# Avoiding DC voltage drop over long distances

I am trying to power my IP camera over a long distance from the power socket (about 10m). The included power adapter, rated for 5V 2A is about 1.5m long.

I cut the cable and extended it using a thicker wire (to prevent voltage drop) but despite that the voltage drops to about 4.5V at the end. The camera does power on, but making any PTZ movements on the camera causes it to reboot.

Any solutions other than buying a new adapter? And if I do buy a new adapter, what the general rule of thumb for voltage drop across x meter of wire when supplying low DC voltage?

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Relevant question. It shows that electrical wires does indeed have a non-zero impedance/resistance and that for longer distance, it is a characteristic to account for. –  Doombot Aug 29 at 13:18
You can buy buck converter modules such as Spehro mentions on ebay amazingly low prices. While they 'should be' [tm] reliable, if they fail short cct they may take your camera with them. A "crowbar" fuse blower may be in order. –  Russell McMahon Aug 29 at 13:31
Why not simply use regulated 5V boost converter on the receiving end? Both Maxim and TI have some sweet ICs that can be used for it... The overall power would drop a bit (sending low voltage over long distance does this anyway), but it still should be simpler than jumbo wires, step-up/step-down converter pairs etc. –  vaxquis Aug 29 at 19:08
Use 7 cables in parallel :-). –  copper.hat Aug 30 at 4:21

If you cut the wire near the adapter and extend it from 1.5m to 10m, you can use wire that has 7 times the cross-sectional area and it will behave similarly to 1.5m of wire.

That means that the copper (not insulation) diameter must be $\sqrt{7} = 2.6$ times the diameter.

An IP camera I happen to have on my desk here has an adapter that is rated 5V/2A and it uses AWG20 wire (marked on the wire). Check your wire to see if it has similar markings and use those rather than the ones I've referred to.

If you look at the above table AWG20 has a cross-sectional area of 0.52mm so, AWG10 wire with a cross-sectional area of 2.588mm should do it (AWG 12, the next thinner common size, is a bit too light).

If you are in a country that uses mm rather than AWG you can use the mm diameter of 2.1 directly. You could use something like speaker wire- for example Belden 5T00UP is about 9mm O.D.

That's pretty fat wire. If it won't fit then you'll have to consider using a higher voltage and putting a regulator at the far end of the cable. A simple adapter alone won't do it unless you have fat wire.

For example, you could use a 12VDC adapter at the far end and an LM2576-based switching regulator to reduce that to 5.0V. Note that if the regulator fails 'on', the camera would likely be destroyed. You may be able to find a module based on such a regulator on eBay etc.

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Skin effect? Either you should use 7 times the diameter, or stranded wire. Or is skin effect only for AC? –  Ben Voigt Aug 30 at 0:04
@BenVoigt Skin effect is only important for relatively high frequency AC or thick wires- even at 60Hz it is not important for wires less than 0.5" copper diameter or so. Resistance (and thus voltage drop for a given current) is inversely proportional to cross sectional area, so 7x the area (not diameter) is all that is required. Remember R= $\frac{\rho L}{A}$ –  Spehro Pefhany Aug 30 at 4:03

Use an extension cord to get the AC the required distance, then use the power adapter.

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Do you not think I would have done that if it was possible? The environment does not allow for an A/C extension as the complete wiring is inside a piping. –  navigator Aug 29 at 5:21
Well you did not specify that and this solution is the smartest, given your constraints. –  Vladimir Cravero Aug 29 at 8:08
@navigator You just need an extension cord with a plug that can be disassembled, then running the wire through the pipe shouldn't be a problem. –  eBusiness Aug 29 at 19:27
@eBusiness, yes that would have been the obvious solution. However, in this case, the camera is located high up in a place where dangling an extension A/C point is neither feasible nor elegant. –  navigator Aug 30 at 2:33
+1 An extension cord with a removable plug is how I get power for lighting through a tube under a walkway. –  Spehro Pefhany Aug 30 at 13:13

Two possible solutions:

(1) MAY work for you: Add a 5.4V capable supercapacitor at the camera end.
This can be 2 x 2.7V caps in series, ideally with balancing, or 1 higher voltage rated (rare).

This will charge to 5V over time and then handle camera surges. With prolonged use it will drop, but still be much better on surges and spikes.

The more capacitance the better.

Here is an example 5.4V. 5F cap.
ESR is 0.067 Ohm suggesting at least 10's of amps of discharge current.

In stock Digikey $US15.30 1's. Data sheet - Bussman, 5.4V, 5F 5F will notionally drop 1V in 5 seconds at 1A drain, or 1V in one second with 5A drain. Which means that surges of say a few amps for fraction of a second will (probably) limit fluctuations to in the order of tenths of a Volt. How well that works for you is TBD. You could use 2 x 2.7V supercaps in series. Net capacitance would ha half the value of 1 cap for two equal caps used. eg 2 x 2.7V, 100 uF in series = 5.4V, 50F equivalent. Here is a table of Digikey supercaps, in stock, priced in 1 quantity and in order of descending capacitance As a guide: 5000F$208!
1200F $49 100F$10
10F $3 (2) 4 x NimH AA batteries will happily settle down to your steady state voltage after a while and provide surge capability of perhaps 5A-10A. 4.5V/4 = 1.125V. A NimH would be better if steady state voltage was higher but may well be OK at this voltage. Knowing your camera's actual draw in various states would help us give better answers. What is the voltage at the adaptor under load? It should not sag much, but may. - Get yourself a 12-24V adapter and a 5V switching buck regulator module. Feed the higher voltage across the line and regulate it at the camera end. This will cause it to draw less current across the line, reducing the voltage (and hence power) loss along the line. - You've said this The camera does power on, but making any PTZ movements on the camera causes it to reboot. A lot of the other answers are telling you how to stop the average voltage drop. I.e. they are telling you how you can use a thick enough wire to supply more power to the camera continuously. I think this is overkill because the PTZ movements are all motors and motors use a lot of current for a very short time when they first start. So it may be that you don't need thicker wires, you just need some power stored at the camera end of the cable to help with this inrush. Russell talks about using batteries or supercaps to solve this problem. That's a good approach but I think it would be a good idea to try to see how much inrush there is before deciding on a particular solution - if you are able to look at the voltage (at the camera end of the cable) on an oscilloscope that would be ideal. A simple cheap aluminum-electrolytic capacitor or 2 may be all you need. - And if I do buy a new adapter, what the general rule of thumb for voltage drop across x meter of wire when supplying low DC voltage? Well, voltage drop depends on current consumption and extension wire resistance. For example if the PTZ motors consume 1A and your wire resistance is 1ohm, by ohm's law you will get a voltage drop of 1Ax1ohm=1volt. If the camera needs 4.5 volt to work, then definetely your cam will reset. If you can't change your load (like in this case), the solution is to use use even a thicker wire to lower the resistance of your extension. For Example, with a 0.3ohm wire you would get only a 0.3v drop and the camera will still work. Edit: I hope a thicker wire fits inside the piping. - My recommendation would be to use some sort of a DC-DC step-up converter at the feed and a step-down converter at the sink. The higher the voltage in the wire is, the smaller are the relative losses per distance. Building and designing efficient DC-DC converters by oneself is a tricky task for experts, I would recommend any novice and anyone wanting it quick to go and buy pre-built ones. - Do you not think I would have done that if it was possible? The environment does not allow for an A/C extension as the complete wiring is inside a piping. – @navigator You just need an extension cord with a plug that can be disassembled, then running the wire through the pipe shouldn't be a problem. – eBusiness 19 hours ago @eBusiness, yes that would have been the obvious solution. However, in this case, the camera is located high up in a place where dangling an extension A/C point is neither feasible nor elegant. Your thinking is too constrained. AC is the best way to transmit power over long distances, that was settled during the War of the Currents back in the 19th century. But "run AC to the camera" doesn't have to mean a Home Depot cord with a duplex outlet and a plug-in wall wart at the end of it. All you need is a couple of transformers. The first one should be mains to something around 30-40V, the second from that voltage to 12v. Then$5 worth of components to produce 5VDC. You should have no problems putting most of the components inside a small tube, the transformer might be a bit bulky.

As soon as you drop below about 60v the electrical code ceases to apply. You will need a suitable enclosure around the first transformer but after that you don't need any specific construction methods, just reasonable ones.

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Building on Enemy Of the State Machine's and paul's suggestions, here's what I think would be the simplest solution:

1. Buy a cheap 12 volt wall wart capable of supplying at least 1A or so. (You might already have one lying around, in which case you can skip this step.)

2. Also buy a small 12 VCD to 5 VCD step-down converter, like this one or this one. These are commonly sold for automotive use, so you should be able to find one cheaply in many stores; the Amazon links are mostly for illustrative purposes:

3. If necessary, cut and replace the cable for the 12 V wall wart and run it up to your camera. Connect your camera to the cable via the step-down converter.

Let's do the math to confirm that this'll work: Your camera draws 2 A at 5 V, for a total of 10 W, and your wire seems to have a resistance of about 0.5 V / 2 A = 0.25 Ω. Assuming a 95% conversion efficiency, the converter will draw 10 W / 0.95 ≈ 10.5 W, or I = 10.5 W / (12 V − U) amps, where U = 0.25 Ω × I is the voltage drop over the wire. Substituting the second equation into the first and solving, we get U ≈ 0.89 A, which is well within the capabilities of a 1 A wall wart.

We also see that the voltage drop over the wire is I ≈ 0.89 A × 0.25 Ω ≈ 0.22 V, leaving 11.78 V for the converter, which it should handle fine. (Car power converters are, necessarily, robust; the one in the picture is specced to take anything from 8 V to 23 V.) The power dissipated over the wire will be 0.89 A × 0.22 V ≈ 0.2 W, or, for a 10 m wire, 0.02 W/m, which also seems reasonable.

If you wanted, you could bump the intermediate voltage up to 24 V or even 48 V to make it even more efficient, but the nice thing about 12 V is that it's so commonly used, so the parts are cheap and easily available, and it's quite sufficient here.

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The way this is normally done is to use a 4-wire power supply which has sense and force wires.

2 thicker wires (force) supply the current and 2 generally thinner wires just sense / measure the voltage (not current carrying) at the camera end. The power supply then regulates to supply the 5V at the target end, automatically compensating for voltage drop along the cables ( obviously need to be thick enough to sensibly supply the current within the regulation range of the PSU)

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