1 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

Semiconductors and PN junction

1. Explain PN junction diode.

When a p-type semiconductor material is suitably joined to n-type semiconductor the contact surface is

called a p-n junction. The p-n junction is also called as semiconductor diode.

•

The left side material is a p-type semiconductor having –ve acceptor ions and +vely charged holes. The

right side material is n-type semiconductor having +ve donor ions and free electrons.

• Suppose the two pieces are suitably treated to form pn junction, then there is a tendency for the free

electrons from n-type to diffuse over to the p-side and holes from p-type to the n-side . This process is

called diffusion.

No external voltage is applied between the terminals of PN junction, hence the pn junction is

called unbiased.

•Due to charge concentration difference (Potential gradient) the free electrons move across the junction

from n-type to p-type, +ve donor ions are formed in . Hence a +ve charge is built on the n-side of the

junction. At the same time, the free electrons cross the junction and form the –ve acceptor ions by

filling in the holes. Therefore a net –ve charge is established on p-side of the junction.

• When a sufficient number of donor and acceptor ions are formed further diffusion is prevented.

At this point the junction is said to have attained an equilibrium.

• Thus a barrier is set up against further movement of charge carriers. This is called potential barrier

or junction barrier or cut in voltage Vo. The potential barrier for si is 0.7 volts and for Ge is 0.3 volts.

Note: outside this barrier on each side of the junction, the material is still neutral. Only inside the

barrier, there is a +ve charge on n-side and –ve charge on p-side. This region is called depletion layer.

The shaded portion on both sides of junction contains only immobile ions and not charge carriers

such as electrons or holes. In other words region is depleted of charge carriers.

2 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

Depletion Region

In the p-type region there are holes from the acceptor impurities and in the n-

type region there are extra electrons.

When a p-n junction is formed, some of the electrons from the n-region which

have reached the conduction band are free to diffuse across the junction and

combine with holes.

Filling a hole makes a negative ion and leaves behind a positive ion on the n-

side. A space charge builds up, creating a depletion region which inhibits any

further electron transfer unless it is helped by putting a forward bias on the

junction.

Energy band diagram for Equilibrium of junction

Potential Barrier

The electric field formed in the depletion region acts as a barrier. Barrier Potential in a PN junction

refers to the potential required to overcome the barrier at the PN junction.

External energy must be applied to get the electrons to move across the barrier of the electric field.

The potential difference required to move the electrons through the electric field is called the barrier

potential.

Barrier potential of a PN junction depends on the type of semiconductor material, amount of doping

and temperature.

This is approximately 0.7V for silicon and 0.3V for germanium.

2. Explain Biasing of PN junction diode.

Connecting a p-n junction to an external d.c. voltage source is called biasing.

1.Forward biasing 2.Reverse biasing

3 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

1.Forward biasing

• When external voltage applied to the junction

is in such a direction that it cancels the

potential barrier, thus permitting current flow

is called forward biasing i.e. P-region is

connected to positive terminal and N-region is

connected to negative terminal of source.

• To apply forward bias, connect +ve terminal

of the battery to p-type and –ve terminal to n-

type as shown in fig. below.

• The applied forward potential establishes the

electric field which acts against the field due to

potential barrier. Therefore the resultant field is weakened and the barrier height is reduced at the

junction as shown in fig.

• Due to negative terminal of the battery free electrons from n side are pushed towards the p-side and

due to positive terminal of the battery holes are pushed towards n-side.

Since the potential barrier voltage is very small, a small forward voltage is sufficient to completely

eliminate the barrier. Once the potential barrier is eliminated by the forward voltage, junction resistance

becomes almost zero and a low resistance path is established for the entire circuit. Therefore current

flows in the circuit. This is called forward current.

Energy band diagram for forward bias

2. Reverse biasing

• When the external voltage applied to the junction is in such a direction the potential barrier is

increased it is called reverse biasing.

• To apply reverse bias, connect –ve

terminal of the battery to p-type and +ve

terminal to n-type as shown in figure

below.

• The applied reverse voltage establishes an

electric field which acts in the same

direction as the field due to potential

barrier. Therefore the resultant field at the

junction is strengthened and the barrier

height is increased as shown in fig. due to movement of electrons and holes away from junction width

of the depletion layer increases.

• The increased potential barrier prevents the flow of charge carriers across the junction. Thus a high

resistance path is established for the entire circuit and hence current does not flow.

4 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

Energy band diagram for Reverse bias

3. Explain Characteristics of PN junction diode.

Forward Bias Characteristics-

P- terminal connected to positive terminal of battery and N-terminal connected to negative terminal of

battery , it is said to be forward biased. Positive terminal of the battery repels majority carriers, holes, in

P-region and negative terminal repels electrons in the N-region and push them towards the junction.

This result in increase in concentration of charge carriers near junction, recombination takes place and

width of depletion region decreases. As forward bias voltage is raised depletion region continues to

reduce in width, and more and more carriers recombine. This results in exponential rise of current.

P-N Junction Diode Equation

I = IS (eV / ɳKBT – 1) ------Here, ɳ = emission co-efficient, which is a number between 1 and 2, KB

– Boltzmann constant and T – temperature.

Is –reverse saturation current

Reverse bias Characteristics

5 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

In reverse biasing P- terminal is connected to negative terminal of the battery and N- terminal to

positive terminal of battery . Thus applied voltage makes N-side more positive than P-side. Negative

terminal of the battery attracts majority carriers, holes, in P-region and positive terminal attracts

electrons in the N-region and pull them away from the junction. This result in decrease in concentration

of charge carriers near junction and width of depletion region increases. A small amount of current

flow due to minority carriers, called as reverse bias current or leakage current. As reverse bias voltage

is raised depletion region continues to increase in width and no current flows. It can be concluded that

diode acts only when forward biased. Operation of diode can be summarized in form of I-V diode

characteristics graph.

For reverse bias diode, V<0, I = IS Where, V = supply voltage I = diode current IS = reverse

saturation current For forward bias, V > 0, I = IS(eV/ ɳKBT - 1)

4. Explain power supply block diagram .

The purpose of a power supply is to take electrical energy in one form and convert it into another.,

supply from 230V, 50Hz AC mains is converted into smooth DC using AC-DC power supply. A power

supply can by broken down into a series of blocks, each of which performs a particular function. A

transformer first steps down high voltage AC to low voltage AC. A rectifier circuit is then used to

convert AC to DC. This DC, however, contains ripples, which can be smoothened by a filter circuit. A

regulated power supply maintains a constant DC output voltage through ‘feedback action’.

Main components of a regulated supply to convert 230V AC voltage to 5V DC are shown below.

6 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

Power supplies are designed to produce as little ripple voltage as possible, as the ripple can cause

several problems. For Example

In audio amplifiers, too much ripple shows up as an annoying 50 Hz or 100 Hz audible hum.

In video circuits, excessive ripple shows up as “hum” bars in the picture. In digital circuits it can

cause erroneous outputs from logic circuits.

5. Explain the working of Half wave rectifier.

A simple Half Wave Rectifier is nothing more than a single pn junction diode connected in series to the

load resistor. Input voltage is given to a step down transformer and the resulting reduced output of

transformer is given to the diode ‘D’ and load resistor RL. The output voltage is measured across load

resistor RL.

The process of rectification is converting alternating current (AC) to direct current (DC).

Working of a Half wave rectifier : During the positive half-cycles of the input ac voltage i.e. when

upper end of the secondary winding is positive w.r.t. its lower end, the diode is forward biased and

therefore conducts current. If the forward resistance of the diode is assumed to be zero (in practice,

however, a small resistance exists) the input voltage during the positive half-cycles is directly applied

7 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

to the load resistance RL, making its upper end positive w.r.t. its lower end. The waveforms of the

output current and output voltage are of the same shape as that of the input ac voltage.

During the negative half cycles of the input ac voltage i.e. when the lower end of the secondary

winding is positive w.r.t. its upper end, the diode is reverse biased and so does not conduct. Thus

during the negative half cycles of the input ac voltage, the current through and voltage across the load

remains zero. The reverse current, being very small in magnitude, is neglected. Thus for the negative

half cycles no power is delivered to the load.

From the figure that the output is not a steady dc, but only a pulsating dc wave. To make the output

wave smooth and useful in a DC power supply, we have to use a filter across the load. Since only half-

cycles of the input wave are used, it is called a half wave rectifier.

Advantages and Disadvantages of Half wave rectifier:

A half wave rectifier is rarely used in practice. It is never preferred as the power supply of an audio

circuit because of the very high ripple factor. High ripple factor will result in noises in input audio

signal, which in turn will affect audio quality.

Advantage of a half wave rectifier is only that its cheap, simple and easy to construct. It is cheap

because of the low number of components involved.

Disadvantages of Half wave rectifier

1. The output current in the load contains, in addition to dc component, ac components of basic

frequency equal to that of the input voltage frequency. Ripple factor is high and filtering is, therefore,

required to give steady dc output.

2. The power output and, therefore, rectification efficiency is quite low. This is due to the fact that

power is delivered only during one half cycle of the input alternating voltage.

3. Transformer utilization factor is low.

4. DC saturation of transformer core resulting in magnetizing current and hysteresis losses and

generation of harmonics.

2. Derive the following values for Half Wave Rectifier .

The following parameters will be explained for the analysis of Half Wave Rectifier:-

1. Peak Inverse Voltage (PIV)

Peak Inverse Voltage (PIV) rating of a diode is important in its design stages. It is the maximum

voltage that the rectifying diode has to withstand, during the reverse biased period.

When the diode is reverse biased, during the negative half cycle, there will be no current flow through

the load resistor RL. Hence, there will be no voltage drop through the load resistance RL which causes

8 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

the entire input voltage to appear across the diode. Thus VSMAX, the peak secondary voltage, appears

across the diode. Therefore,

Peak Inverse Voltage (PIV) of half wave rectifier = VSMAX

2. Average and Peak Currents in the diode

By assuming that the voltage across the transformer secondary be sinusoidal of peak values VSMAX,

instantaneous value of the voltage given to the rectifier can be written as

Instantaneous value of voltage applied to Half Wave Rectifier Assuming that the diode has a

forward resistance of RF ohms and infinite reverse resistance value, the current flowing through

the output load resistance RL is

Current flowing through the diode IMAX = VSMAX / (RF + RL)

3. DC Output Current-- The dc output current is given as

Substituting the value of IMAX for the equation IMAX = VSMAX/(RF + RL), we have

Idc = VSMAX/ = VSMAX/ πRL if RL >> RF

4. DC Output Voltage

Vdc = Idc RL = VSMAX / π (RF + RL)X RL = VSMAX/{1+RF/RL } If RL >> RF, Vdc = VSMAX/π

5. Root Mean Square (RMS) Value of Current

9 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

6. Root Mean Square (RMS) Value of Output Voltage

RMS value of voltage across the load is given as

VLrms = Irms RL = VSMAX RL /2(RF + RL) If RL >> RF, VLrms = VSMAX/2

7. Rectification Efficiency

Rectification efficiency is defined as the ratio between the output power to the ac input power.

Efficiency, Ƞ = DC power delivered to the load / AC input power from the transformer = Pdc/Pac

DC power delivered to the load, Pdc = I2dc RL = (Imax / π)2 RL

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

resistance RL

= I2rms RF + I2

rms RL = {I2MAX/4}[ RF + RL]

So, Rectification Efficiency, Ƞ = Pdc/Pac = {4/π2}[ RL/ (RF + RL)] = 0.406 if RL >> RF,

The maximum efficiency that can be obtained by the half wave rectifier is 40.6%. This is obtained if RF

is neglected.

8. Ripple Factor

It is the ratio of the effective value of the ac components of voltage (or current) present in the output

from the rectifier to the dc component in output voltage (or current).

The effective value of the load current is given as

I2 =I2dc+I2

1+I22+I2

4 = I2dc +I2

ac

So, ripple factor, γ = Iac/ Idc = (I2 – I2dc)/ Idc = {( Irms/ Idc

2)-1} = (Kf2 – 1)

Where Kf is the form factor of the input voltage. For half wave rectifier, form factor is given as

Kf = Irms /Iavg = (Imax/2)/ (Imax/π) = π/2 = 1.57

So, ripple factor, γ = (1.572 – 1) = 1.21

6. Explain cetre tap full wave rectifier.

In the case of centre-tap full wave rectifier, only two diodes are used, and are connected to the opposite

ends of a centre-tapped secondary transformer as shown in the figure below. The centre-tap is usually

considered as the ground point or the zero voltage reference point.

10 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

As shown in the figure, an ac input is applied to the primary coils of the transformer. This input makes

the secondary ends P1 and P2 become positive and negative alternately. For the positive half of the ac

signal, the secondary point D1 is positive, GND point will have zero volt and P2 will be negative. At

this instant diode D1 will be forward biased and diode D2 will be reverse biased. As explained in the

Theory Behind P-N Junction and Characteristics of P-N Junction Diode, the diode D1 will conduct and

D2 will not conduct during the positive half cycle. Thus the current flow will be in the direction P1-D1-

C-A-B-GND. Thus, the positive half cycle appears across the load resistance RLOAD.

During the negative half cycle, the secondary ends P1 becomes negative and P2 becomes positive. At

this instant, the diode D1 will be negative and D2 will be positive with the zero reference point being

the ground, GND. Thus, the diode D2 will be forward biased and D1 will be reverse biased. The diode

D2 will conduct and D1 will not conduct during the negative half cycle. The current flow will be in the

direction P2-D2-C-A-B-GND.

11 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

When comparing the current flow in the positive and negative half cycles, we can conclude that the

direction of the current flow is the same (through load resistance RLOAD). When compared to the

Half-Wave Rectifier, both the half cycles are used to produce the corresponding output. The frequency

of the rectified output voltage is twice the input frequency. The output that is rectified, consists of a

dc component and a lot of ac components of minute amplitudes.

1. Peak Inverse Voltage (PIV) of Centre-Tap Full Wave Rectifier

PIV is the maximum possible voltage across a diode during its reverse biased period. Let us analyze the

PIV of the centre-tapped rectifier from the circuit diagram. During the first half or the positive half of th

input ac supply, the diode D1 is positive and thus conducts and provided no resistance at all. Thus, the

whole of voltage Vs developed in the upper-half of the ac supply is provided to the load resistance

RLOAD. Similar is the case of diode D2 for the lower half of the transformer secondary.

Therefore, PIV of D2 = Vm + Vm = 2Vm PIV of D1 = 2Vm

2. Output Current

3. DC Output Voltage

Vdc = Idc x RL = 2 Im RL / π

4. Root Mean Square (RMS) Value of Current

5. Root Mean Square (RMS) Value of Output Voltage

V rms = Vm / √2

6. Rectification Efficiency

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

Rectification efficiency is defined as the ratio between the output power to the ac input power.

Efficiency, Ƞ = DC power delivered to the load / AC input power from the transformer = Pdc/Pac

DC power delivered to the load, Pdc = I2dc RL = ( 2 Imax / π)2 RL

in load resistance RL

12 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

= I2rms RF + I2

rms RL = {I2MAX/2}[ RF + RL]

So, Rectification Efficiency, Ƞ = Pdc/Pac = {8/π2}[ RL/ (RF + RL)] = 0.812 if RL >> RF,

The maximum efficiency that can be obtained by the full wave rectifier is 81.2%. This is obtained if RF

is neglected.

7. Ripple Factor: Form factor of the rectified output voltage of a full wave rectifier is given as

So, ripple factor, γ = (1.112 – 1) = 0.482

7. Explain Full wave bridge rectifier.

In the circuit diagram, 4 diodes are arranged in the form of a bridge. The transformer secondary is

connected to two diametrically opposite points of the bridge at points A & C. The load resistance RL is

connected to bridge through points B and D.

13 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

During the positive half cycle

During first half cycle of the input voltage, the upper end of the transformer secondary winding is

positive with respect to the lower end. Thus during the first half cycle diodes D1 and D3 are forward

biased and current flows through arm AB, enters the load resistance RL, and returns back flowing

through arm DC. During this half of each input cycle, the diodes D2 and D4 are reverse biased and

current is not allowed to flow in arms AD and BC. The flow of current is indicated by solid arrows in

the figure above. We have developed another diagram below to help you understand the current flow

quickly. See the diagram below – the green arrows indicate beginning of current flow from source

(transformer secondary) to the load resistance. The red arrows indicate return path of current from load

resistance to the source, thus completing the circuit.

14 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

Flow of current in Bridge Rectifier

During the second half cycle

During second half cycle of the input voltage, the lower end of the transformer secondary winding is

positive with respect to the upper end. Thus diodes D2 and D4 become forward biased and current flows

through arm CB, enters the load resistance RL, and returns back to the source flowing through arm

DA. Flow of current has been shown by dotted arrows in the figure. Thus the direction of flow of

current through the load resistance RL remains the same during both half cycles of the input supply

voltage. See the diagram below – the green arrows indicate beginning of current flow from source

(transformer secondary) to the load resistance. The red arrows indicate return path of current from load

resistance to the source, thus completing the circuit.

Path of current in 2nd Half Cycle

1. Peak Inverse Voltage of a Full wave bridge rectifier:

Thus PIV of a bridge rectifier = Vmax (max of secondary voltage)

The different parameters are explained with equations below:

Peak Current --- Imax = Vsmax/(2RF + RL)

2. Output Current

3. DC Output Voltage

15 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

Vdc= Idc RL = 2Imax RL / π

4. Root Mean Square (RMS) Value of Current

5. Root Mean Square (RMS) Value of Output Voltage

Vrms= Irms x RL = Vm / √2

6. Rectification Efficiency

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

Rectification efficiency is defined as the ratio between the output power to the ac input power.

Efficiency, Ƞ = DC power delivered to the load / AC input power from the transformer = Pdc/Pac

DC power delivered to the load, Pdc = I2dc RL = ( 2 Imax / π)2 RL

in load resistance RL

= I2rms RF + I2

rms RL = {I2MAX/2}[ RF + RL]

So, Rectification Efficiency, Ƞ = Pdc/Pac = {8/π2}[ RL/ (RF + RL)] = 0.812 if RL >> RF,

The maximum efficiency that can be obtained by the full wave rectifier is 81.2%. This is obtained if RF

is neglected.

7. Ripple Factor : Form factor of the rectified output voltage of a full wave rectifier is given as

So, ripple factor, γ = (1.112 – 1) = 0.482

16 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

8. Explain the need of filter.

We have learnt in rectifier circuits about converting a sinusoidal ac voltage into its corresponding

pulsating dc. Apart from the dc component, this pulsating dc voltage will have unwanted ac

components like the components of its supply frequency along with its harmonics (together called

ripples). These ripples will be the highest for a single-phase half wave rectifier and will reduce further

for a single-phase full wave rectifier.. Such supply is not useful for driving complex electronic

circuits. For most applications the supply from a rectifier will make the operation of the circuit poor.

Thus, the output of the rectifier has to be passed though a filter circuit to filter the ac components.

The filter is a device that allows passing the dc component of the load and blocks the ac

component of the rectifier output. Thus the output of the filter circuit will be a steady dc voltage.

The filter circuit can be constructed by the combination of components like capacitors, resistors, and

inductors. Inductor is used for its property that it allows only dc components to pass and blocks ac

signals. Capacitor is used so as to block the dc and allows ac to pass.

9. Expalin Capacitor filter.

Filter Circuit – Most of the electronic circuits require a dc voltage that is constant, similar to the voltage

from a battery but the rectifiers cannot provide ripple free dc voltage. They provide a pulsating dc. The

circuit used for filtering or smoothing out the ac variations from the rectified voltage is called as `Filter

circuit’.

Types of Filters –

• 1) Capacitor input filter

• 2) Inductor filter

• 3) RC filter,

• 4) LC filter

17 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

In this type a capacitor C is connected in shunt with load resistor R. Capacitor acts as open circuit for

dc. Therefore direct current passes through the load. Capacitor provides low-reactance path to the ac

components of current. AC gets bypassed to the ground. Only a small part of the ac component passes

through the load producing a small ripple voltage.

When the rectifier output is increasing, the capacitor charges to the peak voltage Vp. When the rectifier

voltage becomes slightly less than Vp, the capacitor starts to discharge through the load (point B). This

prevents the load voltage from falling to zero. The capacitor discharges until the rectifier voltage

becomes more than the capacitor voltage (point C). The capacitor again charges to the peak value Vp.

The current is maintained through the load all the time.

For getting steadier output voltage, the time constant C*R should be large. Then load current is small

and capacitor does not discharge very much and the average load voltage Vdc, is slightly less than the

peak value Vp.

10. Half Wave Rectifier with Capacitor Filter.

We desire a DC power supply with constant output voltage. A constant output voltage from the DC

power supply is very important as it directly impacts the reliability of the electronic device we connect

to the power supply.

We can make the output of half wave rectifier smooth by using a filter (a capacitor filter or an inductor

filter) across the diode. In some cases an resistor-capacitor coupled filter (RC) is also used. The circuit

diagram below shows a half wave rectifier with capacitor filter.

The circuit diagram above shows a half-wave rectifier with a capacitor filter. The filter is applied across

the load RLoad. The output of the RLoad is VLoad, the current through it is ILoad. The current through

the capacitor is Ic.

18 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

During the positive half cycle of the input ac voltage, the diode D will be forward biased and thus starts

conducting. During this period, the capacitor ‘C’ starts charging to the maximum value of the supply

voltage Vsm. When the capacitor is fully charged, it holds the charge until the input ac supply to the

rectifier reaches the negative half cycle. As soon as the negative half supply is reached, the diode gets

reverse biased and thus stops conducting. During the non-conducting period, the capacitor ‘C’

discharges all the stored charges through the output load resistance RLoad. As the voltage across

RLoad and the voltage across the capacitor ‘C’ are the same (VLoad = Vc), they decrease exponentially

with a time constant (C*RLoad) along the curve of the non-conducting period. This is shown in the

graph below.

Half-wave Rectifier with Capacitor Filter - Waveform

The value of the discharge time constant (C*RLoad) being very large, the capacitor ‘C’ will not have

enough time to discharge properly. As soon as the capacitor starts discharging, the time becomes over.

Thus the value of RLoad at the discharge time will also be high and have just a little less value that the

output of RLoad. This is when the positive half cycle repeats again and the diode starts conducting. The

condition to be considered at this stage is that the rectified voltage takes value more than the capacitor

voltage . When the condition occurs the capacitor starts charging to a value of Vsm. The condition

again changes when the negative half cycle comes into pace, and the whole cycle is again repeated to

form the output waveform as shown above. The output shows a nearly constant dc voltage at the load

and that the output voltage is increased considerably.

Ripple factor ---The ripple factor for a Half-wave rectifier with C-filer is given by

1

r = -------------------------

2√3 f CRL

19 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

f-----the line frequency ( HZ)

C-----capacitance ( F)

RL------- Load resistance (Ω)

How to reduce Ripple factor ?

1. By increasing capacitance and resistance

2. Capacitor perform well for high load resistance

Advantages of C-Filter

• low cost, small size and good characteristics.

• It is preferred for small load currents ( upto 50 mA)

• It is commonly used in transistor radio, batteries eliminator etc.

11. The circuit diagram of a full-wave rectifier with capacitor filter is shown below.

The filter capacitor C is placed across the resistance load RLoad. The whole working is pretty much

similar to that of a half-wave rectifier with shunt capacitor explained above. The only difference is that

two pulses of current will charge the capacitor during alternate positive (D1) and negative (D2) half

cycles. Similarly capacitor C discharges twice through RLoad during one full cycle. This is shown in

the waveform below.

20 Mrs. V. S. Patki(Electronics Department) Walchand Institute of Technology, Solapur

Full-wave Rectifier with Capacitor Filter - Waveform

The load current reduces by a smaller amount before the next pulse is received as there are 2 current

pulses per cycle. This causes a good reduction in ripples and a further increase in the average dc load

current.

Ripple factor for full-wave rectifier with C-filter is given by

1

r = -------------------------

4√3 f C RL

12. Full Wave Bridge Rectifier with Capacitor Filter

Ripple factor for full-

wave rectifier with C-

filter is given by

1

r = -------------------------

4√3f C RL