Current Draw
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Posted by Willem on January 28, 2005, 9:18 am
Please log in for more thread options 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 | |||||||||||||||||||||||||||||||
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Posted by John Popelish on January 28, 2005, 4:52 am
Please log in for more thread options 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. -- John Popelish | |||||||||||||||||||||||||||||||
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Posted by Willem on January 28, 2005, 10:00 am
Please log in for more thread options 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 > Willem wrote:
>>
>> 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 >
> 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. > > -- > John Popelish | |||||||||||||||||||||||||||||||
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Posted by John Popelish on January 28, 2005, 2:27 pm
Please log in for more thread options Willem wrote:
>
> 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. 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. > 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. 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. > 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.. This one. > Thanks again for all your help
> > Willem -- John Popelish | |||||||||||||||||||||||||||||||
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Posted by Fritz Schlunder on January 28, 2005, 3:08 pm
Please log in for more thread options
> Hi John,
some
> > 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 > resistance with my amp meter.
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. | |||||||||||||||||||||||||||||||
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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