Dode application

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Diode ApplicationsUnit-1

IntroductionThe primary goal of this chapter is to develop a working knowledgeof the diode in a variety of configurations using models appropriate for the area of application.

SERIES DIODE CONFIGURATIONSThe forward resistance of the diode is usually so small compared to the other series elements of the network that it can be ignoredFor voltages less than 0.7 V for a silicon diode and 0 V for the ideal diode the resistance is so high compared to other elements of the network that its equivalent is the open circuitIn general, a diode is in the on state if the current established by the applied sources is such that its direction matches that of the arrow in the diode symbol, and VD > 0.7 V for silicon, VD > 0.3 V for germanium, and VD > 1.2 V for gallium arsenide.

SERIES DIODE CONFIGURATIONSAn open circuit can have any voltage across its terminals, but the current is always 0 A. A short circuit has a 0-V drop across its terminals, but the current is limited only by the surrounding network.



Determine the resistances for the network when I=200mA

SINUSOIDAL INPUTS; HALF-WAVERECTIFICATIONTheHalf wave rectifieris a circuit, which converts an ac voltage to dc voltage. In theHalf wave rectifiercircuit serves two purposes. It can be used to obtain the desired level of dc voltage (using step up or step down transformers).The diode only conducts when it is forward biased, therefore only half of the AC cycle passes through the diode to the output.

PIV (PRV)For rectifier applications,peak inverse voltage(PIV) orpeak reverse voltage(PRV) is the maximum value ofreverse voltagewhich occurs at thepeakof the input cycle when the diode isreverse-biased.Because the diode is only forward biased for one-half of the AC cycle, it is also reverse biased for one-half cycle.It is important that the reverse breakdown voltage rating of the diode be high enough to withstand the peak, reverse-biasing AC voltage. PIV (or PRV) > VmPIV = Peak inverse voltage PRV = Peak reverse voltage Vm = Peak AC voltage

CLIPPERSClippers are networks that employ diodes to clip away a portion of an input signal without distorting the remaining part of the applied 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 ClipperDiodes 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 clippersThere 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.



Example 2

Solution (continued):- ve region OFF state

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 requiredforward bias voltage.Parallel Clipper

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


Solution (continued)

Solution (continued)

ClampersA 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 circuitMagnitude 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.

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.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.

ExampleDetermine Vo for the below network.

Solutionf=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 Kirchoffs LawVC=20V+5V=25V

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

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

Output waveform