How do you size the correct diodes needed to OR two 12V 1.25A power supplies? What voltage and current rating do the diodes need to be?
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1\$\begingroup\$ The voltage and current is determined by the voltage and current expected on these diodes at extreme conditions multiplied by a safety factor. You do the calculations. \$\endgroup\$– Eugene Sh.Commented May 24, 2016 at 19:07
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\$\begingroup\$ Do not assume that both diodes will have exactly the same Vdrop across them unless it is proven by testing. This means that with diode summing junctions, 1 diode will dominate as the current source, even if by a tiny amount. The power supplies will not put out exactly the same voltage either. This will also cause one diode to dominate as the source of current. \$\endgroup\$– user105652Commented May 24, 2016 at 19:40
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\$\begingroup\$ AT EUGENE : Well, i do actually understand that the diodes need to be sized in relation to voltage and current at the extremes, the question is "How do you size them" and what is a good safety factor? You answer provided zero help. I am looking for a step by step guide which details the "how". If anybody can explain it, it would be appreciated. Professor responses like the above are not helpful at all. \$\endgroup\$– DouglasCommented May 24, 2016 at 19:43
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\$\begingroup\$ At Sparky,256, how much of an issue is it if one diode dominates? and what does that mean long term? ie will one power supply be more likely to burn up before the other? thanks. \$\endgroup\$– DouglasCommented May 24, 2016 at 19:46
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\$\begingroup\$ @Douglas - use the "@" symbol when referring to users such as what I have done with your name. If you do not, they will not be notified that a comment has been made to them. \$\endgroup\$– efox29Commented May 24, 2016 at 19:54
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1 Answer
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Each diode needs to be specified to withstand the highest current thru it and the maximum reverse voltage across it. This really should be obvious.
Keep in mind that the limiting factor for current may be power dissipation, not the maximum current spec of the diode. You have to check both separately. For example, let's say the maximum current you decide the diode will ever have to conduct in steady state is 5 A. You check the datasheet and find that the forward drop at 5 A is 950 mV. That means the diode will dissipate (5 A)(950 mV) = 4.75 W.
Now you look in the datasheet and see if the diode can safely dissipate 4.75 W in free air at your worst case ambient temperature. If not, you need a different diode, a heat sink, or some form of cooling.
answered May 25, 2016 at 12:00