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A conundrum. I've been asked if I can add more current capacity and less voltage drop to a legacy connector, but I have limitations. For the legacy products to work on this same connector it needs to keep the 10pin pinout order the same. This allows a 14pin ribbon cable to simply cut off it's unused pins for those products. But they'd like new products that might draw more current. It's a modular system, so there are unknowns about how much current each module needs. Some modules will draw around 30mA per 3.3v, 5v, +15v, and -15v as were used with this connector. But some modules draw up to 110mA. (For power hungry ones that draw 200mA, there is a beefier power option.) The ribbon cable in question is about 12 inches long and it's 1mm pitch 28AWG (with a 2mm IDC connector on each each end.)
I'd like to consider a whole new pinout, but since this pinout already parallels the power signals in this order -15v Ground +15v +5 +3.3v in that order, perhaps another 4 pins of -15v +15v +5 +3.3v in that order will work OK. I'd rather intersperse ground between signals but they want to only go to 14pins if possible. Should I bother? Perhaps they can simply double up on the connectors for certain modules. The only issue if done that way is that the Clock and Data lines end up with more "stubs" or parallel connections and so more potential for reflections (unless we cut those signal off those secondary cables.) Plus using two power connectors seems a little odd. Schematically it's like this. (P.S. Yes it's got a digital ground for +3.3v and 5v, and an analog ground for audio and +/-15v.)

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  • \$\begingroup\$ Welcome. The wire gauge limits current per wire to 100 mA at most. Voltage rating of ribbon cables is about 150 VDC max. You could raise +15 to as high as +48 volts, the maximum voltage not considered to be a shock hazard, often used in PoE supplies. 100 mA gets you close to 5 watts at 48 volts. Down convert to get 5 volts at 800 mA or 15 volts at 250 mA. Just 1 of many options. \$\endgroup\$
    – user105652
    Commented Jul 4, 2020 at 1:58
  • \$\begingroup\$ I can't change the existing 10pin pinout of +5 +3.3v, +15 and -15v that are getting sent regardless. (Other voltages are not available.) My question is whether anyone sees a problem with that particular pinout for 4 added pins. I can't change the existing 10pin pinout of +5 +3.3v, +15 and -15v that are getting sent regardless. (Other voltages are not available.) My question is whether anyone sees a problem with that particular pinout for 4 added pins. Otherwise I make it the policy that when more current is needed, use a second cable and double the connectors on the module i2c stub be damned. \$\endgroup\$
    – JDeviant
    Commented Jul 6, 2020 at 1:12
  • \$\begingroup\$ Worst case is you need a dedicated 24 VDC power feed (and ground return). 28 ga wire is limited to milliamps of current. \$\endgroup\$
    – user105652
    Commented Jul 6, 2020 at 1:16
  • \$\begingroup\$ BTW the wire I'm talking about is 7 strands of 36AWG wire. I have assumed 7/36 current capacity is then much less than the theoretical 28AWG capacity? Is this where we are getting the 100mA approx. limit from? I find little information on this. I see in one case of an existing module that a potential of 300mA of 5v power is being drawn from a cable that has only 2 pins of 7/36 wire. \$\endgroup\$
    – JDeviant
    Commented Jul 6, 2020 at 17:58
  • \$\begingroup\$ I still need confirmation on where the 100mA limit comes from so I can use this data. Because in theory if tables that I look up are correct I would get over 200mA per 28AWG (or 7x 36AWG at 35 mA). Is this because it's on an IDC connector? \$\endgroup\$
    – JDeviant
    Commented Jul 6, 2020 at 18:28

1 Answer 1

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One approach would be to bring in more current on the +15v line, then add Buck Regulators to convert the +15v to +5v.

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  • \$\begingroup\$ Before I go too far into the rabbit hole on this, I'm trying to get information on where the 100mA limit that Sparky256 comes from so I can use this data for reference. Because in theory if tables that I look up are correct I would get over 200mA per 28AWG (or 7x 36AWG at 35 mA) wire. Is this because it's on an IDC connector? Is it because it's a "practical" limit? \$\endgroup\$
    – JDeviant
    Commented Jul 6, 2020 at 18:31
  • \$\begingroup\$ @JDeviant Just look up "Amp ratings of copper wire". Stranded has a slightly lower rating as it is not a single mass of copper. Actually it is 1.4 amps as individual local signal feeds and 266 mA as a long power feed. The penalty is in the distance or length of the wire. \$\endgroup\$
    – user105652
    Commented Jul 7, 2020 at 23:55
  • \$\begingroup\$ Ah yes, that's what I've found. But we are only talking about 12", so I figured voltage drop was more of an issue than current handling. I've seen the 266mA figure (or something around it) too. But you first wrote "The wire gauge limits current per wire to 100 mA at most." Perhaps you were applying a rule of thumb to divide that by half? \$\endgroup\$
    – JDeviant
    Commented Jul 9, 2020 at 18:26

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