Comparative Safety: 12v v 24v v 48

[CatNewBee]

Yes, there is some truth in there. In all voltage ranges you need the right gear, and the challenges differ.

A cheap fuse in 12V 7.5A that protects a wire that can easily handle 20A without burning to a 60W load is not so sensitive designed. in a 48V environment the same fuse would be used for 240W.

DC sparks are a real thread, even in high current 12V environments, if you have a solenoid or automatic fuse of 500A you better have spark protection or your contacts will melt when connecting / disconnecting under load.

The same load would require in 48V 125A, also with spark protection.

regarding electro-mechanical components: The whole industry runs any size of motors, generators in all voltage ranges DC and AC. So not very different.
You'll need similar amount of copper for 12V, 24V or 48V to crank the same engine, but with 48V your regulators, carbon brushes, wires etc. will be much smaller.

It is always a compromise. Finally you chose what is right for your application.

[Jammer]

Originally Posted by Dockhead

OK, but by the time that happens as the result of a connection fault, the boat will already be in flames, no? I wouldn't think it would take many watts of power to combust a connector weighing a couple of grams -- no?

The extent of the drama is determined by the proximity of the connection to combustible items, the extent to which heat can be moved away by convection or conduction, and the point at which the circuit opens because of the effects of the heat/fire. It does take more than a few watts in every case I've seen. The various electrical safety standards for appliances and electronics have a concept of a "power limited circuit" which is defined by wattage, not volts or amps. Usually the maximum is around 100 watts although it varies from one standard to the next. Any product that operates only on a power-limited circuit has to meet much less stringent safety requirements to meet the standard.


In a circuit that delivers 100 watts from a power supply to a load, a high-resistance connection can't dissipate more than 25 watts. I think in most cases that's a pretty good answer. You can make fire out of 25 watts but it's farfetched with standard wire and connectors.

All this is making me think that it really makes sense to use large and hefty connectors, with large contact surface and significant mass. And higher system voltage? Do you have a view on this, the main topic of this thread?

Using bigger connectors won't help and may create a problem because larger connectors and wires are harder to get right. From a HRC safety standpoint the points of light IMO are:

• Use stranded, tinned wire.
• Avoid splices and unnecessary connections.
• Use high-quality terminations, properly installed. Crimp terminations installed with proper dies and a hydraulic, ratchet, or battery crimper are best. Solder pot terminals installed using flux are probably OK too, I've never seen one fail.
• Make sure connections are clean and bright, use a wire brush or emery cloth as needed.
• Use strain relief close to the terminal so that the wire cannot move and does not put any force on the terminal when the assembly is subject to vibration or movement. Use proper neoprene-cushioned stainless steel cable clamps that are riveted or through-bolted to something solid wherever possible.
• Use internal-tooth star lockwashers (or some other kind of constant-tension lockwasher besides a split ring) to secure nuts in place on bolted connections. Use a torque wrench when tightening. Retighten at least once a few days after installation.
• Use antioxidant grease on the mating, current-conducting portions of connections.
• Make connections to high-current equipment accessible for inspection and service. Do not cover them with adhesive goo. Check them with an infrared thermometer from time to time.
• Avoid wiring approaches that involve placing multiple terminals under one bolt. Tinned copper bar is available and is a better choice for connecting breakers or switches together than daisy-chaining connectors together.
• Aggressively replace switches, relays, and other equipment that is starting to get flakey, looks corroded, or runs hot.

Bit of thread drift, but all this also makes me think that the UK domestic power plugs may be a really good thing -- they are massive, and with a really large contact area compared to the Continental round pin Shuko plugs or the ultra cheap, ultra flimsy American ones. I used to think the UK ones were quite clunky . . .

The UK and USA wiring systems are very different but on the whole have similar safety track records. In the USA or Europe no one cuts their foot stepping on power plugs. The UK system is designed around high-power ring circuits and fused plugs while the USA and European systems are designed around radial circuits that carry much less power.


With any connector system the main problem is that over time corrosion and contamination build up and the springy bits lose their springiness. This is what happens with shore power connectors, for example.

[Dockhead]

Originally Posted by Jammer

. . . The UK and USA wiring systems are very different but on the whole have similar safety track records. In the USA or Europe no one cuts their foot stepping on power plugs. The UK system is designed around high-power ring circuits and fused plugs while the USA and European systems are designed around radial circuits that carry much less power.

I've never owned a land house in the UK, so my experience is somewhat limited, but I've never ever seen a burned plug in the UK, whereas in the U.S. and in Germany it seems to happen regularly . . . And although I have encountered about the same percentage of utter hack marine electricians here, as in the U.S. (why are there so many?), the quality of wiring of English-built boats seems two cuts above either French or American ones, in my experience.

Originally Posted by Jammer

With any connector system the main problem is that over time corrosion and contamination build up and the springy bits lose their springiness. This is what happens with shore power connectors, for example.

That's a separate topic The only boats I ever personally knew to be burned down from electrical fires, burned down because of faulty shore power plugs. I guess that's quite relevant to the topic of this thread. The Marinco/Hubbel ones seem simply awful to me -- I would have a box full of ones I've burned up, if I hadn't thrown them away, and I guess I'm lucky I never had a fire. They are just a fire waiting to happen. I've been using the international 16A blue plug for the last couple of years, which is an improvement on the awful Marinco/Hubbel plug, but I'm still getting some erosion of the contacts. This weekend I will be installing a Smart Plug I brought back from the States.



Probably anyone who cares about electrical safety on a sailboat, should start with the shore power plug.

[CatNewBee]

only important for people living on shore power... [emoji56]

[transmitterdan]

Basically any voltage standard be it 12, 24, 42 or 48 requires standards. And the standards must balance adequate safety, cost and complexity. There is no perfectly safe electrical system anywhere. In fact, simply being alive isn’t “safe” because you can die at any moment.

Safety involves a spectrum and good electrical standards (when adhered to) do not appreciably reduce the expected lifetime of any age group. To wit, the recommendation of “baby proof” outlets for homes with toddlers that came about 20 years ago to better protect that age group.

Any voltage can be designed to be safe. But making things “bigger” is not always “better” or safer. You have to strike the right balance. Obviously 48V systems are “safe enough” else there would not be millions of them installed.

[s/v Jedi]

This topic brings me back out of hibernation:-)

From an engineering standpoint 48V DC is the optimum DC power source for sites exposed to persons who are not certified electrical engineers. This makes it the perfect DC power source for general service on boats as well.

Reasons to use something else: 12V or 24V because many boats have just one battery (economic reasons) up to 144V for electric drive applications.

Arguing for or against 48V is useless when not all participants have expert knowledge and it's futile as well because it has all been done many times before to come to that perfect 48V DC

[transmitterdan]

I agree with Nick (s/v Jedi) that 48V is safe. I think it is pretty clear that 48 (or 42) is more likely to be safer than 12V as cruisers demand ever more power from their DC system. With multi kilowatt inverters and probably DC air conditioners coming one of these days the situation will get worse.

Will it get bad enough that a 40+V standard emerges for rec cruisers? Probably not, because the inertia in the marine recreational industry is slower than ever. Frankly there are not enough customers willing to pay the freight to convince investors they should spend money developing a new standard and associated products. So this topic will likely remain a DIY topic. And the internet is not a good place to get DYI electrical power advice in my experience.

Circuit breakers, fuses and switches for 12V have to be checked if they are compatible with 48V. Most are but some are not. Connectors are most likely ok and there are many commercial types available that are compatible with 48V. Relays for 48V are very common as are battery chargers and solar systems. Alternator will be scarce.

It would not take a lot of book learning to become proficient in designing 48V (or 42) DC systems.

[ramblinrod]

Originally Posted by Dockhead

Someone said that the energy released in a high resistance connection is a function of the square of the voltage, so higher voltage means much greater risk in case of a high resistance connection.

Someone else said it's actually the square of the current, so on the contrary, the higher voltage system will be safer in case of a high resistance connection fault.

These theses can't all be correct! So what is it? Enquiring minds would like to know.

Your answer is in post # 1 and 5.

Watts law: P = E*I (with derivatives E= P/I and I = P/E)...

and using Ohm's Law (E = I*R) to substitute, to produce P = I^2*R and P = E^2/R...

All have to be true.

These are the laws of physics (specifically electro-magnetics) that apply to every boat on the planet without exception.

For an ideal circuit where there is no contact resistance, Watts Law and derivatives can be applied to the supply voltage and load resistance.

This is where everyone, (including myself before post # 1 of this thread) were getting caught and making a mistake.

One cannot use the Watts law solely on the supply voltage and load resistance, as soon as we introduce a CONTACT RESISTANCE.

An increase in contact resistance reduces circuit current whilst increasing the voltage drop across the contact.

A low contact resistance (in relation to the load resistance) has little effect on
circuit current and the voltage drop across the contact is low, and therefore power dissipated small.

But as we increase the contact resistance, it simultaneously impacts circuit current and voltage drop across the contact.

Similarly, when supply voltage is increased, for a small amount of contact resistance, no biggy as the resultant voltage drop across the contact is minimal.

But for a large amount of contact resistance, the voltage drop across the contact, in relation to the load, becomes significant. An increase in supply voltage, has a huge impact on this voltage drop, and hence the power dissipated by the contact.

Maximum Power Transfer Theorum is normally used to describe the impact of a power source internal impedance in relation to a load impedance, especially for a signal from an amplifier to a driven element such as a speaker or fishfinder transducer.

It describes a similar phenomenon to that I am describing here (an impedance in series with a load impedance).

Of course, I am considering an "ideal" power source (no internal resistance), to focus on the impact of the contact resistance in series with the load resistance (so as not to muddy the water even further), but in reality it would also come into play.

[Dockhead]

Originally Posted by ramblinrod

Your answer is in post # 1 and 5.

Watts law: P = E*I (with derivatives E= P/I and I = P/E)...

and using Ohm's Law (E = I*R) to substitute, to produce P = I^2*R and P = E^2/R...

All have to be true.

OK, but if that were true, then these effects would cancel each other out, wouldn't they? For the same power load, doubling the voltage halves the current. So if what you say is true, then the same amount of resistance would produce squared the amount of dissipated power from voltage, but the amount of dissipated power from current would be reduced by the same amount. Not true? The layman's mind gets confused here. This contradicts what CatNewBee wrote, if I understood him correctly.

Originally Posted by ramblinrod

. . . An increase in supply voltage, has a huge impact on this voltage drop, and hence the power dissipated by the contact.. .

I thought resistance would be the same regardless of voltage? I think Jammer and others posted this. What is correct? It's been a long time since I crammed for my amateur radio exams, but I seem to remember that a resistor of x ohms will have the same resistance at any voltage? If it were otherwise, Ohm's Law wouldn't seem to work, no?

[evm1024]

What we are all failing to take into consideration is the reduction in current after the introduction of a series faulty contact.

I have put together a spreadsheet modeling this and it has become obvious that the introduction of a 10 ohm contact resistance in a 12 volt 1000 watt system reduces the total power from 1000 watts to 14 watts.

This is because the 10 ohms is greatly more than the load resistance and thus has a significant effect.

In a 48 volt 1000 watt system 10 ohms is about 4 time the loads resistance and thus we end up with around 5 amps going through that 10 ohm resistance. And thus the significant power dissipation.

This begs the question - what is the typical resistance of a faulty connection?

I have been searching for an answer to that question and have not found an explicit answer.

What I have found indicates that HRC faults are typically less than 1 ohm. There are indications that a typical fault resistance would be in the the 0.5 to 0.25 ohm range. But as I said nothing explicit. (anybody have better info on this?)

When I change the fault to 1 ohm we get a different picture. The resistance in the 12 volt circuit is still greatly limited by the series 1 ohm resistance with a total power dissipation in the fault reaching 110 watts.

For the 48 volt system with 1 ohm the power dissipated in the fault is still high at about 210 watts.

Of interest is that we get a higher power dissipation in the 1 ohm fault in a 24 volt system then either the 12 volt or 48 volt systems. This is a result of the ratio of the fault resistance to the load resistance or so it appears. I should graph this.

Now it gets interesting.

For a series fault of 0.5 ohms the 12 volt systems dissipates 173 watts in the fault and the 48 volt system dissipates 146 watts in the fault. The 48 volt system will have less heating than the 12 volt system. Still, 146 or 173 watts - the difference is moot in my mind.

For a series fault of 0.25 ohms the 12 volt systems fault dissipates 232 watts while the 48 volt system dissipates 88 watts. Better but still a fault.

For a 144 volt system with that same 0.25 ohm series fault we get only 11 watts dissipation in the fault. Looks like higher voltages are better. Of course there are many other hazards to overcome in a 144 volt system.

[s/v Jedi]

For a 1,000W motor at 12V without circuit resistance we have 1,000 / 12 = 83A and 12 / 83 = 0.15 Ohm impedance
For a 1,000W motor at 48V without circuit resistance we have 1,000 / 48 = 21A and 48 / 21 = 2.29 Ohm impedance

Now we introduce a bad contact that results in a 0.1 Ohm resistance

At 12V the circuit impedance becomes 0.15 + 0.1 = 0.25 Ohm. Current 12 / 0.25 = 48A. The motor delivers 48^2 x 0.15 = 345W and the bad contact heats up at 48^2 x 0.1 = 230W

At 48V the circuit impedance becomes 2.29 + 0.1 = 2.39 Ohm. Current 48 / 2.39 = 20A. The motor delivers 20^2 x 2.29 = 916W and the bad contact heats up at 20^2 x 0.1 = 40W

What this means: a bad contact for a 48V motor results in a 84W loss of power which is 8.4%. For a 12V motor, the same bad contact results in a 655W loss of power which is 65% which renders it useless. Also, the heat generation at the bad contact is almost 6 times as high.

There’s no way 12V is safer.

[ramblinrod]

Originally Posted by Dockhead

OK, but if that were true, then these effects would cancel each other out, wouldn't they?

Watts law (P=E*I), ohms law (E=I*R), and all derivatives, have to apply in every case.

Have to.

It's the law. ;-)

[Dockhead]

Originally Posted by ramblinrod

Watts law (P=E*I), ohms law (E=I*R), and all derivatives, have to apply in every case.

Have to.

It's the law. ;-)

That seems very true!


But how then to explain this statement of yours:

Originally Posted by ramblinrod

"But for a large amount of contact resistance, the voltage drop across the contact, in relation to the load, becomes significant. An increase in supply voltage, has a huge impact on this voltage drop, and hence the power dissipated by the contact."

Isn't that a contradiction?


Why would the power dissipated by the contact be more, if the power of the load is the same?


And actually it seems like it's not that simple anyway -- see Post #40, which seems to indicate that in some situations, driving the same load with higher voltage may result in much LESS, rather than "hugely more" power dissipated. Also Post #41 by Jedi (Welcome back, Nick!! Great to see you!).



Obviously I'm in way over my head, but enquiring minds want to know! Since physics are "the law", there must be a single right answer to these questions.

[CatNewBee]

Originally Posted by Dockhead

That seems very true!


But how then to explain this statement of yours:


[I]


Isn't that a contradiction?


Why would the power dissipated by the contact be more, if the power of the load is the same?


And actually it seems like it's not that simple anyway -- see Post #40, which seems to indicate that in some situations, driving the same load with higher voltage may result in much LESS, rather than "hugely more" power dissipated. Also Post #41 by Jedi (Welcome back, Nick!! Great to see you!).



Obviously I'm in way over my head, but enquiring minds want to know! Since physics are "the law", there must be a single right answer to these questions.

that is true, same power, same contact, higher system voltage leads to less voltage drop. Voltage drop is caused by the current trough the conductor. The same conductor has the same resistance, but because on a higher system voltage the current decreases proportional for the same load, so does the voltage drop and finally the loss of power over the contact sinks by ^2.

[ramblinrod]

Originally Posted by s/v Jedi

For a 1,000W motor at 12V without circuit resistance we have 1,000 / 12 = 83A and 12 / 83 = 0.15 Ohm impedance
For a 1,000W motor at 48V without circuit resistance we have 1,000 / 48 = 21A and 48 / 21 = 2.29 Ohm impedance

Now we introduce a bad contact that results in a 0.1 Ohm resistance

At 12V the circuit impedance becomes 0.15 + 0.1 = 0.25 Ohm. Current 12 / 0.25 = 48A. The motor delivers 48^2 x 0.15 = 345W and the bad contact heats up at 48^2 x 0.1 = 230W

At 48V the circuit impedance becomes 2.29 + 0.1 = 2.39 Ohm. Current 48 / 2.39 = 20A. The motor delivers 20^2 x 2.29 = 916W and the bad contact heats up at 20^2 x 0.1 = 40W

What this means: a bad contact for a 48V motor results in a 84W loss of power which is 8.4%. For a 12V motor, the same bad contact results in a 655W loss of power which is 65% which renders it useless. Also, the heat generation at the bad contact is almost 6 times as high.

There’s no way 12V is safer.

Your results are for only one value of contact resistance.

Look at higher values and less power will be dissipated by the contact in the 12 Vdc circuit than in the 48 Vdc circuit.

Hint: Check out my posts # 1 and 5.

So with the greater risk of electric shock, and no real benefit to higher system voltage, everything else being equal, there is NO WAY that 48 Vdc is safer than 12 Vdc.

Increasing system voltage WILL reduce the cable size (to loads requiring greater than 16 AWG at 12 Vdc).

This may result in cost savings IF the sum of the higher voltage components does not exceed the sum of the cable cost savings.

I do review these issues every time I am requested to make a significant change to a vessel electrical system.

Now lets go back to that 37 ft cruising sailboat that started this discussion.

Let's say we have a boat with a 12 Vdc starter battery and 400 A-hr house bank.

Would we change everything to 48 Vdc to put in a 2 kW inverter?

Of course not.

Would we change out the primary system voltage to install an 8 HP (6 kW) thruster.

What about a 1 - 2 kW windlass

Of course not.

Some complain that I repeat the same corrections over and over.

Well, it is just in response to others repeating the same mistakes over and over.