Measuring the power consumption of a battery driven device

[gordon s]

My wife bought a Christmas decoration that requires 4 AAA 1.5v batteries to run.  We put four brand new Duracell batteries in it last night and it worked perfectly.  Switched it on tonight and the batteries have gone flat and it's no longer working.  Two of the batteries had dropped from 1.5v to one volt and the other two were 0.6v.

 

The display has one or two LED's to provide light and some form of pump to blow glitter around inside to give the impression of snow.

 

As someone who loves Christmas, she's quite taken with it.  Of course we can take it back, but I'm curious to know how I can measure the current consumption to see how long batteries should last and then either make up a 6v power pack or just accept it's going to need rechargeable batteries changing every day.

 

I have a multi meter and can measure volts, mA etc, so slid a piece of double sided pub strip between one of the batteries and it's contact point in the unit.  This acted as an insulator and I thought that if I used my multimeter in series by touching the probes to either side of the pub strip that would make the circuit and I would get a measurement.

 

Sadly that doesn't work and the unit doesn't switch on and I'm getting nothing from the meter.  Slide out the pub strip and it all functions normally.

 

Apologies for being a numpty, but what am I doing wrong?

 

How can I measure the current consumption of the circuit?

 

My meter says it is 600V 400mA Max.

[Nearholmer]

You might make a stab at an answer by looking at the battery capacity. They are often 1500mAh or similar. Look to see whether they are in series or parallel, then divide the effective capacity by how many hours they lasted to get an idea of current consumption.

 

Little fans can be quite greedy - we have a bubble blower, and that eats batteries very quickly.

 

Also, did you check for any fault that might be draining the batteries when it is turned off?

[vitalspark]

You measure voltage in parrallel but current in series so placing the meter in series is correct however even if it is set on its lowest current scale the internal resistance in the meter may make it unsuitable for such tiny loads.

 

I usually have to defer to my old but very sensitive analogue AVO for such measurements.

 

On the batteries failing after a day I'm not surprised. These illuminated decorations really need a DC in for a plug in power supply.

 

If theres no facility a small hole in the base and a socket from Maplins soldered in to the outgoing leads from the batteries would be easy enough to do.

 

Either that or have a spare set of rechargeables and as you say swap them every night.

 

Merry Xmas!

 

Dave.

[smokebox]

And don't forget that when measuring current (amps, milliamps etc) most digital meters need to have at least one of the meter leads moved to a different socket on the meter.

[AndyID]

And don't forget that when measuring current (amps, milliamps etc) most digital meters need to have at least one of the meter leads moved to a different socket on the meter.

 

That's a possible explanation. Another is that the meter has a fuse to protect the current shunt and the fuse has blown. Accidentally connecting the meter leads across a power source when it's set on amps instead of volts can pop the fuse.

[gordon s]

And don't forget that when measuring current (amps, milliamps etc) most digital meters need to have at least one of the meter leads moved to a different socket on the meter.

 

 

Thanks, but my Wavetek meter has fixed leads.

 

Thanks to everyone for your input.  Glad to know I was hooking it up correctly but suspect my Wavetek DM9 may not be sensitive enough to measure the current.

 

http://docs-europe.electrocomponents.com/webdocs/013f/0900766b8013f166.pdf

 

I'll see how it goes tonight, but my favoured solution will be to add a 6v supply via a mains charger.

Edited December 15, 2017 by gordon s

[davetheroad]

A 1500mAh battery drawing 400mA should theoretically last 3 hours 45 minutes. You either need a power supply or rechargeables. I use 4 x 850 mAh Duracell NiMh AAA cells in some of my locos and because the manufacturers calculate the capacity in a different way to lipos one of those 850mAH cells is equivalent to a 250mAh lipo. The problem with rechargeables is they can take as long to charge as they take to discharge.

[melmerby]

 

I usually have to defer to my old but very sensitive analogue AVO for such measurements.

 

Merry Xmas!

 

Dave.

No more sensitive than a decent Digital MM.

(I've got an Avo 8 as well as a few old Fluke DMMs)

It's the internal resistance of the current measuring circuit that is the significant factor. Cheap meters are generally high resistance.

 

Keith

Edited December 16, 2017 by melmerby

[Nearholmer]

“Cheap meters are generally high resistance.”

 

I know that to be true, but I’ve never properly understood why.

 

Is it that they contain only one shunt to cover multiple ranges, or something to do with the measurement circuit itself?

[melmerby]

“Cheap meters are generally high resistance.”

 

I know that to be true, but I’ve never properly understood why.

 

Is it that they contain only one shunt to cover multiple ranges, or something to do with the measurement circuit itself?

Judge for yourself.

 

This is my old Fluke 75:

 

This is the circuit diagram:

(Note a custom microprocessor for measurement computation with a separate display driver.)

 

The Fluke has two shunts for current; one for the 10A range and another for the rest and the voltage ranges are auto switched by the processor but it also has a separate low DC (300mV) range

 

The cheaper ones will just have basic switching circuits without the sophistication of these instruments.

Here is a more basic one

 

 

You may notice the very non-linear AC-DC conversion (1N4004s) The Fluke does it in the Chip and is true RMS.

 

.

Keith

Edited December 16, 2017 by melmerby

[Junctionmad]

The key issue when measuring current is the " burden voltage " , typically specified as mV/mA .

 

In the OPs case the burden voltage is given as 1.7mV/mA , hence if the circuit was demanding 400mA , hence 680 mV would be dropped across the meter. With low power battery devices that can be enough to prevent the circuit from working , or where it does to introduce significant error.

[tender]

Assuming the batteries are wired in series to give 6v, hence your comment on replacing with a 6v psu. One has to ask the question as to why two batteries read 1.0v and the other two 0.6v . Wired in series they should have all discharged the same amount so should all read approx the same voltage. Maybe you have some duds. There's a thread on here somewhere about counterfeit duracells.

Edited December 16, 2017 by tender

[AndyID]

Assuming the batteries are wired in series to give 6v, hence your comment on replacing with a 6v psu. One has to ask the question as to why two batteries read 1.0v and the other two 0.6v . Wired in series they should have all discharged the same amount so should all read approx the same voltage. Maybe you have some duds. There's a thread on here somewhere about counterfeit duracells.

 

Batteries discharge due to a chemical reaction. Each battery has slightly different chemistry and they were not all made at the same instant in time. Because of that, some will discharge a bit faster than others.

 

It's a bit like the cells in a 12 volt car battery. When a car battery fails it's usually because one of the six cells konked out even though the other five are still able to hold charge.

 

I suspect Gordon's meter has blown a fuse. Please see my earlier post.

[DCB]

Assuming the batteries are wired in series to give 6v, hence your comment on replacing with a 6v psu. One has to ask the question as to why two batteries read 1.0v and the other two 0.6v . Wired in series they should have all discharged the same amount so should all read approx the same voltage. Maybe you have some duds. There's a thread on here somewhere about counterfeit duracells.

I use Duracell in my camera as rechargables don't give enough voltage to work it and I have never found any appreciable difference between the voltage the batteries which have gone too flat to work it. I reuse mine to operate LEDs in buildings out side, but I suspect your "Duracell" are either the discount version sold in Poundland which have different packaging to the proper long lived versions or are simply counterfeit.

[AndyID]

The key issue when measuring current is the " burden voltage " , typically specified as mV/mA .

 

In the OPs case the burden voltage is given as 1.7mV/mA , hence if the circuit was demanding 400mA , hence 680 mV would be dropped across the meter. With low power battery devices that can be enough to prevent the circuit from working , or where it does to introduce significant error.

 

Gordon's meter is more than sufficiently sensitive to measure the sort of currents we are concerned about here. Either Gordon "coqued-up" the measurement (sorry Gordon) or his meter is not measuring current because the fuse has blown. (Please see above.)

[gordon s]

Thanks for all the info.  Problem solved as I found an old 5v phone charger and once I looked at it more closely found there was a jack socket hidden away in the base.  Plugged it in and everyone is happy.

 

Interested to read about fake Duracell's as my wife buys them from Tesco.  I have to assume they are genuine.  The best before date is 2027!  

 

Jeez, I might not live that long....

[kevinlms]

Thanks for all the info.  Problem solved as I found an old 5v phone charger and once I looked at it more closely found there was a jack socket hidden away in the base.  Plugged it in and everyone is happy.

 

Interested to read about fake Duracell's as my wife buys them from Tesco.  I have to assume they are genuine.  The best before date is 2027!  

 

Jeez, I might not live that long....

At least they're fresh.

 

One of our electronic chains went broke a while back (Dick Smith) and the liquidators reported that the previous owners, in some of the more common sizes (AA & AAA) of their house brand, they had up to 12 years supply on the warehouse shelf!

So dire were their finances, that they had previously sold some of the excess stock, back to the manufacturers! At a significant loss.

Edited December 17, 2017 by kevinlms

[brigo]

I was looking at AA battery life a few months ago for a project my daughter was doing, she needed to run some white LEDs off just two AA cells. I found this site which gives the discharge characteristics for variou types and makes of AA cell.

 

http://www.batteryshowdown.com/index.html

 

There are comments about Duracell batteries, the Poundland ones are Duracell, but have a reduced capacity.

 

She ended up having to use the expensive Energizer Lithium cells because of their flatter discharge characteristics.

 

Brian

[Tricky Dicky]

Tool station do a Duracell Industrial battery in AA & AAA sizes at a fraction of the cost of standard Duracell batteries sold retail. I have used them now for some time and although I have not really noted their longevity I do feel they are as good as the standard Duracell offering.

 

Richard

[dhjgreen]

Tool station do a Duracell Industrial battery in AA & AAA sizes at a fraction of the cost of standard Duracell batteries sold retail. I have used them now for some time and although I have not really noted their longevity I do feel they are as good as the standard Duracell offering.

Richard

JCB batteries 4 AA for 99p, just as good.

[Junctionmad]

Gordon's meter is more than sufficiently sensitive to measure the sort of currents we are concerned about here. Either Gordon "coqued-up" the measurement (sorry Gordon) or his meter is not measuring current because the fuse has blown. (Please see above.)

The issue with burden voltage ( another way of stating resistance ) is that on devices where the battery operating has little room to fall before the device stops , is the meter drops too much voltage to allow the circuit to work.

 

It's not about sensitivity per se

[gordon s]

That's a possible explanation. Another is that the meter has a fuse to protect the current shunt and the fuse has blown. Accidentally connecting the meter leads across a power source when it's set on amps instead of volts can pop the fuse.

 

 

Apologies Andy, I'd missed that one.  Just off to bed, so I'll check that in the morning.  

 

That does make sense, but would it still read other scales with the fuse blown or does that only apply to the current reading element?

 

Voltage, resistance and continuity all work fine.

[AndyID]

Apologies Andy, I'd missed that one.  Just off to bed, so I'll check that in the morning.  

 

That does make sense, but would it still read other scales with the fuse blown or does that only apply to the current reading element?

 

Voltage, resistance and continuity all work fine.

 

Hi Gordon,

 

It depends on the particular meter's design of course but I do know some meters protect the current shunt with a fuse. Other functions will still work properly.

 

I seem to remember I found this out the hard my with my own DVM

 

Andy

[gordon s]

That's a possible explanation. Another is that the meter has a fuse to protect the current shunt and the fuse has blown. Accidentally connecting the meter leads across a power source when it's set on amps instead of volts can pop the fuse.

 

 

Thanks Andy.  You were right...

 

Took the meter apart this morning and the fuse had blown. Something else learned.

 

Pack of 250v 500mA Fast Blow Ceramic fuses on their way to me now. 

[Crosland]

The issue with burden voltage ( another way of stating resistance ) is that on devices where the battery operating has little room to fall before the device stops , is the meter drops too much voltage to allow the circuit to work.

 

The Heisenberg voltage. You can't measure something without affecting it