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.
1
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
( )
[ (
( )
[
)]
]
( )
2
Where
Hence
( )
Flipping the result we have
Note:
(
)
(
)
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.
3
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.
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4
Summary
Type
Equation
Special characteristics
DC or static
resistance
Defined as a point on
the characteristics
AC or
Dynamic
resistance
Defined by a tangent
line at the Q point
|
Graphical
determinations
Defined by a straight
line between limits of
operation
Average AC
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
(
)
(
)
5
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.
6
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.
7
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).
8
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.
9
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.
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10
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
11
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
12
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 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
13
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
[
( )
[
]
]
14
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
15
(
)
.
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’.
16
During the interval t = 0
T/2
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.
During the interval t = T/2
T
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
17
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
(
)
Peak inverse voltage (PIV) or Peak reverse voltage (PRV)
The required PIV rating can be determined from the following figure
18
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.
During the interval t = 0
T/2
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
During the interval t = T/2
T
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
19
When we sketch input ‘Vi’ and the outputs together we get
Peak inverse voltage (PIV) or Peak reverse voltage (PRV)
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
20
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
21
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:
22
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
 Anode is at ground potential.
 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.
 The circuit and the corresponding output waveform is as shown below
Hence applying K.V.L to the loop we have
If ‘Vin = 10Vp – p’ and VK = 0.7V then
24
2) To pass negative peak above some level say (VR + Vk)
During Positive half cycle
 Anode is at ground potential.
 Cathode sees variable positive input voltage from VR to Vm + VR.
 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
 Anode is at ground potential.
 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.
 The circuit and the corresponding output waveform is as shown below
25
Hence applying K.V.L to the loop we have
If ‘Vin = 10Vp – p’, ‘VK = 0.7V’ and ‘VR = 2.3V’ then
3) To pass +ve peak above ‘VK’ level
During Positive half cycle
 Cathode is at ground potential.
 Anode sees variable positive input voltage from 0 to +Vm.
 For complete positive half cycle diode becomes forward biased.
 The circuit and the corresponding output waveform is as shown below
If ‘Vin = 10Vp – p’, ‘VK = 0.7V’
26
During negative half cycle
 Cathode is at ground potential.
 Anode sees variable positive input voltage from 0 to -Vm.
 For complete negative half cycle diode becomes reverse biased and hence Vout = 0V.
 The circuit and the corresponding output waveform is as shown below
4) To pass positive peak above some level (VR + Vk)
During Positive half cycle
 Cathode is at ground potential.
 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.
 The circuit and the corresponding output waveform is as shown below
27
 Hence applying K.V.L to the loop we have
If ‘Vin = 10Vp – p’, ‘VK = 0.7V’ and ‘VR = 2.3V’ then
During negative half cycle
 Cathode is at ground potential.
 Anode sees variable positive input voltage from –VR to – Vm – VR
 For complete negative half cycle diode becomes reverse biased and hence Vout = 0V.
 The circuit and the corresponding output waveform is as shown below
Write the output wave form for the following circuit
A) Series Clippers
1)
28
2)
3)
4)
29
5)
6)
7)
30
8)
9)
10)
31
11)
12)
13)
32
14)
B) Shunt Clippers
1)
2)
33
3)
4)
5)
34
6)
7)
8)
35
9)
10)
36
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
(
)(
)
37
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.
38
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
[
(
)
]
( )
39
 Thus for 25V
At
At
, From equation (3) we have
(
,
,
)
(
)
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
40
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
41
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
42
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
43
 When the input voltage is –ve (Vi < 0) the diode is ‘OFF’ as shown in figure
Applying KVL to the circuit shown in figure
Input /output waveform is as shown below
3) Positive peak clamped at –ve reference level say (-2V)
44
 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
 When the input voltage is –ve (Vi < 0) the diode is ‘OFF’ as shown in figure
Applying KVL to the circuit shown in figure
45
Input /output waveform is as shown below
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
Apply KV to 1st loop
46
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 to the circuit shown in figure
(
)
Input /output waveform is as shown below
5) Negative peak clamped to +ve reference level (say +2v)
47
 When input voltage is –ve (
) the diode is ON applying KVL to the circuit as
shown in figure
Apply KV to 1st loop
Applying KVL to second loop
Hence
 When input is +ve ( >0V) the diode is off applying KVL to circuit in figure
Applying KVL to the circuit shown in figure
48
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
Apply KV to 1st loop
49
Applying KVL to second loop
Hence
 When input is +ve ( >0V) the diode is off applying KVL to circuit in figure
Applying KVL to the circuit shown in figure
Input /output waveform is as shown
50
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 loop2
Case2: During the interval
Applying K.V.L to the loop
(
)
51
Case2: During the interval
Applying K.V.L to the loop1
Applying K.V.L to the loop2
Input /output waveform is as shown
*******END*******
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