# How to extend the DC part of my laptop power cable without experiencing voltage problems?

I use my laptop for extended periods on the mains in two different rooms of the house, and for ease of use I keep a separate power cable in each room. In one of the rooms I need to extend this cable.

Each cable is in two parts. The AC part runs between the mains plug and a two-pin C7 figure-of-eight plug. This then fits into a line lump transformer box, from which it can be removed.

The DC part is moulded into the transformer box and runs to a "universal" laptop-tip female connector at the other end, into which the male tip plugs that fits into the laptop itself.

I want to extend the DC part of the cable. To be clear, I want about 3-4 metres of cable between the transformer box and the laptop, which I want to be able to tie up and leave on a shelf and then untie and bring down to the laptop. I realise I could extend the AC part of the cable, but that's not what I want to do. The extension must be on the DC side, because I do not want to keep moving the transformer box.

The issue is that the DC part of the cable has two small cylindrical boxes, one near each end, which I think are voltage regulators.

It's possible to find DC extension cables with 5.5mm x 2.1mm male and female ends, rated up 24V, but they tend to be described as suitable for CCTV equipment and sometimes other devices too, but not for laptops.

What's the best way to extend the AC part of the cable? Will there be a problem if I have 3-4 metres of cable without any regulators, running from the existing tip to a new tip?

It is hard to extend the DC, because you will have a higher voltage drop on the cable, the voltage will drop with the Current and resistance of the cable, according with Ohm's law, a longer cable, means higher resistance and lower voltage available to your device.

The cylindrical parts are probably not voltage regulators, they are most likely ferrite to filter out High frequency noise, giving you a cleaner DC voltage, so it will not compensate the voltage drop in the extension.

If you add a few meters of cable to the end of the existing DC plug the problem will be: voltage drop, and higher noise pickup.

With the additional voltage drop the supplied voltage might not be enough to power your laptop.

I suggest you to extend the High voltage AC side.

• One quibble: The ferrites are not there to give you "cleaner DC voltage". They are there to prevent RF radiation from the power cable. Otherwise, RF noise from inside the equipment being powered can be picked up by power input cables and then radiated from the unshielded power cable, possibly interfering with radio signals of all sorts and making FCC and similar certification difficult. – Jamie Hanrahan Dec 9 '18 at 21:42
• The increased inductance from lengthening the DC side can be a huge problem. Many laptops are already very close to edge in having sufficient capacitance to hold the supply voltage up during sudden increases in power consumption and any additional inductance in the power cord can push them over the edge. If you do lengthen the DC side, do not ever use the laptop with the lengthened cord and no battery! – David Schwartz Dec 9 '18 at 23:33

It is usually better to design these things instead of guessing.

Let's suppose we want to keep the voltage drop to 0.2 V over 4 m of cable, and let's say the laptop uses a maximum of 5 A. (Most laptop PSUs' output varies by at least that much in normal use.) As for the maximum current consumption, check your power brick's maximum output and use that instead of 5 A.

(The target of a maximum 0.2 V drop is a conservative estimate - it's only 1% of the typical laptop PSU output! In other words most laptops could probably tolerate a considerably greater voltage drop than that. And the 5 A current draw is quite a high estimate especially for many smaller laptops. On the other hand, your power brick's output might already be a little bit low, and other things might be not right on spec, and we're adding some splices too... so a conservative estimate isn't unreasonable here. See the comment discussion.)

The resistive drop in a conductor is simply E = IxR, where E is the voltage drop in volts, I is the current in amperes, and R is the resistance in ohms.

Then R = E / I = 0.2 V / 5 A = 0.04 ohms. You need your round-trip (ie counting both sides) cable resistance to be less than this.

Each side contributes half, so you need the R of each side to be half of this: 0.02 ohms for 4 m of cable.

Most wire tables list wire R in ohms/1000 ft or ohms/km. 1000 m / 4 m = 250, so multiply both sides by 250 to get 5 ohms per 1 km of wire. We want each side of the cable to have R of no more than that.

Consulting a standard wire table, it appears that cable with each conductor using 2mm diameter wire will be close enough (R only a few percent high). In AWG the closest standard is 12 gauge.

That will be some fairly thick cable (in the US we use 12 ga. conductors for 20 A outlets and it is also the heaviest extension cord in anything like common use). And do remember that your splices at each end will likely add some non-insignificant resistance too.

4 m of very thick extension cord does not seem to me to be very convenient to just take down from a shelf and put back up frequently.

If your laptop uses significantly less current than that, or if you think you can tolerate a larger voltage drop, then you can use thinner cable. You now have the formulas and the table to find those numbers.

But I agree with the others - I'd extend the AC.

• But... how reasonable is it to assume that the voltage at the cable is that critical? Wouldn't any critical voltages be regulated onboard? :) – junkyardsparkle Dec 9 '18 at 23:00
• @junkyardsparkle Given that laptop power brick output is usually 20 V or so, it probably isn't that critical. But I won't say "almost certainly". Even if the laptop could run from say 19 V instead of 20 V, its power control circuit might decide not to, and complain "wrong power adapter". So, to be safe, I was using overly conservative inputs. Anyway, my main point here was not in my particular input values, but to show the method. If you or the OP don't like my assumptions, feel free to use others and see what you get. – Jamie Hanrahan Dec 9 '18 at 23:14
• I don't dislike your assumptions... more that I'm starting to question my own. It's quite possible that I give manufacturers too much credit for over-engineering things like that... – junkyardsparkle Dec 9 '18 at 23:21
• Oh... and to answer your other point, yes, of course everything is regulated on-board. But how good that regulation is, is of course yet another unknown. – Jamie Hanrahan Dec 9 '18 at 23:23
• Given that so many electronic circuits are built with 5% resistors... that's good reason to suspect that an input voltage variance of 5% (1 volt over 20 V) should be tolerated... assuming that nothing else in the path is very far off of center. But what if they're not? Suppose your power brick output is already half a volt low? And suppose the laptop's input voltage "tester" is overly fussy? But... that is 2.5x the drop I started with, meaning your wire R can be 2.5x as high. Looks like you'd still need AWG 16 (typical "ordinary lamp cord" in the US) or about 1.5mm wire. – Jamie Hanrahan Dec 10 '18 at 0:15