# Diode Applications (PDF)

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.

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4/15/2012
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```							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

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 : 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

33

```
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