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Description
This article will show the example of diode application such as diode work as rectifiers. Rectification will cover half-wave and full-wave rectifier. After that, this article also show the diode work as clipper and clamper. Lastly, diode work as regulator.
Document Sample
Analogue Electronics I
EAB 2014
Jan 2009
Lecture 2 - Diode
Diode Applications
1
Topics To Cover
Applications of Diode
• Half-Wave and Full-Wave Rectifier
• Clipper
• Clamper
• Regulator (covered in Lecture 1)
• Refer to Chapter 2, Boylestad
2
Sinusoidal Inputs: Half-Wave
Rectification
• Used in ac-to-dc conversion circuit
• The full wave sine wave will be rectified half, becomes a
½ wave rectification
• Also known as Rectifier Diode
with Higher Power & Current Rating
vi
vm
vi R vo
t
Input sine wave ½ wave rectifier circuit
3
Sinusoidal Inputs: Half-Wave
Rectification
Assume Ideal Diode Model:
During positive cycle
vi
vm s/c
vi R vo=vi
t
vo
vm
• Diode Forward-Bias
• Diode “ON”.
• Short-circuit.
t
Vo = Vi
4
Sinusoidal Inputs: Half-Wave
Rectification
Assume Ideal Diode Model:
During negative cycle
vi
- o/c
vi R vo=0V
t
+
vo
vm vm
• Diode Reversed-Bias
• Diode “OFF”.
• Open-circuit. t
vo = 0V
5
Sinusoidal Inputs: Half-Wave
Rectification
v i
vm
Ideal Diode
Model
t
-vm
vo
vm
Vdc = 0.318Vm vdc
t
6
Sinusoidal Inputs: Half-Wave
Rectification
vi
vm
Simplified
Model
vK t
R vo
vi
-vm
vo
Vdc = 0.318 (Vm − VK )
Vm-VT
K
T
vdc
t
7
Full-Wave Rectification
• Used in ac-to-dc conversion circuit
• Improve 100% of the dc level obtained
• 2 common configurations: Bridge & Centre-Tapped Transformer
Bridge Network
vi D1 D2
vm
vo
vi
t R
T/2 T D3 D4
8
Full-Wave Rectification
Assume Ideal Diode Model:
During positive cycle ( t = 0 T/2 )
vi D1 "Off" D2 "On"
vm
vo
vi
t R
T/2 D3 "On" D4 "Off"
• Diode D2 and D3 “On”
• Diode D1 and D4 “Off”
vo = vi ( Ideal Diode) Bridge Network
9
Full-Wave Rectification
Assume Ideal Diode Model:
During negative cycle ( t = T/2 T)
vi D1 "On" D2 "Off"
vo
vi
t R
T/2 T D3 "Off" D4 "On"
-vm
• Diode D1 and D4 “On”
• Diode D2 and D3 “Off”
vo = -vi ( Ideal Diode) Bridge Network
10
Full-Wave Rectification
vi vo
vm vm
vdc
• For the Full-wave rectifier, the dc level is doubled
Vdc = 2(0.318Vm ) = 0.636Vm
• If we use Silicon diode (not ideal diode), Bridge Network
vi − VK − vo − VK = 0
T T
vo = vi − 2VT
K and Vdc = 0.636(Vm − 2VT )
K
11
Full-Wave Rectification
Center-Tapped Transformer
• Use two diodes with a center-tapped transformer
1:2
vi
vm vi
R
vi
t vo
vi
12
Full-Wave Rectification
Assume Ideal Diode Model: Center-Tapped
During positive cycle ( t = 0 T/2 ) Transformer
1:2 D1 "On"
vi
vm vi
R
vi
t vo
T/2 vi
• Diode D1 “On”, D2 “Off” vo = v i D2 "Off"
13
Full-Wave Rectification
Assume Ideal Diode Model:
During negative cycle ( t = T/2 T ) 1:2 D1 "Off"
vi
vm vi
R
vi
t vo
T/2 T
vi
D2 "On"
Diode D2 “On”, D1 “Off”
vo = v i Center-Tapped
Transformer
14
Full-Wave Rectification
Example 1: (Ex. 2.17, Boylestad)
Determine and sketch the output waveform for the network
below. Use the ideal diode model.
vi D1 D2
10V
2kΩ
vi
vo
2kΩ 2kΩ
15
Full-Wave Rectification
During positive cycle ( t = 0 T/2 )
vi
2kΩ
vi
vo
2kΩ 2kΩ
2kΩ vo 1
vi
v o = v i = 5V
2kΩ 2
2kΩ Vdc = 0.636Vm = 0.636(5) = 3.18V
Effect of removing two diodes
16
Full-Wave Rectification
During negative cycle ( t = T/2 T)
vi
2k Ω
vo
vi
2k Ω 2k Ω
10V
2kΩ 1
vo = vi = 5V
vi 2kΩ 2
2kΩ vo
Vdc = 0.636Vm = 0.636(5) = 3.18V
17
Full-Wave Rectification
vo
5V
Vdc = 3.18V
18
Clippers (Wave Shaping)
• Clipper circuit is used to “clip” off a portion of the input signal
without distorting the remaining part
• There are series and parallel clippers
Series Clipper Circuit
• Diode in series with the load
(half-wave rectifier is simplest form of clipper)
vi R vo
19
Clippers
vi vo
• Consider ideal diode
• Try to analyze the ?
following waveform
• Use the series clipper
circuit shown before
vi vo
?
20
Clippers
• Consider ideal diode
Series Clipper Circuit with DC supply
vi
V
Vm
vi R vo
21
Clippers
vi
V
Vm
vi R vo
vo
• Assume diode to be short circuit: “On”
• Use KVL to obtain i/p and o/p relation Vm-V
v i − V − vo = 0 vo = v i − V
• vo will be positive only when vi ≥V
Vi=V
22
Clippers
vi
V
vi R vo vo = 0
vo
• Diode is open circuit: “Off” Vm-V
• When vi <V , diode is open circuit
Vi=V
23
Clippers
Parallel Clipper Circuit
vi vo
• Diode in parallel with the load
R
?
vi vo
vi vo
?
• Consider an Ideal diode
• Observe for each cycle
24
Clippers
Parallel Clipper Circuit with DC supply
vi
vm R
vi vo
V
During positive cycle ( t = 0 T/2 )
• Assume diode to be short circuit
• Apply KVL
• Assume vm > V
25
Clippers
vo = V R
• Only true when vi < V
• When vi > V , diode open circuit: “Off”
vo
vo = v i V
During negative cycle ( t = T/2 T)
• Diode is always short circuit: “On”
vo = V
26
Clippers
vi
vm
V
vo
vm
V
27
Clampers (Wave Shaping)
• Used to “clamp” or attach a signal to a different DC level
• Circuit usually has a capacitor, diode and a resistor
• τ = RC is chosen to be large enough to avoid discharging process
• Assumption: Capacitor will fully discharge and charge in five time
constants (5τ)
C
vi R vo
28
Clampers (Wave Shaping)
Steps for Analysing the Clamper Circuit (pg 91)
1. Start the analysis by considering the signal which forward biases the
diode
2. During the period the diode is in the ‘on’ state, assume C will charge
up instantaneously to a voltage level set by the surrounding network
3. During the period the diode is in the ‘off’ state, assume C holds on
to its established voltage level
4. Throughout the analysis, be aware of the location and polarity for VO
5. Check that the total swing of the output matches that of the input
29
Clampers
Example 2: (Boylestad)
Example : Determine vo
vi f = 1000 Hz
C=1uF
10
vi V R 100kΩ vo
5V
-20
• For clamping circuit, start the analysis when
the diode is forward-biased
30
Clampers C=1uF
From t1-t2 :
vo = 5V vi=20V V R 100kΩ vo
5V
Applying KVL,
−20V + VC − 5V = 0 25V
VC = 25V
From t2-t3 :
Applying KVL, outside loop vi=10V V R 100kΩ vo
10V + 25V − vo = 0 5V
vo = 35V
31
Clampers
Calculate discharge time,
τ = RC = (100k Ω)(0.1μF ) = 10ms
5τ = 5(10ms ) = 50ms vo
Given, 35
f = 1000 Hz; T = 1ms
1/2T= t2-t3 =0.5ms
•Since 1/2T < 5τ, capacitor will not
5
have enough time to discharge
t1 t2 t3 t4
32
Question & Answer
33
