Wire Ratings

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geocom

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#109442 19-Sep-2012 21:41

Hi guys

I hope someone can help me with this one. As i have tried to look and have had nothing but conflicting ideas on the issue.

I am trying to drive an engine underwater. I have measured it to run at 8A@12V.

Now here is where the confusion is. Wire and connectors are rated at Voltages and amps. So if i have a connector that says it is rated for 300V@8A. Some sites say I have an exact limit on the part of 8A where as some say I have a limit of 2400W so at 12V I should be able to run 200A.

So which one is correct?

I think it should be in Watts. But i am in no way a qualified and up till now I have only been working at low voltage and low amps.

Geoff E

Niel

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#688499 19-Sep-2012 23:26

The voltage rating is the breakdown voltage of the wire. If you are to leave the cable in water then it might absorb moisture which will eventually corrode the wire (but will be a long time).

The current rating is irrespective of voltage or cable length. It is the maximum current that will give a known temperature rise under normal ambient conditions for the purpose the cable was designed for (e.g. outdoors rated cable will handle more heat). When you work at extreme temperature then you de-rate the cable accordingly.

There is a 3rd spec which is not a limitation on the cable but a limitation on your application. Wire has a finite resistance which causes a voltage drop so the longer the cable the more the voltage drop. In house wiring the test is to draw maximum current at the furthest socket and measure the voltage drop (between no load and maximum load) which should be less than I think it is 20V (or something like that). If the drop is greater then you need to increase the wire size, if the drop is less then you could have saved money by using thinner cable. But primarily this test checks for poor connections which will heat up and could start a fire, not checking your wire size.

For your application if you get a cable which has a datasheet it will state the resistance per km of cable so you can work out the resistance for your length x2 (because current flows down one wire and returns up the other). Then resistance times current gives you the theoretical voltage drop due to the cable length. I would try to keep it below 5%, or not more than 10% of the applied voltage. Remember 10% voltage drop is 20% power drop.

You might be able to increase the voltage you apply to the cable to compensate for the voltage drop, but remember the voltage drop depends on the load which will not be constant. The ultimate is a cable with a pair of sense wires back to the power supply so it changes the power supply output voltage to always get e.g. 12V at the other end of the cable irrespective of the load.

For short cables of only a few meters this is all academic and just use whatever 8A rated wire you can find. It is in water anyway so will be cooled down much better than if it was in air.

(I assume you measured 8A under load and in the target application, not free running.)

You can never have enough Volvos!

richms

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#688666 20-Sep-2012 11:03

If anything the current will go down with voltage as voltage drop becomes more critical.

That is why 1mm cable is fine for 10A on mains stuff, but would be well undersized in car use etc.

Richard rich.ms

gzt

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#689809 22-Sep-2012 10:38

geocom: Now here is where the confusion is. Wire and connectors are rated at Voltages and amps. So if i have a connector that says it is rated for 300V@8A. Some sites say I have an exact limit on the part of 8A where as some say I have a limit of 2400W so at 12V I should be able to run 200A. So which one is correct?

The relationship above is exactly backwards. Simple example -

10 volts at 10 watts = 1 amp
100 volts at 10 watts = 0.1 amp (100ma)

Taking this a bit further - a connector capable of reliably handing 200A at 12V is huge. Like the lugs on your starter motor or car battery earth.

Niel

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#689924 22-Sep-2012 16:47

Connector (and switch) ratings are done different from wire, because they need to handle a make/break surge (or spec'ed as not capable of). Here you also need to consider AC vs. DC, it uses different plating to handle the make/break current. Specs are also derated with increased ambient temperature. You need to stick within the specifications or else it will not last.

Wire is just wire. The current will cause a temperature rise in the wire. On double the length wire you get double the temperature rise, but spread over double the cable length so the temperature rise at any one point is still the same. But the issue is with length causing a voltage drop so your motor will not get the full voltage. Double the cable length and you get double the voltage loss which translates to 4 times the power loss. The higher the voltage the lower the effective percentage power loss (which is why power lines are run at higher voltage). Suppose you get 5V loss on 240V, it is only 2%. But 5V loss on 10V is 50%. It all depends on your application, there is not exact current rating for cable size.

You can never have enough Volvos!

geocom

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#689937 22-Sep-2012 17:16

gzt:
geocom: Now here is where the confusion is. Wire and connectors are rated at Voltages and amps. So if i have a connector that says it is rated for 300V@8A. Some sites say I have an exact limit on the part of 8A where as some say I have a limit of 2400W so at 12V I should be able to run 200A. So which one is correct?

The relationship above is exactly backwards. Simple example -

10 volts at 10 watts = 1 amp
100 volts at 10 watts = 0.1 amp (100ma)

Taking this a bit further - a connector capable of reliably handing 200A at 12V is huge. Like the lugs on your starter motor or car battery earth.

I dont see how its backwards.

The connector is rated for 300V@8A therefore Watts is Voltage*Amps so 300*8=2400W. If you want to work out the Amp at 12V you take 2400W/12=200A.

Anyway to be safe i am going to up the connector to a 16A connector better to go over than be right on the mark. The cable in this case is able to run at 16A anyway far higher than i will be using and the cable will be shorter than 1/2 a meter.

Geoff E

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#689959 22-Sep-2012 18:17

geocom: I dont see how its backwards. The connector is rated for 300V@8A therefore Watts is Voltage*Amps so 300*8=2400W. If you want to work out the Amp at 12V you take 2400W/12=200A.

Watts is a measure of the power output of the system - the measure of actual work performed in the real world.

If you are getting 2400 watts of output at 300 volts you will see 8 amps flowing in that circuit.

If you are getting 2400 watts of output at 12 volts then you will see 200 amps flowing in that circuit.

The less voltage you are using, the more current (amps) you need to get the same power output.

For cable size the flow of amps in the system is the important part. More amps flowing > bigger cables required.

Does that make things a bit clearer?

Samsung

Shop now on Samsung phones, tablets, TVs and more (affiliate link).

Niel

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#690036 22-Sep-2012 22:29

geocom: The connector is rated for 300V@8A therefore Watts is Voltage*Amps so 300*8=2400W. If you want to work out the Amp at 12V you take 2400W/12=200A.

No, it does not work that way for contact rating. Typically the DC current rating is also lower than the AC current rating, because the changing of polarity in AC helps with self cleaning and reduces the instantaneous power dissipated in the contact when it makes/breaks (or something like that, too late at night to think). You need to refer to the relay/switch datasheet to get the correct rating for the application.

You can never have enough Volvos!

geocom

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#690043 22-Sep-2012 22:40

Niel:
geocom: The connector is rated for 300V@8A therefore Watts is Voltage*Amps so 300*8=2400W. If you want to work out the Amp at 12V you take 2400W/12=200A.

No, it does not work that way for contact rating. Typically the DC current rating is also lower than the AC current rating, because the changing of polarity in AC helps with self cleaning and reduces the instantaneous power dissipated in the contact when it makes/breaks (or something like that, too late at night to think). You need to refer to the relay/switch datasheet to get the correct rating for the application.

Yep fully understand that i though that gzt was saying that my understanding of Wattage was backwards.

Geoff E

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#690096 23-Sep-2012 10:13

geocom: The connector is rated for 300V@8A therefore Watts is Voltage*Amps so 300*8=2400W. If you want to work out the Amp at 12V you take 2400W/12=200A.

geocom: Yep fully understand that i though that gzt was saying that my understanding of Wattage was backwards.

It is. You are taking the amperage rating of a connector at a particular voltage - and then converting that to watts and saying the same connector will handle that number of watts at any voltage/current you want to name.

I don't want to be a bad guy but you made a serious mistake there - and if you do not understand it there will be smoke and fire.

geocom

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#690338 23-Sep-2012 22:52

It's not backwards. Converting the max to wattage is just wrong.

In this case i had conflicting reports from different sites. Neither site I had used in the past. So i had no real ability to tell which one i should believe. Both made sense but conflicted on the main point. After all there are many items that are rated in Watts such as resistors.

So i had some real doubts. Knowing the risks of if the wires get hot enough the plastic will melt which will cause a spark if the 2 wires meet. I decided to ask on a place i trust. At least I now know with 100% certainty which one is correct.

Geoff E