Full Bridge Rectifier


I am applying a sine wave of 60 volts peak to peak and 100KHz to a full bridge diode rectifier made with following diodes

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I connected the resistive load of 300 ohms at the output of the rectifier and got fully rectified 30 volts pulsating DC. Then I connected the capacitor of value 470uF and 100 volts across the load and got the following reading

  1. the input drops from 60 volts peak to peak to 44 peak to peak .
  2. the output looks smooth but the amplitude becomes 20 volts DC.
  3. I have following questions

  4. What did I loose 20 volts peak to peak at the input when I added the capacitor?
  5. Why did the output voltage reduced to 20 volts when it supposed to be 30 volts?
  6. Are these diodes rated for 100KHz signals? Plus what is their switching time?



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john1987 a écrit :

Nothing to do with switching. Have a look at their reverse breakdown voltage...

Reply to
Fred Bartoli

If the diodes were selected for sufficient reverse breakdown voltage, you'll still have to contend with source impedances, though it would have to be pretty high, to show more than capacitive rectification effects with a 300ohm load.

I doubt the parts were zenering, as schottkies don't avalanche tamely; they leak heavily before breakdown and simply short out.


Reply to

You are probably exceeding the diode's reverse voltage capability.


Reply to
John Larkin

"john1987" schreef in bericht news: snipped-for-privacy@p11g2000vbn.googlegroups.com...

According to the datasheet they're rectifiers. It mentions 60Hz halve wave application. They seem to work well at 100kHz and a resistive load. No big deal as they switch near 0V and can take there time. When you connect a capacitor however they have to conduct only for a very short time per period and have to switch much faster. If they do not switch fast enough, you will have a reverse current for some time. You can consider the effect as a part of the capacitor connected directly to the source. This is a pretty low impedance at 100kHz so the internal resistance of the source limits the current and the voltage falls.

As the datasheet does not mention the switching time, the only way to find it is measuring it for yourself :). Guess you'd better go for some fast rectifiers intended for use at 100kHz and beyond.

petrus bitbyter

Reply to
petrus bitbyter

_Schottky_ rectifiers. They're not your grandpa's 1N4001s.

Note the TT and BV parameters from the manufacturer's SPICE model:

*SRC=DFLS130L;DI_DFLS130L;Diodes;Si; 30.0V 1.00A 5.00ns Diodes Inc Low VF Schottky .MODEL DI_DFLS130L D ( IS=7.74u RS=59.7m BV=30.0 IBV=1.00m
  • CJO=225p M=0.333 N=0.832 TT=7.20n )

Best regards, Spehro Pefhany

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Reply to
Spehro Pefhany


** Because the electro cap charges in short bursts of current that average to the load current value.

Your source ( not given ) is unable to supply the peak current needed without dropping voltage - look at the waveform and you will see the peaks are now clipped.

.... Phil

Reply to
Phil Allison

You better check your source of that 60 Volt P-P.. it's not able to supple the current needed for the cap charging, which has the

300 ohm load on it, that is discharging it between cycles.

Caps exert current when charging, just like a cell/battery does.

If there were issues with the bridge not able to operate at that frequency, you'd see a problem before you apply the cap..


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"Spehro Pefhany" schreef in bericht news: snipped-for-privacy@4ax.com...

Ah, did not look at the model. With this numbers, the 100kHz should not be a problem. Remains the internal resistance (impedance) of the voltage source.

petrus bitbyter

Reply to
petrus bitbyter

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