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I noticed when you start running these KBP310 bridge rectifiers in parallel the Vf goes down. From 0.998 V for 1 to 0.900 V for X4 bridge rectifiers together. Why would that be? Also, I was thinking if you run 2 bridge rectifiers in parallel but reverse the A/C leads on the second one would that lower ripple and maybe Vf a little because the rectifiers would be out of phase with each other?

schematic

simulate this circuit – Schematic created using CircuitLab

Null
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allenw
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    these IC rectifiers I don't know what "IC rectifiers" are so include a datasheet. but reverse the A/C leads If it is AC then how is reversing the connections going to change anything? Include a schematic as your question is very vague. – Bimpelrekkie Jan 09 '22 at 14:29
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    if you run 2 chips in parallel but reverse the A/C leads on the second one That's phyisically impossible. Given that the two diodes are in parallel, you cannot connect one diode one way, and the other diode the other way. – Davide Andrea Jan 09 '22 at 14:49
  • KBP310. I don't have a schematic. What I'm saying is having the + and - of both chips in parallel but hooking The A/C wires backwards on the second Rectifier like this =x= the = being the chips A/C legs and the x being the A/C source going from the first chip to the second. – allenw Jan 09 '22 at 14:52
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     – Transistor Jan 09 '22 at 15:03
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    It sounds like you are talking about full wave bridge rectifiers. Using two with the outputs paralleled and the AC reversed on one will not make a difference in the ripple. Bridge rectifiers are symmetrical, you get the same output no matter which way you connect the AC., – GodJihyo Jan 09 '22 at 15:49

2 Answers2

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When you start running these KBP310 bridge rectifiers in parallel the Vf goes down. Anyone know why that would be?

The basic diode (or Shockley) equation explains why: -

enter image description here

Image from here.

So, if the current through a diode is shared between two diodes then \$I_D\$ drops to half. So then, if you re-arrange the formula to solve for \$V_D\$ you will see that \$V_D\$ drops a little bit for a halving the diode current \$I_D\$. You probably thought that all diodes drop a fixed forward voltage irrespective of forward current? Not true. Reality is more complex.

I noticed when you start running these KBP310 bridge rectifiers in parallel the Vf goes down. From .998v for 1 to .900v for \$\color{blue}{\boxed{\text{X4}}}\$ bridge rectifiers together

enter image description here

Image from here.

I was thinking if you run 2 chips in parallel but reverse the A/C leads on the second one would that lower ripple and maybe Vf a little because the rectifiers would be out of phase with each other?

No, it won't make the slightest difference.

Andy aka
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Your data and a linear model of saturated diode with bulk resistance tells you why.

.998v for 1 to .900v for X4

This means 4 Rs drops 98 mV @ 1A shared by 4 equal diodes (assumption) so 4Rs=98 mV/0.25A thus Rs of each diode is 100 mohms @ 0.25A and 1 diode will drop 0.1V @ 1A

Tony Stewart EE75
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