Author Topic: battery drainage (Read 2279 times)

pomprocker · « **on:** March 23, 2009, 05:17:55 PM »

This is kinda non-robotic related but could be some good questions.

I bought this shower clock so that maybe i won't be late in the mornings

The battery that came with it was dead, and I didnt feel like returning it.

The battery recommended on the back case says AG10. I looked it up and that is equivalent to an LR1130. I looked that up and its equivalent to an Energizer 389 1.55vdc.

Anyway the point of the story is, I tested the new battery with my dmm when i got it and it was 1.55vdc, and two days later my clock was dead and the voltage on the battery was less than 1v...approx 0.6v

what the hell? that was a $5 battery!!

Can anyone tell me where things went wrong?

Razor Concepts · « **Reply #1 on:** March 23, 2009, 05:31:17 PM »

Not sure what happened, maybe a poorly designed clock.

For now you can poke around with these 15 cents each batteries
http://www.dealextreme.com/details.dx/sku.2980

Soeren · « **Reply #2 on:** March 23, 2009, 05:44:32 PM »

Hi,

Quote from: pomprocker on March 23, 2009, 05:17:55 PM

Anyway the point of the story is, I tested the new battery with my dmm when i got it and it was 1.55vdc, and two days later my clock was dead and the voltage on the battery was less than 1v...approx 0.6v

what the hell? that was a $5 battery!!

Can anyone tell me where things went wrong?

Somewhere between the desk clerk saying "5 bucks please" and you actally putting out that amount for something that you should get 5 pcs a buck

How much did you pay for the clock?

Measuring the unloaded voltage doesn't really tell all that much. If you want a clear picture of the condition of a cell, measure it with a rated load.
The cell may have been old and only having a "skin-voltage". However, the case is that you allready saw one flat cell in the clock and so, there's a chance the clock is loading the cell harder than it should. Either measure voltage and current with a 1.5V supply (could be from an AA, just make a dummy for the button cell).
Or, measure the load with an Ohmmeter (plus to where the plus pole of the cell goes inside the batt. compartment).

I have had a couple of clocks that just sucked too much current - dropped them into the spare parts bin, since clocks are too cheap to waste more than a few minutes repairing, unless you really like it for some sentimental reason.

I just have to ask... Since I got this image of a clock hosing you down when you're sound asleep... What is a "shower clock"?

pomprocker · « **Reply #3 on:** March 23, 2009, 07:32:32 PM »

This is it

paulstreats · « **Reply #4 on:** March 23, 2009, 07:35:09 PM »

The concept of these things can seem great at the time, being splash proof. But are they steam proof? does the steam get in, then condense and slowly discharge the batteries?

pomprocker · « **Reply #5 on:** March 23, 2009, 09:09:42 PM »

There is a very good rubber seal on the back protecting the battery compartment.

The LEDs use 3 AAs and the clock uses one 1.55v ag10/lr1130/389

however the anti-fog coating on the mirror only lasts one or two showers, you have to clean it with iso alcohol

very poor product...But since I spent $30 on it and i didnt keep the packaging or receipt i'm forced to live with it.

The thing I wanted most was the clock!! i didn't care about the huge mirror and all that.

urrg

pomprocker · « **Reply #6 on:** March 23, 2009, 09:13:59 PM »

Quote from: Soeren on March 23, 2009, 05:44:32 PM

Either measure voltage and current with a 1.5V supply (could be from an AA, just make a dummy for the button cell).
Or, measure the load with an Ohmmeter (plus to where the plus pole of the cell goes inside the batt. compartment).

Can you explain this a little further?

Hawaii00000 · « **Reply #7 on:** March 23, 2009, 09:50:34 PM »

Quote

This is it

I'm guessing thats you in the mirror. lol

pomprocker · « **Reply #8 on:** March 24, 2009, 10:07:04 PM »

Quote from: pomprocker on March 23, 2009, 09:13:59 PM

Quote from: Soeren on March 23, 2009, 05:44:32 PM
Either measure voltage and current with a 1.5V supply (could be from an AA, just make a dummy for the button cell).
Or, measure the load with an Ohmmeter (plus to where the plus pole of the cell goes inside the batt. compartment).

Can you explain this a little further?

measuring ohms at the battery connection points it measured 1.922 and the clock actually powered on from the multimeter. so what does this tell me?

pomprocker · « **Reply #9 on:** March 25, 2009, 09:13:42 PM »

This battery is weird (energizer 389)

After I had originally taken it out when the clock died it measured at around .45v down from 1.55v. I let it sit out for a couple days and measured it again and it was back up to about 1.49v. So i put it back in and the clock worked again. Tonight when i looked the clock was dead again and the battery is reading 0.44v.

I dont get it

SmAsH · « **Reply #10 on:** March 25, 2009, 11:16:46 PM »

is it hot at all when you take it out?

sonictj · « **Reply #11 on:** March 26, 2009, 12:26:18 AM »

Quote

measuring ohms at the battery connection points it measured 1.922

if that is a measurement of ohms not kohms than your current would theoretically be .8A and your power would be 1.24Watts.

I find it hard to believe that you'd have a measurement of 1.922 ohms. So I'm going to assume you measured 1.922 kohms which yields .8mA.

Here are the formulas for calculating how long the battery will last

battery voltage = resistance * current

current* time = capacity

I googled the capcity and found it to be ~68mah

so if I asume you measured 1.922 kohms

68mah/.8ma = 85 hours of continuous use (sounds kinda crappy to me just 3.5 days of use)

if I asume you measured 1.922 ohms

.068Ah/.8Ah = ~ 5 minutes (this would indicate a shot circuit)

I would guess this clock is supposed to be shut off after you take a shower otherwise this thing will essentially die in under week.

pomprocker · « **Reply #12 on:** March 26, 2009, 10:27:14 AM »

Thats lame, well there is no on/off switch :/

Soeren · « **Reply #13 on:** March 26, 2009, 04:18:23 PM »

Hi,

Quote from: pomprocker on March 25, 2009, 09:13:42 PM

This battery is weird (energizer 389)

After I had originally taken it out when the clock died it measured at around .45v down from 1.55v. I let it sit out for a couple days and measured it again and it was back up to about 1.49v. So i put it back in and the clock worked again. Tonight when i looked the clock was dead again and the battery is reading 0.44v.

I dont get it

It's not weird at all. Button cells have a relatively high internal resistance and the behaviour you have observed is typical for a case like, this with a too high current drain.
The battery cell seems flat in a short time due to the load, but chemically it has still got a good deal of its power left, so given a rest, it will seemingly recover.

Too bad you can't return the clock, because if it is due to moisture penetrating the seals and you dried it out (by siccative, by (moderat) heat or any other method), it will redevelop the problem in a short time. If, on the other hand, the problem is by design, no solution exists.

Sorry to be so pessimistic, but in your situation I would cut my losses and dismantle it for parts (hey, a mirror may come in handy in robotics).
Then get a cheap analog clock and put it outside the immediate danger zone.
My S.O. has got a quite ordinary clock in her bathroom ($4 IIRC) and she really steams up the bathroom (semi-scientific fact: Girls need their showers about 10°C hotter than boys) and it has run for about a year. The former clock she had was a bathroom clock who had the same problem as yours (placed in the very same spot) - go figure.

Admin · « **Reply #14 on:** April 08, 2009, 12:39:21 AM »

Hook up a multimeter on current setting to the battery to see if its really a short.

My guess its a el-cheapo 'made in China' clock.

pomprocker · « **Reply #15 on:** April 09, 2009, 11:09:31 AM »

Quote from: Admin on April 08, 2009, 12:39:21 AM

Hook up a multimeter on current setting to the battery to see if its really a short.

My guess its a el-cheapo 'made in China' clock.

To the battery or the battery leads in the back of the clock? and if the battery leads, with or without the battery in place?

SmAsH · « **Reply #16 on:** April 09, 2009, 02:50:16 PM »

to the battery leads in the clock with the battery? and see how much current is going through or i guess you could use the resistance setting to see if its a short.

Trumpkin · « **Reply #17 on:** April 09, 2009, 03:23:15 PM »

you could try getting an ac adapter (same voltage as battery) and solder it to the battery terminals in the clock and then waterproof it with lots of hot glue.

sonictj · « **Reply #18 on:** April 09, 2009, 04:06:26 PM »

to measure current you must place the multimeter leads in series with the power source. SO the battery would have to touch one contact. You would then have to place one test lead to the unconnected side of the battery, and the other lead to the second battery connection.

Essentially BAT_connector+ -> BAT+ and BAT- to multimeter- then multimeter+ to BAT_connector-

this will give you a current measurement. It may be negative but that doesn't matter. Only the magnitude matters.

pomprocker · « **Reply #19 on:** April 09, 2009, 04:34:46 PM »

Thanks I'll try this, but I really don't know what my expected result is. What values should I look out for?

SmAsH · « **Reply #20 on:** April 09, 2009, 05:22:04 PM »

if there is something very high on the current setting or on the resistance a very low reading.

pomprocker · « **Reply #21 on:** April 12, 2009, 01:17:39 PM »

With the DMM leads on the pos and neg leads on the clock w/o battery, The DMM set on 20Kohm, the reading on the DMM was 2.37. What does that tell me?

Soeren · « **Reply #22 on:** April 12, 2009, 08:41:24 PM »

Hi,

Quote from: pomprocker on April 12, 2009, 01:17:39 PM

The DMM set on 20Kohm, the reading on the DMM was 2.37. What does that tell me?

That your AG10 cell, which has a nominal capacity of somewhere between 70 mAh and 90 mAh depending on manufacturer, is loaded with an average current around 0.5 mA (over it's lifetime).
Doing the numbers for the best quality cells says: 90mAh / 0.5mA = 180h (= 7.5 days).
So you should expect the cell to last no more than a week, probably less.

You mentioned 1.9kOhm in a previous post and that only makes it worse.

You should seriously consider, if spending more time on it is worth the effort - you could earn money for a better one, in the same time you're spending fighting this lost cause.

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Author Topic: battery drainage (Read 2279 times)

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