Math, asked by rutujaw2004, 3 months ago

A centre tap rectifier is connected with transformer whose turn ratio is 5:1connected with 230v/50hz main AC supply.find the average DC voltage ,PIV of diode and ripple frequency​

Answers

Answered by JasleenMann
5

During negative half cycle of the input signal sin ( ) i m v V t = ω

The diode D is reverse biased and acts as open.

Hence practically no current flows through the load RL

and no voltage across the load.

• The output of HWR is not a perfect DC, but at least unidirectional.

• For HWR, the typical parameters are:

m

dc

V

V

π

=

2

m

rms

V

V =

Ripple factor=

2

,

1 1.21 r rms rms

dc dc

v v

r

V V

= = − =

The output of HWR has high ripple 121%. The ripple content exceeds the DC content.

This is undesirable. Hence there is need for FWR

Rectification efficiency:

Rectification efficiency dc

ac

Power delivered to the load P

Total input AC power P

η = =

The rectification efficiency of a HWR = 40.6 %

Under best conditions, only 40.6 % of the input power is converted into DC power. The rest

59.4% % remains as AC power in the load. This is highly undesired

PIV: The peak inverse voltage is the maximum reverse voltage the diode should withstand

without breakdown.

In HWR, when the diode D is reverse biased, the maximum voltage across it

would be Vm

o In HWR, the PIV of the diode should be Vm

Note: If asked as long answer question, you need to derive values of Vdc , rms v ripple factor

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