IC Bridge Rectifier vs Diodes

[Lucifre]

Ok i am trying to design ac to dc converter, i have a transformer(NOT center tapped) that will give me 12.6 V rms, what is better? IC full wave rectifier or four diodes? also how can i predict the voltage drop across it?, For diodes i know it is easiest to assume voltage drop of 0.7 per each diode and the output voltage will be 1.4 less than the voltage supplied by the transistor in case of bridge rectifier. how do i determine the optimal dc voltage it can supply?

 

[IanP]

To save on connections use bridge rectifier. Its priciple of operation is exactly the same as 4 individual diodes connected in a bridge configuration.
Also, votage drop will be calculated in the same way for both cases..
Regards,
IanP

 

[ahmad_abdulghany]

You make the very very simple problem as large problem .. no need incase of implementing a DC power supply ( that you called AC to DC converter ) to calculate voltage drops or whatever if you will use regulator .. just i want you to know that voltage out from the transformer is root two its peak value .. so for example if you are speaking about 220/12 volt transformer that means that the dc output from it is 12*2^½ ... if you added a series voltage regulator it's recommended that you make sure that the dc input voltage is greater than output by about 3 volts .. also don't increase it very much as this will cause the regulator to dissipate more power ( heat ) ..

I hope that i helped you ...

Best Regards

 

[Lucifre]

Actually if you have 12V rms transformer, 12 × 2^½ will give you peak value (Vp) of voltage, Dc value will be (2/π)(12 × 2^½) or .636Vp. I like to know voltage drops just to be more accurate. I already figured out how to determine optimal voltage it'll give me.

 

[IanP]

In the "Rectifier Relationships" you will find all possible configurations with voltage-current ratios.
Regards,
IanP

 

[Lucifre]

Correction!! the dc value at secondary of transformer will be zero, the dc value of the rectified output will be (2/π)(12 × 2^½) minus the voltage drop across the diodes. . . Man am i confusing or what? nothin better to do but to create problems.