##### 7.5 EMITTER-COUPLED ASTABLE MULTIVIBRATORS

The main drawback of a collector-coupled astable multivibrator is that the circuit may not necessarily oscillate when *V*_{CC} is switched ON. It is possible that both *Q*_{1} and *Q*_{2} 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 *Q*_{1} switches between cut-off and saturation, and *Q*_{2} switches between the cut-off and active regions.

When *Q*_{1} is OFF, *V*_{CN1} = *V*_{CC1} = *V*_{BN2}. If *Q*_{2} (when ON) is to be in the active region, *V*_{BN2} should be less than *V*_{CN2}. For this, *V*_{CC1} < *V*_{CC2} and to avoid driving *Q*_{1} into saturation, *V*_{BB} < *V*_{CC1}. Therefore, *V*_{BB} < *V*_{CC1} < *V*_{CC2}. The voltages *V*_{BB} and *V*_{CC1} are derived from *V*_{CC2} source using potential divider networks comprising *R*_{1}, *R*_{2} and *R*_{3}, *R*_{4}, respectively. From Fig. 7.16(a) we have:

**FIGURE 7.16(a)** An emitter-coupled astable multivibrator

*R*_{1} and *R*_{2} are chosen such that they do not load the biasing circuit comprising *R*_{3} and *R*_{4}.

To understand the operation of the circuit, assume that initially *Q*_{1} is ON and *Q*_{2} is OFF. The voltage *V*_{EN1} = *V*_{BB} − *V*_{σ} and the capacitor *C* charges from this voltage source, as shown in Fig. 7.16(b).

From Fig. 7.16(b), we have *V*_{EN2} = *V*_{EN1} − *V*_{A}. As the capacitor voltage rises exponentially, *V*_{EN2} falls exponentially to zero, since *V*_{EN1} is constant. When *V*_{EN2} = *V*_{BN2} − *V*_{γ}, *Q*_{2} begins to conduct. As a result, *V*_{EN2} rises. As *E*_{1} and *E*_{2} are connected through a condenser, *V*_{EN1} also rises by the same amount. As *V*_{EN1} rises further, *Q*_{1} comes out of saturation. Therefore, *V*_{CN1} rises and so does *V*_{BN2}. This causes a further increase in the current in *Q*_{2}. Hence, both *V*_{EN1} and *V*_{EN2} rise further. *Q*_{1} then goes into the OFF state and *Q*_{2} into the active region. This process is repeated. To plot these waveforms, the voltages are calculated as illustrated here:

When *Q*_{1} is ON and *Q*_{2} is OFF at *t* = *T*_{1}−:

**FIGURE 7.16(b)** Charging of *C* from *V*_{EN1} source

*V*_{CN1}(*T*_{1}−) = *V*_{BN2}(*T*_{1}−) = *V*_{EN1} + *V*_{CE(sat)} = *V*_{BB} − *V*_{σ} + *V*_{CE(sat)}

As the capacitor C charges, the voltage *V*_{EN2} falls exponentially to zero.

*Q*_{2} begins to conduct when *V*_{EN2} falls to *V*_{CN1}(*T*_{1}−).

At *t* = *T*_{1}+, *Q*_{2} conducts and *Q*_{1} goes into the OFF state because of the regenerative action.

≈ *V*_{CC1}, as *I*_{B2}*R*_{C1} and *V*_{BE2} are small when compared to *V*_{CC1}.

As *Q*_{2} switches into the ON state; *V*_{EN2} rises by *V*_{D}. As *E*_{2} and *E*_{1} are connected through *C*, *V*_{EN1} also rises by *V*_{D}, where,

*V*_{D} = *V*_{EN1}(*T*_{1}+) − *V*_{EN1}(*T*_{1}−) = *V*_{EN2}(*T*_{1}+) − *V*_{EN2}(*T*_{1}−)

Therefore,

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

∴ *V*_{1} ≈ *V*_{CC1}, if the junction voltages are small and *I*_{B2} is small.

Therefore,

*V*_{EN1}(*T*_{1}+) = *V*_{EN2}(*T*_{1}+) = *V*_{1} ≈ *V*_{CC1}

At *t* = *T*_{1}+, *Q*_{2} is ON and the current in

And the current in

At this time *Q*_{1} is OFF. Hence the sum of these two emitter currents must be supplied by *Q*_{2}.

Hence *I*_{E2} ≈ *I*_{C2} = *V*_{CC1}/*R*_{e} and *I*_{B2} = *I*_{C2}/*h*_{FE}, where *R*_{e} is the parallel combination of *R*_{e1} and *R*_{e2}.

In the above circuit, *R*_{e1} = *R*_{e2} and *R*_{e} = *R*_{e1}//*R*_{e2}.

**FIGURE 7.16(c)** Waveforms of the emitter voltages

*V*_{1} = *V*_{CC1} − *I*_{B2}*R*_{C1} − *V*_{BE2} − *V*_{CE(sat)} + *V*_{γ} *V*_{D} = *V*_{1} − *V*_{EN1}(*T*_{1}+)

Also,

*V*_{CN2}(*T*_{1}+) = *V*_{CC2} − *I*_{C2}*R*_{C2}

The variations of *V*_{EN1} and *V*_{EN2} are plotted as shown in Fig. 7.16(c).

The time periods *T*_{2} and *T*_{1} are calculated as illustrated:

*T*_{2} is the time period for which *Q*_{1} is OFF,

∴ *V*_{EN1}(*t*) = *v*_{f} − (*v*_{f} − *v*_{i})*e*^{−t/Re1C}

*v*_{f} = 0, *v*_{i} = *V*_{1}

∴ *V*_{EN1}(*t*) = *V*_{1}*e*^{−t/Re1C}

At *t* = *T*_{1}, *V*_{EN1}(*T*_{1}) = *V*_{BB} − *V*_{σ}

∴ *V*_{BB} − *V*_{σ} = *V*_{1}*e*^{−T1/Re1C}

From which, if *V*_{σ} is small,

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

*V*_{1} = *V*_{CC1}

If this is a symmetric astable multivibrator, *T*_{1} = *T*_{2} = *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 *V*_{CC2} = 18 V, *R*_{1} = 250 kΩ, *R*_{3} = 200 kΩ, *R*_{2} = *R*_{4} = 1 MΩ, *R*_{e1} = *R*_{e2} = 3 kΩ, *R*_{C1} = 0.5 kΩ, *R*_{C2} = 0.2 kΩ, *C* = 0.1 *μF* and *h*_{FE} = 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 *V*_{CC2} = 18 V, *R*_{3} = 250 kΩ and *R*_{4} = 1MΩ

When *Q*_{1} is ON and *Q*_{2} is OFF at *t* = *T*_{1}−

*V*_{CN2}(*T*_{1}−) = *V*_{CC2} = 18 V *V*_{EN1}(*T*_{1}−) = *V*_{BB} − *V*_{σ} = 12 − 0.7 = 11.3 V

*V*_{CN1}(*T*_{1}−) = *V*_{BN2}(*T*_{1}−) = *V*_{BB} − *V*_{σ} + *V*_{CE(sat)} = 11.3 + 0.3 = 11.6 V

*V*_{EN2} exponentially falls to zero.

*V*_{EN2}(*T*_{1}−) = *V*_{BN2}(*T*_{1}−) − *V*_{γ} = 11.6 − 0.5 = 11.1 V

*Q*_{2} begins to conduct when *V*_{EN2} falls to 11.1 V.

At *t* = *T*_{1}+, *Q*_{2} conducts and *Q*_{1} goes into the OFF state because of the regenerative action. *R*_{e} is the parallel combination of *R*_{e1} and *R*_{e2}. In the above circuit,

*I*_{C2}*R*_{C2} = (10 mA)(0.2 kΩ) = 2 V

We have *V*_{γ} = 0.5V, *V*_{BE2} = 0.6 V, *V*_{σ} = 0.7 V, *V*_{CE(sat)} = 0.3 V

∴ *V*_{1} = *V*_{CC1} − *I*_{B2}*R*_{C1} − *V*_{BE2} − *V*_{CE(sat)} + *V*_{γ} = 15 V − 0.165 V − 0.6 V − 0.3 V + 0.5 V = 14.435 V

*V*_{CN1}(*T*_{1}+) = *V*_{CC1} − *I*_{B2}*R*_{C1} = 15 − 0.1655 = 14.8345 V

*V*_{D} = *V*_{1} − *V*_{EN1}(*T*_{1}+) = 14.435 − 11.3 = 3.135 V

∴ *V*_{CN2}(*T*_{1}+) = *V*_{CC2} − *I*_{C2}*R*_{C2} = 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,

*R*_{e2} = 3 kΩ, *C* = 0.1 *μ*F

#### 7.5.1 Advantages of Emitter-coupled Astable Multivibrators

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

- The emitter-coupled astable multivibrator starts oscillating the moment power is switched ON. There is no need to induce oscillations.
- As the second collector is not involved in the regeneration loop, loading is avoided at this collector.
- The first base again is not involved in the regeneration loop.
- The pulses at the collectors will have sharp rising edges.
- The circuit has only one timing capacitance
*C*and thus, it is convenient to adjust the frequency. *Q*_{2}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.*I*_{C1}and*I*_{C2}when*Q*_{1}and*Q*_{2}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:

- The circuit will not function unless the operating conditions are properly adjusted.
- Since a single tuning condenser is used,
*T*_{1}and*T*_{2}are nearly the same. Unsymmetric operation with variable duty cycle is difficult to implement. - This circuit is more complex than a collector-coupled astable multivibrator.
- 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:

- Determine the time period and the frequency of oscillations if
*R*_{1}=*R*_{2}=*R*= 10 kΩ,*C*_{1}=*C*_{2}= 0.01*μ*F. - Determine the time period and the frequency of oscillations if
*R*_{1}= 1 kΩ,*R*_{2}= 10 kΩ,*C*_{1}= 0.01*μ*F,*C*_{2}= 1*μ*F.

*Solution:*

- Given
*R*_{1}=*R*_{2}=*R*= 10 kΩ,*C*_{1}=*C*_{2}= 0.01*μ*F.This is a symmetric astable multivibrator.

- Given
*R*_{1}= 1 kΩ,*R*_{2}= 10 kΩ,*C*_{1}= 0.01*μ*F,*C*_{2}= 0.1*μ*FThis is an un-symmetric astable multivibrator:

*T*= 0.69(*R*_{1}*C*_{1}+*R*_{2}*C*_{2}) = 0.69(1 × 10^{3}× 0.01 × 10^{−6}+ 10 × 10^{3}× 0.1 × 10^{−6}) = 0.69 ms

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

- Find the value of
*C*to provide symmetrical oscillations if*R*= 10 kΩ and*f*= 10 kHz. - Determine the values of capacitors to provide a train of pulses 0.1 ms wide and at a frequency of 1 kHz, if
*R*_{1}=*R*_{2}= 1 kΩ - Find the minimum value of
*R*_{C}in a symmetric astable if*R*= 10 kΩ and*h*_{FE}= 50.

*Solution:*

- Given
*R*= 10 kΩ and*f*= 10 kHz, for a symmetrical astable multivibrator. - Given
*T*_{1}= duration of the pulse = 0.1 ms,*f*= 1 kHz, for the un-symmetric astable multivibrator: - We have
*R*=*h*_{FE}*R*_{C}

*Example 7.5:* For the multivibrator shown in Fig. 7.13(a), *V*_{CC} = 20 V, *V*_{BB} = 10 V, *R*_{1} = *R*_{2} = *R* = 10 kΩ, *C*_{1} = *C*_{2} = *C* = 0.01 *μ*F. Find the time period and the frequency.

*Solution:*

*Example 7.6:* In an astable multivibrator if *R*_{2} = 60 kΩ, *R*_{1} = 40 kΩ, *C*_{1} = *C*_{2} = 2.9 nF, find its frequency and duty cycle.

*Solution:*

*T*_{2} = 0.69 *R*_{1}*C*_{1} = 0.69 × 40 × 10^{3} × 2.9 × 10^{−9} = 80 *μ*s = 0.08 ms

*T*_{1} = 0.69*R*_{2}*C*_{2} = 0.69 × 60 × 10^{3} × 2.9 × 10^{−9} = 120 *μ*s = 0.12 ms

*T* = *T*_{1} + *T*_{2} = 0.08 + 0.12 = 0.2 ms

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

*V*_{CC} = 10 V, *R*_{C} = 1 kΩ, *R* = 10 kΩ, and *C* = 0.01 *μ*F. Silicon transistors with *h*_{FE} = 50 and *r*_{bb′} = 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 *V*_{CC} = 10V, *R*_{C} = 1 kΩ, *R* = 10 kΩ, *C* = 0.01 *μ*F, *r*_{bb′} = 0.2 kΩ and *h*_{FE} = 50.

Assume *V*_{σ} = 0.7V and *V*_{CE(sat)} = 0.2V.

Let *Q*_{2} be ON and in saturation, then *Q*_{1} is OFF. After a time interval *T*_{1}, *Q*_{1} is ON and *Q*_{2} 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 *Q*_{2} switch from OFF to ON and *Q*_{1} 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 *Q*_{2} switches from OFF to ON

**FIGURE 7.18(b)** The circuit when *R* is much larger than *R*_{C}

Let *δ*^{′} and *δ* be the overshoots at the collector of *Q*_{1} and the base of *Q*_{2}.

*V*_{σ} = 0.7 V *V*_{γ} = 0.5 V and *V*_{CE(sat)} = 0.2 V

But, *δ* = *δ*^{′}, since the collector *Q*_{1} and the base of *Q*_{2} 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, *V*_{CC} = 15 V, *R*_{C} = 3 kΩ, *R*_{1} = *R*_{2} = *R* = 30 kΩ, *R*_{3} = 2 kΩ and silicon transistors with *h*_{FE} = 50 are used. *C*_{1} = *C*_{2} = *C* = 0.01 *μ*F. Verify whether the ON device is in saturation or not. Find its frequency.

**FIGURE 7.20** An astable multivibrator with vertical edges

Assume *Q*_{1} is OFF and *Q*_{2} is ON and in saturation. If *Q*_{2} is ON and in saturation

*V*_{C2} = *V*_{CE(sat)} = 0.2V, *V*_{B2} = *V*_{σ} = 0.7 V then *D*_{2} is ON. The collector load is *R*_{3}||*R*_{C}.

Hence, *Q*_{2} is in saturation.

To find, *f*:

For a symmetric astable multivibrator:

##### SUMMARY

- An astable multivibrator is essentially a square-wave generator.
- An astable multivibrator is used as a clock in digital circuits.
- If the time periods
*T*_{1}and*T*_{2}are not equal, the astable multivibrator is called an un-symmetric astable multivibrator. If*T*_{1}=*T*_{2}=*T*/2, then it is called a symmetric astable multivibrator. - The ratio of
*T*_{1}to*T*is called the duty cycle and is expressed as a percentage. - In a symmetric astable multivibrator, the frequency is given by
*f*= 0.7/*RC*cycles/s. - 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.
- An astable multivibrator can be used as a voltage-controlled oscillator (VCO) or a voltage-to-frequency converter (VFC).
- An astable multivibrator can also be used a frequency modulator.
- An emitter-coupled astable multivibrator starts oscillating the moment power is switched ON.

##### MULTIPLE CHOICE QUESTIONS

- An astable multivibrator is a:
- Square-wave generator
- Sweep generator
- Ramp generator
- None of the above

- 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:
- Saturation
- The active region
- The cut-off region
- The reverse-biased condition

- The duty cycle of a symmetric multivibrator is:
- 50%
- 75%
- 25%
- 100%

- The frequency of oscillations of a symmetric astable multivibrator is given as:
- 0.7
*τ*c/s - 2.2
*τ*c/s - None of the above

- 0.7
- To ensure that the ON device is in saturation, the condition that should be satisfied in a symmetric astable multivibrator is given by:
*R*≤*h*_{FE}*R*_{C}*R*≥*h*_{FE}*R*_{C}*R*should be infinity*R*should be zero

- The ratio of
*t*_{rec}/(*T*/2) of a symmetric astable is given by:- 3.2/
*h*_{FE} *h*_{FE}/3.2- 3.2
*h*_{FE} *h*_{FE}

- 3.2/
- An astable multivibrator can be used as a:
- Voltage-controlled oscillator
- Gating circuit
- Ramp generator
- Exponential generator

- An astable multivibrator can also be used as a:
- Frequency modulator
- Gating circuit
- Ramp generator
- Exponential generator

##### SHORT ANSWER QUESTIONS

- Draw the circuit of a collector-coupled astable multivibrator and explain its operation.
- Explain how an astable multivibrator can be modified to operate as a voltage-controlled oscillator.
- Explain why in an astable multivibrator the ON device in the quasi-stable state, is required to be driven into saturation.
- Explain how an astable multivibrator can be used as a frequency modulator.
- What are the advantages of an emitter-coupled astable multivibrator over a collector-coupled astable multivibrator?

##### LONG ANSWER QUESTIONS

- 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. - 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.
- The output of a collector-coupled astable multivibrator does not have sharp rising edges for the voltages at the collectors. Explain why.
- Draw the circuit of a collector-coupled astable multivibrator with vertical edges and explain its operation.

##### UNSOLVED PROBLEMS

- For the multivibrator in Fig. 7.1,
*R*_{1}=*R*_{2}=*R*= 47 kΩ,*C*_{1}=*C*_{2}=*C*= 0.01*μ*F. Find the time period and frequency. - For the astable multivibrator in Fig. 7.1,
*R*_{1}= 20 kΩ,*R*_{2}= 30 kΩ,*C*_{1}=*C*_{2}=*C*= 0.01*μ*F. Find the time period, duty cycle and the frequency. - For the symmetric astable multivibrator that generates square waves with vertical edges shown in Fig. 7.22,
*V*_{CC}= 10 V,*R*_{C}=*R*_{3}= 2 kΩ,*R*_{1}=*R*_{2}=*R*= 20 kΩ,*C*= 0.1*μ*F,*h*_{FE(min)}= 30. Show that the ON device is in saturation. Also find*f*. Assume suitable values for*V*_{CE(sat)}and*V*_{BE(sat)}. Si transistors are used. - Design a symmetric collector-coupled astable multivibrator to generate a square wave of 10 kHz having peak-to-peak amplitude of 10 V where
*h*_{FE(min)}= 30,*I*_{C(sat)}= 2mA. - Design an un-symmetric astable multivibrator having duty cycle of 40 per cent. It is required to oscillate at 5 kHz. Ge transistors with
*h*_{FE}= 40 are used. The amplitude of the square wave is required to be 20 V,*I*_{C}= 5 mA,*V*_{CE(sat)}= 0.1 V and*V*_{BE(sat)}= 0.3 V. - For an un-symmetric astable multivibrator
*R*_{1}= 100 kΩ,*R*_{2}= 100 kΩ,*C*_{1}= 0.02*μ*F,*C*_{2}= 0.01*μ*F. Find the frequency of oscillation and the duty cycle. - Design an unsymmetrical astable multivibrator shown in Fig. 7p.1 using silicon
*n–p–n*transistors having output amplitude of 12 V. Given data,*I*_{C(sat)}= 5 mA,*h*_{FE}(min) = 50,*f*= 5 kHz, duty cycle = 0.6.**FIGURE 7p.1**Un-symmetric astable multivibrator - The voltage-to-frequency converter shown in Fig. 7p.2 generates oscillations at a frequency
*f*_{1}when*V*_{BB}=*V*_{CC}. Find the ratio of*V*_{CC}/*V*_{BB}at which the frequency*f*_{2}= 4*f*_{1}.**FIGURE 7p.2**The given voltage-to-frequency converter