Inverter use calculations (1 Viewer)

[PhilandMena]

I would appreciate some guidance on my logic of calculating inverter usage.
So here is an example of my logic.
I have a sterling 300W pure sine wave inverter which states :
No load Quiescent current 0.95A (I take this to mean when switched on it draws 0.95A an hour from my leisure batteries.
I also have a 240 V plug to charge a device and on that plug it states input 100-240V -50/60Hz -500mA Output 12.0V -1600mA.
Would I be correct in assuming that if I plug this device into my Inverter and switch it on the total draw from my leisure batteries would be.
0.95 + 1.6 =2.55 Amps per hour.

Thanks in advance.

 

[TheCaller]

Nearly right.

Long explanation coming up.

An Amp per hour doesn't really mean very much, although I think I know what you mean.

Amps are a measure of the amount of current flowing at any one instant in time.
Amp Hours are a measure of capacity & indicate for how long a battery could (in theory) supply a particular level of current. An 85AH (AmpHour) battery has the theoretical capacity to supply a current of 1 amp for 85 hours, or 10 Amps for 8.5 hours. (In practice it doesn't work quite like that, but that's another story.)

So the current drawn multiplied by the number of hours it is drawn for, gives us AmpHours (Ah).

You are correct that the quiescent current means that if the inverter is turned on, but nothing is plugged into it, then it will still draw 0.95A all the time. For every hour that it is left like this, it will use 0.95Ah of your battery's capacity.

The figures given for your plugin charger are the maximum it will draw. It doesn't necessarily follow that the device you plug it into will exploit this fully, but it can't draw more than this, so it's not a bad figure to work with.

We are more interested in what goes into the charger than what comes out - after all that's what the inverter is supplying. You have to put more in than you get out because it's not 100% efficient - that's why it feels warm when it's in use.

The higher the voltage, the lower the current needs to be to supply the same power. We know that it never draws more than 500mA, so we can assume that's what it draws when it's being used to maximum capacity on a 100V supply. You are connecting it to a 230V supply, so it will draw somewhere in the 200 - 250mA region. (I must confess that I'm guessing a bit as to how to interpret those Input figures.)

Lets take the higher figure. 250mA (0.25A) at 230V is equivalent to 4.8A at 12V
Why? Well Amps x Volts = Watts (power)
So Amps = Watts/Volts

So 0.25A x 230V = 57.5W
57.5W/12V = 4.8A

Inverters are somewhere around 85% efficient, so the inverter will actually draw around 5.5 - 5.6A to run the plugin charger at maximum output. If you need to run it for 2 hours to charge your device, it would take 5.5 x 2 - 11Ah from your battery.

In reality, the charger won't run at maximum capacity - certainly not for the full charging time. I'll take a stab in the dark & say that it could well be as little as half that. So after all that, we are back pretty close to your original figure.

What this does show is that using an inverter & a 230V charger is an inefficient way to charge a 12V device. If you are limited on habitation battery capacity, it's much better to get a 12V charger for your device if you can.

 

[PhilandMena]

Thanks for that, very helpful.

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[Gromett]

Just a quick question. You are converting 12v DC -> 240V ac, plugging a charger in that converts 240V ac to 12V DC. Each conversion incurs a loss.

Wouldn't it be better to get a small 12V regulator so that the 12-14.5v your battery sees is regulated to a strict 12V. DC-DC conversion is a lot more efficient than DC->AC -> DC.

I just had a quick search on Amazon for something like what I mean... Amazon product ASIN B008A6GCXK I am not recommending this particular one. But something similar will be kinder to your batteries and allow you to run a bit longer.

 

[PhilandMena]

That makes sense! Not sure what a 12 V regulator is ? (I do charge some 12 volt devices through a usb connection from a cigarette 3 way splitter plugged into my TV power point, 12V). However I do have other devices, electric tooth brush & lap top that performs better when receiving 240 volts , and do not have cable/connector for some devices which plug into 240 V and reduced to 12 V.
I will explore ! Thanks again.

 

[Lenny HB]

A quick calculation that gives fairly accurate results is, multiply the 12 v current by 20 and add 10- 15% (depending on inverter)for inverter inefficiency.

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[DBK]

A quick calculation that gives fairly accurate results is, multiply the 12 v current by 20 and add 10- 15% (depending on inverter)for inverter inefficiency.

Divide it by 20.

 

[DBK]

I would appreciate some guidance on my logic of calculating inverter usage.
So here is an example of my logic.
I have a sterling 300W pure sine wave inverter which states :
No load Quiescent current 0.95A (I take this to mean when switched on it draws 0.95A an hour from my leisure batteries.
I also have a 240 V plug to charge a device and on that plug it states input 100-240V -50/60Hz -500mA Output 12.0V -1600mA.
Would I be correct in assuming that if I plug this device into my Inverter and switch it on the total draw from my leisure batteries would be.
0.95 + 1.6 =2.55 Amps per hour.

Thanks in advance.

I think your assumption is about right although because of inefficiencies in the charging device it might be a shade more. A good guess is it might draw up to 3 amps but it isn't possible to be precise because when the inverter is working it may become more efficient and the 0.95 amps of "loss" might be less.

But as already suggested if you can get 12 volt chargers they are generally better. However, you can't always. My Olympus camera can only be charged from the mains so I have a small inverter like you.

 

[funflair]

I think he means multiply the 240v current by 20 plus a bit to give the 12volt draw.

Martin

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