator
4. Exponential generator
1. Draw the circuit of a collector-coupled astable multivibrator and explain its operation.
2. Explain how an astable multivibrator can be modified to operate as a voltage-controlled oscillator.
3. Explain why in an astable multivibrator the ON device in the quasi-stable state, is required to be driven into saturation.
4. Explain how an astable multivibrator can be used as a frequency modulator.
5. What are the advantages of an emitter-coupled astable multivibrator over a collector-coupled astable multivibrator?
1. Draw the circuit of a symmetric collector-coupled astable multivibrator using p–n–p devices and explain its operation. Derive the expression for its frequency of oscillation and plot the waveforms.
2. Explain how an astable multivibrator can be modified to operate as a voltage-controlled oscillator. Derive the expression for its frequency of oscillations and plot the waveforms.
3. The output of a collector-coupled astable multivibrator does not have sharp rising edges for the voltages at the collectors. Explain why.
4. Draw the circuit of a collector-coupled astable multivibrator with vertical edges and explain its operation.
UNSOLVED PROBLEMS
1. For the multivibrator in Fig. 7.1, R1 = R2 = R = 47 kΩ, C1 = C2 = C = 0.01μ F. Find the time period and frequency.
2. For the astable multivibrator in Fig. 7.1, R1 = 20 kΩ, R2 = 30 kΩ, C1 = C2 = C = 0.01μF. Find the time period, duty cycle and the frequency.
3. For the symmetric astable multivibrator that generates square waves with vertical edges shown in Fig. 7.22, VCC = 10 V, RC = R3 = 2 kΩ, R1 = R2 = R = 20 kΩ, C = 0.1μF, hFE(min) = 30. Show that the ON device is in saturation. Also find f. Assume suitable values for VCE(sat) and VBE(sat). Si transistors are used.
4. Design a symmetric collector-coupled astable multivibrator to generate a square wave of 10 kHz having peak-to-peak amplitude of 10 V where hFE(min) = 30, IC(sat) = 2mA.
5. Design an un-symmetric astable multivibrator having duty cycle of 40 per cent. It is required to oscillate at 5 kHz. Ge transistors with hFE = 40 are used. The amplitude of the square wave is required to be 20 V, IC = 5 mA, VCE(sat) = 0.1 V and VBE(sat) = 0.3 V.
6. For an un-symmetric astable multivibrator R1 = 100 kΩ, R2 = 100 kΩ, C1 = 0.02 μF, C2 = 0.01 μF. Find the frequency of oscillation and the duty cycle.
7. Design an unsymmetrical astable multivibrator shown in Fig. 7p.1 using silicon n–p–n transistors having output amplitude of 12 V. Given data, IC(sat) = 5 mA, hFE(min) = 50, f = 5 kHz, duty cycle = 0.6.

FIGURE 7p.1 Un-symmetric astable multivibrator

8. The voltage-to-frequency converter shown in Fig. 7p.2 generates oscillations at a frequency f1 when VBB = VCC. Find the ratio of VCC/VBB at which the frequency f2 = 4 f1.

FIGURE 7p.2 The given voltage-to-frequency converter