Is there an easy way to work out "current draw"?

Submitted: Tuesday, Jun 20, 2006 at 15:34

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Scoey (QLD)

Hi all,

Further to my post 35055 about my Aux battery I am going to have to get myself a new battery and I thought I might also get myself a battery monitor while I'm at it. That's all a bit by the by but.

My question this time is, how do I work out the approximate current draw of a particular appliance? eg my phone charger running through my 300W inverter. I don't have the inverter specs on me but the phone charger has on it:

Input: AC 100-240V, 50-60 Hz, 150mA
Output: DC 5.7V, 800mA

So generally speaking, can I use the specs off the particular appliance to determine the current draw for that appliance so I can estimate battery life?
Also, if that appliance is running through an inverter to add the current draw for the inverter and appliance together or just use the end appliance's draw?

Why I ask is that I can just use the monitor to guage the battery usage but my mind is a "need to know how it works" brain!

Have I completely confused anyone? Apart from myself that is!

Oh, and if it's all to hard, just tell me! ;-) Thanks a lot!

Scoey!

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Reply By: Member - Coyote (SA) - Tuesday, Jun 20, 2006 at 15:45

Tuesday, Jun 20, 2006 at 15:45

Good questions.. I'll be interested to hear too. Have often wondered how to work this stuff out, but always just figured thats why I have a msart solenoid.. if my apliances flatten my reserve battery. no owrries, start the car onthe main and recharge all... but ignorance is no longer bliss, I shold really know to avoid flattening my battery and hence killing it..

AnswerID: 179475

Follow Up By: Scoey (QLD) - Tuesday, Jun 20, 2006 at 15:54

Tuesday, Jun 20, 2006 at 15:54

G'Day Coyote!

Your response has eased my mind! ;-) I thought I might have been asking a stupid question - but if someone else wants to know - then it can't be just me! ;-)

Cheers
Scoey!

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Reply By: hound - Tuesday, Jun 20, 2006 at 16:01

Tuesday, Jun 20, 2006 at 16:01

Given the device _Affordable_Storage_Drawers.aspx 800 milliamp current at 5.7 volts, what is the power requirement of the system? Multiply .80 AMPs by 5.7 to get 4.56 Watts

Think thats correct, only tiny current draw from a phone charger.

Then a 1amp draw fluor light would draw= 1 x 12volt = 12 watts
And cheap 5amp draw fluor would draw= 5 x 12volt= 60 watts

AnswerID: 179478

Follow Up By: Scoey (QLD) - Tuesday, Jun 20, 2006 at 16:07

Tuesday, Jun 20, 2006 at 16:07

So multiply the amps x volts to get watts? Right, got it! ;-)

Can I push the friendship further and then ask how do I then convert watts to amps per hour? If I didn't already mention the fact that I'm an idiot when it comes to this stuff, now is probably a good time to make that statement. I read the article on Battery Power and found the working on Amp/hours idea pretty simple. Just getting the current draw in terms of amp/hours is complicated! :-S haha!

Cheers
Scoey!

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Follow Up By: disco1942 - Tuesday, Jun 20, 2006 at 17:44

Tuesday, Jun 20, 2006 at 17:44

Scoey

You can not convert watts to amps per hour. You can however convert watts per hour to amps per hour. Subtle difference. You can however find out is the amp/hours drawn from your batteries if you know the efficiencies of your battery charger and your inverter. See my follow up to Mike Harding's reply

PeterD

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Reply By: Mike Harding - Tuesday, Jun 20, 2006 at 16:03

Tuesday, Jun 20, 2006 at 16:03

If your phone charger needs 150mA at 240V then it's power consumption is 0.15A x 240V = 36 watts. Your inverter will waste about 6W for the privilege of converting 12V to 240V therefore the overall consumption will be 42W. At 12V that means the current draw from your battery will be 42 / 12 = 3.5 amps. Hope that helps, if not just ask for more info.

Mike Harding

AnswerID: 179479

Follow Up By: Scoey (QLD) - Tuesday, Jun 20, 2006 at 16:10

Tuesday, Jun 20, 2006 at 16:10

Actually, that seems to be a tad more logical! (sorry hound!) So can i then assume that I will be using 3.5 amps...... per hour?? ;-)

Cheers Mike!
Scoey!

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Follow Up By: Mike Harding - Tuesday, Jun 20, 2006 at 16:44

Tuesday, Jun 20, 2006 at 16:44

Exactly. So if you had a 35Ah battery you could power your charger for 10 hours (you couldn't in reality because you'd damage the battery) at which point the battery would be totally knackered and in need of a good lie-down :)

Just work out the power consumption of other items in the same manner and if they are running at the same time add them up to obtain the total current draw. eg. if you are using two phone chargers total current from the battery will be 7A (we'll ingore the inverter loss here) and your 35Ah battery would last for 5 hours.

NB. you should not, usually, take more than 40% or 50% of a battery's capacity without recharging it otherwise you may shorten it's life.

Mike Harding

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Follow Up By: hl - Tuesday, Jun 20, 2006 at 17:19

Tuesday, Jun 20, 2006 at 17:19

Mike,

That 35 Watts for a phone charger is a tad high....It would get bloody hot if it were really dissipating 35 watts. I think that 150mA is running it at 100 Volts.
Be that as it may, running a couple of 240V phone chargers through an inverter is probably the least efficient way to do it. A 12V charger for the phones would make a lot more sense.

Cheers

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Follow Up By: Scoey (QLD) - Tuesday, Jun 20, 2006 at 17:38

Tuesday, Jun 20, 2006 at 17:38

Hey Mike & hl,

Thanks for the info - I only used the phone charger as an example (as that's all I had close to hand in terms of access to specs) but I was essentially after a reasonably simple formula to use to work out current draw, with a reasonable degree of accuracy anyway! I think I have that formula now! ;-)

Thanks again!
Scoey!

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Follow Up By: Mike Harding - Tuesday, Jun 20, 2006 at 17:40

Tuesday, Jun 20, 2006 at 17:40

I agree, but this was really an exercise in how to calculate current consumption rather than the efficiency of a particular device.

Keep in mind that even a 12V phone charger will be a switch mode power supply (just like an inverter) and will have losses - I'll guess the cheap 12V chargers are only about 80% efficient but that is probably better than an inverter running a light load. Fewer wires too.

Mike Harding

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Follow Up By: disco1942 - Tuesday, Jun 20, 2006 at 17:45

Tuesday, Jun 20, 2006 at 17:45

This shows the futility of using an inverter when 12V appliances are available. The input to the charger is 36W (240V x .15A) for an output of 4.56W (5.7V x .8A) – it is only 12.6% efficient.

Scoey – I suggest you purchase a car charger for your phone or a voltage regulator that plugs into the vehicle cigar lighter socket. You will then draw a maximum of 0.8 to 1A whilst your phone is charging. If you do that you can charge the phone whilst driving and save your deep cyclebattery to light your lights and cool your beer.

PeterD

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Follow Up By: Scoey (QLD) - Tuesday, Jun 20, 2006 at 19:20

Tuesday, Jun 20, 2006 at 19:20

G'Day Peter,

Yeah, the phone charger was probably a bad example, but I don't yet have a fridge do get the specs off or anything like that, and yes if I wanted to charge it, I'd use a car charger. The real question to this post was to try and find a basic formula that I could, by using the specs found on each appliance (be it a Waeco, a laptop, or just a simple fluro light), find out how much current it _Affordable_Storage_Drawers.aspx and therefore estimate how long my Aux would last before needing a recharge, or more specifically to try and plan what sort of requirements my Aux will eventually need and try and cater for it now. Does that make sense?? I'm confusing myself again! ;-) hehe!

Cheers for your input though, any advice on how to keep my beer cool for longer is appreciated! ;-)

Scoey!

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Follow Up By: Nifty1 - Tuesday, Jun 20, 2006 at 22:05

Tuesday, Jun 20, 2006 at 22:05

Scoey,

That's about right, except that you will confuse yourself by thinking in terms like "amps per hour". A device will draw a current of so many amps, while it is operating - doesn't matter if it's for a minute, second, or day. Where this becomes relevant is when looking at battery capacity. If a device _Affordable_Storage_Drawers.aspx 3.5 amps, and is run for one hour, then that's 3.5 amp-hours of capacity. (Same as running a device drawing one amp for 3.5 hours.)

As others have pointed out, there is loss in any inverter, and of course if you take 3.5 amp-hours out of a battery, then you need to supply the battery with a bit more than that to make it up again due to the system being less than 100% efficient.

Calculations are all very well. Take the advice, and borrow or buy an ammeter and measure the actual current draw. It's the only way to find out for sure.

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Follow Up By: GaryInOz (Vic) - Tuesday, Jun 20, 2006 at 22:13

Tuesday, Jun 20, 2006 at 22:13

I would guess that the 150 mA nominated would be on the 100 V end of things, and about half that current (62mA) on 240V, given that it would be a small switchmode power supply.

So 15W + 6W = 21W at 14 V = 1.5A

Better more economical way would be to get a 12V car adaprter for the phone, usually ~$20-30, this would use at the most 800 mA, possibly less if it is also a small switchmode (~320-400 mA)

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Reply By: hound - Tuesday, Jun 20, 2006 at 16:21

Tuesday, Jun 20, 2006 at 16:21

Depends if the charger is a cigarette type as they are only run 12volts not 240volts don’t they?

AnswerID: 179483

Follow Up By: Scoey (QLD) - Tuesday, Jun 20, 2006 at 16:24

Tuesday, Jun 20, 2006 at 16:24

Ah yep, sorry! In this case it's 240V mains charger!

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Reply By: Member - Oldplodder (QLD) - Tuesday, Jun 20, 2006 at 16:43

Tuesday, Jun 20, 2006 at 16:43

If concerned, why not install an amp meter?
Even you just hook one up as a temporary measure to check each bit of equipment or use a clamp on type?

AnswerID: 179484

Follow Up By: Scoey (QLD) - Tuesday, Jun 20, 2006 at 17:40

Tuesday, Jun 20, 2006 at 17:40

Yeah I guess I could look at that option, but at this stage I only need to know approximate draw. ie Am I trying to run waaaaay too much on my Aux, about the right amount of gear or have I got heaps more current available?

Thanks for the feedback though, Oldplodder!

Scoey!

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Reply By: luch - Tuesday, Jun 20, 2006 at 17:17

Tuesday, Jun 20, 2006 at 17:17

Your best bet is to speak with a auto electrician who should have the equipment to measure the current draw, then will be able to advise you on which size battery to get (they use this equipment to measure starter motors current AMP draw)

AnswerID: 179491

Follow Up By: Scoey (QLD) - Tuesday, Jun 20, 2006 at 17:41

Tuesday, Jun 20, 2006 at 17:41

Hi Luch,

Probably not a bad option either, but I might hold off on that until I want to get into specifics. Cheers for the info! ;-)

Scoey!

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Follow Up By: disco1942 - Tuesday, Jun 20, 2006 at 17:56

Tuesday, Jun 20, 2006 at 17:56

Bunnings are currently selling a digital multimeter for $10 that will measure up to 10A. (they also have a $5 one but the one I purchased is RS - too cheap.

Most appliances have their input current specified on them. If they only have a wattage on them then divide this figure (watts) by 12 (volts) and you will have the current draw. Multiply this by the number of hours you are going to use the appliance and you have A/H. For car fridges budger on them running for ½ or 1/3 of the total day in summer and 1/3 to ¼ of the day in winter.

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Follow Up By: Craigww2 - Tuesday, Jun 20, 2006 at 21:49

Tuesday, Jun 20, 2006 at 21:49

I have replied to you other post I have a clamp meter to help if needed

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Follow Up By: Scoey (QLD) - Wednesday, Jun 21, 2006 at 08:53

Wednesday, Jun 21, 2006 at 08:53

G'Day Craig,

Just replied to your other post, Thanks a lot for your offer, I might yet take you up on it, but just for the moment I'm gonna have a poke around with the multimeter and see what I can learn/find out on the weekend! ;-)

See how I go! CHeers mate!

Scoey!

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Reply By: Derek from Affordable Batteries & Radiators - Tuesday, Jun 20, 2006 at 22:03

Tuesday, Jun 20, 2006 at 22:03

Hi Scoey,

Where are you based ?

I use a clamp meter to test start up and running current. The old sums don't always work with switch mode chargers and fluorescent lights. Inverters also differ from brand to brand and pure to modified sine wave.

If you are in Brisbane pop into my workshop and I will do a test for you.

Regards Derek.

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Follow Up By: Mike Harding - Wednesday, Jun 21, 2006 at 09:02

Wednesday, Jun 21, 2006 at 09:02

>The old sums don't always work with switch mode chargers and fluorescent lights

Really? Would you explain why that is please Derek?

Mike Harding

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Follow Up By: Scoey (QLD) - Wednesday, Jun 21, 2006 at 09:05

Wednesday, Jun 21, 2006 at 09:05

G'Day Derek,

I'm based just north of Brisbane, so Beauont's a bit of a hike! If I'm down that way I might drop in and say g'day and have a chat!

Cheers
Scoey!

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