##### 7.2 COLLECTOR-COUPLED ASTABLE MULTIVIBRATORS

A collector-coupled astable multivibrator is shown in Fig. 7.1. Here, there is a cross-coupling from the second collector to the first base and also a cross-coupling from the first collector to the second base. This means that the output of one device is the input for the other. While analysing the functioning of this circuit, it is assumed that the circuit is an oscillator; and working backwards, we can justify that it does indeed oscillate.

**FIGURE 7.1** A collector-coupled astable multivibrator

Let us assume that at the given instant of time *Q*_{1} is OFF and *Q*_{2} is ON and saturated. Then *V*_{B2} = *V*_{σ}, *V*_{C2} = *V*_{CE(sat)} and *V*_{C1} = *V*_{CC}. With *Q*_{1} OFF and *Q*_{2} ON, *C*_{1} will try to charge to the supply voltage through the collector resistance *R*_{C1} and the small resistance of *Q*_{2} (≈ *r*_{bb′}), as shown in Fig. 7.2.

However, prior to this condition, *Q*_{2} must have been in the OFF state and *Q*_{1} must have been in the ON state. As a result, *C*_{2} must have been charged through *R*_{C2} and *r*_{bb′} of *Q*_{1}, as shown in Fig. 7.3.

When *Q*_{2} suddenly switches from the OFF state into the ON state, the voltage between its collector and emitter terminals is *V _{CE}* (≈0 V). Hence, the collector of

*Q*

_{2}is at ground potential, i.e., the positive end of the capacitor

*C*

_{2}is at the ground potential and its negative terminal is connected to the base of

*Q*

_{1}. As a large negative voltage is now coupled to the base of

*Q*

_{1}, it is in the OFF state (see Fig. 7.4). However,

*Q*

_{1}will not remain in the OFF state forever. Now, with

*Q*

_{2}ON, the charge on the condenser

*C*

_{2}discharges with a time constant

*τ*

_{2}=

*R*

_{2}

*C*

_{2}.

**FIGURE 7.2** The charging of the capacitor *C*_{1}

**FIGURE 7.3** The charging of capacitor *C*_{2}

**FIGURE 7.4** The discharge of *C*_{2} through *R*_{2}

As a result, the voltage at the base of *Q*_{1} goes on changing as a function of time. Once this voltage reaches *V*_{γ}, *Q*_{1} starts drawing base current. Hence, there is a collector current. In turn, there is voltage drop across *R*_{C1} and the voltage at the collector of *Q*_{1} falls. This voltage was earlier *V*_{CC} and now, it is smaller than *V*_{CC}. Therefore, the negative step at this collector is coupled to the base of *Q*_{2} through *C*_{1}. As the collector of *Q*_{1} and the base of *Q*_{2} are connected through *C*_{1} and the capacitor will not allow any sudden changes in the voltage, the base of *Q*_{2} undergoes changes that are identical to those that have taken place at the first collector. As a result, the base and collector current of *Q*_{2} drop; so, the collector voltage rises. This positive step change is coupled to the base of *Q*_{1}; and thus, its base current increases further. And the collector current increases, the voltage at the collector of *Q*_{1} falls still further. This step change is coupled to the base *Q*_{2} and this process is repeated. Thus, a regenerative action takes place and *Q*_{2} switches to the OFF state and *Q*_{1} goes into the ON state. Hence, the circuit oscillates. As long as the voltage at the base is *V*_{σ}, the voltage at the collector is *V*_{CE(sat)}. Once the voltage at the base is negative, the device switches into the OFF state, giving rise to a voltage *V*_{CC} at its collector. The waveforms at the two bases and the two collectors are shown in Fig 7.5. When a transistor, say *Q*_{1}, switches from the ON state into the OFF state, its collector voltage is required to abruptly rise to *V*_{CC}. Consequently, the waveforms at the collectors should have sharp rising edges and flat tops. But when *Q*_{1} goes into the OFF state and *Q*_{2} into the ON, there is a charging current of the condenser *C*_{1} which prevents the sudden rise of the voltage from *V*_{C1} to *V*_{CC}. Only when this charging current is zero does the collector voltage reach *V*_{CC}, as shown in Fig. 7.6. Hence, there is a rounding off of the rising edge of the pulse.

#### 7.2.1 Calculation of the Frequency of an Astable Multivibrator

To calculate the frequency (*f*), the two time periods *T*_{1} and *T*_{2} are to be calculated. Then:

*T* = *T*_{1} + *T*_{2} and

**FIGURE 7.5** The waveforms of a collector-coupled astable multivibrator

**Calculation of the Time Period, T**: To calculate

*T*, we calculate

*T*

_{1}and

*T*

_{2}. Consider the waveform at the base of

*Q*

_{2}, as shown in Fig. 7.7.

In general, the voltage variation at the base of *Q*_{2} as a function of time is given by the relation:

Here, *v*_{f} = *V*_{CC}, *v*_{i} = *V*_{σ} − *I*_{1}*R*_{C} and the time constant, *τ*_{1} = *R*_{1}*C*_{1}.

**FIGURE 7.6** *V*_{C1} rises to *V*_{CC} only when *i*_{C1} = 0

**FIGURE 7.7** Voltage variation at the base of *Q*_{2}

But, at *t* = *T*_{2}, *v*_{B2}(*t*) = *V*_{γ}

∴ *V*_{γ} = *V*_{CC} − (*V*_{CC} − *V*_{σ} + *I*_{1}*R _{C}*)

*e*

^{−T2/τ1}

Hence,

Similarly,

where, *I*_{1} and *I*_{2} are the currents in *Q*_{1} and *Q*_{2} when ON. However, here, in the expressions for *T*_{1} and *T*_{2}, the currents *I*_{1} and *I*_{2} are present (for all practical purposes these two currents are the same, if the devices have identical parameters).

A transistor can be said to be in the ON state when it is held in the active region or when it is driven into saturation. When the transistor is in the active region, its current (*I*_{1} or *I*_{2}) cannot be maintained at a constant value unless negative feedback is employed to stabilize the current. As there is no such provision in the circuit of Fig. 7.1, *I*_{1} and *I*_{2} are not necessarily stable. Hence, *T*_{1} and *T*_{2} cannot be maintained as constant; and *f*, the frequency of oscillations is also not stable. To ensure that *T*_{1} and *T*_{2} remain constant, the ON device is preferably driven into saturation. Consider the expression for *T*_{2} from Eq. (7.3):

If *Q*_{1}, in the quasi-stable state, is in saturation, then:

But,

Therefore,

Similarly,

*T*_{1} and *T*_{2} are now stable.

In a symmetric astable multivibrator, *τ*_{1} = *τ*_{2} = *τ*

Therefore,

and

If the ON device, in the quasi-stable state, is in the active region, then *T*_{1} and *T*_{2} are dependent of currents *I*_{1} and *I*_{2}, which, in the absence of feedback, are not stable. However, from Eqs. (7.6) and (7.7), it is evident that if the ON device is driven into saturation, *T*_{1} and *T*_{2} are independent of *I*_{1} and *I*_{2}; and hence, will remain stable. We will now derive the condition that ensures that the ON device is saturated as shown in Fig. 7.8(a).

Writing the KVL equations of the base and collector loops, we have:

*V*_{CC} = *I*_{B2}*R* + *V σ* = *I*_{C2}*R*_{C} + *V*_{CE(sat)}

When *Q*_{2} is in saturation, *V*_{σ} ≈ *V*_{CE(sat)}.

Therefore, *I*_{B2}*R* = *I*_{C2}*R*_{C}

Hence, for the ON device to be in saturation, the following condition has to be satisfied:

**Recovery Time.** From Fig. 7.6, it is seen that when a device is switched into the OFF state, the collector voltage will not abruptly rise to *V*_{CC} because of a small charging current associated with the condenser. The time taken for this collector voltage to reach 90 per cent of the steady-state value (*V*_{CC}) is called the recovery time, as shown in Fig. 7.8(b). From the figure it is seen that:

Recovery time *t*_{rec} = *t*_{2} − *t*_{1}

Recovery time

where, *τ*^{′} = ≈ *R _{C}C*

Therefore,

**FIGURE 7.8(a)** The ON transistor is in saturation

**FIGURE 7.8(b)** Recovery time

In a symmetric astable multivibrator: *R*_{1} = *R*_{2} = *R* and *C*_{1} = *C*_{2} = *C*. Hence,

But, *R* ≤ *h _{FE} R_{C}* when the ON transistor is in saturation.

As a specific example, if *h*_{FE} = 20,

Thus, the recovery time is 16 per cent of the half period *T*/2. If, alternately, *h*_{FE} = 80, the recovery time is 4 per cent. Thus, for the recovery time to be small, transistors with large *h*_{FE} may be employed.

**Calculation of the Overshoot.** When suddenly the transistor switches from the OFF state into the ON state, there could be a small overshoot at this base and at the collector of the transistor which switches from the ON state to the OFF state. To understand this, let us see an analogy. Assume that you are asked to come running and stop at a marked line. Now, due to inertia, instead of stopping at the dotted line you would over step it. Similarly, here the voltage at the base of the device which suddenly switches from the OFF state to the ON state will not just be *V*_{σ}, but slightly more. Hence, an overshoot is present at its base and similarly, at the other collector. At the end of the quasi-stable state, let *Q*_{1} go into the OFF state and *Q*_{2} into the ON state and then into saturation. The base-spreading resistance *r*_{bb′}—the resistance seen between the external base lead and the internal base terminal—is the resistance offered to the recombination current and is accounted by the amount of overshoot. This is typically less than 1 *k*Ω, as shown in Fig. 7.9(a).

From Fig. 7.9(a):

Neglecting the current , when compared to , the circuit reduces to that shown in Fig. 7.9(b).

**FIGURE 7.9(a)** *Q*_{2} at the end of the quasi-stable state

**FIGURE 7.9(b)** The simplified circuit of Fig. 7.9(a)

The voltages at *B*_{2} and collector of *Q*_{1} are:

The overshoot *δ* at the second base is the variation over and above *V*_{γ}.

Using Eq. (7.18):

Similarly, the overshoot *δ*^{′} at the first collector is the variation over and above *V*_{CE(sat)}.

Using Eq. (7.19):

As the first collector and the second base are connected through a condenser *C*_{1} which does not allow any sudden changes in voltage, changes, identical to those at the first collector, are required to take place at the second base.

Hence, *δ* = *δ*^{′}.

**FIGURE 7.9(c)** The overshoot at *B*_{2}

The overshoot at *B*_{2} is shown in Fig. 7.9(c).

#### 7.2.2 The Design of an Astable Multivibrator

Let us now try to design the astable multivibrator shown in Fig. 7.10.

We are now required to fix the component values of the astable multivibrator shown in Fig. 7.10.

For this to happen, we have to know *h*_{FE(min)}, *I*_{C(sat)}, *V*_{CE(sat)}, *V*_{BE(sat)} and the frequency *f* at which it is expected to oscillate. The value of *R*_{C} is calculated as:

*I*_{B}(min) is now calculated as:

**FIGURE 7.10** The design of an astable multivibrator

to ensure that the ON device is really in saturation:

For the ON device to be in saturation, Eq. (7.11) needs to be satisfied. If this condition is satisfied, choose the value of *R* as obtained in Eq. (7.28); otherwise choose the value of *R* satisfying Eq. (7.11). For a symmetric astable multivibrator, *f* = 0.7/*RC*. Using this relation, the value of *C* can be fixed. Example 7.1 further elucidates the point.

##### EXAMPLE

*Example 7.1:* Design an astable multivibrator, assuming that silicon devices with *h*_{FE(min)} = 40 are used. Also assume that *V*_{CC} = 10 V, *I*_{C(sat)} = 5 mA. Let the desired frequency of oscillations be 5 kHz. For transistor used, *V*_{CE(sat)} = 0.2 V, *V*_{BE(sat)} = *V*_{σ} = 0.7 V.

*Solution:*

Select *R*_{C} = 2 kΩ

Select *I*_{B(sat)} = 1.5*I*_{B}(min) = 1.5 × 0.125 = 0.187 mA

Select *R* = 47 kΩ

For the ON transistor to be in saturation, the condition is *R* ≤ *h*_{FE}*R*_{C}. Verify whether *R* ≤ *h*_{FE}*R*_{C} or not.

*h*_{FE}*R*_{C} = 40 × 2 kΩ = 80 kΩ

*R* < *h*_{FE}*R _{C}*, hence,

*Q*

_{2}is in saturation.

, for a symmetric astable multivibrator.

We have *f* = 5 kHz

Alternately, if this were to be an un-symmetric astable multivibrator (*T*_{1} ≠ *T*_{2}), then the duty cycle (= *T*_{1}/(*T*_{1} + *T*_{2}) = *T*_{1}/*T*) will have to be specified to fix the component values of *R*_{1}, *R*_{2}, *C*_{1} and *C*_{2}. That is, let *R*_{1} = *R*_{2} = *R* and duty cycle be specified as 40 per cent. If *f* = 5 kHz, then

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

*T*_{1}/(*T*_{1} + *T*_{2}) = *T*_{1}/*T* = 0.4 ms

Hence, *T*_{1} = (0.4)(0.2) = 0.08 ms, *T*_{2} = 0.2 − 0.08 = 0.12 ms.

From Eq. (7.6) we know that:

*T*_{2} = 0.69 *τ*_{1} = 0.69 *RC*_{1} (0.69)(47kΩ)*C*_{1} = 0.08 ms

Hence,

Similarly from Eq. (7.7),

*T*_{1} = 0.69 *τ*_{2} = 0.69*RC*_{2} (0.69)(47kΩ)*C*_{2} = 0.12 ms

The advantages of collector-coupled astable multivibrator in which the ON device is driven into saturation are that it is simple to implement and the dissipation in the ON device (= *V*_{CE(sat)} *I*_{C(sat)}) as well as that in the OFF device (*V*_{CC} *I _{CO}*) is negligibly small. So, devices with smaller dissipation capability can be employed. Yet another advantage is that the time durations are stable.

The reduced switching speed is a drawback with this circuit. As the storage time is large, the circuit takes a longer time delay for the ON device, driven into saturation, to be brought back into the OFF state. Further, as the output of one device is connected as the input to the other device through cross-coupling networks, these cross-coupling networks may cause loading. There is a rounding off of the rising edges (because of recovery transients) for the voltages at the two collectors; with the result; this circuit does not generate square waves with sharp rising edges. These factors also add to its disadvantages.

#### 7.2.3 An Astable Multivibrator with Vertical Edges for Collector Waveforms

The square waves generated by an astable multivibrator are required to have sharp rising edges (zero rise time). In the astable multivibrator shown in Fig. 7.1, the collector waveforms have no sharp rising edges (there is a finite rise time, shown in Fig. 7.5) because of the charging current of the condenser flowing through *R*_{C} of the OFF device.

The collector waveforms with sharp rising edges and flat tops are used in digital circuits to enable some other circuit either at the rising edge or at the falling edge. If needed we could clip these waveforms using a positive-peak clipper, say at a level, *V*, as shown in Fig. 7.11. However, this sacrifices the signal amplitude. If on the other hand, the capacitor is provided with an alternate charging path, the voltage at the collector of the OFF device can abruptly rise to *V*_{CC}, giving rise to pulses with vertical edges. To achieve this, the astable multivibrator circuit shown in Fig. 7.1 is modified as shown in Fig. 7.12.

**FIGURE 7.11** Deriving a square-wave output using a positive-peak clipper

Here, in this circuit, two diodes *D*_{1} and *D*_{2} and two resistors *R*_{3} and *R*_{4} (*R*_{3} = *R*_{4}) are added. If *Q*_{2} switches into the OFF state, *D*_{2} is OFF and the voltage at the collector of *Q*_{2} rises to *V*_{CC} abruptly. At the same time, the charging current of *C*_{2} passes through *R*_{4} and the small input resistance of *Q*_{1} as *Q*_{1} is ON. As this charging current does not pass through *R*_{C} of *Q*_{2}, the voltage at this collector rises abruptly to *V*_{CC}. The amplitude of the square wave is approximately equal to *V*_{CC}.