Wire thickness / amp rating?

[Tim Good]

Ok I'm becoming slowing more proficient with 12v electrics but could someone give me dummies guide to wire ratings in terms of amps? At the moment all I know is that the more amps you expect down a wire then the thicker you want it. But....

- What happens if it is too thin? Assume it will melt?
- If it is too thick does it matter?
- Is there a general chart I can use so I can stock up on various types?

 

[miyagimoon]

Ok I'm becoming slowing more proficient with 12v electrics but could someone give me dummies guide to wire ratings in terms of amps? At the moment all I know is that the more amps you expect down a wire then the thicker you want it. But....

- What happens if it is too thin? Assume it will melt?
- If it is too thick does it matter?
- Is there a general chart I can use so I can stock up on various types?

Try this http://www.aeicables.co.uk/literature/CurrentRatings.pdf

Pages 3, 4 & 5 should give you all you need.

 

[macd]

Ok I'm becoming slowing more proficient with 12v electrics but could someone give me dummies guide to wire ratings in terms of amps? At the moment all I know is that the more amps you expect down a wire then the thicker you want it. But....

1 - What happens if it is too thin? Assume it will melt?
2 - If it is too thick does it matter?
3 - Is there a general chart I can use so I can stock up on various types?

1. In extreme cases it might melt, which is obviously best avoided. But in practice the main issue is voltage drop, which effectively throws away electrical capacity that you have earned with difficulty (cf your thread re wind generators etc). If the voltage drop is great, it can compromise the function of some devices (instruments, fridge, windlass etc).
There are plenty of resources on-line for calculating size/length of cable and voltage drop.

2. Not really, but it adds bulk, weight and expense and can make running cables difficult.

3. Yes, as stated above. Also, given that you're clearing off to distant parts, you'd be well advised to buy Nigel Calder's 'bible', 'Boatowner's Mechanic and Electrical Manual', which will tell you all you need to know and more.

 

[ip485]

Slightly off topic but before I wired in my SSB I had read various stories about just how much power they can draw when transmitting and the potential voltage drops in the cable so without thinking selected cables way over necessary. Several hours later trying to bend, crip and route the cables I rather wished they had not been quite so large and inflexible!

 

[David2452]

- What happens if it is too thin? Assume it will melt?

No it will not, because you will, by the time you get around to actually doing it (I hope) have learned that every circuit should be protected by an over current device suited to it's downstream load and not to use conductors incapable of carrying the required current without inapropriate voltage drop under load.

 

[RobbieW]

The Merlin Wire Sizer, http://www.power-store.com/includes/download.asp?docid=369, is useful if unusable for many of todays whizzy tablet/phone users. Very much an old school Windows program

 

[nigelmercier]

... in practice the main issue is voltage drop ... you'd be well advised to buy Nigel Calder's 'bible', 'Boatowner's Mechanic and Electrical Manual', which will tell you all you need to know and more.

Agree with both, link to the book bottom left of this page

 

[miyagimoon]

With any installation you should be ensuring:
1. The cable current capacity is large enough for the load.
2. The Voltage Drop is within acceptable limits.
3. The protective device is sized correctly to protect the load against overload & short circuit.
4. The protective device is sized correctly to protect the cable feeding the load.
Let’s look at a simple cable installation where we determine size of cable, size of protective device (Fuse / circuit breaker) and volt drop.
Voltage – 12.8v (Using a fully charged battery bank).
Load – For this example something like a lamp rated at 25w.
Length of cable from Fuse Board to Lamp – 8m
Size of cable from Fuse Board to Lamp – TBD (To be determined)
Size of Protective Device – TBD.
From above the load will be Watts / Volts = Amperes = 25/12.8 = 1.95A N.B. When batteries are being charged the current will be less and with partially discharged batteries the current will be more.
Therefore, we will need a cable big enough to carry the load of 1.95A. If we look at page 5 of the AEI cable guide a 1mm², 2 Core cable has a rating of 17A. That’s, more than sufficient for our load. (The method of fixing is irrelevant as the current capacity is always large enough no matter which method is used and the Voltage Drop stays constant).
The 1mm² 2 core, cable has a volt drop per Amp per metre of 46mV, which is 46/1000=0.046v/a/m. So for our load of 1.95A over 8m the voltage drop will be 0.046 x 1.95 x 8 x2 (remember the distance is there and back) = 1.44v.
So at the actual lamp we would get 12.8v – 1.44v = 11.36v.
Now if we use a larger cable let’s say the 1.5mm² the volt drop would be 0.031 x 1.95 x 16 = 0.97v.
The 2.5mm² cable would give a volt drop of 0.59v.
Hence the larger the cable the larger the current capacity and the less the voltage drop.
Now this load and the cable still need protecting from overload and short circuit.
We therefore need to ensure that the size of protective device fitted is large enough to allow the load to operate normally but small enough to ensure it is protected against overload or short circuit. We also need to do the same for the cable.
In this instance a fuse rated at 3Amp will protect the load and the cable.
If no fuse is fitted it is possible for a fault to occur which would result in the cable overheating to the extent that the boat was to catch fire.

 

[pvb]

With any installation you should be ensuring:
1. The cable current capacity is large enough for the load.
2. The Voltage Drop is within acceptable limits.
3. The protective device is sized correctly to protect the load against overload & short circuit.
4. The protective device is sized correctly to protect the cable feeding the load.
Let’s look at a simple cable installation where we determine size of cable, size of protective device (Fuse / circuit breaker) and volt drop.
Voltage – 12.8v (Using a fully charged battery bank).
Load – For this example something like a lamp rated at 25w.
Length of cable from Fuse Board to Lamp – 8m
Size of cable from Fuse Board to Lamp – TBD (To be determined)
Size of Protective Device – TBD.
From above the load will be Watts / Volts = Amperes = 25/12.8 = 1.95A N.B. When batteries are being charged the current will be less and with partially discharged batteries the current will be more.
Therefore, we will need a cable big enough to carry the load of 1.95A. If we look at page 5 of the AEI cable guide a 1mm², 2 Core cable has a rating of 17A. That’s, more than sufficient for our load. (The method of fixing is irrelevant as the current capacity is always large enough no matter which method is used and the Voltage Drop stays constant).
The 1mm² 2 core, cable has a volt drop per Amp per metre of 46mV, which is 46/1000=0.046v/a/m. So for our load of 1.95A over 8m the voltage drop will be 0.046 x 1.95 x 8 x2 (remember the distance is there and back) = 1.44v.
So at the actual lamp we would get 12.8v – 1.44v = 11.36v.
Now if we use a larger cable let’s say the 1.5mm² the volt drop would be 0.031 x 1.95 x 16 = 0.97v.
The 2.5mm² cable would give a volt drop of 0.59v.
Hence the larger the cable the larger the current capacity and the less the voltage drop.
Now this load and the cable still need protecting from overload and short circuit.
We therefore need to ensure that the size of protective device fitted is large enough to allow the load to operate normally but small enough to ensure it is protected against overload or short circuit. We also need to do the same for the cable.
In this instance a fuse rated at 3Amp will protect the load and the cable.
If no fuse is fitted it is possible for a fault to occur which would result in the cable overheating to the extent that the boat was to catch fire.

I'm sure you mean well, but some of this is rather over-complicated, not to mention incorrect.

I imagine the OP just wants a simple way of checking that cable sizes are appropriate. The AEI charts linked to earlier are rather confusing, as you've discovered, not to mention rather off-putting in their complexity.

The prime concern with most boat cabling is to ensure that voltage drop isn't excessive. For general circuits (lights, etc), it's often considered that a 10% voltage drop is acceptable (ie around 1.2v on a 12v circuit), whereas for more sensitive electronics a 3% max drop is often recommended (ie around 0.4v on a 12v circuit).

The simplest way I've found of calculating voltage drop is the "Treble One Twenty" rule:-

* 1 amp going 1 metre along a 1 sq mm wire gives a 20 mv drop.

That’s your basic figure, and all you need do now is change the 20mv figure in proportion to the other changes. So, for example:-

* an increase in current increases the 20mv proportionally

* an increase in length increases the 20mv proportionally

* but an increase in wire size decreases the 20mv proportionally.

In the case you used as an example, the 1.95A load increases the 20mv to a 39mv drop. You’ve then got 16 metres of wire in the circuit (8 metres there on one conductor, and 8 metres back on the other conductor), which further increases the drop by a factor of 16, bringing it to 624mv, or about 0.6v. That’s the theoretical drop you’d experience if you used 1 sq mm wire, and it's perfectly acceptable for a lamp, and it's rather lower than the figure you arrived at. Using 2 sq mm wire would reduce the drop by a factor of 2. Dividing 624mv by 2 means that the 2 sq mm wire will give you a drop of 312mv (0.3v and well within the requirement for sensitive electronics).

This is only a rule of thumb. It’s about 95% accurate, which I reckon is close enough for leisure boats (after all, your 12 volt source probably fluctuates by 15%). But, and this is important, it’s easy to remember, and easy to work out without needing to resort to a calculator (which is why I believe it’s easier to think in millivolts, as there’s less chance of confusion with the decimal point).

 

[macd]

I'm sure you mean well, but...

May I thank, you, pvb, for bringing a note of sanity. It's evident from the opening post that northcave is interested in, but far from savvy with, good practice in his 'lectrics. Techy stuff will not aid him, except insofar as it helps his learning curve, which ohmy stuff will not. At best it will confuse; at worst, put him off for life. He's clearly bright and willing enough to learn (cf his other posts), if not overloaded on the ohms and wattnots.
I'd add that I write this as someone who has a working handle, but no more, on the leccy issues in question. In other words, yr average Joe.

 

[Hydrozoan]

Is it still the case that different tables for cable size selection (from current, length and tolerable voltage drop) may use different conventions for length - some using the 'there and back' length and others using the 'one-way' length?

IIRC it was a US/UK difference. Or has the world now got around to agreeing a single convention?

I believe my old copy of Calder's 'Boatowner's Mechanical and Electrical emphasizes this, but I thought I'd raise it lest the OP finds different tables and gets puzzled.

 

[miyagimoon]

I'm sure you mean well, but some of this is rather over-complicated, not to mention incorrect.

I imagine the OP just wants a simple way of checking that cable sizes are appropriate. The AEI charts linked to earlier are rather confusing, as you've discovered, not to mention rather off-putting in their complexity.

The prime concern with most boat cabling is to ensure that voltage drop isn't excessive. For general circuits (lights, etc), it's often considered that a 10% voltage drop is acceptable (ie around 1.2v on a 12v circuit), whereas for more sensitive electronics a 3% max drop is often recommended (ie around 0.4v on a 12v circuit).

The simplest way I've found of calculating voltage drop is the "Treble One Twenty" rule:-

* 1 amp going 1 metre along a 1 sq mm wire gives a 20 mv drop.

That’s your basic figure, and all you need do now is change the 20mv figure in proportion to the other changes. So, for example:-

* an increase in current increases the 20mv proportionally

* an increase in length increases the 20mv proportionally

* but an increase in wire size decreases the 20mv proportionally.

In the case you used as an example, the 1.95A load increases the 20mv to a 39mv drop. You’ve then got 16 metres of wire in the circuit (8 metres there on one conductor, and 8 metres back on the other conductor), which further increases the drop by a factor of 16, bringing it to 624mv, or about 0.6v. That’s the theoretical drop you’d experience if you used 1 sq mm wire, and it's perfectly acceptable for a lamp, and it's rather lower than the figure you arrived at. Using 2 sq mm wire would reduce the drop by a factor of 2. Dividing 624mv by 2 means that the 2 sq mm wire will give you a drop of 312mv (0.3v and well within the requirement for sensitive electronics).

This is only a rule of thumb. It’s about 95% accurate, which I reckon is close enough for leisure boats (after all, your 12 volt source probably fluctuates by 15%). But, and this is important, it’s easy to remember, and easy to work out without needing to resort to a calculator (which is why I believe it’s easier to think in millivolts, as there’s less chance of confusion with the decimal point).

As such I will agree that in comparison to your "Rule of thumb" it does seem somewhat complicated. Your figure of 0.6v as you say is lower than mine @ 1.44v. But the mV/A/M figure quoted in mine is from a reliable and accurate published document created by the people who manufacture cables. Unless I have made a calculation error I fail to see how it is incorrect and would also believe it to be much more accurate.

 

[prv]

Your figure of 0.6v as you say is lower than mine @ 1.44v.

I haven't read either post in detail, I have a spreadsheet to do these calculations which works well for me. But you're not comparing one-way distance versus there-and-back, by any chance? Online calculators being vague about that distinction is one reason I gave up on them and build the spreadsheet.

Pete

 

[miyagimoon]

I haven't read either post in detail, I have a spreadsheet to do these calculations which works well for me. But you're not comparing one-way distance versus there-and-back, by any chance? Online calculators being vague about that distinction is one reason I gave up on them and build the spreadsheet.

Pete

No Pete, the difference appears to be the mV/A/m figure used. The AEI data shows that as 46mV/A/m whereas pvb is using 20mV/A/m.

 

[pvb]

As such I will agree that in comparison to your "Rule of thumb" it does seem somewhat complicated. Your figure of 0.6v as you say is lower than mine @ 1.44v. But the mV/A/M figure quoted in mine is from a reliable and accurate published document created by the people who manufacture cables. Unless I have made a calculation error I fail to see how it is incorrect and would also believe it to be much more accurate.

You've misunderstood the AEI table. You've taken a figure for twin-core cable, then doubled it up for there-and-back, when the twin-core figure already allows for both cores. As a result, your 1.44v is about double the true figure!

 

[pvb]

No Pete, the difference appears to be the mV/A/m figure used. The AEI data shows that as 46mV/A/m whereas pvb is using 20mV/A/m.

The volt drop for single core 1 sq mm cable is about 20mv/A/m - check out any reference charts.

 

[nigelmercier]

You've misunderstood the AEI table. You've taken a figure for twin-core cable, then doubled it up for there-and-back...

In addition, twin core cable is de-rated WRT single core due to added heating. Thanks for the 1-1-1-20 rule, very handy.

 

[Croftie]

Thanks for the 1-1-1-20 rule, very handy.

+1
just made up a spreadsheet so I can put in different values.

 

[miyagimoon]

You've misunderstood the AEI table. You've taken a figure for twin-core cable, then doubled it up for there-and-back, when the twin-core figure already allows for both cores. As a result, your 1.44v is about double the true figure!

pvb - please accept my humble apologises. School boy error on my part. You are perfectly correct figure should be around 0.7v.

 

[miyagimoon]

In addition, twin core cable is de-rated WRT single core due to added heating. Thanks for the 1-1-1-20 rule, very handy.

Nigel, Surely multi cable derating only affects the current capacity of a cable.