Current Draw

You can increase the current above the pin rating by paralleling, but it is a good idea to derate them a bit to allow for non uniform distribution. Having the junction of those pins as far apart as possible allows the cable resistance to even out the distribution a bit, also. I would go for something like 70% of the pin rating when paralleling, so 4 or 5 1 amp pins in parallel for 3A.

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John Popelish
Reply to
John Popelish
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Hi there,

I'm working on a hobby project which requires a large amount of terminal leads coming into my project box. For this purpose I'm using a 25 pin connector with each pin rated at 1A. The little computer I'm using draws about 3A. Can I simply put 3 or terminals together therefore increasing my current rating? I'm guessing that the path of least resistance will not allow this to work that well. What sort of component would I need to get to draw exactly 1A from each terminal end and then combine this into 3A which I can then deliver to my computer?

I hope I'm making sense.

Thanks

Willem

Reply to
Willem

Hi John,

Thanks for the advise. I tried experimenting with an amp meter and found the non uniform distribution to be quite extensive - but maybe there is some resistance with my amp meter. Just so I understand some of the physics here (its been a while since high school physics - I'm more of a computer guy, but that maybe changing): will my 3A component try to draw from one pin first, whichever one offers the path of least resistance, and as soon as it nears the pin rating of 1A then start drawing from the other pins respectively. Or will the current just flow through all of the pins near or at their respective ratio i.e. 3A draw - 3 pins then 1A each or 4 pins then ..75A each , 5 pins then .6A each etc..

Thanks again for all your help

Willem

Reply to
Willem

There must be a significant (compared to the contact resistance) resistance in your amp meter. You would have to connect similar meters in series with each of the paralleled pins to see what is really going on, and even then the meter resistances would help level out the currants.

The current will divide inversely proportional to the contact resistance. If one contact has 1 milliohm of resistance and one has 2 milliohms of resistance, the first will pass twice as much current as the second. But the heat each contact produces (the thing that actually limits how much current you can pass before the plastic housing melts) is proportional to the current squared times the resistance, so the 1 milliohm contact will produce twice as much heat as the 2 milliohm one does.

This one.

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John Popelish
Reply to
John Popelish

some

This is likely the cause of your non uniform current distribution. A typical digital multi meter will have leads that are about one meter long each and be made of 18AWG wire. Such wire has approximately 20milliohms resistance per meter, and on the 10A (or whatever high current scale your meter has) current shunt usually has something in the range of 10milliohms as well.

My DMM on the 20A current scale has about 44mOhms resistance (7mOhm current shunt, the rest 18AWG test leads). On the mA current scales the resistance is huge. IIRC it is something like one ohm or more, but I could be wrong on that.

Reply to
Fritz Schlunder

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dunno what EXACT you are trying, but if the connector is of a type allowing

1A per connection then as John said taking 4-5 wires should suffice if there are large differences between currents per wire then I'd check out the junctions, since if all are soldered identically ampel difference will exist and no, current will not start with one wire and if reaching 1A begin flowing in the second, all wires of a parallel set will begin conducting at the same time but MAYBE slightly different amount if welding isn't ok
Reply to
peterken

Others have answered your question, but I just wanted to clarify that the old "path of least resistance" business is a common misconception. Current simply flows inversely proportional to resistance. It's like water running in pipes: If you put a fat (low resistance) pipe and a skinny (high resistance) pipe in parallel, the fat pipe gets more of the flow because it has a lower resistance, but it doesn't get

*all* the flow... the skinny pipe still gets its proportional share.

Having said all that, however, there are some situations (and yours isn't one of them) where there is such a thing called "current hogging". That's typically where you have parallel devices like bipolar transistors whose resistance goes down as they heat up. So if one device starts out with a slightly lower resistance, it gets proportionally more current and heats up proportionally more, which makes its resistance go down, which makes it draw still more current, and so on.

Hope this helps.

Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

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Reply to
Bob Masta

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