EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Epaperman on August 28, 2017, 08:46:32 am
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If your power connector pins are rated for 1 amp but you need to load 1.5A.. can you use 2 contacts?
Note.. contacts have a resistance..
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What connector are you using?
Is it hot-plug?
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If you can, use three. At least you would be safe against a single failure. But under design review, it might not pass everywhere.
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I've seen numerous hot swappable power supplies that use multiple pins or edge card connectors in parallel to achieve the power requirements. As Neomys stated, use three instead of two; N+1 redundancy is always preferable to barely getting by.
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Yes, you can see that done in practically every computer on this planet. There are probably 100s of millions of examples of using parallel pins to increase current capacity.
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Very late reply.., sorry guys..
Here is the thing.. Ohms law applies here.
if the resistance even is very low... e.g. 0.01Ohm for contact 1 and 0.02Ohm for contact 2 you must see a big current difference.
Contact 1 will have 2x the current than contact 2...
This can even cascade if that current in contact 1 is more than the rated current.
Any thoughts?
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Doesn't matter. Current ratings of connectors are generally based on heat load. If one contact has exceptionally low resistance such that it hogs most of the current that means it will dissipate less power and get less hot. As long as you don't force unequal current sharing externally, you should be fine. I would still derate somewhat, but don't worry about it too much.
Do watch out for your ratings. Many connectors have lower current/pin ratings in the higher pin models within the same family. So a 4 pin connector might have 10% less capacity/pin than a 2 pin connector.
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I've seen numerous hot swappable power supplies that use multiple pins or edge card connectors in parallel to achieve the power requirements. As Neomys stated, use three instead of two; N+1 redundancy is always preferable to barely getting by.
Hot-swap supplies do it by having a short pin that connects last to enable the supply. That way all the high-current pins are already connected before any load is applied, and the main problem with multiple pins goes away. No sparks.
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Here is the thing.. Ohms lay applies here.
What is "Ohms lay"? Do you mean Ohm's Law?
if the resistance even is very low... e.g. 0.01Ohm for contact 1 and 0.02Ohm for contact 2 you must see a big current difference. Contact 1 will have 2x the current than contact 2...
Maybe if you are designing arc welders or blast furnaces. Is that what you are asking about? thousands of amps?
This can even cascade if that current in contact 1 is more than the rated current.
That can happen in ANY circuit. Even if you are NOT paralleling contacts.
Any thoughts?
You don't appear to understand Ohm's Law well enough to ask a Real World question.
If you are asking about some actual project or design, provide the necessary details and we can help you understand what you can and can't do with parallel contacts.
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Thanks for the reply,
No its not a real world problem..
However i was just wondering what would happen.
In my example i see the contact resistance as a resistor. So when you put these 2 in parallel the current will split according to the values.
I real work you could add a series resistor that is still very low but higher as the contact resistance.
About the heat.. sure when the resistance is very very low. heat will be no issue. but if one contact get more than its rated current it can heat up and the resistance could increase and then the current could transfer to the other contact. (self balancing..?)
anyway thanks again for the feedback.
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A series resistor will have no effect on the division of current in a subsequent parallel resistor situation.
Here's a calculator for your problem. Enter what you consider real-world numbers and see what you get:
https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-circuit-current-divider (https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-circuit-current-divider)
It's already been suggested that 3 1A pins should be sufficient for a 1.5A current. If you're uncomfortable with that for whatever reason, use more pins. 4, 5, 10. Whatever you want. Or use a higher-current connector.