Diode Applications

1
Diode Applications

• CHAPTER 2

2
Load Line Analysis

• The load line plots all possible current (ID)
  conditions for all voltages applied to the diode
  (VD) in a given circuit. E/R is the maximum ID
  and E is the maximum VD.
• Where the load line and the characteristic curve
  intersect is the Q point, which specifies a
  particular ID and VD for a given circuit.

3
How to determine the Q point of a system?

• Identify diode model
• Using Kirchoffs Law
• Set VD 0V (horizontal line)
• Set ID 0A (vertical line)
• Obtain VDQ, IDQ from the graph intersection (
  Q-point)

4
Diode Approximation

• Approximate model notation
• In Forward Bias
• Silicon Diode VD 0.7V
• Germanium Diode VD 0.3V
• In Reverse Bias
• Both diodes act like opens VD source voltage
  and ID 0A

Ideal model notation VD 0V and ID 0A
5
Diode in DC Series Circuit Forward Bias

• The diode is forward bias.
• VD 0.7V (or VD E if E lt0.7V)
• VR E VD
• ID IR VR /R

6
Diode in DC Series Circuit Reverse Bias

• The diode is reverse biased.
• VD E
• VR 0V
• ID IR IT 0A

7

• An open circuit can have any voltage across its
  terminals, but the current is always 0A.
• A short circuit has a 0V drop across its
  terminals, but the current is limited only by the
  surrounding network.
• Source notation

8
Example
9
Solution
10
Example

• Determine ID, VD2 and Vo for the circuit.

Remember, the combination of short circuit in
series with an open circuit always results in
an open circuit and ID0A.
11
Example

• Determine I, V1, V2 and Vo

12
Series Parallel Configurations
Solve this circuit like any Series/Parallel
circuit, knowing VD 0.7V (or up to 0.7V) in
forward bias and as an open in reverse bias. VD1
VD2 Vo 0 .7V VR 9.3V Diodes in parallel
are used to limit current IR E VD 10V
-0 .7V 28mA R
0.33k? ID1 ID2 28mA/2 14mA
13
Example

• Determine the resistance R for the network when
• I200mA.

Si
Si
14
Example

• Determine the currents I1, I2, and ID2 for the
  network

15
Diodes in AC Circuits

• Inputs -Sinusoidal waveform
• -Square wave
• This circuit is called half-wave rectifier, which
  generate waveform vo that will have an average
  value of particular use in the ac-to-dc
  conversion process.
• The diode only conducts when it is in forward
  bias,
• therefore only half of the AC cycle passes
  through the diode.

16
Half-Wave Rectification

• The diode that employed in the rectification
  process is typically
• referred to rectifier.
• The diode only conducts for one-half of the AC
  cycle. The remaining
• half is either all positive or all negative.
  This is a crude AC to DC
• conversion.
• The DC Voltage out of the diode
• Vdc 0.318Vm
• where Vm the peak voltage

17
Peak Inverse Voltage (PIV)

• Because the diode is only forward biased for
  one-half of the AC cycle, it is then also off for
  one-half of the AC cycle. It is important that
  the reverse breakdown voltage rating of the diode
  be high enough to withstand the peak AC voltage.
• PIV (PRV) gt Vm
• PIV Peak Inverse Voltage PRV Peak Reverse
  Voltage
• Vm Peak AC Voltage

18
Full-Wave RectificationBridge Network

• The dc level obtained from a sinusoidal input can
  be improved 100 using a process called full-wave
  rectification.
• The most familiar network is bridge configuration
  with 4 diodes.

Vdc 0.636 Vm
19
Operation of the Bridge Rectifier Circuit
For the positive half of the AC cycle
For the negative half of the AC cycle
20
Determining Vo for silicon diodes in the bridge
configuration
The effect of using a silicon diode with VD0.7
is demonstrated in below figure. The dc level has
change to
21
Example
Determine the output waveform for the network
below and calculate the output dc level.
22
Solution
Conduction path for the ve region
23
Solution
Conduction path for the -ve region
24
Animation of full bridge rectifier
25
Full-Wave RectificationCenter-Tapped Transformer

• A second popular full-wave rectifier with only
  two diodes but
• requiring a center-tapped transformer to
  establish the input
• signal across each section of the secondary of
  the transformer.

Two diodes and a center-tapped transformer are
required. VDC 0.636(Vm) for ideal diode Note
that Vm here is the transformer secondary voltage
to the tap.
26
Operation of the CenterTapped Transformer
Rectifier Circuit
For the positive half of the AC cycle
During the positive cycle of vi applied to the
primary of the Transformer the network will
appear as shown in figure. D1 assumes the
short-circuit equivalent and D2 the open-circuit
equivalent, as determined by the secondary
voltages and the resulting current directions.
27
For the negative half of the AC cycle
During the negative cycle of vi, reversing the
roles of the diodes (D2 is short-circuit) but
maintaining the same polarity for the voltage
across the load resistor R.
28
Animation of center-tapped transformer rectifier
29
Example
Show the voltage waveform across the secondary
winding and across R when an input sinusoidal is
applied to the primary winding.
30
Solution
The transformer turns ratio 0.5. The total
peak secondary voltage is,Vp(sec) nVp(pri)
0.5(100)50V. There is a 25 V peak across each
of the secondary with respect to ground.
31
Rectifier Circuit Summary
Note Vm peak of the AC voltage. Be careful, in
the center tapped transformer rectifier circuit
the peak AC voltage is the transformer secondary
voltage to the tap.
32
Clippers

• Clippers or diode limiting is a diode network
  that have the ability to clip(cut short/crop)
  off a portion on the input signal without
  distorting the remaining part of the alternating
  waveform.
• Clippers are used to eliminate amplitude noise or
  to fabricate new waveforms from an existing
  signal.
• Simplest form of diode clipper- one resistor and
  a diode
• Depending on the orientation of the diode, the
  positive or negative region of the applied signal
  is clipped off.
• 2 general of clippers
• a) Series clippers
• b) Parallel clippers
• Series Clippers
• The series configuration is defined as one where
  the diode is in series with the load.
• A half-wave rectifier is the simplest form of
  diode -clipper-one resistor and diode.
• Parallel Clippers
• The parallel configuration has the diode in a
  branch parallel to the load.

33
Series Clipper

• Diodes clip a portion of the AC wave.
• The diode clips any voltage that does not put
  it in forward bias. That would be a reverse
  biasing polarity and a voltage less than 0.7V for
  a silicon diode.

Any type of signals can be applied to a clipper
34
Analysis steps for series clippers

• There is no general procedure for analyzing
  series clippers
• network but there are some things one can do to
  give the
• analysis some direction.
• Take careful note of where the output voltage is
  defined.
• Try to develop an overall sense of the response
  by simply noting the pressure established by
  each supply and the effect it will have on the
  conventional current direction through the diode.
• Determine the applied voltage (transition
  voltage) that will result in a change of state
  for the diode from the off to the on state.
• It is often helpful to draw the output waveform
  directly below the applied voltage using the same
  scales for the horizontal axis and the vertical
  axis.

Series clipper with dc supply examples
35
Series clipper with dc supply
By adding a DC source to the circuit, the voltage
required to forward bias the diode can be
changed.
36
Series clipper example

• Positive region of Vi - turn the diode ON.
• Negative region of Vi - turn the diode OFF.
• Vi gt V to turn ON the diode.
• In general, diode is open circuit (OFF state)
  and short circuit (ON state)
• For Vi gt V the Vo Vi V
• For Vi V the Vo 0 V

37
Example 1
Determine the output waveform for the network
below
38
Solution (continued)
39
Example 2
Repeat previous example for the square-wave input.
40
Solution (continued) - ve region ? OFF state
41
Parallel Clipper

• By taking the output across the diode, the
  output is now the
• voltage when the diode is not conducting.
• A DC source can also be added to change the
  diodes required
• forward bias voltage.

Parallel clipper example
42
Example 2
Determine the Vo and sketch the output waveform
for the below network
43
Solution
44
Solution (continued)
45
Solution (continued)
46
Example 2
Repeat the previous example using a silicon diode
with VD0.7 V
Solution
47
Solution (continued)
For input voltages greater than 3.3 V the diode ?
open circuit and VoVi. For input voltages less
than 3.3 V the diode ? short circuit and the
network result as shown below
48
Series Clippers (Ideal Diode) Summary
49
Parallel Clippers (Ideal Diode) Summary
50
Clampers

• A clamper is a network constructed of a diode,
  resistor, and a capacitor that shifts a waveform
  to a different dc level without changing the
  appearance of the applied signal.
• Clamping networks have a capacitor connected
  directly from input to output with a resistive
  element in parallel with the output signal. The
  diode is also in parallel with the output signal
  but may or may not have a series dc supply as an
  added element.

51

• Element of the clamper circuit
• Magnitude of R and C must be appropriate to
  ensure ?RC where the time constant is large
  enough and capacitor may not discharge during the
  time interval while diode is not conducting.
• We will assume that all practical purposes the
  diode will fully charge or discharge in 5 time
  constant.

A diode in conjunction with a capacitor can be
used to clamp an AC signal to a specific DC
level.
52

• The input signal can be any type of waveform
• - sine, square, triangle wave, etc.
• You can adjust the DC camping level with a DC
  source.

Clampers example
53

• There is a sequence of steps that can be applied
  to
• help make the analysis straight forward.
• Start the analysis by examining the response of
  the portion of the input signal that will forward
  bias the diode. If the diode is reverse bias,
  skip the analysis for that interval time, and
  start analysis for the next interval time.
• During the period that the diode is in the on
  state, assume that the capacitor will charge up
  instantaneously to a voltage level determined by
  the surrounding network.
• Assume that during the period when the diode is
  in the off state the capacitor holds on to its
  established voltage level.
• Throughout the analysis, maintain a continual
  awareness of the location and defined polarity
  for vo to ensure that the proper levels are
  obtained.
• Check that the total swing of the output matches
  that off the input.

54
Clampers Summary
55
Example

• Determine Vo for the below network.

56
Solution

• f1000Hz, so a period of 1ms or interval 0.5ms
  between each level.
• Define the period that the diode is start to
  conduct (t1t2), which is the V05V.
• Determine VC from the Kirchoffs Law
• VC20V5V25V
• When in the positive input, we will find
  V035V(outside loop)
• Time constant, ? RC 10ms, total discharge time
  50ms where is large enough before the capacitor
  is discharge during interval t2t3

57
Output waveform
58
Zener Diode

• The state of the diode must be determined
  followed by a
• substitution of the appropriate model and a
  determination of
• the unknown quantities of the network. For the
  off state as a
• defined by a voltage less than Vz but greater
  than 0V. The
• Zener equivalent is the open circuit.

59
Vi and R Fixed

• The applied dc voltage is fixed, as the load
  resistor.
• The analysis
• Determine the state of the Zener diode by
  removing it from the network and calculating the
  voltage across the resulting open circuit.

60

• 2. Substitute the appropriate equivalent circuit
  and solve for the desired unknowns.
• - For the on state diode, the voltages across
  parallel elements must be the same.
• VLVZ
• The Zener diode current is determined by KCL
• IZ IR IL
• The power dissipated by the Zener diode is
  determined by
• PZ VZ IZ
• - For the off state diode, the equivalent
  circuit is open-circuit.

61
Fixed Vi, Variable RL

• Due to the offset voltage Vz, there is a specific
  range of resistor values (and therefore load
  current) which will ensure that the Zener is in
  the on state.
• Too small RL ? VL lt Vz ? Zener diode will be in
  the off state
• To determine the min RL that will turn the Zener
  diode on
• Any load resistance value greater than the RL min
  will ensure that the Zener diode is in the on
  state and the diode can be replaced by its Vz
  source equivalent.
• The max IL

62

• Once the diode is in the on state, the voltage
  across R remains fixed at
• Iz is limited to IZM as provided on the data
  sheet, it does affect the range of RL and
  therefore IL.

63
Fixed RL, Variable Vi

• For fixed values of RL, the voltage Vi must be
  sufficiently large to turn the Zener diode on.
  The min turn-on voltage ViVi min
• The max value of Vi is limited by the max Zener
  current IZM.
• IRmaxIZMIL
• Since IL is fixed at VZ/RL and IZM is the max
  value of IZ, the max Vi is defined by
• Vi max VRmaxVz
• Vi maxIRmaxRVz

64
Zener Diode Examples
65
Example
Determine the network to find the range of RL and
IL to maintained VRL at 10V.
66
Solution