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I'm going to transmit about 7 Watt DC over 4 meters long twisted pair Ethernet cable. I'm thinking of using Buck and Boost converters to step-up to 30 V and step-down to 5 V the voltage before and after the transmission to reduce path loss.

My question is can DC-DC conversion cause to more power loss than the path loss?

EDIT

Another requirement is ~4A @ 5V and the cable is 10m long. Then ~20W has to be transmitted.

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Yes, and in your case Boost/Buck certainly will be less efficient unless you are using extremely low voltage/high current for your application.

DC-DC convertors tend to be no more than about 90-95% efficient, so placing one at each end of your 4 m cable will result in approximately 10% loss.

In addition, the Ethernet cable has a resistance, so there will be an additional loss in the cable. The fact that you boosted and bucked the voltage does not eliminate these losses, however conversion to a higher voltage will reduce the I^2R losses.

7 W @ 5 V is about 700 mA. Typical Ethernet cable is about 200m Ohm/m (loop), so voltage drop will be about 0.15 V/m. If you can tolerate 150 mV/m drop in your 5 V (about 3%), then you don't convert. If you need regulated 5 V, then start with say 6 V and put a low dropout regulator at the application end.

EDIT:

If you want to consume 5 V @ 4 A at the remote end, then it does make sense to boost/buck. Resistive losses at 5 V would be close to 3.2 V .....so not possible.
If you boost to 30 V (about 700 mA current for 20 W) then the voltage loss will be about 0.56 V.
You then Buck down to 5 V at the remote end. Likely you will have about (let's be conservative ) 90% efficiency in each convertor, so about 80% efficiency overall. You need to consume 20 W but need to supply 25 W of input power to allow for losses.

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  • \$\begingroup\$ Thank you for the answer. I have edited the question with another requirement. I think in that case conversion is efficient. \$\endgroup\$ – TRiNE Oct 7 '17 at 15:02
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    \$\begingroup\$ 7 W @ 5V is 1.4 amps, by my calculator. \$\endgroup\$ – Peter Bennett Oct 7 '17 at 15:30
  • \$\begingroup\$ @PeterBennett. Read the update to the question. The op updated to 4A @ 5 V \$\endgroup\$ – Jack Creasey Oct 7 '17 at 15:31
  • \$\begingroup\$ Yeahbut your calculation in your fourth paragraph is wrong. \$\endgroup\$ – Peter Bennett Oct 7 '17 at 15:34
  • \$\begingroup\$ @PeterBennett Fixed. \$\endgroup\$ – Jack Creasey Oct 7 '17 at 16:02
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Easiest method to check this is simply to measure the resistance of the wire you're planning on using. Your power loss over the wire is then easily calculated based on that measured resistance and the current you'd be using.

Assuming your power consumption is 7 watts at 5V, we can compute how much current that involves:

$$I = \frac P V = \frac 7 5 = 1.4A$$

We can then work out what the power loss is, based on resistance:

$$P_\mathrm{loss} = I^2R = 1.4^2R$$

You can then turn this into a percentage efficiency by dividing your loss power by your total power:

$$\mathrm{Efficiency} = \frac {P_\mathrm{loss}} {P_\mathrm{total}} \times 100$$

In this case your total power is the sum of your load power (7W) and the loss power.

Now for the regulator option, you need to keep in mind that you're dealing with the efficiencies of both regulators, so we need to consider the product of them both:

$$\mathrm{Total\space efficiency} = \mathrm{Boost\space efficiency} \times \mathrm{Buck\space efficiency}$$

A reasonable estimate for DC-DC regulator efficiency is 80-90%.

By comparing the two efficiencies you can see exactly which solution has the better efficiency. For short distances I'd bet my money on just using the wire.

Efficiency isn't the only consideration though. While the wire-only solution is likely to be cheaper and more efficient, there will be a voltage drop across that wire proportionate to the current:

$$\Delta V = I \times {R_\mathrm{wire}}$$

We computed the current above as 1.4A, so using your measured wire resistance you can work out how big a voltage drop there will be. You can then increase your input voltage by that amount in order to negate the drop. As an example, if your wire resistance measures 0.2Ω then your drop is 1.4*0.2=0.28V. Assuming you want 5V at the load end of the wire, you'd want 5.28V on input.

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  • \$\begingroup\$ Thank you for the answer. I know the calculations you did. But what I don't know is power loss of conversion. \$\endgroup\$ – TRiNE Oct 7 '17 at 15:18
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    \$\begingroup\$ A really important factor to make sure gets included is that you have to include the resistance of the wire both to and from the remote end when delivering power over a cable. It would be best to incorporate that into your answer so it does not slide into the cracks between the floor boards [or chassis cover :-) ] \$\endgroup\$ – Michael Karas Oct 7 '17 at 15:59
  • \$\begingroup\$ Which brings up another point. If the voltage drop along the return wire is significant, it will change the effective logic/reference levels for any other signals being carried on the cable. \$\endgroup\$ – Trevor_G Oct 7 '17 at 16:10

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