##### 9.4 SCHMITT TRIGGERS

An emitter-coupled bistable multivibrator is also called a Schmitt trigger, named after the designer of the vacuum tube version. In addition to being used as a bistable multivibrator, it has some more important applications.

In the Schmitt trigger circuit shown in Fig. 9.12(a), it is seen that the output of the first transistor is connected to the input of the second transistor through a potential divider network comprising *R*_{1} and *R*_{2}. This is simply an attenuator circuit. Normally *R*_{1} and *R*_{2} are reasonably large resistors so as to avoid loading the collector of *Q*_{1}. Further, the emitter resistance *R _{E}* stabilizes the currents and voltages. Note that the second collector and the first emitter are not involved in the regenerative loop (there is no cross-coupling from the second collector to the first base). So, when used as a bistable multivibrator, there is no loading on the second collector and the trigger is applied at the first base and the output is taken from the second collector.

As long as the battery voltage *v _{i}* is small,

*Q*

_{1}is OFF. The voltage at this collector is approximately

*V*. This voltage is coupled to the second base through

_{CC}*R*

_{1}and

*R*

_{2}. As a result,

*Q*

_{2}can conduct. If

*Q*

_{2}conducts, it can operate in the active region or it may be driven into saturation.

Let it be assumed that *Q*_{2} is in the active region. The base current *I*_{B2} and collector current *I*_{C2} flows through *R _{E}*. Therefore, a voltage

*V*=

_{EN}*V*

_{EN2}is developed in

*R*. As

_{E}*V*

_{BE1}=

*V*

_{BN1}−

*V*

_{EN2}and if

*V*

_{BE1}reverse-biases the emitter diode of

*Q*

_{1}, then as assumed

*Q*

_{1}is OFF.

If *V*_{BN1} (*v _{i}*) is increased, at a value (

*V*

_{EN2}+

*V*

_{γ1}),

*Q*

_{1}begins to conduct. As a result, the voltage at the second base decreases, hence the base current of

*Q*

_{2}decreases, its collector current also reduces and consequently the voltage at the second collector rises. If the input is increased further,

*Q*

_{1}goes into the ON state and

*Q*

_{2}into the OFF state. If the loop gain is less than unity (this condition can be satisfied by reducing the collector load of

*Q*

_{1}), there exists a region of linearity in the transfer characteristic. In this region an incremental change at the input Δ

*v*will cause a proportional change in the output, Δ

_{i}*v*as shown in Fig. 9.12(b). If the loop gain is made equal to 1 by adjusting

_{o}*R*

_{C1}and

*R*

_{C2}, the transfer characteristic is as shown in Fig. 9.12(c).

If on the other hand, the loop gain is made greater than 1, the transfer characteristic is an S-shaped characteristic, as shown in Fig. 9.12(d). When the input is increased from 0 to a larger value, at a voltage *V*_{1}, the output suddenly jumps from a smaller value to *V _{CC}* as shown in Fig. 9.13(a). Even if the input is increased further, the output remains at

*V*.

_{CC}If on the other hand, the input is decreased, then at a voltage *V*_{2} the output falls from *V _{CC}* to a smaller value, as shown in Fig. 9.13(b). If these two curves are combined together, the resultant transfer characteristic, that gives the relation between the input and the output, is shown in Fig. 9.13(c).

**FIGURE 9.12(a)** The Schmitt trigger

*V*_{1} is called the upper trip point (UTP) and *V*_{2} is called the lower trip point (LTP). The closed loop in Fig. 9.13(c) is termed the hysteresis loop and the difference in voltages *V*_{1} and *V*_{2} is called the hysteresis voltage, *V _{H}*. Thus,

*V*=

_{H}*V*

_{1}−

*V*

_{2}.

It is seen from the above discussion that a Schmitt trigger exhibits hysterisis, i.e., when the input *v _{i}* is increased to reach a voltage

*V*

_{1}it is required to first pass through a point,

*V*

_{2}at which the reverse transition takes place. Similarly, when the input now is reduced to reach

*V*

_{2}it has to pass through the point

*V*

_{1}. This is called hysteresis. This characteristic of the Schmitt trigger is used to an advantage in waveshaping applications.

**FIGURE 9.12(b)** The transfer characteristic of a Schmitt trigger when the loop gain *<* 1

**FIGURE 9.12(c)** The transfer characteristic of a Schmitt trigger when the loop gain is 1

**FIGURE 9.12(d)** The transfer characteristic of a Schmitt trigger when the loop gains *>* 1

**FIGURE 9.13(a)** The output rises suddenly to *V _{CC}*

**FIGURE 9.13(b)** When the input is decreased the output falls from *V _{CC}*

**FIGURE 9.13(c)** The transfer characteristic of a Schmitt trigger

#### 9.4.1 Calculation of the Upper Trip Point (*V*_{1})

**FIGURE 9.14** The Schmitt trigger circuit

In the Schmitt trigger circuit, shown in Fig. 9.14, when the input is increased, till *V*_{1} is reached *Q*_{1} is OFF and *Q*_{2} is ON. As a result, *I*_{B2} and *I*_{C2} flow through *R _{E}* developing a voltage

*V*in

_{EN}*R*. Now, if the input is such that its value is (

_{E}*V*+

_{EN}*V*

_{γ1}) =

*V*

_{1}(UTP),

*Q*

_{1}switches into the ON state and

*Q*

_{2}switches into the OFF state.

If the circuit show in Fig. 9.14 is Thévenised at the base of *Q*_{2}, the Thévenin voltage source is,

And its internal resistance *R*^{′} is given by the relation:

The resultant simplified circuit is shown in Fig. 9.15(a).

**FIGURE 9.15(a)** The circuit that enables computation of *V*_{1}

The total current in *R _{E}* is

*I*

_{B2}(1 +

*h*) and the current in

_{FE}*R*

^{′}is

*I*

_{B2}. As far as

*I*

_{B2}is concerned,

*R*is seen to have increased by a factor (1 +

_{E}*h*). The net voltage in the base loop is, (

_{FE}*V*

^{′}−

*V*

_{BE2}) and is equal to the sum of the voltage drops across

*R*

^{′}and (1 +

*h*)

_{FE}*R*, as shown in Fig. 9.15(b).

_{E}**FIGURE 9.15(b)** The circuit to calculate *V*_{EN2}

In Eq. (9.58), as:

*R*^{′} << *R _{E}*(1 +

*h*

_{FE}) R^{′}+

*R*(1 +

_{E}*h*≅

_{FE})*R*(1 +

_{E}*h*

_{FE})Then Eq. (9.58) reduces to:

From Fig. 9.15(a), using Eq. (9.60):

The calculation of *V*_{1} is made based on the assumption that *Q*_{2} is in the active region. Now, to verify whether *Q*_{2} is in the active region, *V*_{CB2} is calculated and checked if this reverse-biases the collector diode by a reasonable voltage or not. If it does, the device *Q*_{2} is indeed in the active region.

From the circuit in Fig. 9.15(a), we have:

*V*_{CB2} = *V*_{CE2} − *V*_{BE2} and *V*_{CE2} = *V _{CC}* −

*I*

_{C2}

*R*

_{C2}−

*V*

_{EN2}

To calculate *V*_{CB2} using Eq. (9.62), we have to find *I*_{C2}.

where,

Substituting Eq. (9.63) in Eq. (9.61) we get:

where is given by Eq. (9.64). From Eq. (9.63) *I*_{C2} is given as:

These calculations were made based on the assumption that *Q*_{2} is in the active region. Having made the calculations, we once again verify whether *Q*_{2} is really in the active region or not, to justify the validity of the calculations made. *V*_{CB2} is calculated using Eq. (9.62). If the base−collector diode is reverse-biased, then *Q*_{2} is in the active region as assumed.

#### 9.4.2 Calculation of the Lower Trip Point (*V*_{2})

At the voltage *V*_{1} (UTP), *Q*_{1} is ON and *Q*_{2} is OFF. Now if the input is reduced, till voltage *V*_{2} is reached, *Q*_{1} is ON and *Q*_{2} is still OFF. However, when the voltage at the input is *V*_{2} then the voltage at the second base is *V*_{EN1} + *V _{γ}* 2.

*Q*

_{2}again switches into the ON state and

*Q*

_{1}into the OFF state.

Consider the Schmitt trigger circuit shown in Fig. 9.14. The Thévenin voltage source at the first collector is:

And its internal resistance *R _{t}* is,

The first collector and the second base are connected through *R*_{1} and *R*_{2}, as shown in Fig. 9.16.

where

The circuit that enables us to calculate *V*_{2} is shown in Fig. 9.17.

Writing the KVL equation for the base loop of *Q*_{2}:

*αV*_{CN1} = *V*_{γ2} + (*I*_{B1} + *I*_{C1}) *R _{E}*

*V*

_{CN1}=

*V*−

_{t}*I*

_{C1}

*R*

_{t}**FIGURE 9.16** The coupling network from the first collector to the second base

**FIGURE 9.17** The circuit to calculate *V*_{2}

Let:

However, we have:

Therefore, from Eq. (9.73)

Using Eqs. (9.72), (9.74) and (9.75), *V*_{2} is calculated. To understand the method of calculation for *V*_{1} and *V*_{2}, let us consider Example 9.5.

##### EXAMPLE

*Example 9.5:* For the Schmitt trigger circuit shown in Fig. 9.18(a), calculate *V*_{1} and *V*_{2}.

**FIGURE 9.18(a)** The Schmitt trigger with components mentioned

a) Calculation of *V*_{1}:

Consider the Schmitt trigger circuit, shown in Fig. 9.18(a). From Eq. (9.55):

*R*^{′} the internal resistance of this Thévenin source, as given by Eq. (9.56), is:

The resultant circuit is shown in Fig. 9.18(b).

**FIGURE 9.18(b)** The circuit that enables computation of *V*_{1}

From Eq. (9.58):

If *Q*_{2} is in the active region, typically, for silicon *V*_{BE2} = 0.6 V and let *h _{FE}* = 50,

*R _{E}*(1 +

*h*= 3(1 + 50) = 153 kΩ

_{FE})Therefore,

Therefore,

*V*_{1} = *V*_{EN2} + *V*_{BE2} = 4.76 + 0.6 = 5.36 V.

The calculation of *V*_{1} is made based on the assumption that *Q*_{2} is in the active region. To find out whether *Q*_{2} is in the active region or not, we calculate *V*_{CB2}.

*V*_{CB2} = *V _{CC}* −

*I*

_{C2}

*R*

_{C2}−

*V*

_{EN2}−

*V*

_{BE2}.

From Eq. (9.72)

Hence

*V*_{CB2} = 10 − (1.56 × 1) − 4.76 − 0.6 = 10 − 6.92 = 3.08 V

As the collector of *Q*_{2} is positive with respect to the base by 3.08 V the collector diode is reverse-biased. Hence, *Q*_{2} is in the active region, as assumed.

(b) Calculation of *V*_{2}:

The circuit that enables us to calculate *V*_{2} is shown in Fig. 9.18(c). From the circuit values:

∴ *V*_{2} = 0.6 V + (1.05 mA)(3.06 kΩ)= 0.6 V + 3.22 V = 3.82 V

Hence, for the given Schmitt trigger:

*V*_{1} = 5.36 V *V*_{2} = 3.82 V *V _{H}* =

*V*

_{1}−

*V*

_{2}= 5.36 − 3.82 = 1.54 V

**FIGURE 9.18(c)** The circuit to calculate *V*_{2}

#### 9.4.3 Methods to Eliminate Hysteresis in a Schmitt Trigger

It is evident from Fig. 9.13(c), that hysteresis is present in a Schmitt trigger because the loop gain is not exactly unity, but is greater than 1. Hysteresis is needed when a schmitt trigger used as a histable multitrigger and also when converting a time-varying signal into a square wave. But when it used as a comparator, hysteresis needs to be elimimated. The following schemes can be implemented to eliminate hysteresis in a Schmitt trigger:

*V*_{1}and*V*_{2}are made to coincide with the proper choice of*R*_{1}and*R*_{2}. However, by this method, though, it is not possible to make*V*_{1}identical to*V*_{2}, they can be brought close to each other in practice.- Another method to eliminate hysteresis is by introducing a resistance
*R*_{e2}in series with the emitter terminal of*Q*_{2}, as shown in Fig. 9.19(a).

*R*_{e2} will change *V*_{1}, but has no effect on *V*_{2}. Therefore, by including *R*_{e2} in series with the emitter of *Q*_{2}, it is possible to reduce *V*_{1} to the level of *V*_{2},(*V*_{1} is made equal to *V*_{2}), thereby reducing hysteresis. The method to calculate *R*_{e2}, for the circuit shown in Fig. 9.19(b), calculate *V*^{′} and *R*^{′} as illustrated in Section 9.4.1, as shown in Fig. 9.15(a).

From Fig. 9.19(b), we have:

**FIGURE 9.19(a)** An alternate method to eliminate hysteresis

**FIGURE 9.19(b)** The circuit to calculate *R*_{e2}

To eliminate hysteresis, *V*_{1} should be made equal to *V*_{2}, which means that in Eq. (9.78) *V*_{1} is replaced by *V*_{2}

*R*_{e2} that eliminates hysteresis is calculated using Eq. (9.79). To further understand the procedure let us consider Example 9.6.

##### EXAMPLE

*Example 9.6:* From the problem in Example 9.5, we have *V*_{1} = 5.36 V, *V*_{2} = 3.82 V, *V ^{′}* = 5.45 V,

*R*= 2.73 kΩ,

^{′}*R*= 3kΩ,

_{E}*h*= 50,

_{FE}*V*

_{BE2}= 0.6 V. Find the value of

*R*

_{e2}that ensures

*V*

_{1}=

*V*

_{2}= 3.82 V and eliminates hysteresis.

*Solution:* We have from Eq. (9.79):

Therefore,

**FIGURE 9.19(c)** Another method to eliminate hysteresis

(iii) Another method to eliminate hysteresis is to introduce a resistance *R*_{e1}in series with the emitter terminal of *Q*_{1}, as shown in Fig. 9.19(c).

*R*_{e1} is not going to change *V*_{1} but will only influence *V*_{2}. To eliminate hysterisis, *R*_{e1} is chosen such that *V*_{2} is increased to the level of *V*_{1}. To achieve this, we must ensure that *V*_{2} plus the voltage drop across *R*_{e1} is *V*_{1}.

From Eq. (1):

Substituting Eq. (3) in Eq. (2), we get:

To understand the procedure to calculate *R*_{e1}, let us consider an example.

##### EXAMPLE

*Example 9.7:* From the problem in Example 9.5, we have *V*_{1} = 5.36 V, *V*_{2} = 3.82 V, *I*_{C1} = 1.05 mA, *h _{FE}* = 50. Find the value of

*R*

_{e1}.

*Solution:*

Using Eq. (9.83),

#### 9.4.4 Applications of a Schmitt Trigger

The following are some applications of a Schmitt trigger:

(a) An emitter-coupled bistable multivibrator is called the Schmitt trigger. Hence, a Schmitt trigger can be used as a bistable multivibrator. Consider the transfer characteristic, shown in Fig. 9.20(a).

To use this circuit as a bistable multivibrator, the first device *Q*_{1} is biased to have a voltage *V* at its base. Initially, let the output be at 0 level (*V _{CC}* −

*I*

_{C2}

*R*

_{C2}). To change this to 1(

*V*) apply a positive pulse at the base of

_{CC}*Q*

_{1}, whose magnitude is more positive than (

*V*

_{1}−

*V*) as shown in Fig. 9.20(b).

To once again change the output to a 0 level, apply a pulse at the base of *Q*_{1}, which is negative with respect to *V* and whose magnitude is more negative than (*V*_{2} − *V*) as shown in Fig. 9.20(c).

(b) A Schmitt trigger can be used as an amplitude comparator. In an amplitude comparator, the amplitude of a time varying signal is compared with a reference and it tells us the time instant at which the input has reached this set reference level. For example, consider the diode comparator circuit shown in Fig. 9.21(a). As long as *v _{i} < V_{R}, D* is OFF and

*v*=

_{o}*v*. When

_{i}*v*≥

_{i}*V*is ON and

_{R}, D*v*=

_{o}*V*.

_{R}**FIGURE 9.20(a)** The transfer characteristic of a Schmitt trigger

**FIGURE 9.20(b)** The trigger to change the output from 0 to 1 level

**FIGURE 9.20(c)** The trigger to change the output from 1 to 0 level

**FIGURE 9.21(a)** A diode comparator

The output follows the input till *t* = *t*_{1} and then the slope of the output abruptly changes. This is called the break point, and *t*_{1} is the time instant at which *v _{i}* has reached

*V*.

_{R}Now consider the Schmitt trigger (in which hysteresis is eliminated) as a comparator with input and output shown in Fig. 9.21(b). A relatively small dc voltage is there at the output till *V*_{1}(= *V*_{2}) is reached at the input. The moment the input is *V*_{1}, the output abruptly jumps to *V _{CC}*. The slope of the input has no relation to the slope of the signal at the output. Thus, a Schmitt trigger can be used as a better amplitude comparator.

(c) A Schmitt trigger can be used as a waveshaping circuit (or a squaring circuit). It can be used to convert any arbitrarily time varying signal into a square-wave output. The only condition to be satisfied is that the input signal has amplitude more than *V*_{1} and also less than *V*_{2}. Consider the input for which the output is plotted as shown in Fig. 9.22.

**FIGURE 9.21(b)** A Schmitt trigger as comparator

**FIGURE 9.22** The Schmitt trigger as a squaring circuit

**FIGURE 9.23(a)** A Schmitt trigger circuit

#### 9.4.5 The Design of a Schmitt Trigger

In this section, the procedure to design a Schmitt trigger is presented. For designing the Schmitt trigger shown in Fig. 9.23(a), UTP (*V*_{1}) and LTP (*V*_{2}) values, *h _{FE}*, desired

*I*and

_{C}*V*are to be specified.

_{CC}(a) Till UTP is reached *Q*_{1} is OFF and *Q*_{2} is ON and in saturation. Just at *V*_{1} (UTP), *Q*_{1} goes ON and *Q*_{2} goes OFF.

Therefore, *V*_{1} = UTP = *V*_{BN2} and *I _{E}* =

*I*is specified. Therefore:

_{C}*R _{E}* is chosen using Eq. (9.80). If

*Q*

_{2}is in saturation,

*V*=

_{CE}*V*

_{CE(sat)}. From the circuit shown in Fig. 9.23(a), we have

*I*

_{C2}

*R*

_{C2}=

*V*−

_{CC}*V*

_{CE(sat)}−

*V*

_{EN2}

Therefore,

*R*_{C2} is calculated using Eq. (9.81).

Assume that

Using Eq. (9.82):

Using the value of *h _{FE}* specified, we can now calculate

*I*

_{B2(min)}as:

For *Q*_{2} to be in saturation *I*_{B2(sat)} must be larger than *I*_{B2(min)}, choose

Using Eq. (9.82) and (9.85), we can calculate (*I*_{2} + *I*_{B2}).

From the circuit shown in Fig. 9.23(a):

*R* is calculated using Eq. (9.86).

*R*_{1} is chosen using Eq. (9.87).

**FIGURE 9.23(b)** The circuit at LTP

(b) At LTP = *V*_{2}, consider the circuit shown in Fig. 9.23(b).

*V*_{BN2} = *V*_{BN1} = LTP = *V*_{2}

Let *I*_{1} be the current in *R*_{2},

Writing the KVL equation of the outer loop consisting of *R*_{C1}, *R*_{1} and *R*_{2}:

*V _{CC}* = (

*I*

_{C1}+

*I*

_{1})

*R*

_{C1}+

*I*

_{1}(

*R*

_{1}+

*R*

_{2})

Using Eq. (9.87):

*V _{CC}* = (

*I*

_{C1}+

*I*

_{1})

*R*

_{C1}+

*I*

_{1}(

*R*−

*R*

_{C1}+

*R*

_{2})

Therefore,

From Eq. (9.90),

*R*_{C1} is calculated using Eq. (9.95),

*R*_{1} = *R* − *R*_{C1}

All the components of the Schmitt trigger are fixed. The procedure to design a Schmitt trigger is presented in the Example 9.8.

##### EXAMPLE

*Example 9.8:* Design a Schmitt trigger shown in Fig. 9.24(a) with UTP of 6 V and LTP of 3 V. Ge transistors with *h*_{FE(min)} = 50 and *I _{C}* = 4 mA are used. The supply voltage is 15 V.

**FIGURE 9.24(a)** The Schmitt trigger circuit

*Solution:*

Till UTP is reached, *Q*_{1} is OFF and *Q*_{2} is ON and in saturation. Just at *V*_{1} (UTP) *Q*_{1} goes ON and *Q*_{2} goes OFF.

Therefore, *V*_{1} = UTP = *V*_{BN2} = 6 V

*I _{E}* =

*I*= 4mA

_{C}Choose *R _{E}* = 1kΩ

If *Q*_{2} is in saturation *V*_{CE(sat)} = 0.1 V, *V _{σ}* = 0.3 V, so

*I*_{C2}*R*_{C2} = *V _{CC}* −

*V*

_{CE(sat)}−

*V*

_{EN2}

Therefore,

Choose *R*_{C2} = 2.2 kΩ

*R*_{1} = 17.31 kΩ − *R*_{C1}

**FIGURE 9.24(b)** The circuit at LTP

At LTP = 3 V, consider the circuit shown in Fig. 9.24(b).

*V*_{BN2} = *V*_{BN1} = 3 V = LTP = *V*_{2}

Let *I*_{1} be the current in *R*_{2},

Writing the KVL equation of the outer loop:

*V _{CC}* = (

*I*

_{C1}+

*I*

_{1})

*R*

_{C1}+

*I*

_{1}(

*R*

_{1}+

*R*

_{2}) = (

*I*

_{C1}+

*I*

_{1})

*R*

_{C1}+

*I*

_{1}(17.31 −

*R*

_{C1}+

*R*

_{2})

*V _{CC}* =

*I*

_{C1}

*R*

_{C1}+

*I*

_{1}(17.31 +

*R*

_{2})

*R*_{C1} = 3kΩ *R*_{1} = 17.31 − *R*_{C1} = 17.31 − 3 = 14.31 kΩ

Choose *R*_{1} = 15 kΩ.

The designed Schmitt trigger circuit is shown in Fig. 9.24(c) with all the component values.

**FIGURE 9.24(c)** The designed Schmitt trigger

##### SOLVED PROBLEMS

*Example 9.9:* Design a fixed-bias bistable multivibrator shown in Fig. 9.25 using *p–n–p* Ge transistors having *h*_{FE(min)} = 50, *V _{CC}* = −10 V,

*V*= 10 V,

_{BB}*V*

_{CE(sat)}= −0.1,

*V*

_{BE(sat)}= −0.3 and

*I*

_{C(sat)}= −5 mA. Assume

*I*= −5 mA.

_{C}**FIGURE 9.25** The fixed-bias bistable multivibrator

*Solution:*

*R _{C}* = 1.98 kΩ ≈ 2.2 kΩ(standard resistance).

If *Q*_{2} is in saturation,

*I*_{B2} = 1.5*I*_{B2(min)} = −0.15 mA *I*_{1} = *I*_{2} + *I*_{B2} = −0.5 mA − 0.15 mA = −0.65 mA

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

*R*

_{1}) −

*R*= 14.92 − 1.98 = 12.94 kΩ

_{C}Choose *R*_{1} ≈ 13 kΩ

Note: Choose the nearest standard values.

*Example 9.10:* For a fixed-bias bistable multivibrator using Si transistors, shown in Fig. 9.26(a), *h _{FE}* = 20,

*V*

_{CE(sat)}= 0.3 V,

*V*= 0.7 V.

_{σ}- Calculate the stable-state currents and voltages.
*I*_{CO}=*I*_{CBO}= 15*µ*Aat 25°C.*I*_{CBO}gets doubled for every 10°C rise in temperature. The OFF transistor remains OFF only till the net voltage at its base is 0 V. Find the maximum value of*I*_{CBO}and the temperature up to which the OFF transistor will remain OFF and the multivibrator operates normally.- If the commutating condenser is 100 pF, find the maximum switching speed of the bistable multivibrator
- The maximum switching frequency is 50 kHz. Find the value of the commutating condenser.

**FIGURE 9.26(a)** The given fixed-bias bistable multivibrator

(a)

(i) If *Q*_{2} is ON and in saturation, to find its base current *I*_{B2}, consider the base circuit shown in Fig. 9.26(b).

From Fig. 9.26(b),

**FIGURE 9.26(b)** The circuit to calculate *I*_{B2}

(ii) To calculate *I*_{C2} consider the collector circuit of *Q*_{2}, shown in Fig. 9.26(c).

From Fig. 9.26(c),

*I*_{C2} = *I*_{3} − *I*_{4} = 7.35 − 0.13 = 7.22 mA

Given *h _{FE}* = 20.

Therefore,

*I*_{B2(sat)} = 1.5 *I*_{B2(min)} = 1.5 × 0.361 = 0.54 mA

**FIGURE 9.26(c)** The circuit to calculate *I*_{C2}

As *I*_{B2} = 0.558 mA, *Q*_{2} is in saturation.

Hence, *Q*_{1} is OFF.

*V*_{C1} = *V _{CC}* −

*I*

_{1}

*R*= 15 − (0.715)2 = 13.57 V

_{C}(b) *I*_{CO} = *I*_{CBO} = 15 *µ*Aat 25°C.

To calculate the positive voltage that *I _{CBO}* develops at

*B*

_{1}, the base of the OFF device, short

*V*

_{CE(sat)}and

*V*sources, shown in Fig. 9.26(d). We see that

_{BB}*I*flows through the parallel combination of

_{CBO}*R*

_{1}and

*R*

_{2}. If

*R*

_{B}is the effective resistance of

*R*

_{1}and

*R*

_{2}in parallel, then,

*V*_{B1} was calculated earlier.

*V*_{B1} = −2.034 V

When *I _{CBO} R_{B}* =

*V*

_{B1}, the net voltage at

*B*

_{1}is zero. Till that instant

*Q*

_{1}is OFF. If

*I*

_{CBO}R_{B}> V_{B1}, the voltage at B

_{1}becomes positive and

*Q*

_{1}will not be in the OFF state. Therefore, for

*Q*

_{1}to be OFF,

*R _{B} I*

_{CBO(max)}=

*V*

_{B1}

Therefore,

**FIGURE 9.26(d)** The circuit to calculate *R _{B}* and

*I*

_{CBO(max)}

where, *T*_{2} is the temperature at which *I _{CBO}* =

*I*

_{CBO(max)}.

As *n* =

Δ*T* = *T*_{2} – 25 *T*_{2} – 25 = 31.5 *T*_{2} = 31.5 + 25 = 56.5°C

(c)

Given *C* = 100 pF.

Therefore,

(d) Given *f*_{(max)} = 50 kHz.

*Example 9.11:* The self-bias transistor bistable multivibrator shown in Fig. 9.27(a) uses a *n–p–n* Ge transistor. Given that *V _{CC}* = 15 V,

*V*

_{CE(sat)}= 0.1 V,

*V*= 0.3 V,

_{σ}*R*= 2kΩ,

_{C}*R*

_{1}= 30 kΩ,

*R*

_{2}= 10 kΩ,

*R*= 500 Ω.

_{E}Find:

- Stable-state currents and voltages and the
*h*needed to keep the ON device in saturation_{FE} - The value of
*C*_{1}needed to ensure a resolution time of 0.02 ms. - The maximum value of
*I*that will still ensure one device OFF and the other ON._{CBO} - The maximum temperature up to which the multivibrator can work normally if
*I*at 25 °C = 10_{CBO}*µ*A.

**FIGURE 9.27(a)** The self-bias bistable multivibrator

*Solution:* (a)

**FIGURE 9.27(b)** The circuit to calculate *V*_{thb} and *R*_{thb} of *Q*_{2}

(i) To calculate *I*_{B2}, consider the base circuit of *Q*_{2}, shown in Fig. 9.27(b).

From Fig. 9.27(b),

(ii) To calculate *I*_{C2}, consider the collector circuit of *Q*_{2}, Fig. 9.27(c).

Now let us draw the base and collector circuits of *Q*_{2}, shown in Fig. 9.27(d).

**FIGURE 9.27(c)** The circuit to calculate *V*_{thc} and *R*_{thc} of *Q*_{2}

**FIGURE 9.27(d)** The circuit to calculate *I*_{B2} and *I*_{C2}

Writing the KVL equations of the input and output loops:

Eqs. (1) and (2) are simplified as:

Solving Eqs. (3) and (4) for *I*_{B2} and *I*_{C2} we get,

*I*_{B2} = 0.212 mA *I*_{C2} = 4.18 mA

Therefore, *h _{FE}* = 4.18/0.212 = 19.7.

The *h _{FE}* that keeps the ON device in saturation is 20.

*V*_{EN2} = (*I*_{B2} + *I*_{C2}) *R _{E}* = (4.18 + 0.212) 0.5 = 2.196 V

*V*_{CN2} = *V*_{EN2} + *V*_{CE(sat)} = 2.196 + 0.1 = 2.296 V

*V*_{BN2} = *V*_{EN2} + *V _{σ}* = 2.196 + 0.3 = 2.496 V

*V*_{BE1} = *V*_{BN1} − *V*_{EN2} = 0.765 − 2.196 = −1.431 V

Hence, *Q*_{1} is OFF.

Therefore, *V*_{CN1} should be *V _{CC}*, but actually it is less than

*V*.

_{CC}We have from Fig. 9.27(b),

*V*_{CN1} = *V _{CC}* −

*I*

_{1}

*R*= 15 − (0.379)(3) = 13.86 V

_{C}(b)

(c) *V*_{BE1} was calculated as −1.431 V. This voltage exists at the base of *Q*_{1} to keep *Q*_{1} OFF. Till such time the voltage at *B*_{1} of *Q*_{1} is 0 V, let us assume that *Q*_{1} is OFF, as shown in Fig. 9.27(e). To calculate *R*_{B} and hence *I*_{CBO}*R*_{B}, short *V _{EN}* (though

*I*

_{E1}= 0, there exists a voltage

*V*at the first emitter) and

_{EN}*V*

_{CE(sat)}sources. From Fig. 9.27(e), it is seen that

*R*is the parallel combination of

_{B}*R*

_{2}and (

*R*

_{1}+

*R*).

_{E}Until *I*_{CB0(max)}*R*_{B} = *V*_{BE1}, *Q*_{1} will be OFF.

Therefore,

**FIGURE 9.27(e)** The circuit to calculate *I*_{CBO}*R*_{B}

d) *I*_{CB0} at 25°C = 10 µ*A*

*Example 9.12:* Consider the Schmitt trigger circuit designed, shown in Fig. 9.28(a). *R*_{e2} is now included to eliminate hysteresis, as shown in Fig. 9.28(a). For this circuit *V*_{1} = 8 V and *V*_{2} = 4 V and *h _{FE}* = 40. Calculate

*R*

_{e2}such that

*V*

_{1}=

*V*

_{2}= 4 V.

*Solution:*

From Eq. (9.79), we have:

143.5*R*_{e2} = 301.81 *R*_{e2} = 2.10 kΩ

**FIGURE 9.28(a)** *R*_{e2} is connected to eliminate hysteresis

##### SUMMARY

- A bistable multivibrator has two stable states. Initially if the multivibrator is in one of the stable states (say
*Q*_{1}is ON and*Q*_{2}is OFF), change of state (*Q*_{1}is OFF and*Q*_{2}is ON) occurs only after the application of an external trigger. - A bistable multivibrator is also known by many names, namely, binary, flip-flop, scale-of-two circuit and Eccles–Jordan circuit.
- A bistable multivibrator can be used for storing and counting of binary information. It can also be used for generating pulsed output.
- The output of a bistable multivibrator, if connected to some other circuit, could cause loading on the bistable multivibrator, which in turn will reduce the output swing.
- To ensure that the output of a bistable multivibrator does not fall below a specified threshold, the collector of the OFF transistor is clamped to a dc voltage using collector catching diodes.
- The time taken for conduction to transfer from one device to the other is called the transition time.
- Once conduction is transferred from one device to the other, an additional time, known as the settling time, is required to elapse before the voltages across the commutating condensers interchange. Only then do we say that the multivibrator has settled down in its new state completely.
- Commutating condensers help in reducing the transition time.
- The resolution time of a bistable multivibrator is the minimum time interval required between successive trigger pulses to be reliably able to drive the multivibrator from one stable state to the other.
- The reciprocal of the resolution time is the maximum switching speed of the bistable multivibrator.
- Unsymmetric triggering is a method of pulse triggering in which one trigger pulse taken from one source is applied at point in the circuit. The next trigger pulse is taken from a different source and is applied at a different point in the circuit to cause a change of state in both the devices. This method of triggering is used to generate a gated output; the duration of gate being dependent on the time interval between these trigger pulses.
- Symmetric triggering is a method of pulse triggering in which successive trigger pulses taken from the same source and applied at the same point in the circuit will cause a change of state in either direction. This method of triggering is used in counters.
- An emitter-coupled bistable multivibrator is called a Schmitt trigger.
- When the loop gain is greater than 1, the Schmitt trigger exhibits hysteresis.
- Hysteresis in a Schmitt trigger can be eliminated by including a suitable resistance in series with either the first emitter (
*R*_{e1}) or the second emitter (*R*_{e2}). - A Schmitt trigger can also be used as a comparator and squaring circuit in addition to being used as a bistable multivibrator.
- The input voltage at which the output voltage of a Schmitt trigger goes high is called the upper trip point (UTP).
- The input voltage at which the output voltage of a Schmitt trigger goes low is called the lower trip point (LTP).

##### MULTIPLE CHOICE QUESTIONS

- Unless an external trigger is applied, the state of a bistable multivibrator:
- Remains unaltered
- Changes automatically
- Goes into the quasi-stable state
- None of the above

- When a trigger is applied to a bistable multivibrator, conduction is transferred from one device to the other. The time taken for conduction to transfer from one device to the other is called:
- Delay time
- Rise time
- Transition time
- Fall time

- The time taken for the voltages across the commutating condensers to interchange is called:
- Transition time
- Rise time
- Recovery time
- Settling time

- Resolution time of a bistable multivibrator is the sum of the:
- Rise time and fall time
- Delay time and rise time
- Transition time and settling time
- Storage time and fall time

- The reciprocal of the resolution time of the bistable multivibrator is called:
- Maximum switching speed of the bistable multi-vibrator
- Minimum switching speed of the bistable multi-vibrator
- Frequency of oscillations of the bistable multivibrator
- Gate width

- If the commutating condensers are large:
- Transition time decreases and the settling time increases
- Transition time increases and the settling time decreases
- Both, transition time and settling time increase
- Both transition time and settling time decrease

- An emitter-coupled bistable multivibrator is also called as:
- Astable multivibrator
- Monostable multivibrator
- Schmitt trigger
- None of the above

- A Schmitt trigger can be used as a:
- Comparator
- Astable multivibrator
- Monostable multivibrator
- None of the above

- If the loop gain is greater than 1, a Schmitt trigger exhibits:
- Oscillations
- Hysteresis
- Instability
- None of the above

- When symmetric pulse triggering is used in a bistable multivibrator, its application is in:
- Astable multivibrators
- Monostable multivibrators
- Counters
- Schmitt trigger

*R*_{e2}connected in series with the second emitter in a Schmitt trigger influences:*V*_{2}but not*V*_{1}*V*_{1}but not*V*_{2}- Both
*V*_{1}and*V*_{2} - None of the above

*R*_{e1}connected in series with the first emitter in a Schmitt trigger influences:*V*_{2}but not*V*_{1}*V*_{1}but not*V*_{2}- Both
*V*_{1}and*V*_{2} - None of the above

##### SHORT ANSWER QUESTIONS

- What is a bistable multivibrator? What are the other names by which it is called? Explain, in short, how a bistable multivibrator can be used as a memory element?
- What is meant by loading in a bistable multivibrator? How can you make the output swing constant?
- What is the purpose served by a collector catching diodes?
- What is the use of commutating condensers in a bistable multivibrator?
- Define transition time? Suggest how to minimize it.
- What is meant by the resolution time of a bistable multivibrator? Suggest simple methods to improve it.
- What do you understand by unsymmetric triggering? What is its application?
- What do you understand by symmetric triggering? What is its application?
- What is a Schmitt trigger? Mention some of its applications.
- Suggest simple methods to eliminate hysteresis in a Schmitt trigger.
- A Schmitt trigger can be used as a regenerative comparator – justify.

##### LONG ANSWER QUESTIONS

- With the help of a neat circuit diagram explain the working of a fixed-bias bistable multivibrator. Derive the expression for the resolution time and maximum switching speed of a bistable multivibrator.
- With the help of a neat circuit diagram explain the working of a self-bias bistable multivibrator. List the advantages of this circuit over a fixed-bias bistable multivibrator.
- With the help of a suitable circuit diagram, explain the methods of symmetric and unsymmetric triggering of a bistable multivibrator.
- Draw the circuit of a Schmitt trigger and explain its operation. Derive the expressions for (i) UTP and (ii) LTP.
- Explain with the help of waveforms how a Schmitt trigger can be used as a:
- Bistable multivibrator;
- Squaring circuit; and
- Amplitude comparator.

##### UNSOLVED PROBLEMS

- Design a fixed-bias bistable multivibrator using Ge transistors having
*h*_{FE((min)}= 50,*V*= 10 V and_{CC}*V*= 10 V,_{BB}*V*_{CE(sat)}= 0.1 V,*V*_{BE(sat)}= 0.3 V,*I*_{C(sat)}= 5 mA and assume*I*_{B(sat)}= 1.5*I*_{B(min)}. - For a fixed-bias bistable multivibrator shown in Fig. 9p.1 using
*n–p–n*Ge transistor*V*= 10 V,_{CC}*R*= 1 kΩ,_{C}*R*_{1}= 10 kΩ,*R*_{2}= 20 kΩ,*h*_{FE(min)}= 40,*V*= 10 V._{BB}

Calculate:- Stable-state currents and voltages assuming
*Q*_{1}is OFF and*Q*_{2}is ON and in saturation. Verify whether*Q*_{1}is OFF and*Q*_{2}is ON or not. - The maximum load current.

FIGURE 9p.1 The fixed-bias bistable multivibrator

- Stable-state currents and voltages assuming
- Design a self-bias bistable multivibrator shown in Fig. 9p.2 with a supply voltage of −12 V. A
*p*−*n*−*p*silicon transistor with*h*_{FE(min)}= 50,*V*_{CE(sat)}= −0.3 V,*V*_{BE(sat)}= −0.7 V and*I*_{C2}= −4 mA is used.FIGURE 9p.2 The given self-bias bistable multivibrator

- A self-bias bistable multivibrator uses Si transistors having
*h*_{FE(min)}= 50.*V*= 18 V,_{CC}*R*_{1}=*R*_{2},*I*_{C(sat)}= 5 mA. Fix the component values*R*_{E}, R_{C}, R_{1}and*R*_{2}. - For the Schmitt trigger in Fig 9p.4 using
*n–p–n*silicon transistors having*h*_{FE(min)}= 40, the following are the circuit parameters:*V*= 15 V,_{CC}*R*= 0,_{S}*R*_{C1}= 4kΩ,*R*_{C2}= 1kΩ,*R*_{1}= 3kΩ,*R*_{2}= 10 kΩ and*R*= 6kΩ. Calculate_{E}*V*_{1}and*V*_{2}. - The self-bias transistor bistable multivibrator shown in Fig. 9p.3 uses
*n–p–n*Si transistors. Given that*V*= 15 V,_{CC}*V*_{CE(sat)}= 0.2 V,*V*= 0.7 V,_{σ}*R*= 3kΩ,_{C}*R*_{1}= 20 kΩ,*R*_{2}= 10 kΩ,*R*= 500 Ω._{E}Find:

- Stable-state currents and voltages and the
*h*needed to keep the ON device in saturation._{FE} *f*_{(max)}, if*C*_{1}= 100 pF.- The maximum value of
*I*that will ensure one device is OFF and the other is ON._{CBO} - The maximum temperature up to which the multivibrator can work normally if
*I*at 25°_{CBO}*C*= 20 µA.

FIGURE 9p.3 The given self-bias bistable multivibrator

- Stable-state currents and voltages and the
- Design a Schmitt trigger shown in Fig. 9p.4 with UTP of 8 V and LPT of 4 V. Si transistors with
*h*= 40 and_{FE}*I*= 5 mA are used. The supply voltage is 18 V. The ON transistor is in the active region for which_{C}*V*= 0.6 V,_{BE}*V*= 2.0 V. (b) Calculate_{CE}*R*_{e1}for eliminating hysteresis.FIGURE 9p.4 The Schmitt trigger circuit

- Design a Schmitt trigger in Fig. 9p.5 with UTP of 8 V and LTP of 4 V. Si transistors with
*h*= 40 and_{FE}*I*= 4 mA are used. The supply voltage is 12 V. The ON transistor is in saturation for which_{C}*V*= 0.7 V,_{BE}*V*_{CE(sat)}= 0.2 V.(ii) Calculate

*R*_{e1}for eliminating hysteresis.(iii) Find

*R*_{e2}to eliminate hysteresis.FIGURE 9p.5 The given Schmitt trigger circuit