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PEAK INVERSE VOLTAGE Using the ideal diode model, the PIV of each diode in the bridge rectifier is equal to V2.This is the same voltage that was applied.

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Presentation on theme: "PEAK INVERSE VOLTAGE Using the ideal diode model, the PIV of each diode in the bridge rectifier is equal to V2.This is the same voltage that was applied."— Presentation transcript:

1 PEAK INVERSE VOLTAGE Using the ideal diode model, the PIV of each diode in the bridge rectifier is equal to V2.This is the same voltage that was applied to the diode in the full-wave center-tapped rectifier. Following fig helps us to illustrate this point .Fig in next slide , two things have been done :

2 PEAK INVERSE VOLTAGE

3 PEAK INVERSE VOLTAGE 1) The conducting diodes (D1 and D3 ) have been replaced by straight wires. Assuming that the diode are ideal they will have the same resistance as wire; therefore replacement is valid. 2) The positive side of the secondary has been labeled A and the negative side has been labeled B.

4 PEAK INVERSE VOLTAGE Connecting the common A point along a straight line and doing the same with B points gives us the circuit shown in the fig b .With the equivalent circuit ,you can see that two reverse biased diodes and the secondary of the transformer are all in parallel. Since parallel voltages are equal ,the PIV across each diode is equal to V2.The same situation will exist for D1 and D3 when they are reverse-biased.

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6 FILTERS Filters are used in power supplies to reduce the variations in the rectifier output signal. Since our goal is to produce a constant dc output voltage, it is necessary to remove as much of the rectifier output variations as possible.

7 FILTERS The overall result of using a filter is illustrated

8 CLIPPERS There are variety of diode networks called clippers that have the ability to clip off a portion of the input signal without distorting the remaining part of the alternating waveform. The half wave rectifier studied earlier is a simplest form of diode clipper.

9 CLIPPERS Depending on the orientation of the diode, the positive or negative region of the input signal is “clipped” off.

10 IMPORTANT POINTS FOR CLIPPERS
Make a sketch in your mind about the response of the network. Determine the applied voltage (Transition Voltage) that causes change in the diode bias. Be continuously aware of the defined terminal and polarity of Vo.

11 IMPORTANT POINTS FOR CLIPPERS
(4) Sketch the input signal on the top and the output at the bottom to determine the output at instantaneous points of the input.

12 TYPES OF CLIPPERS There are two general categories of the clippers.
Series Parallel

13 SERIES CLIPPERS The series configuration is defined as one where the diode is in series with the load as half wave rectifier.

14 EXAMPLES

15 ADDITION OF A BATTERY IN THE SERIES CLIPPER CIRCUIT
In the fig. the direction of the diode suggests that the signal Vi must be positive to turn it on. The dc supply further requires that the voltage Vi be

16 ADDITION OF A BATTERY IN THE SERIES CLIPPER CIRCUIT
greater than V volts to turn the diode on. The negative region of the input signal is pressuring the diode into off region.

17 ADDITION OF A BATTERY IN THE SERIES CLIPPER CIRCUIT
Now we determine the applied voltage that will cause a change in state for the diode. The ideal diode transition occur at the point on the characteristic where Vd= 0 and Id = 0

18 ADDITION OF A BATTERY IN THE SERIES CLIPPER CIRCUIT
In this case transition will occur at Vi = V

19 ADDITION OF A BATTERY IN THE SERIES CLIPPER CIRCUIT
When the diode is short circuited as shown in the fig. The out put voltage V0 can be calculated bu applying KVL Vi – V - V0 = 0 V0 = Vi – V

20 ADDITION OF A BATTERY IN THE SERIES CLIPPER CIRCUIT
In this case V0 = Vm - V

21 EXAMPLE Determine the output waveform for the network shown

22 SOLUTION The equivalent circuit will be V0 = Vi + 5V
Transition voltage =5V

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24 EXAMPLE Determine the output for the square wave input shown in the fig. lec39_m

25 SOLUTION In positive half cycle. The diode is in the short circuit condition and by KVL V0= = 25V

26 SOLUTION For the negative half cycle Vi = -10V the result in placing the diode in the reverse condition.

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