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 Q&D Automated load test
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kcarring
Moderator


Canada
1057 Posts

Posted - November 20 2012 :  23:57:16  Show Profile Send kcarring a Private Message  Reply with Quote
I have been considering a way to do a quick and dirty and hopefully cheap automated load testing circuit.

It occurred to me that a low voltage disconnect circuit obtainable, like this one:

You must be logged in to see this link.

Then it also occurred to me, that if a mechanical wall clock is set to 12:00 (or whatever), and a relay opens connection to it's batteries...
When the low voltage disconnect disables the relay, the clock stops, and displays the runtime...

At a given load, you then have your amp-hours. Well, approximate, as the voltage is always changing, but, good enough for the .... that i ....


~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!

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kcarring
Moderator



Canada
1057 Posts

Posted - November 21 2012 :  00:33:41  Show Profile Send kcarring a Private Message  Reply with Quote
quote:
Originally posted by SD3T

you could try one of the timer clocks light they use on lighting instead of photo cell.


Not sure what you mean, really, SD. What Im after is, a circuit, that when I'm done conditioning a battery through its cycle, I can turn on, and walk away, and when I come back, I know how many amp hours the battery gave at C20. So in this case, I set the disconnect to say 11.55-11.6V. For a 150 aH trojan 1275, for example, on goes the 8A load. And I walk away. When I return, the battery drops to 11.58V (ideally/approximately) THEN SHUTS OFF THE LOAD & THE CLOCK. Whenever I get home, is fine. I now have the runtime. 11.58V is 30% capacity, but, the battery may recover to 11.75V. I use the chart I presented in the capacity of a used battery thread to approximate what percentage I have discharged the battery, based on it's "recovered" voltage. You then calculate the number of amp hours that would have been left (theoretically) had it have been dropped right to 10.5V. Adding that to the amp-hours it actually ran, you then compare to what it ought to do, ideally, new.

Most importantly, you can see how many hours it ran the C20 load (on the clock), and arrive home (anytime you want) and not destroy the battery. Based on when you left, what the clock says, you know how long it's had to rest. without needing to be too technical, at very least, you can do one run, pulse condition, then do another - and compare.

cheers

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!

Edited by - kcarring on November 21 2012 00:35:35
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kcarring
Moderator



Canada
1057 Posts

Posted - November 21 2012 :  01:22:04  Show Profile Send kcarring a Private Message  Reply with Quote
Yes the timer would work, too, especially for say a "known to be reasonably good battery" in it's 2nd or 3rd cycle and on... but from my experience when u get a junk bat, the first time around... voltage alone means next to nothing, so lets say you put a 5A load on it, cause it SAYS 12.6, and you take off... you really dont know if that will run for the expect 12 min. hours, 20 hrs or maybe its gonna be hitting 9V by the time you are eating lunch at work... who knows. this circuit would cure that dilemna. and, when you get back the number and hours and minutes it ran to a set cutout voltage, is right there on the clock... justa cheap way of doing an automated sustained c20 load test basically

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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kcarring
Moderator



Canada
1057 Posts

Posted - November 21 2012 :  01:41:59  Show Profile Send kcarring a Private Message  Reply with Quote
One might think, ok no problem use the above circuit, and instead of having the LED light, have a relay, that disconnects the load

BUT

There is an inherent problem of hysteresis with that relay under load. The relay might pop, then relatch (I've read)... so maybe it's not quite as simple as I thought...

This one looks pretty good:

You must be logged in to see this link.

Given that it is load testing, these simplified circuits should do the trick, but they are too "power thirsty" for a LV disconnect in "all device", i.e. an LED lamp. you don't want a low voltage disconnect switch running at 300mA for a 300mA LED circuit LOL

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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kcarring
Moderator



Canada
1057 Posts

Posted - November 21 2012 :  01:56:06  Show Profile Send kcarring a Private Message  Reply with Quote
This is such a beast, but a bit spendy at $36
You must be logged in to see this link.

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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kcarring
Moderator



Canada
1057 Posts

Posted - November 21 2012 :  02:27:26  Show Profile Send kcarring a Private Message  Reply with Quote


660-CF1/4C511J 510 OHM RESISTOR
660-CF1/4C102J 1.0 K RESISTOR
660-CF1/4C472J 4.7K RESISTOR
660-CF1/4C822J 8.2K RESISTOR
660-CF1/4C332J 3.3K RESISTOR
511-STP55NF06L 60 VOLT 55 AMP MOSFET
31VA401-F 10K 24 MM POT
512-2N3906TA PNP SMALL SIGNAL TRANSISTOR
511-TL431ACZ ADJUSTABLE PRECISION SHUNT REGULATOR
103-1012-EVX N.O. PUSHBUTTON SWITCH


Courtesy:
You must be logged in to see this link.

User Dan Baldwin on the forum, stated:

quote:
In a circuit using a tl431 where there are two resistors setting the cut off voltage such that one goes to the positive rail (resistor A), one goes to the negative rail (resistor B), and the junction of the two is connected to the reference lead of the TL431
cut off voltage = ((resistor A + resistor B)/ resistor B) * 2.5

Or

cut off voltage = ((resistor A / resistor B) * 2.5) + 2.5

Both formulas should give the same answer.



~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!

Edited by - kcarring on November 21 2012 02:29:23
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - November 21 2012 :  05:07:11  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Kyle

I would go with the extremecircuits circuit and do like you said , add a suitably size relay.

ron
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kcarring
Moderator



Canada
1057 Posts

Posted - November 21 2012 :  13:57:18  Show Profile Send kcarring a Private Message  Reply with Quote
ron_ I am going to try it, but my "thought" is, that - hysteresis could wreak havoc in that circuit, with a relay, switching a heavy load, causing the circuit to open, then relatch closed. We shall see. The latter circuit looks more usable for say a lamp (regular usage) - in that relays, add a fair bit of power consumption (if large) ... so I thought I'd toss both of the circuits out there for people to look at / play with.

Of course I could be totally wrong, wouldn't be the first time (today).

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!

Edited by - kcarring on November 21 2012 14:13:47
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - November 21 2012 :  17:26:05  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Kyle

The circuit " as is " , would have the relay switched on until the battery voltage falls and the circuit switches the relay off.
If you power the circuit via the relay contacts ( use a by pass switch across the relay contacts to initially switch the circuit/ relay on ) once the relay switches off , everythings off , so no relay chatter / spikes / hysteresis problems

ron
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