Current Draw

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.

 

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

 

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

 

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
www.daqarta.com

 

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.

 

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.