Diode Circuits               
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One way of obtaining DC power is to transform, rectify, filter and regulate AC line voltage. The principles of rectification, filtering and regulation are treated in the following sections.

Block diagram representation of a power supply:
fig9.gif (8480 bytes)


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Half-Wave Rectifier

In the following figures, let's imagine that a diode is added to the circuit at t=0.

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Left-hand side of the above two figures show the voltage across RL with no diode while the right-hand side show the rectified signal. The diode only allow current to flow in the direction of the arrow.

There is a voltage drop of 0.6V across the diode when it is conducting (The graph of VL versus t  on the right-hand side of the figures above  is therefore actually 0.6V lower than that on the left-hand side.)


Filter capacitor in the rectifier circuit
fig11.gif (2006 bytes)
 

The ripple in the signal can be filtered with a capacitor, the rectified signal charges the capacitor to the peak voltage, when the diode is conducting. When the diode is not conducting, the capacitor discharges through the resistor. The larger the capacitor the better the filter.
 

Voltage across load in the above circuit:fig12.gif (2336 bytes)
 

The magnitude of DV:
For a half-wave rectifier, the relationship of ripple voltage £GV, current i drawn by the load, the frequency f and the filter capacitance C is estimated to be:

eqt1.gif (1512 bytes)

The ripple factor, a commonly quoted specification of a power supply, is defined as,

eqt3.gif (1512 bytes)

This factor is sometimes measured as the rms value of the AC signal component divided by the DC component,

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Example 1

A power supply consisting of a 24V transformer, a single diode and a capacitor of 10000£gF is to supply a maximum current of 0.5A. What is the approximate ripple voltage and the ripple factor at maximum load?

Solution:

Since the rectification is half wave, the frequency of the rectified signal is 50Hz. Then,

£GV= 0.5A / (10000*10-6F*50Hz) = 1V

For a 24V rms signal the peak voltage is 34V. The DC voltage can be approximated by 34V-1V/2=33.5V. Then the ripple factor will be

r = 1V/ 33.5V = 0.03 = 3%

Full wave Rectifier
An improvement over the half-wave rectifier.

fig13.gif (5146 bytes)fig13b.gif (4096 bytes)

Full wave rectifier action

This configuration includes a centertapped (CT) transformer and two diodes. The rectified signal is doubled in frequency from that of the half-wave rectifier. With all other parameters being the same, the ripple voltage is halved.

Bridge rectifier

The bridge rectifier, shown in Fig14, produces The same waveform without the need for a center-tapped transformer. It requies two more diodes.

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Fig14 Bridge rectifier action

Split Power Supplies

The power supply that provides negative voltage as well as positive voltage.

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Two split power supplies

Voltage Multipliers

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Voltage doubler circuit

Zener Diode

An important feature of the operation of a diode: the avalanche or reverse breakdown process.

Breakdown voltage: for voltages more negative than the breakdown voltage, the voltage is virtually independent of the current.

The breakdown, or Zener, voltage of a diode can be controlled in the manufacturing process, and these special diodes, called Zener diodes, are available in values from 3.9V to several hundred volts.

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Zener diode behavior. Note that the units for positive current are mA and those for negative current are  £gA

The electronic characteristics of the Zener diode make the device suitable as a voltage regulator.

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Fig19 Simple Zener diode voltage regulator circuit

Example 2

With an input voltage of 7.0V, design a 5.1V power source capable of supplying 10mA.

Solution:

First we choose a Zener diode which has a breakdown voltage of 5.1V and typically a value of

IZmax = 50mA(1N4733A diode)

If we design the circuit such that the current through the diode is 20% of its maximum rated value, then
R = (Vin-VZ) / (0.2IZmax  + Iout), or

R = (7.0-5.1) / ((0.2)(0.5) + 0.01) = 95£[

To estimate the power rating of the resistor, substitute values for the input current, Iin = 0.2IZmax  + Iout, into the equation: PR = Iin2R =  (Vin-VZ)2 / R=38mW. This means that the power rating of the resistor must be greater than 38mW.  A 100£[ 1/4W resistor would be suitable.
To estimate the power rating of the Zener diode, we have PZ = IZVZ = (Iin- Iout) VZ = [((Vin-VZ)/ R ) - Iout]VZ=5.1mW. The power rating of the diode must be greater than 5.1mW.