Inverters - how do they work? Something for nothing?

Robert Wilson

Robert Wilson

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I can't get my head round "12 volts becoming 240 volts."
Electricity baffles (and scares) me, so I have always let the professionals do all work, bar very basic stuff.
It seems to me that an inverter will take 12 volts from boat batteries and produce enough power to run hairdriers, fridges, TVs, tools etc etc.
HOW? :confused:

Please don't get too scientific, but what is the basic principle, and how long does a small boat domestic battery give so much "power" to run 240 volt stuff?
I am not talking about plain old "shore power" connected to 13amp sockets.
Or am I missing something very basic? Most likely I am.:o

Thanks to the wise ones.
 
at its most basic

take the 12 volt dc and have a circuit that swaps the out put polarity about regularly

so now you have 12 volt ac

put that in a transformer, on the out put take off the 240 ac.


Obviously there is lots more to get a smooth ac wave form.

equally the dc side needs very thick wires to get enough energy in to convert.

for a fuller explanation see here, from a solar panel connection perspective


http://www.solar-facts.com/inverters/how-inverters-work.php


for the how long does the boat battery last, depend on how big the battery is. If the conversion was totally efficient a mains 2400 w hair drier draws 10 amps. At 12v the current will be in the order of 200Amps. so vastly more than normal domestic load, and about 1/3 of engine cranking!
 
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volts arent power.

Let me try to put it into simpler terms you might understand using water. Volts are equivalent to water pressure - you can feel that pressure putting your thumb over the end of the tap. Amps are the rate of flow of the water - lot more flow out of a big hole than a small one. Power is the combination of the two so a river which flows slowly but flows a lot of water is producing a lot of power. And a pressure washer which flows not a lot of water but does it at very high pressure, is also flowing a lot of power.

in simple maths pressure x volume flow = power just as volts x amp = power.

now here is the basic law of physics - the power remains constant. So if you take the 12v and change it up to 240v with a lot of lectrickery, the power remains the same apart from a bit of waste in doing the process. so it follows that if you increase the volts then you decrease the amps in proportion.

its all a bit more complicated as you can see from the post above but dont worry about that. all you need to remember is that with power as with everything else in this life, you dont get something for nothing. Leaving aside a bit of waste which appears as heat, the power going into your inverter is the same as the power coming out of it. So if the volts go up, the amps go down. If you have the power to run a hairdyer for swbmo, then thats also the amount of power you are taking out of the batteries.

not surprisingly its the same with your battery charger. When it drops the 240v AC coming into it by a factor of 20 down to 12v DC going into the battery, then the current goes up by a factor of 20. Simples.

P.S. For the benefit of any engineers coming afterwards, yes I do know that the above is a gross simplification but the OP wants and needs a simple answer not someone wittering on about phases and losses etc
 
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To keep it simple an inverter will take a constant current (DC) and switch it on and off for equal periods of time. This will produce a square wave form. Draw it on a bit of paper, a line at 0 volts and one at 12 volts above it. Switch on for 1 second and off for 1 second. The voltage will "Alternate" between 0 and 12.
If you then add a "Step up Transformer" with a ratio of 20-1 you will get 240 volts Alternating current (ac) out. This will still be a square wave. Some basic inverters output this wave form. More sophisticated ones will "Smooth it, or, round it off to provide something more like a sinewave. They will also have some form of voltage stabilisation. You must remember also that 1Kw at 240 volts will draw 4amps, approx and at 12 volts the current will be 83 amps. approx. If you allow for inefficiency probably more like 86 amps.
That is all very basic, no doubt someone will come along and query the maths etc. but I hope it at least lets you understand the principle.
 
Spot on, thank you.

That is what I call a gentleman's approach to explanation. Well done, Wotayottie.

So, am I correct in thinking that running e.g. a hairdryer in that manner would kn**ker my domestic battery in short (no pun intended) time, or do the lecky-magicals actually give me something for nothing?
I understand that the amperage would be well down, so am I correct in thinking that heavy-drain stuff (kettle??) would either not work/struggle or blow fuses?

By the way, I look forward to the physicists getting their teeth into this one - it's a horrid wet n windy day up here and I'm not leaving the house (poor dog :D)
 
Alex_Blackwood
To keep it simple an inverter will take a constant current (DC) and switch it on and off for equal periods of time. This will produce a square wave form. Draw it on a bit of paper, a line at 0 volts and one at 12 volts above it. Switch on for 1 second and off for 1 second. The voltage will "Alternate" between 0 and 12.
If you then add a "Step up Transformer" with a ratio of 20-1 you will get 240 volts Alternating current (ac) out. This will still be a square wave. Some basic inverters output this wave form. More sophisticated ones will "Smooth it, or, round it off to provide something more like a sinewave. They will also have some form of voltage stabilisation. You must remember also that 1Kw at 240 volts will draw 4amps, approx and at 12 volts the current will be 83 amps. approx. If you allow for inefficiency probably more like 86 amps. That is all very basic, no doubt someone will come along and query the maths etc. but I hope it at least lets you understand the principle.

So, the amp drain on the battery would be enormous?
Or am I missing that point completely?
 
So, the amp drain on the battery would be enormous?
Or am I missing that point completely?[/QUOTE]

As Woty said the drain on the battery will be approx. 20 times what you take out. So what ever the current required by your hair drier/kettle/CD player. the current drawn by the battery wil be X 20.
There is no such thing as a "Free lunch" so you will not get something for nothing. No piece of Electrical or Mechanical equipment is 100% efficient so you must have system losses somewhere.
 
So, the amp drain on the battery would be enormous?
Or am I missing that point completely?[/QUOTE]

No, you have it it: To get 1 amp mains (around 240 watts at mains voltage) you need to put in the region of 20 amps in at 12volts, So your 500w hairdryer would be discharging the 12v battery at around 40 - 50 amps. Thats why usually they are only used fro running low power mains things like laptops or Tv, and even then they can be drawing 5 amps, and often a lot more from the 12v battery.
 
oldharry said:
So, the amp drain on the battery would be enormous?
Or am I missing that point completely?
Click to expand...

No, you have it it: To get 1 amp mains (around 240 watts at mains voltage) you need to put in the region of 20 amps in at 12volts, So your 500w hairdryer would be discharging the 12v battery at around 40 - 50 amps. Thats why usually they are only used fro running low power mains things like laptops or Tv, and even then they can be drawing 5 amps, and often a lot more from the 12v battery.[/QUOTE]

Great thread this, for replenishing knowledge of simple electrics, thanks guys.
 
Your battery bank is a bit like a fuel tank. If you want it to supply fuel longer, you need a bigger tank. Same with batteries. Rafiki has 3x 110 ah batteries, so 330 ah, will last longer than 1x 110 battery when supplying power.
 
Something not mentioned (or asked) is that the elektricky is needed becauae a transformer will not work off DC. It needs the cyclic volts(AC) to induce the magnetism in the transformer that then converts to the higher voltage.
 
Another thing to bear in mind is that roughly a typical fully charged good condition 110Ah 12v battery has 1000 watt/hr of usable energy. Some would argue less as that still takes it to quite a deep discharge. That implies that you could run a 500w light etc. for 2 hours, since that would take (500w/hr*2)

However in reality, the higher the load placed on the battery the less apparent energy it is able to supply before it is discharged. The opposite is also true. Draw a tiny current for a long period of time, and it will appear to have more energy available.

This is because of the batteries own ability to provide energy efficiently. Put a very heavy load on a battery (inverters 1-2Kw are easily able to do this) for any period of time and the battery will begin to heat up, wasting stored energy it has available to you. This is caused by 'internal resistance' or ESR (equivalent series resistance), the lower this value, the better the ability of the battery to supply high currents (and usually the better the battery)

I will always try to keep maximum current draw across a battery to around 1/10th its stated capacity for medium term loads, 1/20th for long term loads. Very short intermittent loads (of seconds) can of course be much higher (engine cranking), but shouldn't be done a the same time as with devices like mains inverters sharing the same supply, as it can cause those to drop out/reset.

HTH :)
 
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Robert Wilson said:
That is what I call a gentleman's approach to explanation. Well done, Wotayottie.

So, am I correct in thinking that running e.g. a hairdryer in that manner would kn**ker my domestic battery in short (no pun intended) time, or do the lecky-magicals actually give me something for nothing?
I understand that the amperage would be well down, so am I correct in thinking that heavy-drain stuff (kettle??) would either not work/struggle or blow fuses?

By the way, I look forward to the physicists getting their teeth into this one - it's a horrid wet n windy day up here and I'm not leaving the house (poor dog :D)
Click to expand...

You never ever get something for nothing. If your hairdryer uses a kilowatt of energy, then it takes that out of your battery. Exactly the same with your kettle. But as for would it kn**cker your batteries, that all depends on how big your batteries are ( ie how much energy is inside them) and how long you use the hairdryer for. In my case even at 1 minute per hair, it wouldnt be running for very long. ;)

I am a physicist. Perhaps better say "was". In 1967.
 
I know I shouldn't let it get to me but I cringe when I read some of the above. The "1000 watts of usable energy" in the latest missive has pushed me over the edge. Aaaagh!!!!

... and now another "kilowatt of energy" from someone who thinks he's a "physicist". Ye gods!
 
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Why? Prove me wrong. 110Ah x 12v nominal. = 1320watt/hr. Drain a battery to 0% SOC and it will die in short order, taking 1kw/hr as a basis to work from is a nice round number that will take into account margin for error/and sub optimal battery condition.

People without an advance understanding of these topics don't care that written text isn't of such quality of some published white paper. They care about reading it in a way in which makes sense to them.
 
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It can be surprising how quickly the current drain mounts up.

We have a small inverter which plugs into the 12V "cigar" socket on the boat. It's rated at 200W or similar and my wife uses it for her hair tongs but it's not powerful enough for the hairdryer. A couple of years ago she tried to use the tongs whilst one of the boys was running his laptop off the inverter at the same time. It immediately blew the 10 amp fuse in the boat wiring loom!

Lesson learned - tongs always go solo!

Richard
 
Why needlessly complicate things for people who don't care to know the difference? Granted I could have made the point clearer, but equally you could have refined it further yourself.

Your mentality intrigues me, your probably about twice my age, and you behave as above? Where did it go wrong? I'd rather not make the same mistakes...
 
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