C hapter 2 diode applications

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TOPIC OUTLINES
 LOAD LINE ANALYSIS
SERIES DIODE CONFIGURATION
PARALLEL AND SERIES-PARALLEL CONFIGURATION
AND/OR GATES
HALF WAVE RECTIFICATION
FULL WAVE RECTIFICATION
CLIPPERS
CLAMPERS
ZENER DIODES
VOLTAGE MULTIPLIER CIRCUITS


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.


How to determine the Q point of a
system?
• Identify diode model

• Using Kirchoff’s Law :
– Set VD = 0V (horizontal line)
– Set ID = 0A (vertical line)
– Obtain VDQ, IDQ from the graph intersection ( Q-point)


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

Ideal model notation
VD = 0V and ID = 0A


Diode in DC Series Circuit:
Forward Bias

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

Diode in DC Series Circuit:
Reverse Bias

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


• 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 :


Example
For the series diode configuration below, employing the diode
characteristics of figure below, determine VDQ, IDQ and VR.


Solution
Step1: Find the maximum ID. VD = 0V→ ID = IR= E/R
Step 2: Find the maximum VD. ID =0A → E = VD + IDR
Step 3: Plot the load line
Step 4 : Find the intersection between the load line and the
characteristic curve. This is the Q-point
Step 5: Checking !!!!


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 ID=0A.


Example
Determine I, V1, V2 and Vo


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


Example
Determine the resistance R for the network when
I=200mA.

Si
Si


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


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.

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


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) > Vm

• PIV = Peak Inverse Voltage; PRV = Peak Reverse Voltage
• Vm = Peak AC Voltage


Full-Wave Rectification:
Bridge 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

Operation of the Bridge Rectifier
Circuit
For the positive half of the AC cycle:

For the negative half of the AC cycle:


Determining Vo for silicon diodes in the bridge
configuration

The effect of using a silicon diode with VD=0.7 is
demonstrated in below figure. The dc level has change to:


Example
Determine the output waveform for the network below and
calculate the output dc level.


Solution
Conduction path for the +ve region


Solution
Conduction path for the -ve region


Full-Wave Rectification:
Center-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.


Operation of the Center–Tapped 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.


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.


Animation of center-tapped
transformer rectifier


Example
Show the voltage waveform across the secondary winding
and across R when an input sinusoidal is applied to the
primary winding.


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.


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.


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.


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


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.
1. Take careful note of where the output voltage is defined.
2. 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.
3. Determine the applied voltage (transition voltage) that will
result in a change of state for the diode from the “off” to the
“on” state.
4. 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


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


Series clipper example

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


Example 1
Determine the output waveform for the network below:


Solution (continued)


Example 2
Repeat previous example for the square-wave input.


Solution (continued):
- ve region  OFF state


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 diode’s required
forward bias voltage.

Parallel clipper example

Example 2

Determine the Vo and sketch the output waveform for the
below network


Solution


Example 2
Repeat the previous example using a silicon diode with
VD=0.7 V

Solution


Solution (continued)

For input voltages greater than 3.3 V the diode  open
circuit and Vo=Vi.
For input voltages less than 3.3 V the diode  short circuit
and the network result as shown below


Series Clippers (Ideal Diode) Summary


Parallel Clippers (Ideal Diode) Summary


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.


Element of the clamper circuit
• Magnitude of R and C must
be appropriate to ensure
г=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.

•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

There is a sequence of steps that can be applied to
help make the analysis straight forward.
1. 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.
2. 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.
3. Assume that during the period when the diode is in the
“off” state the capacitor holds on to its established
voltage level.
4. Throughout the analysis, maintain a continual awareness
of the location and defined polarity for vo to ensure that
the proper levels are obtained.
5. Check that the total swing of the output matches that off
the input.

Clampers Summary


Example
Determine Vo for the below network.


Solution
• f=1000Hz, so a period of 1ms or
interval 0.5ms between each level.

• Define the period that the diode is
start to conduct (t1~t2), which is
the V0=5V.

• Determine VC from the Kirchoff’s
Law
– VC=20V+5V=25V

• When in the positive input, we will
find V0=35V(outside loop)

• Time constant, г =RC = 10ms,
total discharge time = 50ms
where is large enough before the
capacitor is discharge during
interval t2~t3


Output waveform


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.


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


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.
VL=VZ
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.


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 < 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


• 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.


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
Vi=Vi min :

• The max value of Vi is limited by the max Zener current IZM.
IRmax=IZM+IL

• Since IL is fixed at VZ/RL and IZM is the max value of IZ, the
max Vi is defined by:
Vi max =VRmax+Vz

Vi max=IRmaxR+Vz

Zener Diode Examples


Example
Determine the network to find the range of RL and IL to
maintained VRL at 10V.


Solution:


Voltage Multiplier Circuit

Half Wave voltage doubler



Full Wave voltage doubler



Voltage Tripler and Quadrupler


Practical Application

• Rectification
• Protective Configuration
• Polarity Insurance
• Controlled Battery-Powered Backup
• Polarity Detector
• AC Regulator and Square Wave
Generator


End of Chapter 2

Thank you


C hapter 2 diode applications

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