# 9.4 Schmitt Triggers – Pulse and Digital Circuits

##### 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 R1 and R2. This is simply an attenuator circuit. Normally R1 and R2 are reasonably large resistors so as to avoid loading the collector of Q1. Further, the emitter resistance RE 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 vi is small, Q1 is OFF. The voltage at this collector is approximately VCC. This voltage is coupled to the second base through R1 and R2. As a result, Q2 can conduct. If Q2 conducts, it can operate in the active region or it may be driven into saturation.

Let it be assumed that Q2 is in the active region. The base current IB2 and collector current IC2 flows through RE. Therefore, a voltage VEN = VEN2 is developed in RE. As VBE1 = VBN1VEN2 and if VBE1 reverse-biases the emitter diode of Q1, then as assumed Q1 is OFF.

If VBN1 (vi) is increased, at a value (VEN2 + Vγ1), Q1 begins to conduct. As a result, the voltage at the second base decreases, hence the base current of Q2 decreases, its collector current also reduces and consequently the voltage at the second collector rises. If the input is increased further, Q1 goes into the ON state and Q2 into the OFF state. If the loop gain is less than unity (this condition can be satisfied by reducing the collector load of Q1), there exists a region of linearity in the transfer characteristic. In this region an incremental change at the input Δvi will cause a proportional change in the output, Δvo as shown in Fig. 9.12(b). If the loop gain is made equal to 1 by adjusting RC1 and RC2, 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 V1, the output suddenly jumps from a smaller value to VCC as shown in Fig. 9.13(a). Even if the input is increased further, the output remains at VCC.

If on the other hand, the input is decreased, then at a voltage V2 the output falls from VCC 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

V1 is called the upper trip point (UTP) and V2 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 V1 and V2 is called the hysteresis voltage, VH. Thus, VH = V1V2.

It is seen from the above discussion that a Schmitt trigger exhibits hysterisis, i.e., when the input vi is increased to reach a voltage V1 it is required to first pass through a point, V2 at which the reverse transition takes place. Similarly, when the input now is reduced to reach V2 it has to pass through the point V1. 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 VCC

FIGURE 9.13(b) When the input is decreased the output falls from VCC

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

#### 9.4.1 Calculation of the Upper Trip Point (V1)

FIGURE 9.14 The Schmitt trigger circuit

In the Schmitt trigger circuit, shown in Fig. 9.14, when the input is increased, till V1 is reached Q1 is OFF and Q2 is ON. As a result, IB2 and IC2 flow through RE developing a voltage VEN in RE. Now, if the input is such that its value is (VEN + Vγ1) = V1(UTP), Q1 switches into the ON state and Q2 switches into the OFF state.

If the circuit show in Fig. 9.14 is Thévenised at the base of Q2, 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 V1

The total current in RE is IB2 (1 + hFE) and the current in R is IB2. As far as IB2 is concerned, RE is seen to have increased by a factor (1 + hFE). The net voltage in the base loop is, (VVBE2) and is equal to the sum of the voltage drops across R and (1 + hFE) RE, as shown in Fig. 9.15(b).

FIGURE 9.15(b) The circuit to calculate VEN2

In Eq. (9.58), as:

R << RE(1 + hFE)     R + RE(1 + hFE)RE(1 + hFE)

Then Eq. (9.58) reduces to:

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

The calculation of V1 is made based on the assumption that Q2 is in the active region. Now, to verify whether Q2 is in the active region, VCB2 is calculated and checked if this reverse-biases the collector diode by a reasonable voltage or not. If it does, the device Q2 is indeed in the active region.

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

VCB2 = VCE2VBE2   and   VCE2 = VCCIC2RC2VEN2

To calculate VCB2 using Eq. (9.62), we have to find IC2.

where,

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

where is given by Eq. (9.64). From Eq. (9.63) IC2 is given as:

These calculations were made based on the assumption that Q2 is in the active region. Having made the calculations, we once again verify whether Q2 is really in the active region or not, to justify the validity of the calculations made. VCB2 is calculated using Eq. (9.62). If the base−collector diode is reverse-biased, then Q2 is in the active region as assumed.

#### 9.4.2 Calculation of the Lower Trip Point (V2)

At the voltage V1 (UTP), Q1 is ON and Q2 is OFF. Now if the input is reduced, till voltage V2 is reached, Q1 is ON and Q2 is still OFF. However, when the voltage at the input is V2 then the voltage at the second base is VEN1 + Vγ 2. Q2 again switches into the ON state and Q1 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 Rt is,

The first collector and the second base are connected through R1 and R2, as shown in Fig. 9.16.

where

The circuit that enables us to calculate V2 is shown in Fig. 9.17.

Writing the KVL equation for the base loop of Q2:

αVCN1 = Vγ2 + (IB1 + IC1) RE     VCN1 = VtIC1Rt

FIGURE 9.16 The coupling network from the first collector to the second base

FIGURE 9.17 The circuit to calculate V2

Let:

However, we have:

Therefore, from Eq. (9.73)

Using Eqs. (9.72), (9.74) and (9.75), V2 is calculated. To understand the method of calculation for V1 and V2, let us consider Example 9.5.

##### EXAMPLE

Example 9.5: For the Schmitt trigger circuit shown in Fig. 9.18(a), calculate V1 and V2.

FIGURE 9.18(a) The Schmitt trigger with components mentioned

Solution:

a) Calculation of V1:

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 V1

From Eq. (9.58):

If Q2 is in the active region, typically, for silicon VBE2 = 0.6 V and let hFE = 50,

RE(1 + hFE) = 3(1 + 50) = 153 kΩ

Therefore,

Therefore,

V1 = VEN2 + VBE2 = 4.76 + 0.6 = 5.36 V.

The calculation of V1 is made based on the assumption that Q2 is in the active region. To find out whether Q2 is in the active region or not, we calculate VCB2.

VCB2 = VCCIC2RC2VEN2VBE2.

From Eq. (9.72)

Hence

VCB2 = 10 − (1.56 × 1) − 4.76 − 0.6 = 10 − 6.92 = 3.08 V

As the collector of Q2 is positive with respect to the base by 3.08 V the collector diode is reverse-biased. Hence, Q2 is in the active region, as assumed.

(b) Calculation of V2:

The circuit that enables us to calculate V2 is shown in Fig. 9.18(c). From the circuit values:

V2 = 0.6 V + (1.05 mA)(3.06 kΩ)= 0.6 V + 3.22 V = 3.82 V

Hence, for the given Schmitt trigger:

V1 = 5.36 V   V2 = 3.82 V   VH = V1V2 = 5.36 − 3.82 = 1.54 V

FIGURE 9.18(c) The circuit to calculate V2

#### 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:

1. V1 and V2 are made to coincide with the proper choice of R1 and R2. However, by this method, though, it is not possible to make V1 identical to V2, they can be brought close to each other in practice.
2. Another method to eliminate hysteresis is by introducing a resistance Re2 in series with the emitter terminal of Q2, as shown in Fig. 9.19(a).

Re2 will change V1, but has no effect on V2. Therefore, by including Re2 in series with the emitter of Q2, it is possible to reduce V1 to the level of V2,(V1 is made equal to V2), thereby reducing hysteresis. The method to calculate Re2, 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 Re2

To eliminate hysteresis, V1 should be made equal to V2, which means that in Eq. (9.78) V1 is replaced by V2

Re2 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 V1 = 5.36 V, V2 = 3.82 V, V = 5.45 V, R = 2.73 kΩ, RE = 3kΩ, hFE = 50, VBE2 = 0.6 V. Find the value of Re2 that ensures V1 = V2 = 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 Re1in series with the emitter terminal of Q1, as shown in Fig. 9.19(c).

Re1 is not going to change V1 but will only influence V2. To eliminate hysterisis, Re1 is chosen such that V2 is increased to the level of V1. To achieve this, we must ensure that V2 plus the voltage drop across Re1 is V1.

From Eq. (1):

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

To understand the procedure to calculate Re1, let us consider an example.

##### EXAMPLE

Example 9.7: From the problem in Example 9.5, we have V1 = 5.36 V, V2 = 3.82 V, IC1 = 1.05 mA, hFE = 50. Find the value of Re1.

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 Q1 is biased to have a voltage V at its base. Initially, let the output be at 0 level (VCCIC2RC2). To change this to 1(VCC) apply a positive pulse at the base of Q1, whose magnitude is more positive than (V1V) as shown in Fig. 9.20(b).

To once again change the output to a 0 level, apply a pulse at the base of Q1, which is negative with respect to V and whose magnitude is more negative than (V2V) 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 vi < VR, D is OFF and vo = vi. When viVR, D is ON and vo = VR.

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 = t1 and then the slope of the output abruptly changes. This is called the break point, and t1 is the time instant at which vi has reached VR.

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 V1(= V2) is reached at the input. The moment the input is V1, the output abruptly jumps to VCC. 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 V1 and also less than V2. 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 (V1) and LTP (V2) values, hFE, desired IC and VCC are to be specified.

(a) Till UTP is reached Q1 is OFF and Q2 is ON and in saturation. Just at V1 (UTP), Q1 goes ON and Q2 goes OFF.

Therefore, V1 = UTP = VBN2 and IE = IC is specified. Therefore:

RE is chosen using Eq. (9.80). If Q2 is in saturation, VCE = VCE(sat). From the circuit shown in Fig. 9.23(a), we have IC2RC2 = VCCVCE(sat)VEN2

Therefore,

RC2 is calculated using Eq. (9.81).

Assume that

Using Eq. (9.82):

Using the value of hFE specified, we can now calculate IB2(min) as:

For Q2 to be in saturation IB2(sat) must be larger than IB2(min), choose

Using Eq. (9.82) and (9.85), we can calculate (I2 + IB2).

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

R is calculated using Eq. (9.86).

R1 is chosen using Eq. (9.87).

FIGURE 9.23(b) The circuit at LTP

(b) At LTP = V2, consider the circuit shown in Fig. 9.23(b).

VBN2 = VBN1 = LTP = V2

Let I1 be the current in R2,

Writing the KVL equation of the outer loop consisting of RC1, R1 and R2:

VCC = (IC1 + I1)RC1 + I1(R1 + R2)

Using Eq. (9.87):

VCC = (IC1 + I1)RC1 + I1(RRC1 + R2)

Therefore,

From Eq. (9.90),

RC1 is calculated using Eq. (9.95),

R1 = RRC1

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 hFE(min) = 50 and IC = 4 mA are used. The supply voltage is 15 V.

FIGURE 9.24(a) The Schmitt trigger circuit

Solution:

Till UTP is reached, Q1 is OFF and Q2 is ON and in saturation. Just at V1 (UTP) Q1 goes ON and Q2 goes OFF.

Therefore, V1 = UTP = VBN2 = 6 V

IE = IC = 4mA

Choose RE = 1kΩ

If Q2 is in saturation VCE(sat) = 0.1 V, Vσ = 0.3 V, so

IC2RC2 = VCCVCE(sat)VEN2

Therefore,

Choose RC2 = 2.2 kΩ

R1 = 17.31 kΩ − RC1

FIGURE 9.24(b) The circuit at LTP

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

VBN2 = VBN1 = 3 V = LTP = V2

Let I1 be the current in R2,

Writing the KVL equation of the outer loop:

VCC = (IC1 + I1)RC1 + I1(R1 + R2) = (IC1 + I1)RC1 + I1(17.31 − RC1 + R2)

VCC = IC1RC1 + I1(17.31 + R2)

RC1 = 3kΩ   R1 = 17.31 − RC1 = 17.31 − 3 = 14.31 kΩ

Choose R1 = 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 hFE(min) = 50, VCC = −10 V, VBB = 10 V, VCE(sat) = −0.1, VBE(sat) = −0.3 and IC(sat) = −5 mA. Assume IC = −5 mA.

FIGURE 9.25 The fixed-bias bistable multivibrator

Solution:

RC = 1.98 kΩ ≈ 2.2 kΩ(standard resistance).

If Q2 is in saturation,

IB2 = 1.5IB2(min) = −0.15 mA     I1 = I2 + IB2 = −0.5 mA − 0.15 mA = −0.65 mA

R1 = (RC + R1) − RC = 14.92 − 1.98 = 12.94 kΩ

Choose R1 ≈ 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), hFE = 20, VCE(sat) = 0.3 V, Vσ = 0.7 V.

1. Calculate the stable-state currents and voltages.
2. ICO = ICBO = 15 µAat 25°C. ICBO 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 ICBO and the temperature up to which the OFF transistor will remain OFF and the multivibrator operates normally.
3. If the commutating condenser is 100 pF, find the maximum switching speed of the bistable multivibrator
4. The maximum switching frequency is 50 kHz. Find the value of the commutating condenser.

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

Solution:

(a)

(i) If Q2 is ON and in saturation, to find its base current IB2, consider the base circuit shown in Fig. 9.26(b).

From Fig. 9.26(b),

FIGURE 9.26(b) The circuit to calculate IB2

(ii) To calculate IC2 consider the collector circuit of Q2, shown in Fig. 9.26(c).

From Fig. 9.26(c),

IC2 = I3I4 = 7.35 − 0.13 = 7.22 mA

Given hFE = 20.

Therefore,

IB2(sat) = 1.5 IB2(min) = 1.5 × 0.361 = 0.54 mA

FIGURE 9.26(c) The circuit to calculate IC2

As IB2 = 0.558 mA, Q2 is in saturation.

Hence, Q1 is OFF.

VC1 = VCCI1RC = 15 − (0.715)2 = 13.57 V

(b) ICO = ICBO = 15 µAat 25°C.

To calculate the positive voltage that ICBO develops at B1, the base of the OFF device, short VCE(sat) and VBB sources, shown in Fig. 9.26(d). We see that ICBO flows through the parallel combination of R1 and R2. If RB is the effective resistance of R1 and R2 in parallel, then,

VB1 was calculated earlier.

VB1 = −2.034 V

When ICBO RB = VB1, the net voltage at B1 is zero. Till that instant Q1 is OFF. If ICBO RB > VB1, the voltage at B1 becomes positive and Q1 will not be in the OFF state. Therefore, for Q1 to be OFF,

RB ICBO(max) = VB1

Therefore,

FIGURE 9.26(d) The circuit to calculate RB and ICBO(max)

where, T2 is the temperature at which ICBO = ICBO(max).

As n =

ΔT = T2 – 25     T2 – 25 = 31.5     T2 = 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 VCC = 15 V, VCE(sat) = 0.1 V, Vσ = 0.3 V, RC = 2kΩ, R1 = 30 kΩ, R2 = 10 kΩ, RE = 500 Ω.

Find:

1. Stable-state currents and voltages and the hFE needed to keep the ON device in saturation
2. The value of C1 needed to ensure a resolution time of 0.02 ms.
3. The maximum value of ICBO that will still ensure one device OFF and the other ON.
4. The maximum temperature up to which the multivibrator can work normally if ICBO at 25 °C = 10 µA.

FIGURE 9.27(a) The self-bias bistable multivibrator

Solution: (a)

FIGURE 9.27(b) The circuit to calculate Vthb and Rthb of Q2

(i) To calculate IB2, consider the base circuit of Q2, shown in Fig. 9.27(b).

From Fig. 9.27(b),

(ii) To calculate IC2, consider the collector circuit of Q2, Fig. 9.27(c).

Now let us draw the base and collector circuits of Q2, shown in Fig. 9.27(d).

FIGURE 9.27(c) The circuit to calculate Vthc and Rthc of Q2

FIGURE 9.27(d) The circuit to calculate IB2 and IC2

Writing the KVL equations of the input and output loops:

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

Solving Eqs. (3) and (4) for IB2 and IC2 we get,

IB2 = 0.212 mA     IC2 = 4.18 mA

Therefore, hFE = 4.18/0.212 = 19.7.

The hFE that keeps the ON device in saturation is 20.

VEN2 = (IB2 + IC2) RE = (4.18 + 0.212) 0.5 = 2.196 V

VCN2 = VEN2 + VCE(sat) = 2.196 + 0.1 = 2.296 V

VBN2 = VEN2 + Vσ = 2.196 + 0.3 = 2.496 V

VBE1 = VBN1VEN2 = 0.765 − 2.196 = −1.431 V

Hence, Q1 is OFF.

Therefore, VCN1 should be VCC, but actually it is less than VCC.

We have from Fig. 9.27(b),

VCN1 = VCCI1RC = 15 − (0.379)(3) = 13.86 V

(b)

(c) VBE1 was calculated as −1.431 V. This voltage exists at the base of Q1 to keep Q1 OFF. Till such time the voltage at B1 of Q1 is 0 V, let us assume that Q1 is OFF, as shown in Fig. 9.27(e). To calculate RB and hence ICBORB, short VEN (though IE1 = 0, there exists a voltage VEN at the first emitter) and VCE(sat) sources. From Fig. 9.27(e), it is seen that RB is the parallel combination of R2 and (R1 + RE).

Until ICB0(max)RB = VBE1, Q1 will be OFF.

Therefore,

FIGURE 9.27(e) The circuit to calculate ICBORB

d) ICB0 at 25°C = 10 µA

Example 9.12: Consider the Schmitt trigger circuit designed, shown in Fig. 9.28(a). Re2 is now included to eliminate hysteresis, as shown in Fig. 9.28(a). For this circuit V1 = 8 V and V2 = 4 V and hFE = 40. Calculate Re2 such that V1 = V2 = 4 V.

Solution:

From Eq. (9.79), we have:

143.5Re2 = 301.81     Re2 = 2.10 kΩ

FIGURE 9.28(a) Re2 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 Q1 is ON and Q2 is OFF), change of state (Q1 is OFF and Q2 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 (Re1) or the second emitter (Re2).
• 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
1. Unless an external trigger is applied, the state of a bistable multivibrator:
1. Remains unaltered
2. Changes automatically
3. Goes into the quasi-stable state
4. None of the above
2. 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:
1. Delay time
2. Rise time
3. Transition time
4. Fall time
3. The time taken for the voltages across the commutating condensers to interchange is called:
1. Transition time
2. Rise time
3. Recovery time
4. Settling time
4. Resolution time of a bistable multivibrator is the sum of the:
1. Rise time and fall time
2. Delay time and rise time
3. Transition time and settling time
4. Storage time and fall time
5. The reciprocal of the resolution time of the bistable multivibrator is called:
1. Maximum switching speed of the bistable multi-vibrator
2. Minimum switching speed of the bistable multi-vibrator
3. Frequency of oscillations of the bistable multivibrator
4. Gate width
6. If the commutating condensers are large:
1. Transition time decreases and the settling time increases
2. Transition time increases and the settling time decreases
3. Both, transition time and settling time increase
4. Both transition time and settling time decrease
7. An emitter-coupled bistable multivibrator is also called as:
1. Astable multivibrator
2. Monostable multivibrator
3. Schmitt trigger
4. None of the above
8. A Schmitt trigger can be used as a:
1. Comparator
2. Astable multivibrator
3. Monostable multivibrator
4. None of the above
9. If the loop gain is greater than 1, a Schmitt trigger exhibits:
1. Oscillations
2. Hysteresis
3. Instability
4. None of the above
10. When symmetric pulse triggering is used in a bistable multivibrator, its application is in:
1. Astable multivibrators
2. Monostable multivibrators
3. Counters
4. Schmitt trigger
11. Re2 connected in series with the second emitter in a Schmitt trigger influences:
1. V2 but not V1
2. V1 but not V2
3. Both V1 and V2
4. None of the above
12. Re1 connected in series with the first emitter in a Schmitt trigger influences:
1. V2 but not V1
2. V1 but not V2
3. Both V1 and V2
4. None of the above
1. 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?
2. What is meant by loading in a bistable multivibrator? How can you make the output swing constant?
3. What is the purpose served by a collector catching diodes?
4. What is the use of commutating condensers in a bistable multivibrator?
5. Define transition time? Suggest how to minimize it.
6. What is meant by the resolution time of a bistable multivibrator? Suggest simple methods to improve it.
7. What do you understand by unsymmetric triggering? What is its application?
8. What do you understand by symmetric triggering? What is its application?
9. What is a Schmitt trigger? Mention some of its applications.
10. Suggest simple methods to eliminate hysteresis in a Schmitt trigger.
11. A Schmitt trigger can be used as a regenerative comparator – justify.
1. 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.
2. 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.
3. With the help of a suitable circuit diagram, explain the methods of symmetric and unsymmetric triggering of a bistable multivibrator.
4. Draw the circuit of a Schmitt trigger and explain its operation. Derive the expressions for (i) UTP and (ii) LTP.
5. Explain with the help of waveforms how a Schmitt trigger can be used as a:
1. Bistable multivibrator;
2. Squaring circuit; and
3. Amplitude comparator.
##### UNSOLVED PROBLEMS
1. Design a fixed-bias bistable multivibrator using Ge transistors having hFE((min) = 50, VCC = 10 V and VBB = 10 V, VCE(sat) = 0.1 V, VBE(sat) = 0.3 V, IC(sat) = 5 mA and assume IB(sat) = 1.5IB(min).
2. For a fixed-bias bistable multivibrator shown in Fig. 9p.1 using n–p–n Ge transistor VCC = 10 V, RC = 1 kΩ, R1 = 10 kΩ, R2 = 20 kΩ, hFE(min) = 40, VBB = 10 V.
Calculate:
1. Stable-state currents and voltages assuming Q1 is OFF and Q2 is ON and in saturation. Verify whether Q1 is OFF and Q2 is ON or not.

FIGURE 9p.1 The fixed-bias bistable multivibrator

3. Design a self-bias bistable multivibrator shown in Fig. 9p.2 with a supply voltage of −12 V. A pnp silicon transistor with hFE(min) = 50, VCE(sat) = −0.3 V, VBE(sat) = −0.7 V and IC2 = −4 mA is used.

FIGURE 9p.2 The given self-bias bistable multivibrator

4. A self-bias bistable multivibrator uses Si transistors having hFE(min) = 50. VCC = 18 V, R1 = R2, IC(sat) = 5 mA. Fix the component values RE, RC, R1 and R2.
5. For the Schmitt trigger in Fig 9p.4 using n–p–n silicon transistors having hFE(min) = 40, the following are the circuit parameters: VCC = 15 V, RS = 0, RC1 = 4kΩ, RC2 = 1kΩ, R1 = 3kΩ, R2 = 10 kΩ and RE = 6kΩ. Calculate V1 and V2.
6. The self-bias transistor bistable multivibrator shown in Fig. 9p.3 uses n–p–n Si transistors. Given that VCC = 15 V, VCE(sat) = 0.2 V, Vσ = 0.7 V, RC = 3kΩ, R1 = 20 kΩ, R2 = 10 kΩ, RE = 500 Ω.

Find:

1. Stable-state currents and voltages and the hFE needed to keep the ON device in saturation.
2. f(max), if C1 = 100 pF.
3. The maximum value of ICBO that will ensure one device is OFF and the other is ON.
4. The maximum temperature up to which the multivibrator can work normally if ICBO at 25°C = 20 µA.

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

7. Design a Schmitt trigger shown in Fig. 9p.4 with UTP of 8 V and LPT of 4 V. Si transistors with hFE = 40 and IC = 5 mA are used. The supply voltage is 18 V. The ON transistor is in the active region for which VBE = 0.6 V, VCE = 2.0 V. (b) Calculate Re1 for eliminating hysteresis.

FIGURE 9p.4 The Schmitt trigger circuit

8. Design a Schmitt trigger in Fig. 9p.5 with UTP of 8 V and LTP of 4 V. Si transistors with hFE = 40 and IC = 4 mA are used. The supply voltage is 12 V. The ON transistor is in saturation for which VBE = 0.7 V, VCE(sat) = 0.2 V.

(ii) Calculate Re1 for eliminating hysteresis.

(iii) Find Re2 to eliminate hysteresis.

FIGURE 9p.5 The given Schmitt trigger circuit