CHAPTER 1 DIODE CIRCUITS - VTURESOURCE - Homevturesource.weebly.com/uploads/3/7/4/0/37406139/chapter1.pdf · CHAPTER 1 DIODE CIRCUITS ... Repeat the above problem for Approximate

1

CHAPTER 1

DIODE CIRCUITS

Resistance levels

Semiconductor act differently to DC and AC currents

There are three types of resistances

1. DC or static resistance

The application of DC voltage to a circuit containing a semiconductor diode will

result in an operating point on the characteristics curve that will not change with time. i.e.

For specified applied voltage the diode will have a specified current ‘Id’ and specified

resistance ‘RD’.

The resistance of the diode depends on applied voltage ‘VD’ and can be found by

The D.C resistance levels at the knee and below will be greater than the

resistance levels obtained for the vertical section of the charecterstics. The resistance

levels in the reverse bias region will naturally quite high. Also higher the current through

the diode the lower is the resistance level.

2. AC or dynamic resistance

The varying input will move the instantaneous operating point up and down a

region of characteristics.

2

Without AC signal applied to the diode the operating point ‘Q’ which is fixed not

moving hence it is called Quiescent point or ‘Q’ – point.

A straight line drawn tangent to the curve through the ‘Q’ point as shown in the

figure will define particular change in voltage and current that can be used to determine

AC or dynamic resistance for this region of diode characteristics.

The steeper the slope, the lower is the value of for the same change in ,

and there by lowering the value of resistance.

The A.C resistance in the vertical-rise region of the characteristics is quite small,

Whereas A.C resistance is much higher at low current levels. The lower the ‘Q’ point of

operation, the higher is the A.C resistance.

Mathematical expression of AC resistance

We have

( )

[ (

)]

( )

[

]

( )

3

Where

Hence

( )

Flipping the result we have

( )

Substituting n =1 and ‘VT = 26mV’

We have

Observation

The equation can be used directly by substituting ‘Q’ point diode current ‘ID’ to find the

AC resistance.

At lower values of ID, ( ) and the value of obtained must be multiplied

by factor of ‘2’. For small values of ‘ ’ below the knee of the curve, the equation becomes

inappropriate.

Every diode has some resistance due to resistance of semiconductor material which is

called body resistance and the resistance introduced by the connection between the

semiconductor material and external metallic conductor called contact resistance. These can be

included in the equation

i.e.

( )

Note:

4

Average AC resistance

If the input signal is sufficiently large to produce a broad swing as indicated in figure the

resistance associated with the device for this region is called average AC resistance.

The average AC resistance is determined by a straight line drawn between the two

intersections established by the maximum and minimum values of the input signal voltages.

|

5

Summary

Type Equation Special characteristics Graphical

determinations

DC or static

resistance

Defined as a point on

the characteristics

AC or

Dynamic

resistance

Defined by a tangent

line at the Q point

Average AC

|

Defined by a straight

line between limits of

operation

Diode Equivalent Circuits

An equivalent circuit is combination of elements properly chosen to best represent the

characteristics of a device in a particular operating region.

Piecewise linear equivalent circuit

The equivalent circuit for a diode can be obtained by approximating the characteristics

of the diode by straight line segment which is as shown in the figure

( )

( )

6

The barrier potential which is equal in magnitude to the cut in voltage is represented as a

battery of e.m.f ‘VK’

The resistance ‘rav’ offered by the forward bias diode is shown as an external resistance

in series with the ideal diode.

Simplified equivalent circuit

For most application the resistance ‘rav’ is quite small compared to the other elements in

the network. So removing ‘rav’ from the equivalent circuit the characteristics of the diode is as

shown in the figure, along with reduced equivalent circuit.

7

Ideal equivalent circuit

Let us ignore both resistance and cutoff voltage VK = 0.7V. In the case the equivalent

circuit will be reduced to that of an ideal diode as shown in figure

Transition and diffusion capacitance

Every electronic or electrical device is frequency sensitive. In the p-n semiconductor

diode, there are two capacitive effects to be considered

1. Transition capacitance

2. Diffusion capacitance

The basic equation for the capacitance of a parallel – plate capacitor is defined as

Where

Transition capacitance

In the reverse bias region there is a depletion region that behaves essentially like an

insulator between the layers of opposite charge. As the voltage is increased the depletion region

widens and hence its capacitances decreases.

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Diffusion capacitance

In forward bias region the width of the depletion region is much reduced resulting in the

increased levels of current that will result in increased level of diffusion current.

Reverse Recovery time

The diode conducts rapidly when forward biased and blocks conduction when reverse

biased. If the applied voltage is reversed to establish a reverse bias situation we would ideally

like to see the diode change instantaneously from the conduction state to non conduction state,

however the diode current will simply reverse as shown in figure and stay at this level for the

period of time ‘ts’ (storage time) required for minority carrier to move into other region, after

this reverse current decreases exponentially for a time tt (transition time).

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Load line analysis

Let us consider a diode ‘D’ in series with a load resistance ‘R’. The diode characteristics

are placed on the same set of axis as a straight line defined by the parameters of network.

The intersection on the vertical axis is defined by applied load ‘R’. Hence the analysis is

therefore called load line analysis

Applying KVL to the circuit

( )

Hence When VD = 0, we have

| ( )

When ID = 0, we have E ( )

A straight line drawn between the two points will define load line which is as shown in

the figure

Change the value of ‘R’ and the intersection on the vertical axis will change. The result

will be a different point of intersection between the load line and device characteristics.

10

The point of operation is called Quiescent point which gives current through diode for

given load and input voltage

By mathematically calculate we have from equation (1)

And

( )

Example

1. For the series diode configuration of fig (a) employing the diode characteristics of figure (b)

Determine I) II)

Fig. a

Fig.b

I.

|

11

From the resulting load line analysis we find

II.

( )( )

2. Repeat the above problem for Approximate equivalent model

we have

The level of remains same

12

3. Repeat the above problem for for ideal diode model

We have

Rectification

Rectification is the process of converting alternating current into direct current.

Rectifier

Rectifier is device that converts AC current into DC current (pulsating DC)

Half wave rectifier

The following figure shows the half wave rectifier circuit. An alternating input voltage is

applied to the diode connected in series with load resistance ‘R’

During the interval t = 0 T/2

The polarity of ‘Vi’ makes the diode to turn ‘ON’ this provides short – circuit

equivalence for the ideal diode. The output signal is exact replica of the applied signal which

is as shown in figure

13

During the interval t = T/2 T

The polarity of ‘Vi’ makes the diode to turn off. This provides open – circuit

equivalence for the ideal diode. The result is absence of path which is as shown in the figure

When we sketch input ‘Vi’ and the outputs together then we get

The D.C output waveform is expected tobe a straight line but the half wave rectifier gives output in the form of positive sinusoidal pulses hence output is called pulsating D.C

The D.C output waveform is expected to be a straight line but the halfwave rectifier

gives output in the form of positive sinusoidal pulses. The average value of D.C voltage is given

by

14

Since we are using silicon diode we have a cut-in voltage ‘VK’ of 0.7V. Hence the diode

will be in off state when ‘Vi’ is less than 0.7V hence output will be zero which is as shown in

figure. The net effect is reduction in area which reduces the average DC voltage hence

( ) ( )

Note:

One complete cycle 0 to 2Π. To find he average value of alternating waveform, we have

to determine the area under the curve one complete cycle and then dividing it by the base i.e 2 Π

Im peak value of load current

∫ ( )

∫ ( )

2Π Π current flows only in positive half cycle

[ ( ) ]

[ ]

15

Average D.C load voltage

( )

Neglecting the internal resistance we have

Example

1. Sketch the output ‘Vo’ and determine the DC level of the output for the network of

figure (a)

2. Repeat part (a) if the ideal diode is replaced by a silicon diode

3. Repeat part (a) and (b) if ‘Vm’ is increased to 200V and compare solutions. Using

equation (1) and (2).

Solution

1. The diode will conduct for negative half cycle hence the output is as shonwn in figure

16

( )

.

2. For silicon diode

( )

( )

3. ( )

( )

( )

( )

Peak inverse voltage (PIV) or Peak reverse voltage (PRV)

The peak inverse voltage rating of the diode is of primary importance in the design of

rectification system. It is the voltage rating that must not be exceeded in the reverse bias region.

The required PIV rating can be determined from the following figure.

Full wave reactivation

The following figure shows the full-wave bridge rectifier circuit with four diodes in a

bridge configuration. An alternating in out is applied to diodes with a load resistance ‘R’.

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The polarity of input ‘Vi’ makes the diodes ‘D2’ and ‘D3’ to turn ‘ON’ whereas ‘D1’ and

‘D3’ to turn off which is as shown in figure. Since the diodes are ideal, the load voltage is Vo =

Vi.


The polarity of input ‘Vi’ makes the diodes ‘D1’ and ‘D4’ to turn ‘ON’ where as ‘D2’ and

‘D3’ to turn off which is as shown in figure. Since the diodes are ideal the polarity across the

load resistor ‘R’ is same as shown in figure (a) thus establishing a second positive pulse as

shown in figure (b)

When we sketch input ‘Vi’ and the outputs together we get

18

The D.C output waveform is expected to be a straight line but the full wave rectifier

gives output in the form of positive sinusoidal pulses. The average value of D.C voltage is given

by

( )

If we use silicon diode as shown in the figure the application of KVL around the

conduction path result in

The peak output voltage is therefore

The DC voltage of silicon diode

( )


The required PIV rating can be determined from the following figure

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Center tapped transformer

The following figure shows the full wave rectifier circuit with only two diodes, but

requiring a center tapped transformer to establish the input signal across each section of the

secondary of the transformer.


The positive portion of ‘Vi’ applied to primary of the transformer. The polarity of ‘Vi’

makes the diode ‘D1’ to turn ‘ON’ where as ‘D2’ to turn off which is as shown in figure


The negative portion of ‘Vi’ applied to primary of the transformer. The polarity of ‘Vi’

makes the diode ‘D2’ to turn ‘ON’ where as ‘D1’ to turn off which is as shown in the figure

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When we sketch input ‘Vi’ and the outputs together we get


The required PIV rating can be determined from the following figure

Example

Determine the output waveform for the network of figure and calculate the output DC level and

the required PIV of each diode.

Solution

The network appears as shown in the figure below for the positive region of the input

voltage

21

Redrawing the above circuit we have

Applying voltage divider rule we have

Hence

( )

For the negative part of the input the roles of the diodes are interchanged and Vo appears

as shown below

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Clippers

The circuits which are used to clip off unwanted portion of the waveform by making use

of diodes without disturbing the remaining part of the waveform are called clippers.

The half wave rectifier is an example of the simplest form of diode clippers which

contains one resistor and one diode. Clippers are also called limiters or slicers.

Depending on the orientation of the diode the positive or the negative region of the

applied signal is clipped off.

Clippers are classified into two types

1) Series clippers

2) Parallel clipper

1. Series clipper:

In series clipper the diode is connected in series with the load as shown in figure

Figure (a) shows a negative series clipper circuit, because it passes only the positive half

cycle of an alternating waveform and clips off the other half cycle. A diode series clipper is

simply a half wave rectifier circuit

Output:

Figure (b) shows a positive series clipper circuit because it passes only the negative half

cycle of an alternating waveform and clips off the positive half cycle.

Output:

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The zero level output from a series clipper circuit is not exactly zero. The reverse

saturation current (IR) of the diode produce a voltage drop across resistor. This is almost so

small that it can be neglected.

2. Shunt clippers:

In shunt clipper the diode is connected in parallel with the load as shown in figure

The figure shows positive shunt clipper circuit because it passes only the negative half

cycle of an waveform and clips off the other half cycle.

The figure above shows negative shunt clipper circuit because it passes only the positive

half cycle of an waveform and clips off the other half cycle

Examples

Series clipper

1) Positive clipper or to pass negative peak above Vk level

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During Positive half cycle

Anode is at ground potential.

Cathode sees variable positive input voltage from o to +Vm.

For complete positive half cycle diode becomes reverse biased and hence Vout = 0V.

The circuit and the corresponding output waveform is as shown below

During negative half cycle


Cathode sees variable negative input voltage from o to -Vm.

For complete negative half cycle diode becomes forward biased and hence the output

is given as Vout = -Vm + Vk.


Hence applying K.V.L to the loop we have

If ‘Vin = 10Vp – p’ and VK = 0.7V then

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2) To pass negative peak above some level say (VR + Vk)



Cathode sees variable positive input voltage from VR to Vm + VR.

For complete positive half cycle diode becomes reverse biased and hence Vout = 0V.




Cathode sees variable negative input voltage from VR to -Vm + VR.

When the diode is reverse biased and ‘Vo = 0V’

When the diode is forward biased.


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If ‘Vin = 10Vp – p’, ‘VK = 0.7V’ and ‘VR = 2.3V’ then

3) To pass +ve peak above ‘VK’ level


Cathode is at ground potential.

Anode sees variable positive input voltage from 0 to +Vm.

For complete positive half cycle diode becomes forward biased.


If ‘Vin = 10Vp – p’, ‘VK = 0.7V’

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Anode sees variable positive input voltage from 0 to -Vm.

For complete negative half cycle diode becomes reverse biased and hence Vout = 0V.


4) To pass positive peak above some level (VR + Vk)



Anode sees variable positive input voltage from -VR to +Vm – VR

When the diode is reverse biased and ‘Vo = 0V’

When the diode is forward biased.


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If ‘Vin = 10Vp – p’, ‘VK = 0.7V’ and ‘VR = 2.3V’ then



Anode sees variable positive input voltage from –VR to – Vm – VR

For complete negative half cycle diode becomes reverse biased and hence Vout = 0V.


Write the output wave form for the following circuit

A) Series Clippers

1)

29

2)

3)

4)

30

5)

6)

7)

31

8)

9)

10)

32

11)

12)

13)

33

14)

B) Shunt Clippers

1)

2)

34

3)

4)

5)

35

6)

7)

8)

36

9)

10)

37

1) Sketch the output waveform ‘Vo’ to the time scale

Solution

Case: 1 for

Diodes D1 and D2 are off therefore Vo = Vi.

Case: 2 When

Diode ‘D1’ conducts and diode ‘D2’ is open circuited, Which is as shown in the figure

below

Applying K.V.L to loops we have

( )( )

38

Case: 3 When

Diodes ‘D1’ and ‘D2’ are off therefore Vo = Vi.

Case: 4 When

Diode ‘D2’ conducts and hence

Hence the input and output waveform along with transfer characteristics is as shown

2) The input voltage Vi to the two level clipper circuit as shown in the figure caries linearly

from 0 – 150V. Sketch the output voltage to time scale. Assume the diodes are ideal.

39

Solution:

Case: 1 for

Diode ‘D1’ and ‘D2’ are off. The circuit is as sown in figure, Hence

Case: 2 for 25V

Diodes ‘D1’ is off and ‘D2’ is on. The circuit is as shown below

Applying K.V.L to the loop we have

( )

( )

( ) ( )

( )

Substituting (1) in and (2) we have

[

( )]

( )

40

Thus for 25V , From equation (3) we have

At , ( )

At , ( )

Case: 3 for 100V

Both the diodes becomes on. The circuit is as shown below

Apply K.V.L to the above circuit

( )

( )

( )

( )( )

Hence the input and output waveform along with transfer characteristics is as shown

41

Clamping circuit

Clampers are networks or circuits that changes the input signal to a different DC level

but the peak to peak swing of applied signal will remain the same.

It adds a DC voltage to the A.C signal hence it is also called as DC restorer DC inserter

circuits. The capacitor diode and resistance are the three basic elements of a clamper circuit

The clamper circuit relies on a change in the capacitors time constant current path with

changing input voltage. The magnitude of R and C are chosen so that T = RC is large enough

that the voltage across the capacitor does not discharge significantly during the diodes non

conduction state

Design:

If f = 1KHz

T = 1ms

( ) ( )

( )

Let C = 1μF ( )

1. Positive peak clamped at reference level

Capacitor charges only when diode conducts in above circuit diode gets conducted

during positive half cycle.

When the input voltage is +Ve (Vi > 0) the diode is ‘ON’ and capacitor charges to peak

value of input signal as shown in figure

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Apply KV to 1st loop

We know that

,

Applying K.V.L to 2nd

loop

When the input voltage is –ve (Vi < 0) the diode is ‘OFF’ as shown in figure

Applying KVL

Input /output waveform is as shown below

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2) Positive peak clamped at +ve reference (+2V)

When diode – ON and capacitor starts charging at a very low time constant

which is as shown in the figure.

Applying KVL to the circuit shown in figure

Applying KVL to the second loop

Hence

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3) Positive peak clamped at –ve reference level say (-2V)

45

When diode – ON and capacitor starts charging at a very low time constant

Which is as shown in the figure.


( )

Applying KVL to the second loop

( )

Hence



46


4) Negative peak clamped level

Diode starts conducting during negative half cycle. During this capacitor starts charging

to the peak value of the input voltage which is as shown in the figure


47

Applying K.V.L to 2nd

loop

When the input voltage is +ve (Vi > 0) the diode is ‘OFF’ as shown in figure


( )


5) Negative peak clamped to +ve reference level (say +2v)

48

When input voltage is –ve ( ) the diode is ON applying KVL to the circuit as

shown in figure


Applying KVL to second loop

Hence

When input is +ve ( >0V) the diode is off applying KVL to circuit in figure


49

Input /output waveform is as shown

6) Negative peak clamped at –ve reference level (-2V)

When the input voltage is negative (Vin < 0) the diode is ON which is as shown in the figure


50

Applying KVL to second loop

Hence

When input is +ve ( >0V) the diode is off applying KVL to circuit in figure



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1) Determine ‘Vo’ for the network as shown in the figure below (ideal diode)

Solution

Case1: During the interval

Diode is ON capacitor charges to peak value of the input signal which is ass shown in

the figure

Applying K.V.L to the loop1



Applying K.V.L to the loop

( )

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CHAPTER 1 DIODE CIRCUITS - VTURESOURCE - Homevturesource.weebly.com/uploads/3/7/4/0/37406139/chapter1.pdf · CHAPTER 1 DIODE CIRCUITS ... Repeat the above problem for Approximate