Chapter
C
apte 2
DIODE
Part ７
Sinusoidal Inputs:
Half-Wave Rectification
Chapter 2 DIODE, Part 7 Half-Wave Rectification
1
AC INPUTS (SINUSOIDAL INPUTS) ; HALF-WAVE RECTIFICATION
The diode analysis to include time-varying functions such as Sinusoidal and Square waveform
Example of sinusoidal input
rectifier
T:p
period ((one full cycle)
y )
Vm : amplitude
t : time
vi : input
vo : output
t t
Half-wave rectifier
Chapter 2 DIODE, Part 7 Half-Wave Rectification
2
t = 0 ÆT/2
The polarity of the applied voltage vi is such as to “establish” pressure in the direction
indicated and “turn on” the diode
diode.
short circuit:Ideal diode
Conduction region(0 ÆT/2)
t = T/2 ÆT
Th polarity
The
l it off th
the applied
li d voltage
lt
vi is
i such
h as tto “t
“turn off”
ff” the
th di
diode.
d
open circuit:Ideal diode
Nonconduction region(T/2 ÆT)
v0 = iR = (0)R = 0 V
Chapter 2 DIODE, Part 7 Half-Wave Rectification
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Sketch of input,vi and output,vo for half-wave rectifier
Ideal Diode ((no effect of Si diode))
The output signal v0 has a net +ve
area above the axis over the a full
period and an average value determined
by
Eq.(1)
The process of removing one-half the
input signal to establish a dc level is
called half-wave rectification.
Chapter 2 DIODE, Part 7 Half-Wave Rectification
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Effect of Si Diode
Si Diode
The applied signal must have at least 0.7V (VT) before the diode can “turn on”.
Nonconducting region: For vi less than 0
0.7V,
7V the diode is in an “open
open circuit
circuit” state and v0 =0V
0V.
Conducting region: v0 = vi - VT (VT =0.7V).
The net effect is a reduction in area above the axis, which naturally reduces the resulting
d voltage
dc
l
llevel.l
an average value (Vm >> VT)
Eq. (2)
If Vm is sufficiently larger than VT, eq.(1) is often applied as approximation for Vdc.
Chapter 2 DIODE, Part 7 Half-Wave Rectification
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Example (1):
(a) Sketch the output v0 and determine the dc level of the output. Use ideal diode model.
Solution: In this situation the diode will conduct during the –ve part of input.
- 6.36 V
Vdc
0 318 Vm = -0
0.318
318 (20V) = - 6.36
6 36 V
d = - 0.318
The –ve sign indicates that the polarity of the output is
opposite to the defined polarity.
Chapter 2 DIODE, Part 7 Half-Wave Rectification
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(b) Replace the ideal diode with a silicon diode.
Solution
Vdc = - 0.318 (Vm - VT )= - 0.318 (20 – 0.7 )= -0.318 (19.3 V) = - 6.14 V
The resulting drop in dc level
is 0.22 V (6.36 - 6.14 V)
or about 3.5%.
(c) Vm is increased to 200 V and compare solutions using Eq.(1) and Eq.(2)
Solution
E (1) Vdc = - 0.318
Eq.(1)
0 318 Vm = - 0.318
0 318 (200 )=
) - 63.6
63 6 V
Eq.(2) Vdc = - 0.318 (Vm - VT )= - 0.318 (200 – 0.7 )= -0.318 (199.3 V) = - 63.38 V
The difference is very small that can be ignored.
Chapter 2 DIODE, Part 7 Half-Wave Rectification
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Peak Inverse Voltage (PIV) : maximum reverse-bias before entering Zener avalanche region.
Requirement for the diode to behave as rectifier
Vm ≤ PIV
If Vm ≥ PIV , there will a short circuited diode
or the diode will enter zener avalanche region.
This will make the diode to loose its function
as rectifier.
Chapter 2 DIODE, Part 7 Half-Wave Rectification
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AC INPUTS (SINUSOIDAL INPUTS) ; FULL-WAVE RECTIFICATION
4 diodes
t = 0 Æ T/2
Full-Wave Bridge rectifier
(a) Positive Region
D2 & D3 : ON ; D1 & D4 : OFF
Ideal diode
Polarityy of R
Conduction path for +ve region of vi
Chapter 2 DIODE, Part 7 Full-Wave Rectification
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(b) Negative Region
Polarity of R
Ideal diode
Conduction path for -ve region of vi
D2 & D3 : OFF ; D1 & D4 : ON
The important result is that the polarity across the load resistor R is the same as
+ve region,
g
establishing
g a second p
positive p
pulse.
Chapter 2 DIODE, Part 7 Full-Wave Rectification
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Input and output of full wave rectifier for one full cycle.
Since th
Si
the area above
b
th
the axis
i ffor one ffullll cycle
l iis now ttwice,
i
th
the d
dc llevell h
has also
l b
been
doubled.
(1))
Effect of Si diode
KVL of conduction path
Conduction path
Chapter 2 DIODE, Part 7 Full-Wave Rectification
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Peak Inverse Voltage (PIV) : maximum reverse-bias before entering Zener avalanche region.
Requirement for the diode to behave as rectifier
Vm ≤ PIV
If Vm ≥ PIV , there will a short circuited diode
or the diode will enter zener avalanche region.
This will make the diode to loose its function
as rectifier.
Chapter 2 DIODE, Part 7 Full-Wave Rectification
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CLIPPERS
Clipper has the ability to “clip”
clip off a portion of the input signal without distorting the
remaining part of the alternating waveform. Half-wave rectifier is an example of the
simplest form of diode clipper ; one resistor and one diode.
Depending on the orientation of the diode, the +ve and –ve region of the input signal is
“clipped” off.
S i clipper
Series
li
Example of clipped input
Chapter 2 DIODE, Part 7 Clipper
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Procedure for analyzing Networks
Additional dc supply
(1)
Make a mental sketch of the response of the network based on the direction of the
diode and the applied voltage levels
The direction of the diode suggests that the signal vi must be +ve to turn it on.
The voltage vi must be greater than V volts to turn the diode “ON”.
The –ve
ve region of the input signal is “pressuring”
pressuring the diode into the “OFF”
OFF state supported
further by dc supply.
Here, we can be quite sure that the diode is an open circuit for –ve region of the input signal.
Chapter 2 DIODE, Part 7 Clipper
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(2) Determine the applied voltage (transition voltage) that will cause a change in state
for the diode.
For the ideal diode the transition between the states will occur at the point of
vd = 0 V and id = 0 A.
The level of vi that will cause a
transition in state is
vi = V
An input greater than V volts the diode is in
the short
short-circuit
circuit state
state.
An input less than V volts the diode is in
the open-circuit state
(3) Be continually aware of the defined terminals and polarity of v0
Example: Short-circuit state
The output v0 can be determined by KVL in
clockwise direction
Chapter 2 DIODE, Part 7 Clipper
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(4) It can be helpful to sketch the input signal above the output and determine the output
at instantaneous values of the input.
Keep in mind that at an instantaneous value of vi the input
can be treated as a dc supply of that value and the
corresponding dc value (the instantaneous value) of the output
determined.
Example: vi = Vm
Vm > V ; the diode is in short
short-circuit
circuit, then v0 = Vm – V.
V
vi = V ; the diode change state
vi = - Vm, v0 = 0 V
Determining v0 when vi = Vm
Sketching v0
Chapter 2 DIODE, Part 7 Clipper
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Example (2): Determine the output waveform.
Series clipper
The diode will be in the “ON” state for
the positive region of vi. Note here
that V=5V also will aid to this effect.
vo = vi + 5V
Substituting ideal diode.
The transition level (vd=0 & id=0)
vo = vi + 5V
Here, vo = 0V
vi = -5 V
For vi more –ve than – 5V the diode
will enter its open-circuit state,
For vi more +ve than – 5V the diode
is in the short-circuit state.
Chapter 2 DIODE, Part 7 Clipper
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Example (3): Determine the output waveform.
Circuit with square-wave input is easier to analyze than with
sinusoidal wave input.
(a) vi = 20V (0ÆT/2)
The diode is in shortcircuit state.
v0 = 20V + 5V = 25V
(b) vi = -10V (T/2 ÆT)
The di
Th
diode
d iis iin opencircuit state.
v0 = iRR = ((0)R
) = 0V
The clipper not only clipped
off the input signal but raised
h d
dc llevell off the
h signal
i
lb
by
the
5V.
Output Voltage
g
Chapter 2 DIODE, Part 7 Clipper
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Parallel clipper
Example of response to a parallel clipper
Chapter 2 DIODE, Part 7 Clipper
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Example (4): Determine the output waveform.
The polarity of the dc supply and the
Direction of the diode strongly suggest
That the diode will be in “ON” state for
-ve region of the input signal.
Use ideal diode (vd=0 & id=0)
The transition level is
vo = V = 4V
vi = V = 4V
Since the dc supply is “pressuring” the
diode to stay in short-circuit state,
th iinputt voltage
the
lt
mustt be
b greater
t than
th
4V for the diode to be in the “OFF”
state.
Anyy input
p less than 4V will
result in a short-circuited diode.
Short-circuit
Short
circuit
Chapter 2 DIODE, Part 7 Clipper
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Determining v0 for open state.
v0 = vi
open-circuit
Input & Output
Voltage
Chapter 2 DIODE, Part 7 Clipper
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Effect of Si diode (VT = 0.7V) in parallel clipper
The transition voltage is determined at id = 0A & vd = 0.7V
KVL in the clockwise direction
For input greater than 3.3V the diode will be an open
circuit. v0 = vi
For input less than 3.3V the diode will be in the
“ON” state.
t t
The diode in the “ON” state
Chapter 2 DIODE, Part 7 Clipper
The only effect
off was to drop
d
the
h
transition level to
3.3 V from 4V.
22
SUMMARY : CLIPPER
Chapter 2 DIODE, Part 7 Clipper
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Chapter 2 DIODE, Part 7 Clipper
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Center-Tapped Transformer
This full-wave rectifier appears with two diodes but requiring a center-tapped (CT)
transformer to establish the input signal across each section of the secondary of the
transformer.
(a)
Consider the +ve portion of vi applied to primary of the transformer.
D1 assumes the short-circuit equivalent and D2 the open-circuit equivalent, as
determined by the secondary voltages and the resulting current directions.
Chapter 2 DIODE, Part 7 Full-Wave Rectification,CT
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(b) Consider the –ve portion of the input applied to primary of the transformer.
Now, D1 assumes the open-circuit equivalent and D2 the short-circuit equivalent, as
determined by the secondary voltages and the resulting current directions
directions.
The voltage across the load resistor R maintain to the same polarity. The net effect is
the same output with the same dc levels as appearing in 4 diode bridge full-wave rectifier.
(c) PIV
Chapter 2 DIODE, Part 7 Full-Wave Rectification,CT
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Example (5): Determine the output waveform and calculate the output dc level and the
required PIV of each diode.
Solution
(a) +ve region of the input voltage.
Redrawing the circuit.
Here, v0 = ½ vi or vomax= ½ vimax =1/2(10) = 5 V.
Chapter 2 DIODE, Part 7 Full-Wave Rectification
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(b) -ve region of the input voltage.
Th roles
The
l off th
the di
diodes
d will
ill b
be iinterchanged.
t h
d
Output waveform for +ve and –ve region of input
(c) -ve region of the input voltage.
Vdc = 0.636 ((5 V)) = 3.18 V
The effect of removing two diodes from the bridge configuration was therefore to reduce
the available dc level.
Chapter 2 DIODE, Part 7 Full-Wave Rectification
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CLAMPERS
The clamping circuit is one that will “clamp” a signal to a different dc level.
The circuit must have a capacitor
capacitor, a diode and a resistive element and it can also
have an independent dc supply to introduce an additional shift.
(a) 0 Æ T/2
v0 = 0 V
(b) T/2 Æ T
KVL
Chapter 2 DIODE, Part 7 Clampers
The resulting output
waveform.
29
The output
p signal
g
is clamped
p to 0V for the interval
0 to T/2 but maintains the same total swing (2V) as
the input.
For the clamping circuit
circuit, the total swing of the output
Is equal to the total swing of the input signal.
The resulting output
waveform.
waveform
Chapter 2 DIODE, Part 7 Clampers
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Example (6): Determine vo for the input indicated.
C=0.1μF
Solution:
Frequency is 1000Hz, resulting in a period of 1 ms (T
(T=1
1 / f ) and an interval of 0.5ms
between levels.
KVL around the input loop
t1Æt2 :the diode is in its
short circuit state
state.
The output,v0 is across R
but it is also directly across
5V battery.
y The result is
v0 = 5V.
The capacitor will charge
up to 25V
charging circuit
Chapter 2 DIODE, Part 7 Clampers
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t2Æt3 :the diode is in its open
circuit state.
The open circuit equivalent will
remove the 5 V battery from
having any effect on v0 .
KVL around the outside loop
discharging circuit
The time constant of the discharging
For all practical purposes, the capacitor will fully charge and discharge in five time
constants.
The total discharge time is
5 τ = 5(10 ms) = 50 ms
Chapter 2 DIODE, Part 7 Clampers
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The resulting output waveform
Note that the output swing of 30V matches the input swing.
Chapter 2 DIODE, Part 7 Clampers
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Example (7): Repeat example (6) using silicon diode with VT = 0.7V.
KVL in the output section
(a) Short Circuit state (t2Æt3)
KVL in the input section
(b) Open Circuit state (t2Æt3)
The resulting output waveform
Chapter 2 DIODE, Part 7 Clampers
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Example of Clamping Circuit with square wave input (Considering ideal diode)
Chapter 2 DIODE, Part 7 Clampers
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Example of Clamping Circuit with sinusoidal wave input (Considering ideal diode)
Chapter 2 DIODE, Part 7 Clampers
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