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VT
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Inverter Current

08/04/2009 12:31 AM

Greetings CR4'ers,

If I have an 800W, 12Vdc to 240Vac, power inverter hooked up to a 105Ah deep cycle (12V) battery, how will the current (dc) drawn by the inverter (from the battery) vary with increasing ac loads (from the usual home appliances)? Can I measure this variation by simply series-connecting an ammeter between the positive terminal of the battery and the red (positive) cable of the inverter?

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PWSlack
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#1

Re: Inverter current

08/04/2009 3:25 AM

Yes, that would do it. It should be linear, and almost proportional.

Expect the internal circuitry of the inverter to shut the thing down if the aggregate of the connected AC loads exceeds 800W.

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Nah
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#2

Re: Inverter Current

08/04/2009 2:12 PM

DC power in = AC power out + 10%.
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eng.hmk
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#4
In reply to #2

Re: Inverter Current

08/17/2009 3:06 AM

do you have any reference for this equation
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Andy Germany
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#5
In reply to #2

Re: Inverter Current

08/19/2009 11:54 AM

A really well designed one may get to over 80%, but I simply cannot believe 90%, sorry!

A bought (cheap?) one will probably be 60% efficient at best......

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lighthasmass
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#6
In reply to #5

Re: Inverter Current

08/19/2009 9:14 PM

The Outback inverters I install are rated at 96% at full output. The key is "full output". Efficiency drops as the load does.

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Andy Germany
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#7
In reply to #6

Re: Inverter Current

08/20/2009 2:20 PM

Excellent point (I did not dream that even that was possible, thanks for the info).....But take an average for a small user and what do you get? Lets say 70% on average? and that would be excellent......

Have a great day.

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ashoktoshniwal
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#3

Re: Inverter Current

08/05/2009 1:15 AM

Losses in the inverter are

(a) Fixed loss, which includes power consumed by the control circuit and MORE IMPORTANT the standby current & therefore power "consumed" by transformer & filter choke ( if any ). I presume the inverter is not high frequency switching type, rather a more conventional one using 50/60 Hz power transformer.

(b) Variable loss, which is the power loss in the power circuit ( IGBTs, MOSFETs. etc.) and copper loss in the transformer & a very marginal increase in the magnetising current of transformer.

Thus the current drawn from battery will not be directly proportional.

Thanks and regards

Ashok Toshniwal, Bangalore, India
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Andy Germany (2); ashoktoshniwal (1); eng.hmk (1); lighthasmass (1); Nah (1); PWSlack (1)

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