# 13. Electronic Circuits – Basic Electrical and Electronics Engineering

## Electronic Circuits

TOPICS DISCUSSED
•       Half-wave and full-wave rectifier circuits
•       Analysis of rectifiers
•       Filters
•       Clipping and clamping circuits
•       Amplifiers
•       Oscillators
•       Integrator and Differentiator
•       multivibrator
•       schmitt Trigger

#### 13.1.1 introduction

A rectifier is a device that converts ac supply into dc using diodes. Rectification can be done by halfwave or full-wave rectification circuits. All electronic circuits need a dc voltage for their operation. The supply voltage available is from the ac mains which is 230 V, 50 Hz supply. A rectifier will first step down the ac supply voltage to the required level by using a step-down transformer. A single diode can be used to rectify the ac voltage into half-wave rectified dc voltage. since the rectified voltage is a unidirectionally changing dc, filter circuits are used to get steady dc output.

In this section we will discuss half-wave and full-wave rectifier circuits.

#### 13.1.2 Half-wave Rectifier

A half-wave rectifier circuit consisting of a transformer, a diode, and a load represented by a resistor has been shown in Fig. 13.1. The input and output wave shapes have been shown. The diode is forward biased during the positive half cycle of the applied voltage and reverse biased during the negative half cycle.

Figure 13.1 Half-wave rectifier circuit with input and output voltage waveforms

During the positive half cycle, i.e., from time 0 to π, voltage is positive, and hence terminal A of the transformer, T is positive. Diode is forward biased, and hence current will flow through the diode and through the load resistance, RL. If a step down transformer is used, the magnitude of output voltage will be reduced. If however, step down of voltage is not required, a transformer having an equal turn ratio, i.e., N2/N1 will be used. The function of the transformer will be to electrically isolate the dc output circuit from the input circuit.

During the negative half cycle of the input voltage wave, the terminal A of the transformer will be negative and terminal B will be positive. The diode will be reverse biased and virtually no current will flow through the diode and the load.

This sequence of allowing the positive half cycle of current through the diode and blocking the negative half cycle will continue for each and every cycle of power supply. As a result, a half-wave rectified current will flow through the output circuit. The voltage drop across the load resistance, RL will be the output voltage, Vo which will be a half-wave rectified voltage. When the diode is reverse biased, the maximum or peak voltage of the negative half cycle of the input will be appearing across the diode terminals. This is the peak of the reverse voltage or peak-inverse voltage (PIV) which is getting applied to the diode in every alternate half cycles.

It is seen that a diode works as a closed switch during the positive half cycle of the input voltage and works as an open switch during the negative half cycle. The output voltage appears across the load during the positive half cycle only. The load voltage and load current although positive all the time (i.e., unidirectional) are fluctuating dc as its magnitudes changing. Our aim will be to obtain steady dc at the output.

The average value of this fluctuating dc as also its rms values can be calculated as and , respectively.

The performance parameters of a half-wave rectifier output are calculated in terms of the output dc current (i.e., the average value of the rectified wave), the rms value of the output current, the output voltage, ripple factor, peak inverse voltage, etc.

#### 13.1.3 Analysis of Half-wave Rectifier

The equation for a sinusoidal voltage is written as v = Vm sin θ and for a sinusoidal current,

i = Im sinθ,

where i is the instantaneous value, Im is the maximum value, and θ = ωt.

Figure 13.2 Half-wave rectified current

The average value is calculated by integrating the current for a period θ = 0 to θ = π and averaging it for the entire cycle, i.e., θ = 0 to q = 2π.

Idc is the average value of the rectified current.

The RMS value, I is calculated by first squaring the current, then taking its mean, and then taking its root as

or

or

or

Ripple Factor: The output of a half-wave rectifier is a pulsating dc. If we analyse, we will see that it has a steady dc component and an ac component. The ac component is called the ripple. Ripple factor is defined as the ratio of the RMS value of ac component to the value of dc component. The ripple factor indicates the level of fluctuation of the output voltage from its steady value. Ripple is an undesired effect and should be minimized. The ripple factor, r for a half-wave rectifier is calculated as

Again,I2rms=I2dc+I2rms+I22+I2244+...=I2dc+I2ac

where I1, I2, I4, etc., are the fundamental and harmonics of the ac component

or

or

where Iac is the RMS value of the ac component of the output current.

Idc is the dc component of the output current and Irms is the RMS value of the output current.

Substituting

Considering and a half-wave rectifier,

ripple factor,

=1.21

Output voltage, Vdc across the load is

Rectifier efficiency: It is calculated as the ratio of output power to input power.

AC input power, Pac = Power dissipated in diode junction + Power dissipated in the load

The forward resistance, Rf of the diode is very small, and hence I2rmsRf can be neglected in comparison with I2rmsRL.

Therefore, rectifier efficiency

This value of efficiency is considered low and ripple factor is considered very high. A filter circuit has to be used to minimize the ripples.

Peak inverse voltage (PIV): As mentioned earlier, PIV is the maximum value of reverse voltage that appears across the diode when it gets reverse biased. Here

PIV = Vm

Rectifier diodes are specified for their average forward current-carrying capacity and their reverse voltage capacity, i.e., their PIV capacity. For example, low-power rectifier diode series, IN 4000 to IN 4007 are rated for forward current of 1000 mA and maximum reverse voltage of value varying from 50 V to 1000 V.

Voltage Regulation of the rectifier is calculated using the relation,

The difference between no-load voltage and full-load voltage is the voltage drop in the transformer winding and across the diode. The value of voltage regulation, which is generally expressed in percentage should be low.

Example 13.1 A half-wave diode rectifier has a forward voltage drop, i.e., voltage drop across the diode when conducting is 0.7 V. The load resistance is 600 Ω. The rms value of the ac input is 28.87 V. Calculate Idc, Irms, PIV, and form factor.

Solution:

Figure 13.3

Vi(RMS)=28.27V

= 1.414 × 28.27

i.e., Vm = 40V

PIV = Vm = 40V

Example 13.2 A half-wave rectifier produces a maximum load current (peak value) of 50 mA through a 1200 Ω resistor. Calculate the PIV of the diode. The diode is of silicon material.

Solution:

Assuming a voltage drop of 0.7 V across the silicon diode, the peak value of current, Im is

Im is given as 40 mA.

Therefore,

or                                      Vm – 0.7 = 1200×40×10-3 = 48v

or                                           Vm = 48+0.7 = 48v

PIV = Vm = 48.7 V

Example 13.3 A half-wave rectifier circuit has been made using a step-down transformer of turn ratio 10:1. The input voltage is v = 325 sin ωt the diode forward resistance is 25 Ω. A load resistance of 1.2 kΩ has been connected in the circuit. Assuming a secondary winding resistance of the transformer as 1Ω, calculate the following: (a) Rms value of load current (b) rectification efficiency, and (c) ripple factor.

Solution:

Input voltage, v = 325 sin ωt

Input, Vm = 325

Transformer has a turn ratio of 10:1

where R2 is the secondary winding resistance, RF is the forward resistance of the diode and RL is the load resistance.

Figure 13.4 Half-wave rectifier

=26.5 mA

Since it is a half-wave rectifier circuit,

output dc power=I2dcRL=(8.44×10-3)2×1200

= 85.48 mW

AC input power = (Irms)2 [R2 + RF + RL]

= (13.25 10-3)2×1226

= 0.215 W = 215 mW

#### 13.1.4 Full-wave Rectifier

Full-wave rectifiers can be made using two diodes and a centre-tapped transformer. Full-wave rectifiers are also made using a two-winding transformer and four diodes. Such rectifiers are called bridge rectifiers. These are discussed as follows.

Two-diode full-wave rectifier: here ac input voltage is supplied from the secondary of a centre-tapped transformer. The circuit consists of the transformer, two diodes and the load resistance. The circuit is essentially the summation of two half-wave rectifiers as shown in Fig. 13.5.

For the positive half cycle of the input voltage, diode D1 will conduct. This is because terminal A is positive and the diode D1 is forward biased. As terminal B is negative, diode D2 will not conduct. In the negative half cycle, terminal B is positive and terminal A is negative. Hence, diode D2 will conduct and diode D1 will be reverse biased. This way in each half cycle one of the two diodes will conduct and current will flow through the load resistance, RL. The output current and the output voltage across the load will be a full-wave rectified current and voltage, respectively. The output wave form is a series of consecutive positive half cycles of sinusoidal wave form. The current through the load resistance is unidirectional but its magnitude is fluctuating as shown in Fig. 13.5. The PIV is the maximum voltage that would appear across a diode when it is reverse biased. Here, when D1 is conducting, D2 is reverse biased and vice versa. When D1 is conducting, the voltage that would appear across diode D2 is the sum of the voltage across the lower half of the transformer secondary winding and the voltage appearing across the load. PIV of the diode is equal to 2 Vm.

Figure 13.5 Full-wave rectifier using two diodes and a centre-tapped transformer

#### 13.1.5 Full-wave Bridge Rectifier

A bridge rectifier circuit uses four diodes connected in the form of a bridge. The various ways the four diodes in the bridge circuit are drawn have been shown in Fig. 13.6 (a) and (b). As shown in Fig. 13.6 (a), the arrow head symbols of all the diodes are pointing towards the positive terminal of the output, i.e., the load.

Figure 13.6 A bridge rectifier circuit for full-wave rectification using four diodes and a transformer

During the positive half cycle of the input voltage, terminal A of the transformer is positive. Current will flow from the positive terminal through diode D1, load RL and diode D4, and back to the negative terminal, B of the transformer. The direction of current through the load has been shown to be from C to D, i.e., from top to bottom. The polarities of the load terminals have been shown. During this period diodes D2 and D3 are reverse biased.

During the negative half cycle of the input voltage diodes D3 and D2 are forward biased while diodes D1 and D4 are reverse biased. Current through the load will flow in the same direction, i.e., from terminal C to D. During both positive and negative half cycles of the input voltage, current will pass through the load in the same direction. The output voltage wave shape is a series of positive half cycles of the sinusoidal voltage. This is dc output but having a varying magnitude. The transformer provides isolation of the dc output from the supply ac input.

The important parameters of a full-wave bridge rectifier are determined as follows.

#### 13.1.6 Analysis of Full-wave Rectifiers

Average value or dc value of load current, Idc: The average value or dc value will be the same if calculated for a period 0 to π or 0 to 2π

Rms value of the laod current, Irms

Output voltage, Vdc

Vdc=IdcRL

where RL is the load resistance, RF is the forward resistance of the diode, and R2 is the secondary winding resistance of the transformer.

Rectifier efficiency, η

=81.2 percent

Ripple factor,

=0.48 percent

We had earlier calculated the ripple factor for a half-wave rectifier as 1.21. For a full-wave rectifier, the ripple factor is reduced to 0.48. This indicates that the fluctuation of dc output is reduced. PIV for a bridge rectifier Vm.

It is now possible to compare the performance parameters of a half-wave rectifier with a bridge rectifier. This has been shown in Table 13.1. Although the performance parameters of a bridge rectifier are superior than a half-wave rectifie, the quality of output voltage is still not acceptable and has ripple content which must be further minimized. This is achieved by using filters. Filters are discussed in the section that follows.

#### 13.1.7 Comparison of Half-wave and Full-wave Rectifiers

The performance of half-wave and full-wave rectifiers with respect to certain parameters has been compared as in Table 13.1.

Table 13.1 Comparison of Half-wave and Full-wave Rectifiers Against Their Salient Parameters

Example 13.4 The input to a bridge rectifier is through a step-down transformer of turn ratio 10:1. The supply voltage is 230 V at 50 Hz. The load resistance is 1.2 kΩ secondary winding resistance of the transformer is 4Ω diode forward resistance is 2 Ω. Calculate the efficiency of the bridge rectifier.

Solution:

Figure 13.7 Bridge-rectifier circuit

Given Vi (RMS) = 230 V

RF = 2 Ω, R2 = 4 Ω, RL = 1200 Ω

The RMS value of the emf in transformer secondary

=23 V percent

Peak secondary voltage, Vm is

=32 5. V percent

Current through the load will flow from the transformer secondary via two diodes. Therefore, following the current path during the positive half cycle

For a bridge rectifier,

DC power output, Pdc=I2dcRL=(17×10-2)2×1200

= 346.8 mW

AC power input,

Pac = (Irms)2 (RL + 2RF + Rs)

= 432 mW

Example 13.5 Determine for the bridge circuit the peak value of load current when Vi = 15 V, R1 = 600Ω and the forward voltage drop of the diode is 0.7 V. Also calculate the average value of the output current

Figure 13.8 Bridge rectifier circuit

Solution:

Vi = 15V

Maximum value of voltage appearing across the load will be Vi (max) – 2 VF.

This is because two diodes are involved in the current flow through the load at any point of time

Vo(max) = Vi(max) − 2 VF
= 21.21 − 2 × 0.7
= 19.81V
##### 13.2 TRANSIST OR AMPLIFIER

One of the main applications of the transistors involves amplification of the input signal. The amplifier is used to tune the low power input signal into a higher power signal, its output frequency is same as that of the input. Here, the transistor is working as an amplifier unit it should be biased to operate in the active region. The input signal to the amplifier is in sinusoidal form then the output will also remain in sinusoidal form.

Three main classifications of the transistor amplifiers: common emitter, common collector and common base amplifiers. In each of the three configurations, one of the three nodes is permanently grounded and the other two nodes are either an input or output of the amplifier.

#### 13.2.1 The Common Emitter Amplifier Circuit

Figure 13.9 shows the basic circuit for a common emitter amplifier, it uses potential divider biasing. This type of biasing arrangement uses two resistors as a potential divider network across the supply with their centre point supplying the required base bias voltage to the transistor. The potential divider network formed by the two resistors, R1, R2 and the power supply voltage Vcc will hold the base bias at a constant steady voltage level allowing for best stability.

Figure 13.9 The practical common emitter amplifier

The circuit for the common emitter amplifier consists of three capacitors C1, C2, CE. The input capacitor C1 allows the ac signal for amplification and it blocks the dc component in the signal. So it is also known as blocking capacitor. The input is biased to the base of the transistor through this capacitor. Blocking capacitor is used to provide good stability to the circuit by maintaining constant bias condition.

CE is an emitter bypass capacitor and it is connected parallel to the resistor RE. If it is not inserted in the circuit, the amplified output voltage drop will occur.

Coupling capacitor C2 will connect the output of the amplifier to the load. It will block the dc component and passes only the amplified Ac signal.

#### 13.2.2 Common Collector Amplifier Circuit

The basic circuit for common collector configuration is shown in Fig. 13.10. In this configuration, the source and the load shares the collector lead as a common connection point. Input of the common collector configuration is applied to base-collector terminal of the transistor and the output is taken from the emitter-collector lead. Load is placed in series with the emitter, which receives both base and collector current. The heavy current through the emitter lead, the amplifier will have a high current gain approximately equal to unity.

Figure 13.10 Common collector amplifier

The dc biasing for the circuit is provided by the resistors R1, R2 and RE. When the input voltage (Vi) is applied to the base terminal, the voltage developed across the Resistor RE, VE = VB − VBE, where VBE remains constant, so the emitter voltage follows the base voltage. Here, emitter is the output terminal, so it can be specified that output voltage is same as that of its input voltage. In other words, in base voltage appears an equal change across the load at the emitter; hence the common collector amplifier is also known as emitter follower.

#### 13.2.3 Common Base Amplifier Circuit

The basic circuit for the common base amplifier is shown in Fig. 13.11. It is called so, because the source and the load share the base of the transistor as a common connection point. This is a complex amplifier configuration compared to others and they are rarely utilized because of its strange characteristics.

When the input is applied to the emitter terminal, the emitter voltage will increase. But the base voltage remains constant, so the forward bias of the emitter base junction will get reduced. This reduces the base current Ib and hence IC will get reduced. The drop across the resistor RC will also get reduced. This will cause increase in output voltage V0 = VCC – ICRC. When positive going input is given, output also develop positive going signal and negative going input will also develop negative going output without any phase shift.

Figure 13.11 Common base amplifier

##### 13.3 OSCILLATORS

An electronic oscillator is an electronic circuit that produces a periodic, oscillating electronic signal, often a sine wave or a square wave. The oscillators produce sinusoidal as well as non sinusoidal waveforms from very low frequencies up to very high frequencies. It is also called as a waveform generator which incorporates both active and passive elements. They are widely used in many electronic devices.

#### 13.3.1 Classification of Oscillators

Based on the output, oscillators may be classified as

1. Sinusoidal oscillator or Harmonic Oscillator
1. Tuned circuit oscillators
2. RC oscillators
3. Crystal oscillators
4. Negative resistance oscillator
2. Non-sinusoidal oscillator

#### 13.3.2 Sinusoidal Oscillator or Harmonic Oscillator

These types of oscillators generate a sine wave output and are characterised by amplitude and frequency of oscillations. Sine wave oscillators are employed to generate sinusoidal high frequency sine waves in wide ranging applications in electronic equipment and computers. For example, these are used to generate carrier (high-frequency sinusoidal) waves in tuning stages in radio and television receivers. These are extensively employed in radars.

Tuned circuit oscillators

These oscillators use a tuned circuit consisting inductors (L) and capacitors (C) and are used to generate high frequency signals. Thus, they are also known as radio frequency (RF) oscillators. Such oscillators are Hartely, Colpitts, Clapp oscillators etc.

RC oscillators

These oscillators use resistors and capacitors and are used to generate low frequency signals. Thus they are also known as audio frequency (AF) oscillators. Such oscillators are Phase Shift Oscillator and Wien Bridge Oscillator.

Crystal oscillators

These oscillators use quartz crystal and are used to generate highly stabilized output signal with frequency up to 10 MHz. For example, piezo oscillator.

Negative resistance oscillator

These oscillators use negative resistance characteristics of the device such as tunnel diodes. For example, tuned diode oscillator.

#### 13.3.3 Non-sinusoidal Oscillator

As the name suggests, the output wave in these types of oscillators is non-sinusoidal, such as rectangular, square, triangular, or saw-tooth. Figure 13.12 shows different forms of output waveforms oscillators. Non sinusoidal oscillators are also known as relaxation oscillators.

Figure 13.12 Waveforms generated by oscillator

#### 13.3.4 Basic Functional Blocks of Oscillator Circuit

The primary function of an oscillator is to convert DC power into a periodic signal or AC signal at a very high frequency. An oscillator does not require any external input signal to produce sinusoidal or other repetitive waveforms of desired magnitude and frequency at the output and even without any use of mechanical moving parts. The basic functional blocks of an oscillator circuit is shown in Fig. 13.13

Figure 13.13 Block diagram of an oscillator circuit

Amplifier circuit

An amplifier circuit converts the dc power of the battery into ac power of required frequency and supplies it to the oscillatory circuit via the feedback circuit to meet the losses.

Feedback circuit

A feedback circuit supplies a part of the output energy to the oscillatory circuit and ensures that the input to it satisfies the Barkhausen criteria of = 1 and βAv = 180° to produce sustained oscillations.

Oscillatory circuit

An oscillatory circuit is a tuned LC or RC circuit that generates the desired oscillations, which, in turn, constitute the input to the transistor amplifier circuit. Due to the amplifying characteristics of the transistor, the output obtained is an amplified version of the input oscillations.

#### 13.3.5 Barkhausen Criterion or Conditions for Oscillation

The main statement of the oscillator is that the oscillation is achieved through positive feedback which generates the output signal without input signal. Also, the voltage gain of the amplifier increases with the increase in the amount of positive feedback.

In order to understand this concept, let us consider a non-inverting amplifier with a voltage gain ‘A’ and a positive feedback network with feedback gain of β as shown in Fig. 13.14.

Figure 13.14 Block diagram of a feedback system

Let us assume that a sinusoidal input signal Vs is applied at the input. Since the amplifier is noninverting, the output signal Vo is in phase with Vs. A feedback network feeds the part of Vo to the input and the amount of Vo feedback depends on the feedback network gain β.

No phase shift is introduced by this feedback network and hence the feedback voltage or signal Vf is in phase with Vs. A feedback is said to be positive when the phase of the feedback signal is same as that of the input signal.

The open loop gain ‘A’ of the amplifier is the ratio of output voltage to the input voltage, i.e.,

A = Vo/VI

By considering the effect of feedback, the ratio of net output voltage Vo and input supply Vs called as a closed loop gain Af (gain with feedback).

Af = Vo/Vs

Since the feedback is positive, the input to the amplifier is generated by adding Vf to the VS,

Vi = Vs + Vf

Depends on the feedback gain β, the value of the feedback voltage is varied, i.e.,

Vf = βVo

Substituting in the above equation,

Vi = Vs + βVo
Vs = Vi – βVo

Then the gain becomes

Af = Vo/ (Vi – βVo)

By dividing both numerator and denominator by Vi, we get

Af = (Vo / Vi)/ (1 – β) (Vo / Vi)
Af = A/ (1 – Aβ) Since A = Vo/Vi

Where Aβ is the loop gain and if Aβ = 1, then Af becomes infinity. From the above expression, it is clear that even without external input (Vs = 0), the circuit can generate the output just by feeding a part of the output as its own input.

And also closed loop gain increases with increase in amount of positive feedback gain. The oscillation rate or frequency depends on amplifier or feedback network or both.

The circuit will oscillate when two conditions, called as Barkhausen’s criteria are met. These two conditions are

1. The loop gain must be unity or greater.
2. The feedback signal feeding back at the input must be phase shifted by 360 degrees (which is same as zero degrees). In most of the circuits, an inverting amplifier is used to produce 180 degree phase shift and additional 180 degree phase shift is provided by the feedback network.

At only one particular frequency, a tuned inductor-capacitor (LC circuit) circuit provides this 180 degrees phase shift.

#### 13.3.6 LC Tuned Oscillators

Hartley oscillator

The Hartley oscillator is one of the classical LC feedback circuits used to generate high frequency waveforms or signals. The circuit diagram of a Hartley oscillator is shown in the Fig. 13.15. The tank circuit consists of two coils L1 and L2 and a capacitor C. The coil L1 is inductively coupled to the coil L2 and the combination works as an auto transformer. A coil called Radio Frequency Choke (RFC) is connected between the oscillator and VCC supply. The feedback between the input and output is accomplished through auto transformer action, which also introduces a phase shift of 180°. The phase reversal between the output and input voltage occurs because they are taken from the opposite ends of the coils (L1 and L2) with respect to tap which is grounded.

Since, the transistor also introduces a phase shift of 180°, therefore, the total phase shift is 360° and hence the feedback is positive. Ignoring the loading effects of the base, the feedback fraction is given by the relation,

β = L1/L2

Also since

Aβ ≥ 1

Figure 13.15 Hartley oscillator

The resistors R1 and R2 and Re are used to provide DC bias to the transistor. When the circuit is energised, switching on the supply, the collector current flows. The oscillations are produced because of positive feedback from the tank circuit.

Colpitts oscilator

The basic configuration of the Colpitts oscillator resembles that of the Hartley Oscillator but the difference is that the centre tapping of the tank sub-circuit is made up of a “capacitive voltage divider” network instead of a tapped autotransformer type inductor as in the Hartley oscillator as shown in Fig. 13.16. The tank circuit is made up of two capacitors C1 and C2 connected in series with each other across a fixed inductor (L). The resistors R1, R2, RE, and RF choke have the same function as mentioned in Hartley oscillator. The advantage of capacitive circuit configuration is less self and mutual inductance within the tank circuit; hence the frequency stability of the oscillator is improved along with a more simple design.

The feedback between the output and input circuit is accomplished by the voltage developed across Capacitor C2. Ignoring the loading effect of the base, the feedback fraction,

β = C1/C2

Also since

Aβ ≥ 1

Figure 13.16 Colpitts oscilator

Thus, to start the oscillation, the voltage gain (Av) must be greater than (C1/C2). The frequency of oscillation (neglecting mutual inductance) is given by the relation,

where

C=C1 C2 /( C1 +C2)

The configuration of the transistor amplifier is of a Common Emitter Amplifier with the output signal 180° out of phase with regard to the input signal. The additional 180° phase shift required for oscillation is achieved by the fact that the two capacitors are connected together in series but in parallel with the inductive coil resulting in overall phase shift of the circuit being zero or 360°.

Colpitts oscillator is widely used in commercial signal generator ranging from 1MHz to 500 MHz. Frequency of oscillation is varied by gang tuning the capacitor C1 and C2.

#### 13.3.7 RC Oscillators

RC phase shift oscillator

Basic principle of phase shift oscillator is that a fraction of the output single-stage amplifier is passed through a phase shift network, before feeding back to input. The phase-shift network gives another phase-shift of 180° in addition to the phase-shift of the 180° introduced by the amplifier. Thus, there is a total phase shift of 360°, which is also equal to 0°.

Circuit arrangement of phase shift oscillator is shown in Fig. 13.17. There are three R-C combinations forming feedback network. Each R-C combination provides a phase shift of 60°. Hence the net phase shift produced by three RC networks is 180° and another 180° phase shift is provided by the transistor itself. Thus, total 360° phase shift is produced between the input and output signal.

Figure 13.17 RC phase shift oscillator

The phase shift φ, given by each R-C section is,

By varying one or more of the resistors or capacitors in the phase-shift network, the frequency can be varied and generally this is done by a 3, R-C combination.

If all the resistors, R and the capacitors, C in the phase shift network are equal in value, then the frequency of oscillations produced by the RC oscillator is given as:

Where

ƒ is the Output Frequency in Hertz

R is the Resistance in Ohms

C is the Capacitance in Farads

N is the number of RC stages. (N = 3)

Wien Bridge oscillator

The Wien Bridge oscillator is so called because the circuit is based on a frequency-selective form of the Wheatstone bridge circuit. The Wien Bridge oscillator is a two-stage RC coupled amplifier circuit that has good stability at its resonant frequency, low distortion and is very easy to tune making it a popular circuit as an audio frequency oscillator but the phase shift of the output signal is considerably different from the previous phase shift RC oscillator.

Figure 13.18 Wein bridge oscillator

The Wien Bridge Oscillator uses a feedback circuit consisting of a series of RC circuit connected with a parallel RC of the same component values producing a phase delay or phase advance circuit depending upon the frequency. At the resonant frequency ƒ, the phase shift is 0°.

The Wien bridge oscillator circuit is as shown in Fig. 13.18. A second stage of amplifier is used for producing another 180° phase shift in addition to the phase shift of 180° produced by the first stage. Thus, there is a total phase shift of 360°, which is the basic requirement. A fraction of output from the second stage is fedback to the input of the first stage without producing any further phase shift.

The Wien bridge oscillator consist of two transistor (CE configuration) which, provides an approximately, 360° or 0° phase shift so the feedback network has no need to introduce any additional phase shift where R5, R6, R7, and R8 are biasing resistors.

The feedback network consists of C1-R1 and C2-R2 (called a lead-lag network) and R3-R4 (called a voltage divider). The lead-lag network provides a positive feedback to the output of the first stage and the voltage divider provides the negative feedback to the emitter of the transistor Q1.

##### 13.4 Filters

We have seen that the wave form of the rectified voltage is a series of positive half cycles of the input voltage wave form either of equal or of reduced magnitude. For a half-wave rectifier we get a series of positive half cycles with one missing in between. Our objective is to get a steady-value dc output. To convert the fluctuating output voltage into a steady dc, smoothing circuits called filters must be used. The simplest filter is a capacitor which is connected across the load. Fig. 13.19 shows a capacitor C connected across the load resistance RL in a half-wave rectifier. The effect of the use of a capacitor on the output voltage wave has been shown.

During the positive half cycle of the input voltage the diode D1 is forward biased. Current flows through the diode and the load resistor, RL. At the same time the capacitor, C gets charged upto the peak value, Vm of the input voltage.

Figure 13.19 Half-wave rectifier circuit with a capacitor filter

After attaining the pick value, the input voltage starts reducing, its value becoming less and less than Vm. But the capacitor has been charged to a voltage Vm. Thus, the potential of terminal B becomes higher than the potential of terminal A. As a result, diode D1 gets reverse biased but the capacitor voltage remains close to Vm. With the diode D1 reverse biased, the changing of the capacitor stops. The capacitor now starts getting discharged through the load resistor RL. The voltage across the capacitor, VC starts falling, as has been shown in Fig. 13.19, through a thick horizontally inclined line. The diode, D1 remains reverse biased throughout the rest of the positive half cycle and also during the negative half cycle, and further to the next positive half cycle until at θ2 when the input voltage starts becoming higher than the capacitor voltage, VC once again. At this point the diode becomes conducting supplying current to the load as also charging the capacitor once again. This process of charging and discharging of the capacitor continues in every cycle and an output voltage waveform, as shown by a thick line, is achieved. This wave shape of the output voltage is superior to the wave shape of the output voltage obtained when no capacitor was connected. The ripple of the output voltage is now reduced and a near-steady dc output voltage obtained.

Amplitude of ripple voltage and selection of capacitor

The half-wave rectified voltage with a capacitor filter has been shown again in Fig. 13.20. Vr represents the peak to peak ripple voltage. The time of discharge of the capacitor is represented by t1 as shown in Fig. 13.20. The output voltage fluctuates between V0(min) to V0(max).

Peak to peak, Vr = V0 (max)–V0 (min).

The capacitor C gets discharged during the time t1 when the voltage across it drops by Vr causing a load current, IL to flow through the resistance RL. So we can write

Q = CVr = IL × t

Calculation of the value of the capacitor to be used depends on the allowable ripple voltage, the average output voltage, the load resistance, and the supply frequency. The approximate value is calculated using the procedure illustrated through an example. The standard manufacture’s list is then consulted to select the next higher value of the capacitor available in the market.

Figure 13.20 Half-wave rectified voltage with a capacitor filter

Example 13.6 A half-wave capacitor filter rectifier has maintained an average output voltage of 15 V with a peak to peak ripple of not more than 3 V. The load resistance is 100 Ω Calculate the value of the capacitor filter. The ac supply frequency is 50 Hz.

Solution:

The time of discharge of the capacitor t1 can approximately by considered equal to T. Therefore,

t1 = T = 20ms

using the relation,

CVr = ILt1

Capacitors of 100 μF are available as can be checked from the manufacturer’s list.

Similar to the capacitor filter used in half-wave rectifiers, capacitor filters are used in full-wave rectifier also as shown in Fig. 13.21. The circuit works exactly the same way as has been explained in the case of half-wave rectifier with a capacitor filter. From the dc output voltage wave shape it is observed that the ripple is minimized to a very small level.

The ripple voltage that appears across the capacitor can further be reduced by use of another resistor and a capacitor. The resistor is connected in series while the capacitors are in parallel making a π formation. Such filters are called R–C filters or simply π filters. Similarly, we can use L–C filters also to further smoothen the output voltage. An inductor in series with the load also works as a filter as the inductor allows dc current to flow and opposes ac current flow.

Figure 13.21 Bridge rectifier with a smoothing capacitor

#### 13.5.1 Integrator

The Integrator is a type of low pass filter circuit that converts a square wave input signal into a triangular waveform output. Why it is known as low pass filter because it allows only the low frequency signal to pass through it. The basic circuit for an RC integrator is shown in the Fig. 13.22

Figure 13.22 Basic integrator circuit

The AC input from the signal generator with voltage Vin (t), is given to the RC series circuit. And the output is taken across the capacitor. Here the time constant of the circuit should be very large. As the input square wave is applied, during the positive half cycle the voltage across capacitor increases from zero, to the maximum (peak value of applied voltage). During the negative half cycle, the capacitor starts to discharge and comes to zero. This process repeats for the remaining cycles and a triangular wave is obtained.

Considering the signals with high frequency ω >> 1/RC, the charging up time for the capacitor is very less and its voltage will be small, so the input voltage is approximately equal to voltage across the resistor.

So, the input voltage equation,

But                                          ωC >> (1/R), so Vin ≈ IR

So,

#### 13.5.2 Differentiator

A differentiator gives the derivative of input voltage as output. It can convert a square wave input signal into high frequency spikes at its output. A differentiator using passive components such as resistors and capacitors is a high pass filter. The basic circuit is RC differentiator shown in the Fig. 13.23. It acts as a differentiator only when the time constant is too small.

Figure 13.23 The Basic circuit of differentiator

The input AC signal from the signal generator with voltage Vin(t), is applied to an RC series circuit. Here, the output is the voltage across the resistor. This time, we consider only low frequencies ω << 1/RC, so that the capacitor has time to charge up until its voltage almost equals that of the source. The voltage at output is proportional to the current through the capacitor. The current through the capacitor can be expressed as C dv/dt. The output is taking across the resistor. So output will be RC dv/dt. Thus differentiation of input takes place.

So the input signal to the differentiator,

But R >> (1/ωC), so Vin = (I/ωC)

For frequency ω << (1/RC), Vin ≈ Vc

##### 13.6 APPLICAT IONS OF DIODES IN CLIPPING AND CLAMPING CIRCUITS

A clipping circuit removes, i.e., clips off a certain portion of the input voltage wave form. Clipping circuits are used when it becomes necessary to protect a circuit or a device that might get destroyed due to large amplitude signal. In fact, the half-wave rectifier described earlier is a clipper circuit. It clips off the negative half cycle and allows only the positive half cycle. A clipping circuit, or also called a clipper is used to clip off certain unwanted portions of the wave form which may lead to noise, and deteriote the performance of the device through which such currents pass. Let us see one noise clipper circuit as shown in Fig. 13.24. Here the noise level is lower than the forward voltage drop VF of the diodes i.e., VF = 0.7 V. Two diodes, D1 and D2 clear both the halves of the main signal but do not allow the noise signal to pass.

Figure 13.24 Noise clipping circuit

#### 13.6.1 Negative and Positive Series Clippers

When the input is positive the diode is forward biased, and hence the positive half cycle is passed to the output as shown in Fig. 13.25 (a). During the negative half cycle the diode is reverse biased, and hence the output will remain zero. This way we can say that the circuit cuts off the negative half cycles of the input voltage.

In Fig. 13.25 (b), the positive half cycles are clipped off as the diode is reverse biased during the positive half cycles of the input. The input voltage wave in these two clippers are sinusoidal. However, the input wave form shapes could be of any other form like square, rectangular, triangular, etc.

#### 13.6.2 Shunt Clippers

Like series clippers, shunt clippers can also clip off the positive half cycle or the negative half cycle of the input.

Figure 13.25 Negative and positive series clippers

Figure 13.26 Positive shunt clipper

Figure 13.26 shows a positive shunt clipper circuit where the diode is connected in parallel, i.e., in shunt with the load resistance RL.

When the input voltage is positive the diode is forward biased and is in conducting mode. The voltage drop across the diode VF, which is only 0.7 V for a silicon diode will appear across the load. Thus, effectively the positive half cycle of the input voltage is clipped off and the output voltage will be nearly zero. When the input voltage is negative the diode is reverse biased. Current flows through the load resistance RL and the series resistance R. The voltage drop across R is IL R which is small as compared to the voltage drop across the load resistance RL. The output voltage V0 is nearly equal to the negative input voltage as has been shown. Resistance R is connected in the circuit to limit the diode current when it is forward biased.

A negative shunt clipper can be made by reversing the diode terminal connections.

#### 13.6.3 Biased Clippers

The level of clipping can be adjusted by introducing a biased voltage in the circuit. Fig. 13.27 shows a biased positive clipper circuit where a biased voltage, VB has been connected in series with the diode.

The biased voltage, VB is kept lower than Vm. When the input voltage is less than VB, the diode is not forward biased, and hence the whole input is passed on to the load and a load current flows through RL.

Figure 13.27 Biased shunt clipper

Figure 13.28 Positive clamper’s wave shape

When the input voltage level crosses the voltage VB, the diode gets forward biased and starts conducting, and no further increase in the output current is possible. Thus, the current through the load is resistricted by the bias voltage level. A combination biased clipping circuit can be made by using another diode together with a bias voltage. The bias voltage for the two diodes can be made different to achieve the clipping of the positive and negative half cycles at different voltage levels. Biased clipping circuits are used to protect circuits and devices from over voltages. We have also observed that clipper circuits can change the output voltage wave shape also.

#### 13.6.4 Clamping Circuits

A clamping circuit essentially adds a dc component to the ac signal in either direction. As shown in Fig. 13.28 a dc voltage has been added to the sinusoidal voltage. The signal voltage equation is say, v = Vm sin ωt = 5 sin ωt. If we add +5 V dc, the equation of the resultant voltage will be v = 5 + 5 sin ωt. The clamper circuit has added a dc voltage of +5 V to the signal and pushed the signal upwards without changing its wave shape. This is called a positive clamper. If a negative dc voltage is added, the signal will be brought downwards, and such a clamper will be called a negative clamper. A negative clamper circuit has been explained in Fig. 13.28.

Assume that the input voltage is a square one. During the positive half cycle of the input voltage the diode is forward biased and will work like a short circuit (neglecting the forward voltage drop across the diode). The capacitor will be charged to input voltage level. The voltage across the diode, and hence across the load resistor RL will be zero.

During the negative half cycle of the input the diode will be reverse biased and will be like an open switch (no current through it can flow). The voltage across the load, i.e., V0 can be calculated in the circuit of Fig. 13.29 (b) by applying Kirchhoff’s voltage law as

− Vi − Vc − V0 = 0

or,                                              V0 = − Vi − Vi (as Vc = Vi)

or,                                                          V0 = −2Vi

This shows that the input signal is negatively clamped by a dc voltage equal to the magnitude of the input voltage.

Figure 13.29 A negative clamper circuit

##### 13.7 MULTIVIBRATORS

A Multivibrator is an electronic circuit that generates square, rectangular, pulse waveforms, also called nonlinear oscillators or function generators. Multivibrator is basically a two amplifier circuit arranged with regenerative feedback.

There are three types of Multivibrators:

• Astable Multivibrator: Circuit is not stable in either state—it continuously oscillates from one state to the other. (Application in Oscillators)
• Monostable Multivibrator: One of the states is stable but the other is not. (Application in Timer)
• Bistable Multivibrator: Circuit is stable in both the state and will remain in either state indefinitely. The circuit can be flipped from one state to the other by an external event or trigger. (Application in Flip flop)

#### 13.7.1 Astable Multivibrator

A multivibrator circuit in which there is no single stable state is known as an astable multivibrator, as the name suggests. Basically, the circuit keeps switching from one state to another, and that theoretically sounds undesirable, there are actually many practical applications for it, namely, relaxation oscillators. It consists of two amplifying stages connected by coupling networks, with a positive feedback loop. The actual amplifying elements used range from vacuum tubes to op-amps and many more, including field effect transistors and bipolar junction transistors. The circuit diagram for the astable multivibrator is shown in Fig. 13.30.

Figure 13.30 Astable multivibrator

The circuit keeps switching from one state to another because of the positive feedback, and has only one amplifying element running at a time. The charge cycle is much faster when compared to the discharge cycle, and is facilitated by the coupling capacitors, since voltage change in a capacitor cannot be instantaneous. Just like other circuits, it requires some time to begin conducting, which depends on the amplifying elements used.

Applications

• Used in applications where low clock frequency clock pulse train is required
• Relaxation oscillators, which are parts of vehicle indicator lights, disco strobe lights, early oscilloscopes and television receivers
• Timing signals

#### 13.7.2 Monostable Multivibrator

A monostable multivibrator, as the name suggests, is a circuit with a single stable state. Conceptually and practically, it can be termed as a half astable multivibrator. While an astable multivibrator uses two resistive-capacitive networks, a monostable multivibrator uses one resistive-capacitive and one simple resistive network, hence it is being called as half astable multivibrator. It outputs a perfect square waveform because of the absence of a loaded capacitor, and is used in precise applications.

The circuit, when triggered by an input pulse, switches to the unstable state for some time, and then returns to the stable state. Depending on the application, the circuit can be tweaked to stay in the unstable state for as long as required, even sometimes requiring multiple input pulses to keep it in the unstable state. This is also called a retriggerable monostable multivibrator. Consequently, if the trigger pulses don’t affect the time spent in the unstable state, it is called a non-retriggerable monostable multivibrator.

Applications

• Used as pulse generators
• Used to produce time delay in circuits
• Used to regenerate pulses in old telecommunication systems
• Used to reduce pulse distortion in computer systems
• Used as gated circuits

The circuit which is shown in Fig. 13.31, when triggered by an input pulse, switches to the unstable state for some time, and then returns to the stable state.

Figure 13.31 Monostable multivibrator

#### 13.7.3 Bistable Multivibrator

A bistable multivibrator is a circuit that has two stable states. Instead of using two resistive-capacitive networks like an astable multivibrator, a bistable multivibrator uses two resistive networks only. Therefore, it uses only direct or resistive coupling, and the latch type has no charge or discharge time because of the lack of any capacitors. Switching between the states can be done by two terminals called as set and reset. Bistable multivibrators are among the most important components of digital computer systems, because they’re used to store data. It is also the basic storage element in sequential logic, and can be clocked or unclocked. Just like other types of multivibrators, bistable multivibrators also use multiple control elements, such as vacuum tubes and bipolar junction transistors. In complex circuits, multiple bistable multivibrators are cascaded as well. Instead of using two resistive-capacitive networks like an astable multivibrator, a bistable multivibrator uses only two resistive networks as shown in Fig. 13.32.

Figure 13.32 Bistable multivibrator

Bistable multivibrators, particularly flip-flops, are subject to something called metastability, which happens when both inputs change at the same time. It can cause unforeseen problems like extra time to settle into a state, unintended oscillations and the like. In digital computer systems, this can lead to data corruption and crashes because of the inconsistent state change. This is fixed by setting proper time constraints and cascading multiple flip-flops.

Applications

• Used in counting circuits
• Used in memory storage units
• Used as frequency dividers
• Used in pulse generation circuits.
##### 13.8 SCHMITT TRIGGER

Schmitt Trigger is a wave shaping circuit, used for generation of a square wave from a sine wave input. It is a bistable circuit in which two transistors switches are connected regeneratively.

Figure 13.33 and 13.34 shows the circuit of the Schmitt trigger with the input and output waveforms. It consists of two identical transistors Q1 and Q2 coupled through an emitter resistor RE. Resistors R1 and R2 form a voltage divider across VCC and ground. These resistors provide a small forward bias on the base of transistor Q2. When the supply is switched on, transistor Q2 starts conducting. The flow of its current through resistor RE produces a voltage drop across it. This voltage drop acts as a reverse bias across the emitter junction of transistor Q1 due to which it cuts-off. As a result of this, the voltage across the collector will rise up to VCC. This rising voltage is coupled to the base of transistor Q2 through the resistor R1. It increases the forward bias at the base of transistor Q2 and therefore it reaches the saturation. At this instant, the collector voltage level are VC1 = VCC and VC2 = VCE(sat) as shown in Fig 13.34.

Consider an AC supply is given to the input of transistor Q1. As the voltage increases above the VUT, transistor Q1 starts conducting. The point at which the Q1, starts conducting is the Upper Triggering Point (UTP). As transistor Q1 conducts, the collector voltage falls below VCC. Hence the forward bias to Q2 is reduced, which in turn reduces the current of Q2 and voltage drop across RE. As a result, the reverse bias of transistor Q1, is reduced and it conducts more which drives Q2 nearer to cut-off. This process continues till Q1 is driven into saturation, and Q2 is cut off. At this instant, the collector voltage levels are VC1 = VCE(sat) and VC2 = VCC.

Figure 13.33 Schmitt Trigger

Figure 13.34 Input and output waveforms of Schmitt trigger

Transistor Q1 will continue to conduct till the input voltage crosses the VLT. When the input voltage becomes equal to VLT, the emitter base junction of Q1, become reverse biased. Hence its collector voltage starts rising towards VCC. This forward biases Q2 and it starts conducting. The point at which Q2 starts conducting is Lower Trigger Point (LTP). Then Q2 is very quickly driven into saturation and Q1 is cut off. At this instant the collector voltage levels are VC1 = VCC and VC2 = VCE(Sat). The difference between UTP and LTP is known as Hysteresis voltage (VH).

Applications

1. Schmitt trigger is used for wave shaping circuits.
2. It can be used as a Voltage Comparator.
3. The hysteresis in Schmitt trigger is valuable when conditioning noisy signals for using digital circuits. The noise does not cause false triggering and so the output will be free from noise.
##### 13.9 REVIEW QUESTIONS

1. Draw a half-wave rectified circuit and show the input and output voltage waveforms.
2. List the performance parameters of a halfwave rectifier and explain this significance.
3. Draw the circuit diagram for a centre-tap fullwave rectifier and explain its operation with the help of input and output voltage wave forms.
4. Draw and explain the circuit for a bridge rectifier and draw the input and output voltage waveforms.
5. Derive the expression for the following in the case of a half-wave rectifier:

(a) rectifier efficiency (b) ripple factor (c) average value of output dc current (d) output voltage, Vdc.

6. Compare the performance parameters of a half-wave rectifier and a full-wave bridge rectifier.
7. What is meant by ripple factor and what is its significance. Calculate its value for a halfwave rectifier and a full-wave rectifier.
8. What are the disadvantages of a half-wave rectifier?
9. What is meant by rectifier efficiency? What is its value for a full-wave and a half-wave rectifier?
10. What is the purpose of using a filter circuit in a rectifier and how does it work?
11. Draw the circuit for a capacitor filter and explain how does it half reduce the ripple factor?
12. Sketch the nature of ripple in the output of a capacitor filter used in a full-wave bridge rectifier.
13. Draw and explain the half-wave rectifier circuit with a capacitor filter. Also draw the input and output voltage wave form.
14. Draw and explain a bridge rectifier with a smoothing capacitor. Draw the output voltage wave shape.
15. What is a clipping and a clamping circuit? Where are they used?
16. Draw and explain negative and positive series clipper circuits with their input and output voltage waveforms, respectively.
17. What is a shunt clipper circuit? Draw the input and output voltage waveforms.
18. Draw the circuit for a biased clipper. Also draw the input and output voltage waveforms.
19. What are clamping circuits and what are their applications?
20. Draw and explain positive and negative clamper circuits.
21. What are the basic configurations of transistor amplifiers?
22. State the Barkhausen criterion for oscillators.
23. Why the integrator is known as a Low Pass Filter?
24. Why the differentiator is known as a High Pass Filter?
25. State the applications of astable multivibrator.
26. What are the advantages of bistable multivibrator?
27. State the applications of Schmitt trigger.

B. Multiple Choice Questions

1. The average or dc value of output current for a half-wave rectifier is
 a. b. c. d. .
2. The RMS value of load current for a half-wave rectifier is
 a. b. c. d. .
3. The output voltage Vdc for a half-wave rectifier is
 a. b. c. d. .
4. Rectifier efficiency of a half-wave and full-wave rectifier circuit, respectively are
 a. 40.6% and 81.2% b. 20.3% and 81.2% c.40.6% and 91.2% d. 20.3% and 40.6%.
5. Ripple factor for a half-wave and full-wave rectifier circuit, respectively are
 a. 0.48 and 1.21 b. 0.48 and 0.121 c. 4.8 and 1.21 d. 8.21 and 0.48.
6. Peak inverse voltage is defined as the
1. Minimum voltage that appears across the diode when it is forward biased
2. Minimum voltage that appears across the diode when it is reverse biased
3. Maximum voltage that appears across the diode when it is reverse biased
4. Minimum voltage that appears across the diode when it is reverse biased.
7. In a capacitor filter circuit,a capacitor is connected
1. In series with the load and it allows dc current to flow through it
2. In parallel with the load and it allows ac but blocks dc
3. In parallel with the load and it allows dc but blocks ac
4. In series with the load and it allows ac and blocks dc.
8. A clamping circuit
1. adds a dc component to an ac signal in either direction
2. adds an ac component to a dc signal in either direction
3. adds a dc component to an ac signal in positive direction only
4. adds a dc component to an ac signal in negative direction only.