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I'm going to use a RS485 modbus connection which requires a twisted pair. Now I also need power, that will be a 5V line and a ground. The distance will be about 3 meters max. Can i use a twisted pair for the 5V and gnd as well? Or will that result in a voltage drop?

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    \$\begingroup\$ You will always get a volt drop down cable if you are taking current. \$\endgroup\$ – Andy aka May 29 '18 at 11:12
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    \$\begingroup\$ Check this link \$\endgroup\$ – Pryda May 29 '18 at 11:17
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    \$\begingroup\$ The company I work at designs gas detection systems for car parks. The detection nodes use modbus RTU or canbus. We use 3x2x0,8mm² STP cables to couple the nodes to the loop. To save on cabling and avoid the hassle of bringing different types of cables to the project, we use one pair of the cable to pass the 24V power supply to the nodes. Our detection loops are generally over 500 meters long and work without problems. \$\endgroup\$ – Bart May 29 '18 at 11:26
  • \$\begingroup\$ @Bart thanks for verifying with an excisting application. I guess it won't be a problem at all then :) The 2 nodes on the RS485 modbus RTU get their own power (one only gnd) and have a minimum of 3.3V signal I believe. I will be using like 24AWG wiring tho, but still.. \$\endgroup\$ – Matthijs van Hest May 29 '18 at 11:34
  • \$\begingroup\$ We did not however make a theoretical analysis of this, but commissioning tests never revealed problems. Of course your device needs proper decoupling from the power supply, and you need to keep the voltage drop low. Our sensors (CO and LPG) sink about 300 mA each from the 24V supply, which is doubly fed into the loop. For very long loops we use 2 wire pairs for the power supply. \$\endgroup\$ – Bart May 29 '18 at 11:42
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There are a couple things to take note of to know if this will work:

  1. Peak current draw of your load: $$I_p$$
  2. The acceptable voltage drop your load can accept: $$V_d$$
  3. The resistance of your cable: $$R_C$$

If:

$$I_p \times R_c \leq V_d$$

Then you're good to go.

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  • \$\begingroup\$ Supply is 2.4A max. Do'nt expect to draw that much tho. Voltages above 3.3V are usually good enough. So 1.7 volt drop. The cable i'm thinking of is 0.5 ohm per 5 meter. 2.4 x 0.5 = 1.2. Which is lower than 1.7. These are maximum values. Should be ok haha.. But thanks for the formula! \$\endgroup\$ – Matthijs van Hest May 29 '18 at 11:51
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    \$\begingroup\$ For short current peaks, proper decoupling should reduce the magnitude of voltage drops. For the average current draw, yes, the cable resistance should be sufficiently low as you indicate. \$\endgroup\$ – Bart May 29 '18 at 11:53
  • \$\begingroup\$ @MatthijsvanHest if possible, I would put a DC/DC converter at both ends of the cable. Step up to something like 12V or 24V at the supply end, and step back down to 5V at the sink. That way the current down the cable will be lower (higher volt = lower current) so lower losses, and you'll get a regulated 5V at the other end. \$\endgroup\$ – Tom Carpenter May 29 '18 at 11:53
  • \$\begingroup\$ @TomCarpenter that would be very reliable yes. It's going to be enclosed, and therefor I don't like to use dc converters (heat) too much. Also there's a budget, so I rather have a very very very small voltage drop than more components which can break and such. But thanks for the advice, will keep it in mind ;) \$\endgroup\$ – Matthijs van Hest May 29 '18 at 11:56
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    \$\begingroup\$ @MatthijsvanHest 95% efficient DC/DCs are readily available and quite inexpensive nowadays. The trouble with the calculation is the current flows through both VCC and GND cables, so you actually have 6m of cable (3m there, 3m back), so you'll end up with about 3.5V, assuming no other source of loss (e.g. is your 5V supply exactly 5V? or does it drop to 4.8V or maybe lower under load). Further, 1.5V is not a "very very very small voltage drop" - it equates to about 3.5W dissipated in your cable, far more than a DC-DC converter would dissipate. \$\endgroup\$ – Tom Carpenter May 29 '18 at 12:38

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