Simple (I hope) voltage drop/resistan over distance question

[nadimaj]

Hi,

So, it's been a while since elementary electronics and I now have an issue I desperately need to resolve other wise I'm going to have to rip through a wall.

I'm working on a project where I'm installing a USB based touch screen monitor (yes 100% USB based) that is situated 3m away from a PC. USB, if you're in the know delivers 5v along with the data connection and as expected over 3m it doesn't work too well, barely being able to power up the device. The device also comes with an additional 5v DC jack and AC adapter that one could use if extra power is required. I was able to forsee this, so along with the wall embedded USB cable, also pre-wired an additional (telephone grade) 2-core cable that takes 5v from a distribution panel and delivers it to the unit via a DC jack. The wiring from the distribution panel is about 5m in length if I am to include all the bends and twists.

Despite all this I cannot get the monitor to power up using this 3m USB lead + approx 5m of cable carrying the additional 5v DC. The voltage reading at the DC jack end is delivering a pure 5v and the distributor can deliver up to 3A (the unit requires 0.5).

However, using the same wall embedded USB lead and instead of using the distributor supplied 5v, if I plug in the supplied AC adapter into a socket near the unit (supplied AC adapter cable is 3m in length), it all works just fine. The supplied voltage coming from the AC adapter is identical to the voltage provided by my distributor (measured using a multimeter), although the max current from the adapter is about 1A.

Clearly the wall embedded 5v carrying cable is not able to deliver the same voltage/current as the AC adapter, but I find it difficult to believe that this is due to the additional 2m length. I have alarm cable running around the entire house in some cases delivering 5v and 12v up to 25m away!

So here's where I guess I'm demonstrating my ignorance. Using a multi-meter, why am I able to measure a pure and consistent 5v from the wall embedded power cable (exactly the same as the AC adapter)? What (and how do I measure it) can I do in order to determine why the wall embedded cable is not able to deliver the same voltage/current as the supplied AC adapter? Is there something I can do with a multi-meter?

Thanks for trying to understand my problem![/u]

 

[flameport]

The multimeter has a very high impedance, and as such doesn't put any load on the supply. Therefore effects such as voltage drop won't be apparent.

For any meaningful measurements, you need to measure the voltage with the load (the screen) connected.

 

[Mursal]

Yes, always leave the load connected and measure the voltage across it, to get a true voltage reading. True USB will not (as far as I know) supply 500mA (0.5A) constant supply.
Probably overloading and going into current limiting mode to self protect the output.
Unless you have an old fashioned bad connection? Just place the voltmeter across any suspect connections (with the load connected) the meter should read 0V. I wouldn't think its anything to do with wire length

Let us know how you get on?

 

[ericmark]

A 7805 is cheap so I would send over 5 volt and reduce to 5 volt at the screen.

I do hate this 5v, 12v marking there must be a tolerance but they never seem to give it.

 

[FDJ]

an additional (telephone grade) 2-core cable

That's the problem.
There is little scope for voltage drop with just a 5v supply. Really needs a cable with a much higher cross-sectional area.
Can you pull one through with the existing "telephone" cable?
Frank

 

[nadimaj]

Thanks everyone for your input and suggestions.

There is little scope for voltage drop with just a 5v supply. Really needs a cable with a much higher cross-sectional area.
Can you pull one through with the existing "telephone" cable?

If that is true, how do you explain something like alarm wiring (very similar to telephone wiring) that runs up to 25m from the power distribution point?

The latter suggestion is a good one but wont work as there is now unfortunately plaster and coving in the way not giving me any free run either way.

A 7805 is cheap so I would send over 5 volt and reduce to 5 volt at the screen

- http://electronics.howstuffworks.com/digital-electronics4.htm - Excellent idea, will work on that.

 

[SimonH2]

If that is true, how do you explain something like alarm wiring (very similar to telephone wiring) that runs up to 25m from the power distribution point?

Several factors :
1) They use 12V (or sometimes 24V)
2) The sensors take very little power
So low current = low volt drop. Low volt drop against a higher initial voltage is less (% wise) loss.

3) The devices are designed for this and I'd imagine have a reasonable tolerance for their supply voltage (and, I'd imagine, a fairly good tolerance to noise/interference picked up in the wiring).


Back to your original question ...
CW1308 cable (assuming it's the standard 0.5mm^2 size) is a tad under 100R/km/conductor. So you've about 10m total (5m in each of 2 conductors) which means about 1 ohm of resistance. Try and put 0.5A through that, and you'll get about 0.5V drop. That puts your supply at the equipment down to about 4.5V which is way below the tolerance for a lot of 5V electronics.

As suggested, given your constraints and understandable reluctance to rip up the decor, you'd be best using a higher voltage supply (say 9V to 12V) and a 5V regulator by the screen.

 

[nadimaj]

Thanks Simon, what you say makes perfect sense. I thing I might have had the solution all along (you jogged my memory), going to try it out today.

Regards

 

[JohnW2]

CW1308 cable (assuming it's the standard 0.5mm^2 size) is a tad under 100R/km/conductor.

Lest there be any confusion, although what you say about the resistance of CW1308 cable (hence your subsequent calculations and conclusions) is correct, CW1308 conductors are 0.5 mm diameter (about 0.2 mm² CSA), not 0.5 mm².

Kind Regards, John

 

[SimonH2]

Oops

 

[nadimaj]

Done it!

Had an old DC 12v to 5v convertor, so sent 12v down the telephone cable and converted to 5v nearer the unit. All working.

Thanks