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CHAPTER-2
Diode Applications
In general, approximate model of diode is used in applications because of non-ideal real
life conditions (tolerance, temperature effect, etc) never allow an ideal case to be applied
Load Line Analysis. The load of a circuit determines the point or the region of operation
of a diode (or device). The method: A line is drawn on
the characteristic of the device. The intersection point gives the point of operation
Load Line Analysis: In the above figure, E has to be in the forward bias direction and >
VT of the diode, in order for the current to flow. Using KVL:
-E + VD + VR = 0
E = VD + VR = VD + IDR
Notice: Variables VD, ID are same in the above equation. For VD = 0: ID = E/R
For ID = 0: E = VD + ID R
Load Line Analysis: The intersection point is called Q point. Same solution can be
found by using nonlinear diode equation. We can avoid heavy math using load line
analysis.
Exp from notes 1
Exp from notes 2
Exp from notes 3
Example 1
For below figure, determine VDQ, IDQ and VR
Solution:
Refer to notes for the solution
1
Example 2
Repeat the previous example using approximate model of the Si diode.
Solution: Refer to notes and see the next figure
Solution to Example 2
Example 3
Repeat the same example using ideal model of the diode. Solution: Refer to notes and see
next figure
Solution to example 3 using ideal model
Diode Approximations
As an engineer, we will generally use approximate models to avoid extensive
mathematical calculations. This is achieved by using approximate model of a device
whenever it is possible. Approximate model of diodes are given in Table 2.1 Also see
Figure 2.11 Series Diode Config with DC Inputs. When connected to voltage sources in
series, the diode is ON if the applied voltage is in the direction of forward-bias and it is
greater than the VT of the diode. When a diode is on, we can use the approximate model
for the on state. See next two figures.
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Using equivalent model of the diode in the forward-bias region
Series Diode Config. with DC Inputs. Here, VD = VT, VR = E – VT. ID = IR = VR / R.
When the diode is in the off state, the model for the off state is used. See two figures
Here, VD = E, VR = 0, ID = 0. Keep in mind that KVL has to be satisfied under all
conditions. Exp from notes 1, 2, 3, 4, 5
Example 4
For the figure below, determine VD, ID, and VR. Refer to notes for the solution
Example 5
For the figure below, determine VD, ID, and VR. Refer to notes for the solution
3
Example 6
Determine VD2, ID and V0 for the figure below. Refer notes for the solution
Example 6
Determine I1, V1, V2 and V0 for the figure below. Refer to notes for the solution
Parallel Diode Configurations
Determine I1, VD1, VD2 and V0 for the parallel diode circuit in below figure. Refer to
notes for the solution
Determining unknown quantities
Example 9
Determine the current I for the circuit below. Refer to notes for the solution
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Example 9
Determine I1, I2, and ID for the figure below. Refer to notes for the solution
Example 9
Determining the unknown quantities for the above example
Sinusoidal Inputs: Half-wave Rectifier. We expand our analysis to include time varying
signals. Such a network is shown as in the next figure. This circuit is called half-wave
rectifier. For the positive and negative cycles, the circuit is approximated as in below. See
following 2 figures
Half-wave Rectifier
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Half-wave Rectifier
The total effect of diode on the output signal is given in below
Half-wave Rectifier
For the half-wave rectified signal: Vdc = 0.318 Vm. If the effect of VT is also considered,
the output of the system will as below. Vdc = 0.318 (Vm- VT). See next Figure Effect of
VT on half-wave rectified signal
PIV rating of Half-wave
Rectifiers: PIV rating is very important consideration for rectifier circuits. For the halfwave rectifier: PIV ≥ Vm Full-wave Rectifiers Bridge Networks. The dc level obtained
from a sinusoidal input to the half-wave rectifier can be improved to 100% using fullwave rectifiers. Bridge networks are used for this purpose. See Figure 2.54. For positive
and negative cycles, network acts as below. See Figures 2.55, 2.58
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Center-Tap Full-Wave Rectifier
Diode Applications
Transfer Characteristic
Rectification ("frequency shifting")
 Typical power supply applications

Half-Wave
Rectification
"Figure shows a half-wave rectifier circuit. The signal is exactly the top half of
the input voltage signal, and for an ideal diode does not depend at all on the size
of the load resistor.
"The rectified signal is now a combination of an AC signal and a DC component.
Generally, it is the DC part of a rectified signal that is of interest, and the un-welcomed
AC component is described as ripple. It is desirable to move the ripple to high
frequencies where it is easier to remove by a low-pass filter.
"When diodes are used in small-signal applications - a few volts - their behaviour is not
closely approximated by the ideal model because of the PN turn-on voltage. The
equivalent circuit model can be used to evaluate the detailed action of the rectifier under
these conditions. During the part of the wave when the input is positive but less than the
PN turn-on voltage, the model predicts no loop current and the output signal voltage is
therefore zero. When the input exceeds this voltage, the output signal becomes
proportional to, or about 0.6 V lower than the source voltage."
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(source)
Op Amp solution to PN turn-on problem
Half-wave rectifier with filter capacitor or peak detector

Full-Wave Rectification
o Version 1 - Center-Tap Full-Wave Rectifier
o
Version
2
Bridge
Full-Wave
Rectifier
"The diode bridge circuit shown Ö is a full-wave rectifier. The diodes act
to route the current from both halves of the AC wave through the load
resistor in the same direction, and the voltage developed across the load
resistor becomes the rectified output signal. The diode bridge is a
commonly used circuit and is available as a four-terminal component in a
number of different power and voltage ratings."
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(source)
Go to Diode Bridge Modules for a collection of pdf data sheets on many integrated
diode bridges.
Op Amp solution to PN turn-on
Problem
(source)
Split Power Supply
"Often a circuit requires a power supply that provides negative voltage as well as positive
voltage. By reversing the direction of the diode and the capacitor (if it is polarized), the
half-wave rectification circuit with low-pass filter provides a negative voltage. Similarly,
reversing the direction of the diodes and capacitor in the full-wave rectified supply
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produces a negative voltage supply. A split power supply is shown in figure Ö"
(source)
A Variety of Other Applications:

Clamp
Also called a "dc restorer" in Sedra & Smith
"When a signal drives an open-ended capacitor the average voltage level on the output
terminal of the capacitor is determined by the initial charge on that terminal and may
therefore be quite unpredictable. Thus it is necessary to connect the output to ground or
some other reference voltage via a large resistor. This action drains any excess charge
and results in an average or DC output voltage of zero.
"A simple alternative method of establishing a DC reference for the output voltage is by
using a diode clamp as shown in figure Ö. By conducting whenever the voltage at the
output terminal of the capacitor goes negative, this circuit builds up an average charge on
the terminal that is sufficient to prevent the output from ever going negative. Positive
charge on this terminal is effectively trapped."
(source)
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(source)
 Clippers/Limiters
"A diode clipping circuit can be used to limit the voltage swing of a signal. Figure Ö
shows a diode circuit that clips both the positive and negative voltage swings to
references voltages."
11
(source)
Limiting with ordinary diodes:
(source)
Limiting with Zener diodes:
 Multiplier

Doubler - Version 1: A dc restorer followed by a peak detector
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(source)

Doubler - Version 2
"A voltage multiplier circuit is shown in figureÖ. We can think of it as two halfwave rectifier circuits in series. During the positive half-cycle one of the diodes
conducts and charges a capacitor. During the negative half-cycle the other diode
conducts negatively to charge the other capacitor. The voltage across the
combination is therefore equal to twice the peak voltage. In this type of circuit we
have to assume that the load does not draw a significant charge from the
capacitors."
(source)
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(source)

Tripler
(source)

For more on multipliers see: (Geometry she bs
Cockroft-Waton Diode Voltage Multipliers
 Diode Logic
"To the left (above) you see a basic Diode Logic OR gate. We'll assume that a logic 1 is
represented by +5 volts, and a logic 0 is represented by ground, or zero volts. In this
figure, if both inputs are left unconnected or are both at logic 0, output Z will also be held
at zero volts by the resistor, and will thus be a logic 0 as well. However, if either input is
raised to +5 volts, its diode will become forward biased and will therefore conduct. This
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in turn will force the output up to logic 1. If both inputs are logic 1, the output will still be
logic 1. Hence, this gate correctly performs a logical OR function.
"To the right (above) is the equivalent AND gate. We use the same logic levels, but the
diodes are reversed and the resistor is set to pull the output voltage up to a logic 1 state.
For this example, +V = +5 volts, although other voltages can just as easily be used. Now,
if both inputs are unconnected or if they are both at logic 1, output Z will be at logic 1. If
either input is grounded (logic 0), that diode will conduct and will pull the output down to
logic 0 as well. Both inputs must be logic 1 in order for the output to be logic 1, so this
circuit performs the logical AND function."
(source)
 Standby Voltage
(source)
 Gate
(source)
Mixer Circuits
Consider the following circuit:
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Using the Shockley Diode Equation we can write
This is a pretty complicated expression, but we can get a reasonable and useful result
with a bit algebra and analysis. Let us assume that there are dc biases on the signals
(usual case) so that
where
are fluctuations around the bias values
With much algebra we can show that
and
where
The important point is that the output has components
This page was prepared and is maintained by R. Victor Jones
Comments to: [email protected].
Last updated October 25, 2001
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