##### 8.3 CALCULATION OF THE VOLTAGES TO PLOT THE WAVEFORMS

To plot the waveforms at the two collectors and the two bases, the voltages in the circuit are to be calculated; when the multivibrator is in the stable state, when it is driven into the quasi-stable state and finally when it returns to the initial stable state. Consider the collector-coupled monostable multivibrator as shown in Fig. 8.1.

#### 8.3.1 In the Stable State (*t* < 0)

Here, the assumption is that *Q*_{2} is ON and in saturation while *Q*_{1} is OFF. In this situation, the need is to verify whether *Q*_{2} is really in saturation or not and whether *Q*_{1} is in the OFF state.

**To Verify that Q_{2} is in Saturation.** For this we need to calculate

*I*

_{C2}and

*I*

_{B2}and then verify whether

*I*

_{B2}is significantly larger than

*I*

_{B2}(min) or not. If

*I*

_{B2}>>

*I*

_{B2}(min), then

*Q*

_{2}is really in saturation. To verify this, consider the circuit shown in Fig. 8.7(a). From Fig. 8.7(a),

**FIGURE 8.7(a)** In the stable state *Q*_{1} is OFF and *Q*_{2} is ON

For *Q*_{2} to be in saturation, *I*_{B2} should be at least 1.5 *I*_{B2} (min). If *I*_{B2} >> *I*_{B2} (min), *Q*_{2} is really in saturation, as per the assumption made. Hence *V*_{C2} = *V*_{CE(sat)} and *V*_{B2} = *V*_{BE(sat)} = *V*_{σ}.

**To Verify that** *Q*_{1} **is OFF.** To show that *Q*_{1} is OFF, the voltage *V*_{B1} at *B*_{1} is to be found out and then checked whether the base–emitter diode is reverse-biased or not. If this diode is reverse-biased, then *Q*_{1} is OFF. To calculate the voltage *V*_{B1} at *B*_{1}, consider the circuit shown in Fig. 8.7(b). The voltage *V*_{B1} is due to the two sources; the *V*_{BB} source and the *V*_{CE(sat)} source. Use the superposition theorem to calculate *V*_{B1}, considering one source at a time. Shorting the *V*_{BB} source, the resultant circuit is as shown in Fig. 8.7(c).

Now shorting the *V*_{CE(sat)} source, the resultant circuit is as shown in Fig. 8.7(d).

Combining Eqs. (8.14) and (8.15), the net voltage *V*_{B1} at *B*_{1} due to the two sources *V*_{CE(sat)} and −*V*_{BB} is:

If the base of *Q*_{1} is negative with respect to the emitter, the base–emitter diode is reverse-biased. Therefore, *Q*_{1} is OFF, as assumed. Hence, *V*_{C1} = *V*_{CC}. The voltage across the capacitor terminals is,

#### 8.3.2 In the Quasi-stable State (*t = 0+*)

At *t* = 0, a negative pulse of proper amplitude is applied at the base of *Q*_{2} that drives *Q*_{2} into the OFF state. As a result *Q*_{1} goes into the ON state and into saturation as shown in Fig. 8.7(e).

**FIGURE 8.7(b)** The circuit for calculating *V*_{B1}

**FIGURE 8.7(c)** The circuit to calculate *V*_{B1} due to *V*_{CE}(sat) source

**FIGURE 8.7(d)** The circuit to calculate *V*_{B1} due to – *V*_{BB} source

**FIGURE 8.7(e)** In the quasi-stable state *Q*_{1} is ON and *Q*_{2} is OFF

First let us verify whether *Q*_{1} is really in saturation or not. Let us calculate *I*_{C1}.

To calculate *I*_{R}, write the KVL equation of the loop consisting of *R*_{C}, *C* and *R*.

Using Eqs. (8.19) and (8.20), we get *I*_{1} and *I*_{R}.

Therefore,

**To Calculate I_{B1}.** Considering Fig. 8.7(f):

If *I*_{B1} >> *I*_{B1(min)}, then *Q*_{1} is in saturation,

In the quasi-stable state, only *V*_{B2} changes and all other voltages remain unaltered. At *t* = *T*, when *V*_{B2} = *V*_{γ}, the quasi-stable state ends and the multivibrator returns to the initial stable condition of *Q*_{1} and *Q*_{2} in the OFF and ON states, respectively.

#### 8.3.3 At the End of the Quasi-stable State (at *t* = *T* +)

At the end of the quasi-stable state, *Q*_{1} goes OFF and *Q*_{2} goes ON and into saturation. In this process, overshoots (increase over and above the expected value) can occur at the base of *Q*_{2} and at the collector of *Q*_{1}, because the voltage change occurs abruptly. The amount of overshoot is accounted for by the base spreading resistance *r*_{bb′} which is the resistance seen between the external base lead and the internal base terminal and is the resistance offered to a recombination current. This is typically less than 1 kΩ, as shown in Fig. 8.7(g).

From Fig. 8.7(g),

As

Neglecting the current *I*_{R} when compared to the circuit reduces to as shown in Fig. 8.7(h).

The voltages at *B*_{2} and *C*_{1} are

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

Using Eq. (8.30):

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

Therefore,

Using Eq. (8.31):

**FIGURE 8.7(f)** Circuit that is used to calculate *I*_{B1}

**FIGURE 8.7(g)** The circuit at the end of the quasi-stable state

**FIGURE 8.7(h)** The simplified circuit of Fig. 8.7(g)

The first collector and the second base are connected through a condenser *C* and as the condenser will not allow any sudden changes in the voltages, whatever is the change that takes place at the first collector an identical change is required to take place at the second base. Hence, *δ* = *δ*^{′}.

The waveforms can now be plotted. To plot the waveforms of a collector-coupled monostable multivibrator with specific component values mentioned, consider the following example.

##### EXAMPLE

*Example 8.2:* Consider the circuit shown in Fig. 8.8(a), which uses an *n*−*p*−*n* silicon transistors with the following specifications: *V*_{CC} = 10 V, *V*_{BB} = 10 V, *R*_{C} = 1 kΩ, *R*_{1} = 10 kΩ = *R*, *R*_{2} = 100 kΩ, *h*_{FE(min)} = 30, *r*_{bb} = 0.2 kΩ, *V*_{CE(sat)} = 0.3 V, *V*_{BE(sat)} = *V*_{σ} = 0.7 V. Calculate all the current and voltages and then plot the waveforms.

*Solution:* (i) In the stable state (*t* < 0)

The assumption made is *Q*_{2} is ON and in saturation and *Q*_{1} is OFF. To verify that *Q*_{2} is ON and in saturation, *I*_{C2} and *I*_{B2} of *Q*_{2} are to be calculated. Further check whether *I*_{B2} >> *I*_{B2} (min) or not. If *I*_{B2} >> *I*_{B2} (min), then *Q*_{2} is really in saturation. Consider the circuit shown in Fig. 8.8(b).

(a) To verify if *Q*_{2} is ON and in saturation:

**FIGURE 8.8(a)** Practical collector-coupled monostable multivibrator

**FIGURE 8.8(b)** In the stable state, *Q*_{1} is OFF and *Q*_{2} is ON

∴ *I*_{C2} = *I*_{2} − *I*_{3} = 9.7 mA − 0.0934 mA = 9.61 mA

For *Q*_{2} to be in saturation, *I*_{B2} should be at least 1.5 *I*_{B2} (min). Hence, *I*_{B2} should be selected to keep *Q*_{2} in saturation as

*I*_{B2} = 1.5 × 0.32 mA = 0.48 mA.

As *I*_{B2}(0.93 mA) >> 1.5*I*_{B2(min)}(0.48 mA), as per the assumption made *Q*_{2} is really in saturation. Hence,

*V*_{C2} = 0.3 V, and *V*_{B2} = 0.7 V.

b) To verify that *Q*_{1} is OFF:

To verify whether *Q*_{1} is in the OFF state or not, *V*_{B1} = *V*_{BE1} is calculated and seen if it reverse-biases the base–emitter diode. The voltage *V*_{B1} is due to two sources—the *V*_{BB} source and the *V*_{CE(sat)} source, as shown in Fig. 8.8(c). Use the superposition theorem to calculate *V*_{B1}, considering one source at a time. Considering the *V*_{CE(sat)} source and shorting the *V*_{BB} source, the resultant circuit is as shown in Fig. 8.8(d).

**FIGURE 8.8(c)** The circuit for calculating *V*_{B1}

**FIGURE 8.8(d)** The circuit to calculate *V*_{B1} due to *V*_{CE(sat)} source

Now shorting the *V*_{CE(sat)} source, the resultant circuit is as shown in Fig. 8.8(e).

Therefore, the net voltage *V*_{B1} at *B*_{1} due to the two sources *V*_{CE(sat)} and −*V*_{BB} is

*V*_{B1} = 0.27 − 0.91 = −0.64 V.

This explains that the base of *Q*_{1} is negative with respect to the emitter by 0.64 V. Hence, the base–emitter diode is reverse-biased. Therefore, *Q*_{1} is OFF, as assumed. Hence, *V*_{C1} = *V*_{CC} = 10 V. The voltage across the capacitor terminals is

*V*_{A} = *V*_{C1} − *V*_{B2} = *V*_{CC} − *V σ* = 10 V − 0.7 V = 9.3 V

In the stable state, the voltages are *V*_{B1} = −0.64 V, *V*_{C1} = 10 V, *V*_{B2} = 0.7 V, *V*_{C2} = 0.3 V, *V*_{A} = 9.3 V.

ii) In the quasi-stable state (*t* = 0+)

In the quasi-stable state, *Q*_{2} is driven into the OFF state, by the application of a trigger. Consequently, *Q*_{1} goes into the ON state and into saturation as shown in Fig. 8.8(f).

**FIGURE 8.8(e)** Circuit to calculate *V*_{B1} due to −*V*_{BB} source

**FIGURE 8.8(f)** In the quasi-stable state, *Q*_{1} is ON and *Q*_{2} is OFF

a) To verify if *Q*_{1} is ON and in saturation or not

To verify whether *Q*_{1} is really in saturation or not, calculate *I*_{C1}.

To calculate *I*_{R} the KVL equation of the loop consisting of *R*_{C}, *C* and *R* is,

*I*_{R}*R* = *I*_{1}*R*_{C} + *V*_{A} = 9.7 V + 9.3 V = 19 V

Therefore,

*I*_{C1} = *I*_{1} + *I*_{R} = 9.7 mA + 1.9 mA = 11.6 mA

To calculate *I*_{B1}, consider Fig. 8.8(g).

*I*_{B1} = *I*_{3} − *I*_{4}

*I*_{B1} = 0.84 − 0.11 = 0.73 mA

**FIGURE 8.8(g)** The circuit that is used to calculate *I*_{B1}

As already calculated, *I*_{B1(min)} = 0.39 mA. Thus, *I*_{B1} >> *I*_{B1(min)}.

Hence, *Q*_{1} is in saturation.

∴ *V*_{C1} = 0.3 V, *V*_{B1} = 0.7 V,

*V*_{C2} = *V*_{CC} − *I*_{3}*R*_{C} = 10 − (0.84)(1) = 9.16 V (But for the current, *I*_{3}, *V*_{C2} should have been 10 V)

*V*_{B2} = *V*_{CC} − *I*_{R}*R* = 10 − (1.9)(10) = −9 V

In the quasi-stable state, except *V*_{B2}—which changes exponentially as a function of time—all other voltages remain unaltered. At *t* = *T*, when *V*_{B2} = *V*_{γ}, the quasi-stable state ends and the multivibrator returns to its initial stable condition. The voltages at the beginning of the quasi-stable state are *V*_{C1} = 0.3 V, *V*_{B1} = 0.7 V, *V*_{C2} = 9.16 V, *V*_{B2} = −9 V initially and varies exponentially.

iii) At the end of the quasi-stable state (at *t* = *T* +)

At the end of the quasi-stable state *Q*_{1} goes OFF and *Q*_{2} goes ON and into saturation, resulting in overshoots at the base of *Q*_{2} and at the collector of *Q*_{1}. The overshoots are calculated using Fig. 8.8(h).

Neglecting the current *I*_{R} when compared to the circuit reduces to Fig. 8.8(i).

**FIGURE 8.8(h)** The circuit that is used to calculate *I*_{B1}

**FIGURE 8.8(i)** The simplified circuit of Fig. 8.8(h)

**FIGURE 8.8(j)** Waveforms of the collector-coupled monostable

Using Eq. (8.34):

= 7.91 mA

= *V _{CC} − R_{C}* = 10 − (7.91)(1)

= 2.09 V

= *r _{bb′}* +

*V*

_{σ}

= (7.91)(0.2) + 0.7 = 1.58 + 0.7

= 2.28 V

At *t* = *T*+, the voltages are *V*_{C2} = 0.3 V, *V*_{B2} = 2.28 V, *V*_{C1} = 10 V, *V*_{B1} = −0.64 V. The waveforms are plotted as shown in Fig. 8.8(j).

#### 8.3.4 The Design of a Collector-coupled Monostable Multivibrator

Let us design a collector-coupled monostable multivibrator of Fig. 8.1, having a gate width *T*. From the circuit of Fig. 8.1 we have,

Select *I*_{B(sat)} = 2 × *I*_{B(min)}

For ON device to be in saturation:

On verifying whether the condition in Eq. (8.40) is satisfied or not; the value of *R* is accepted (if satisfying). If not, the value of *R* is changed suitably, to satisfy the Eq. (8.40).

Assuming that the current in *R*_{2} is *I*_{2}:

When *Q*_{1} is ON, if *I*_{1} is the current in (*R*_{C} + *R*_{1}):

*R*_{1} = (*R _{C}* +

*R*

_{1}) −

*R*

_{C}

Using the relation, *T* = 0.69*RC*, the value of *C* is calculated. To understand the design procedure let us consider an example.

##### EXAMPLE

*Example 8.3:* Design a collector-coupled monostable circuit of Fig. 8.1 to generate a pulse of width 100 *µ*s. Silicon devices with *h*_{FE(min)} = 50 are used. ON device is in saturation.

Given that:

*V*_{CC} = 12 V, *V*_{CE(sat)} = 0.2 V, *V*_{BE(sat)} = 0.7 V, *I*_{B(sat)} = 2 *I*_{B(min)}, *V*_{BB} = 12 V, *I*_{C(sat)} = 2 mA, *T* = 100 *µ*s.

*Solution:*

- Let
*Q*_{2}be ON and*Q*_{1}be OFF.*I*_{B2(sat)}= 2 ×*I*_{B2}(min) = 2 × 40 = 80*µ*A*h*= 50 × 6 = 300 kΩ_{FE}R_{C}For ON device to be in saturation,

*R*≤*h*_{FE}R_{C}Hence, the condition is verified.

- When
*Q*_{1}is ON, let*I*_{2}be the current in*R*_{2}.Let

*I*_{1}be the current in*R*_{C}+*R*_{1}.*I*_{1}=*I*_{B1}+*I*_{2}= 0.08 + 0.2 = 0.28 mA*R*_{1}= (*R*+_{C}*R*_{1}) −*R*_{C}= 40 − 6 = 34 kΩ*T*= 0.69*R*C 100 × 10^{−6}*s*= 0.69 × 141 × 10^{3}×*C*