# 7.2 Collector-coupled Astable Multivibrators – Pulse and Digital Circuits

##### 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 Q1 is OFF and Q2 is ON and saturated. Then VB2 = Vσ, VC2 = VCE(sat) and VC1 = VCC. With Q1 OFF and Q2 ON, C1 will try to charge to the supply voltage through the collector resistance RC1 and the small resistance of Q2 (≈ rbb), as shown in Fig. 7.2.

However, prior to this condition, Q2 must have been in the OFF state and Q1 must have been in the ON state. As a result, C2 must have been charged through RC2 and rbb of Q1, as shown in Fig. 7.3.

When Q2 suddenly switches from the OFF state into the ON state, the voltage between its collector and emitter terminals is VCE (≈0 V). Hence, the collector of Q2 is at ground potential, i.e., the positive end of the capacitor C2 is at the ground potential and its negative terminal is connected to the base of Q1. As a large negative voltage is now coupled to the base of Q1, it is in the OFF state (see Fig. 7.4). However, Q1 will not remain in the OFF state forever. Now, with Q2 ON, the charge on the condenser C2 discharges with a time constant τ2 = R2C2.

FIGURE 7.2 The charging of the capacitor C1

FIGURE 7.3 The charging of capacitor C2

FIGURE 7.4 The discharge of C2 through R2

As a result, the voltage at the base of Q1 goes on changing as a function of time. Once this voltage reaches Vγ, Q1 starts drawing base current. Hence, there is a collector current. In turn, there is voltage drop across RC1 and the voltage at the collector of Q1 falls. This voltage was earlier VCC and now, it is smaller than VCC. Therefore, the negative step at this collector is coupled to the base of Q2 through C1. As the collector of Q1 and the base of Q2 are connected through C1 and the capacitor will not allow any sudden changes in the voltage, the base of Q2 undergoes changes that are identical to those that have taken place at the first collector. As a result, the base and collector current of Q2 drop; so, the collector voltage rises. This positive step change is coupled to the base of Q1; and thus, its base current increases further. And the collector current increases, the voltage at the collector of Q1 falls still further. This step change is coupled to the base Q2 and this process is repeated. Thus, a regenerative action takes place and Q2 switches to the OFF state and Q1 goes into the ON state. Hence, the circuit oscillates. As long as the voltage at the base is Vσ, the voltage at the collector is VCE(sat). Once the voltage at the base is negative, the device switches into the OFF state, giving rise to a voltage VCC at its collector. The waveforms at the two bases and the two collectors are shown in Fig 7.5. When a transistor, say Q1, switches from the ON state into the OFF state, its collector voltage is required to abruptly rise to VCC. Consequently, the waveforms at the collectors should have sharp rising edges and flat tops. But when Q1 goes into the OFF state and Q2 into the ON, there is a charging current of the condenser C1 which prevents the sudden rise of the voltage from VC1 to VCC. Only when this charging current is zero does the collector voltage reach VCC, 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 T1 and T2 are to be calculated. Then:

T = T1 + T2    and

FIGURE 7.5 The waveforms of a collector-coupled astable multivibrator

Calculation of the Time Period, T: To calculate T, we calculate T1 and T2. Consider the waveform at the base of Q2, as shown in Fig. 7.7.

In general, the voltage variation at the base of Q2 as a function of time is given by the relation:

Here, vf = VCC, vi = VσI1RC and the time constant, τ1 = R1C1.

FIGURE 7.6 VC1 rises to VCC only when iC1 = 0

FIGURE 7.7 Voltage variation at the base of Q2

But, at t = T2, vB2(t) = Vγ

Vγ = VCC − (VCCVσ + I1RC) eT2/τ1

Hence,

Similarly,

where, I1 and I2 are the currents in Q1 and Q2 when ON. However, here, in the expressions for T1 and T2, the currents I1 and I2 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 (I1 or I2) 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, I1 and I2 are not necessarily stable. Hence, T1 and T2 cannot be maintained as constant; and f, the frequency of oscillations is also not stable. To ensure that T1 and T2 remain constant, the ON device is preferably driven into saturation. Consider the expression for T2 from Eq. (7.3):

If Q1, in the quasi-stable state, is in saturation, then:

But,

Therefore,

Similarly,

T1 and T2 are now stable.

Therefore,

In a symmetric astable multivibrator, τ1 = τ2 = τ

Therefore,

and

If the ON device, in the quasi-stable state, is in the active region, then T1 and T2 are dependent of currents I1 and I2, 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, T1 and T2 are independent of I1 and I2; 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:

VCC = IB2R + V σ = IC2RC + VCE(sat)

When Q2 is in saturation, VσVCE(sat).

Therefore,       IB2R = IC2RC

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 VCC 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 (VCC) is called the recovery time, as shown in Fig. 7.8(b). From the figure it is seen that:

Recovery time trec = t2t1

Recovery time

where, τ = ≈ RCC

Therefore,

FIGURE 7.8(a) The ON transistor is in saturation

FIGURE 7.8(b) Recovery time

In a symmetric astable multivibrator: R1 = R2 = R and C1 = C2 = C. Hence,

But, RhFE RC when the ON transistor is in saturation.

As a specific example, if hFE = 20,

Thus, the recovery time is 16 per cent of the half period T/2. If, alternately, hFE = 80, the recovery time is 4 per cent. Thus, for the recovery time to be small, transistors with large hFE 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 Vσ, 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 Q1 go into the OFF state and Q2 into the ON state and then into saturation. The base-spreading resistance rbb—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 kΩ, 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) Q2 at the end of the quasi-stable state

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

The voltages at B2 and collector of Q1 are:

The overshoot δ at the second base is the variation over and above Vγ.

Using Eq. (7.18):

Similarly, the overshoot δ at the first collector is the variation over and above VCE(sat).

Using Eq. (7.19):

As the first collector and the second base are connected through a condenser C1 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 B2

The overshoot at B2 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 hFE(min), IC(sat), VCE(sat), VBE(sat) and the frequency f at which it is expected to oscillate. The value of RC is calculated as:

IB(min) is now calculated as:

FIGURE 7.10 The design of an astable multivibrator

Now select:

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 hFE(min) = 40 are used. Also assume that VCC = 10 V, IC(sat) = 5 mA. Let the desired frequency of oscillations be 5 kHz. For transistor used, VCE(sat) = 0.2 V, VBE(sat) = Vσ = 0.7 V.

Solution:

Select RC = 2 kΩ

Select IB(sat) = 1.5IB(min) = 1.5 × 0.125 = 0.187 mA

Select R = 47 kΩ

For the ON transistor to be in saturation, the condition is RhFERC. Verify whether RhFERC or not.

hFERC = 40 × 2 kΩ = 80 kΩ

R < hFERC, hence, Q2 is in saturation.

, for a symmetric astable multivibrator.

We have f = 5 kHz

Alternately, if this were to be an un-symmetric astable multivibrator (T1T2), then the duty cycle (= T1/(T1 + T2) = T1/T) will have to be specified to fix the component values of R1, R2, C1 and C2. That is, let R1 = R2 = R and duty cycle be specified as 40 per cent. If f = 5 kHz, then

T = T1 + T2 = 0.2 ms

T1/(T1 + T2) = T1/T = 0.4 ms

Hence, T1 = (0.4)(0.2) = 0.08 ms, T2 = 0.2 − 0.08 = 0.12 ms.

From Eq. (7.6) we know that:

T2 = 0.69 τ1 = 0.69 RC1      (0.69)(47kΩ)C1 = 0.08 ms

Hence,

Similarly from Eq. (7.7),

T1 = 0.69 τ2 = 0.69RC2    (0.69)(47kΩ)C2 = 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 (= VCE(sat) IC(sat)) as well as that in the OFF device (VCC ICO) 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 RC 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 VCC, 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 D1 and D2 and two resistors R3 and R4 (R3 = R4) are added. If Q2 switches into the OFF state, D2 is OFF and the voltage at the collector of Q2 rises to VCC abruptly. At the same time, the charging current of C2 passes through R4 and the small input resistance of Q1 as Q1 is ON. As this charging current does not pass through RC of Q2, the voltage at this collector rises abruptly to VCC. The amplitude of the square wave is approximately equal to VCC.