1

This may be a simple and beginner question but I am kinda lost how else I could solve my problem.

I decided to make 2 boards and stack them via a board connector. Now usually these board connector have current ratings per connection. On the bottom board I have my power regulation (24V -> 12VDC, 10A max.) that needs to get to the top board. If I now take 2x 5A connector, would that work as the current should split up equally between the pins or do I think wrong here?

Thanks.

Daniel Do
  • 191
  • 10
  • 2
    Can you afford three? That should address any current imbalance due to slight variations in resistance and would run the pins below rated power so there would be a lot less heat. \$ P = I^2R \$ so reducing the current in a pin by 1/3 results in a contact heating reduction to \$ (\frac 2 3)^2 = \frac 4 9 \$ of the 100% rating. – Transistor May 12 '21 at 10:47

3 Answers3

4

When you consider current rating of connectors, they are usually tested with conservative criteria: maximum voltage and a target operating life that is sometimes 10^5 operation, sometimes much less (100-200 operations). This target operating life depends o the intended use of the connector, so this may give you additional margin or none at all.

This said, for the typical Harting-like board-to-board pin headers, you may use more than the strict maths, as @Transistor says. Also, take into account the return (so + and -), possibly separating a bit the pin pairs (1 + always with its 1 -) to help dissipation and avoid hot spots. Since such pins are from EMC and signal integrity viewpoint a "ground", they help the shielding of signal lines, normalizing signal transmission and reducing crosstalk.

So, with 10A flowing between boards, allow let's say 3 pairs (+ & -): 2 may be placed at the extremities of the connector and one pair in the middle of the pin array (assuming 2 rows, so + and - facing each other).

andrea
  • 1,589
  • 8
  • 13
  • Thanks for that answer. Ontop of this question maybe you can answer this one too. I want to also include a 50MHz clock onto either of the boards but booth would need it. Can I use a connector for high freq. signals ? I heard that I should keep the length short but nothing about connectors. If you know that would be a great help. ! – Daniel Do May 12 '21 at 13:31
  • I think that a normal pin header 2-row connector will work, puttin the clock with a ground reference next to it. The, maybe if you equalize the trace length on each board to have the same delay (approx), that would be even better. Good signal connectors and cables are needed when you go between boards far away, not in a stack, or your clock is really fast (then you need an interposer designed for GHz ... I tested some of Smiths Connectors, they are really good stuff) – andrea May 12 '21 at 13:48
0

Yes, you can see a connector as a resistor that's almost zero. So adding some in parallel will divide the current through them.

Although I'd add one or two more to not have to operate on maximum specs.

Swedgin
  • 723
  • 6
  • 20
  • But "almost zero" is not zero, and one-half of "almost zero" is still almost zero but will take more than half of the current. You are implicitly assuming that the connections will have the **same** resistance, which does not seem to be a conservative design practice to me. – Elliot Alderson May 12 '21 at 11:29
  • @ElliotAlderson I meant that its resistance is negligible and I did not asume they are all the same resistance. Draw the circuit (voltage source, x-number of sub 10 mOhm (does not need to be the same) in parallel), and you'll see the current drop across the resistors as you add mode resistors. -> `So adding some in parallel will divide the current through them.` – Swedgin May 12 '21 at 13:08
  • Though feel free to delete as @andrea made a better answer – Swedgin May 12 '21 at 13:09
0

First of all, you didn't say if the 5A per connector value is a derated number or a rating (max, maybe) from the manufacturer.

We always start off by derating the manufacturer's current rating by 50%, so that a 1 amp rated pin (or connector) would only be used with a max current of 0.5 amps.

Then, there's another derating we apply when multiple pins are used to carry higher currents. We take the total current, divided by the derated-current-per-pin, then add another 25% to the number of pins calculated.

So lets say we need to carry 3 amps, with a manufacturer's connector whose pins are rated for 1 A. The number of pins used would be (3/0.5)*1.25 = 7.5 pins, rounded up to 8.

SteveSh
  • 9,672
  • 2
  • 14
  • 31