Whilst in storage, your battery should be on float charge.
For a typical unloaded and fully charged and good condition wet lead-acid battery at normal room temperature, recharging is generally recommended at no less than 3 monthly intervals. (Though a mate has a great wet cell battery that stood uncharged for 2 years! There are always freaks and exceptions...)
For any battery with a load, sulfation builds. And soft sulfates turn hard after about one week.
A loaded battery with infrequent charging and maintenance (equalisation etc) may last a season or year etc, but not the expected 4-6 years.
LOL. If "275 out of 340" is it's capacity, 275/340 = 80.9%...
80% (of original capacity as opposed to "rated" capacity) is the "normal" criterion used for an end of life battery. Sorry - I just LOL'd at the coincidence. (FYI - I retired my last crankers at an estimated 10% remaining capacity.)
And I'm glad that they replaced it. IMO they should have upon you first flattery experience... (Imagine being stranded somewhere... Sue the pricks!)
And BTW - don't buy batteries from automotive dealers, find a reputable battery shop.
But to answer your previous post - and possibly save a premature battery failure....
Re the PIC, again, it depends on your setup. The PIC could sequentially power up a few devices and can sense various inputs, whether voltage, or diode-connected etc interrupts or state changes, etc.
The 8-pin (leg) PIC08 has 6 inputs/outputs, but bigger PICs have more, however I was only considering the PIC08M2 for its $3-$5 price and small size (especially if SMD). For more inputs etc, I'd go CPU but only because my many I/O applications IMO are easier with a CPU and their expandable memory (rather than single wire RAM etc) and other peripherals. But the PICs have many of those peripherals internally (counters/timers, ADCs, DAC(s), etc).
For a 5V supply, a resistor would feed a 5V Zener (or whatever Zener is close; 4.7V, 5.2V etc).
The resistor drops the voltage from the supply (eg, 8V-16V) but must carry the current required by the load.
Its Wattage is its voltage drop time the current thru it. EG, for 16V in and 5V out, it drops 11V. For 0.5A, that's 11V x 0.5A = 5.5W, hence a 10W resistor though 5W might do if the voltage is rarely above (say) 14.5V (hence 9.5V x 0.5A = 4.75W hence 5W is ok if the resistor wattage is conservative and it doesn't get too hot (for practical purposes) OR the normal load is under 0.5A OR additional heatsinking is added.
(And the resistor value: for a 11V drop at 0.5A, V=IR hence R=V/I = 11V/0.5A = 22 Ohms which is coincidentally one of the preferred resistor values, so a 22 Ohm 10W resistor.
The Zener must be able to "dump" the max current through the resistor - eg, 0.5A. For a 5V Zener, that's 5V x 0.5A - 2.5W, hence a 5W Zener. (Zeners are often 400mW, 1W & 5W. Bigger than that gets rare and expensive.)
Hence Zener regulators are suited only for small current applications, and the 0.5A example above is probably too high current wise for practicality.
Besides, an LM317 is only a $few and has internal thermal limitation, regulates very accurately, and there is no trickiness involved in selecting its resistor. (2 low-power resistors required where one is 120 Ohms to ensure the minimum required load of 10mA for a stable output). Zener voltages vary a bit depending on their current (as with all diodes).
They are both linear and waste about the same amount of power, eg, "burn" up to 11V @ 0.5A in (5.5W) for an output of [email protected](2.5W), hence an efficiency of 2.5/(2.5+5.5) or ~30% at 16V in; 50% at 10V in etc.
A regulator for the Arduino etc:
For power conservation, a dc-dc converter is recommended. They are almost 100% efficient (typically 95% or more), and hence for 12V to 5V will more than half the current as used for a linear regulator.
Mind you, that depends on their overhead (or standby) current. If that's ~equal to the average load current, then overall efficiency is ~50%. If it's higher than the load current, then linear regulators may be more efficient.
IOW, dc-dc converters are NOT used for low current applications if power conservation is critical (eg, uA and maybe mA loads etc).
In your case, for an Arduino battery, maybe a CurrentLogic PCB-mounting CLD5-12S05 (9~18VDC input; 5V 1A (5W) output) for ~USD$10.50 ($9.50 + $1 for 5W output).
[ Note - unfortunately they do not state their standby currents, but assuming "typical efficiency >80%" applies to 50% load and higher, that could mean up to 20% inefficiency x 50% load x 5W = 10% x 5W = 0.5W or 0.5/12V = 40mA overhead, though I'd expect far less. You'd have to confirm with them. Maybe >80% efficiency covers a mere 10% of rated load etc. (Max efficiency is usually at ~80% of rated load and is hence easier to calculate, though how much is current-dependent circuit loses as opposed to mere no-load "idling" overhead/drain is just as tricky...) So ignoring its idling overhead... ]
Hence a normal 12V battery whose AH rating you can probably calculate (based say on max 20% discharge; I'd recommend the Yuasa NP range.
[ Quick quesstimate: (24+9)mA = ~35mA => ~15mA @ 12V => 15mAH => 20x15 = 300mAH for 20hours (C20) x 1/.2 for 2-% discharge => 1.5AH. Hence an NP1.2-12 for ~20 hours to 80% capacity, or NP4-12 for a guesstimated 3 days, or NP7-12 (usually the best bang for bucks (~$25-$35)) for 6 days etc. But please check my guesstimates - I could be wildly wrong!. And those reserve times multiplied 1/3 to 1.2 for linear regulators. ]
And an isolating relay. (For small batteries, current limiting is desirable, but that might not be a problem if not too discharged.)
Capacitors - forget it. Not for mA loads for more than a few seconds or minutes.
If they hadn't backed off of the cost I would have purchased it elsewhere. After the warranty wears out and I need a new one (my 2k3 mx-5 still had the stock battery when I sold it last year but it was a gel), I'll buy elsewhere.
Originally Posted by OldSpark
I hadn't thought of a battery like those Yuasa. I can probably get one from a buddy who has a bunch. I just really don't want another car battery sized thing.
I'm not sure I follow the need for the regulator. There is already one built into the Arduino. I'm not up on the dual battery setups and will need to read some but would think you need a relay that would hold open when the computer(or ignition) is on and slam immediately there after to keep from draining the main crank battery. Sounds like you suggest a current limiter to keep from smoking the little guy when it starts charging? I assume it would just charge like normal off of the charging system. I would also guess a GOOD diode is in order.
A regulator is needed if the load does not handle the typical 8-16V range of a car's electrical system. At least that is my own traditional "design" spec plus some high voltage (spike) suppression, though some cars may not exceed 15V etc and good batteries or low current starters may not cause dips as low as 8V. (The M2 & M4 dc-dc PSUs are spec'd down to 6V!)
If the Arduino handles car voltages, then fine. But if it is merely a "12V input", then it probably only handles ~11.5V to maybe 14V or 14.5V - the typical range for a battery with bench charger - if that (ie, it may require regulated 12V). I know the Arduino can be powered via USB, but I don't remember its higher voltage input spec.
The isolating relay is to prevent batteries being in parallel when not being charged (otherwise 2 parallel batteries can fail FOUR times as often as a single battery; one failing battery drains the other battery, etc etc).
The relay is not to open during cranking - that would require a 2nd power source, ie, a 2nd battery.
But normally, battery isolators are open during cranking - they should only be closed when the system is charging. (Hence one possible flaw with voltage sensing or smart isolators - if they have a minimum connect time with no immediate disconnect when the vehicle cranks (however that may be determined!), then the 2nd battery also powers the starter, and pop go the one or both of the 2 interlink fuses or circuit-breakers (else the inter-wiring or relay if not protected).
Batteries have a specified maximum recharge current which is typically 10% to 20% of their AH rating - eg, a C10 or C20 rated 40AH car battery might have spec'd max 10% or 20% of "40" - ie, 4A or 8A.
But that is for warranty purposes and optimal life but is otherwise usually a semi-long term spec to prevent battery overheating and demise.
My 38AH cranker (a Yuasa UXH38-12 UPS battery) typically takes 40-45A after a very long crank or headlight discharge (maybe 10% discharged???), but that drops to under 10A within ~30-60 seconds though that is still above its likely 20% = 8A "max" charge current (I'd have to check its spec). And being AGM, it has no fluid to boil off to keep cool.
However, despite it being a 13 year old battery with a 10-year UPS design life - replaced from the UPS after 35 years (a typical preventative measure) and used for 3 years in my vehicle (which IMO is a totally unsuitable application for it!) - it is still behaving ok. Maybe that's because I am underestimating its spec (compared to other UPS batteries), or that that Yuasa battery is so bluddy good. (I liked Yuasa from my early biking days.) (Did I mention my mother's house alarm? It had a Yuasa NP4-12 that lasted 23 years before failing its yearly reserve test. As I recall, they have a ~4 or 5 year design-life spec.)
If my UXH38 translates to a "small"(?) 7AH NP7-12, it means a ~10A initial charge current acceptance, hence at least a 10A diode (and I'd suggest self-resetting circuit breakers) and such diodes can get expensive and uncommon - especially if Shottky types.
But the diode voltage drop is undesirable anyway. IE - if the charging system suits the main battery which is also lead-acid whether wet cell, AGM or gel, then it is NOT optimal after a diode's voltage drop.
Besides, a flat 2nd battery can still drain the main battery though the diode - it only needs to drop to a voltage below that of the main battery minus the diode's forward voltage drop, and a collapsed cell means a voltage drop of ~2V. And charging a collapsed AGM battery at several Amps can cause thermal runaway and (or) a possible fire or explosion. (Same for wet cells, but they boil off their water and acid before thermal runaway. Not that its surroundings like its acid vapor!)
I did initially include a diode as an option to a relay, but I deleted that (I think!) for those reasons. Likewise, diode isolators that were once not uncommon for vehicle dual-battery installations have since virtually ceased to exist. (Or they should have! Some installations expressly forbid them due to battery life considerations as well as safety issues.)
POST EDIT: Commercial diode isolators for dual batteries use TWO diodes (a diode between the alternator and each battery) hence one batter will not discharge into the other (nor share any loads) unless a diode shorts. I was referring a single diode used for dual-battery isolation, however the same problem of different battery charge-voltages occur (diode voltage drops vary with current), and that the alternator must sense one of the batteries (hence zero wire or "single wire" D+ only alternators are unsuitable without appropriate modifications. /end POST EDIT
And don't confuse the (Zener) diode regulator with a battery isolation diode - that's a totally different application.
The regulator on the Arduino is to convert 4-12vdc to 5vdc for the Atmega chip and other components on the board. The pixaxe board would have the same thing. You should be able to put the Arduino into sleep mode though, and have it use a fraction of the power... and then just watch for some signal (Ignition line/wifi rx pin, whatever) to turn itself back on.
From what I read it pulls 10ma for the regulator alone so sleep is ill advised. Have you put yours to sleep? The only difference would be you could possibly choose a more efficient regulator for the picaxe than is used on the arduino. (oh and the article I read from: http://arduino.cc/playground/Learning/ArduinoSleepCode)
Originally Posted by malcom2073
Input voltage for the arduino is listed as 6-20v
My house alarm battery is years old and still goes. Hey, I think my brother in law has a system that all the copper was cut off before he bought his house. I wonder if he has the battery :) I once ran my entire house alarm on the battery including 4 smoke alarms for about a week or 2 when I unplugged it for a reason and forgot to plug it back in.
Ideally regulator current is proportional to voltage difference AND current draw. If you're drawing no current, the regulator will be wasting no energy. That being said, it's entirely possible that something on that board is perpetually connected, so at the minimum (in sleep mode) it still draws 10mA, but I highly doubt that. I've not put mine to sleep, since I've never really had to deal with low power situations, but in a month when I get home I'm definitely going to give it a shot :-D. You've got me curious about if it's even possible to reduce power significantly.
Something else, if you're using an uno, the USB-Serial chip is another atmega, one which I do not believe you can put to sleep. That could account for the slight current draw off the regulator... but I highly doubt it's pulling that much power. 10mA from a regulator at 12V, means it's only using about 12mA of current at 5V in a linear regulator. (OldSpark correct me if I'm wrong, but linear regulators waste the voltage drop as pure heat don't they?) The Atmega 383 uses 7mA when active, and assuming the Atmega doing the USB-Serial probably uses about the same you'd get 14-15mA under normal operation (Or about 13mA at 12v, again assuming I'm correct about linear regulators, OldSpark?). Assuming you put the 383 in sleep mode, it will only use 25uA (microamps). So that should cut your usage in half, bringing you down to a potential of 7mA at 12V. Of course... if you were seeing 28mA from the Arduino, then you'll probably only see 14mA. There's more on that board that is likely taking up power. I know I/O takes up power if say, you leave the pullup resistors on.
So with sleep mode you could cut your power usage in half. Well worth it i think.
Could be right. I think you are right about the linear regulator as well but definitely would count on oldspark for that info.
I'm working with a nano. It uses an FTDI 1133-C chip and what looks like the regulator has 10AK N00A written on it. I have a spare for testing and the like so I'll try putting it to sleep tonight and see what kind of differences there are. I know I messed with sleep very briefly before but I think you have to import a library to get it to work and I just decided it wasn't worth the time. I'll take another crack I guess.
OK, let the fun begin. So I spent about 30 minutes and figured it out. With a very basic sketch that basically counts to 10 then goes into standby until pin 2 is brought high (pulled down with 10k) I get:
I also tried another suggestion where they suggested setting all unused pins to input and pulling them up as this will save power. I'm not convinced:
7.9ma sleeping <-oops!
I noticed that the LED for 13 was about half lit so I set it to output and low to get it out:
7.8ma sleeping <-better but still not best.
For somebody running on a watch battery, I would suggest doing a hybrid where during running, they are pulled up and during sleep they float. In my case... I don't care about running as there will be an alternator going and a ma won't matter. For me only on sleep does it matter.
There may be more that can be done to save power. As for me, I'm going to rewrite my sketch to go to sleep when no power is coming from the ignition circuit. This (plus doing something about the obd2 scanner's 9ma draw) will reduce me about 14 and 9 for 23ma.
Thanks for the motivational kick in the arse.
Input and high is a pullup, which causes extra current draw since it's applying +5V through a 10k resistor on every single input! Set them to OUTPUT not input and they won't do this.
That's awesome that sleep mode actually saves that much power, certainly an improvement! Thanks for doing the experimentation, I'm really happy that you got those results and I'm sure I'll find that useful in the future :)
Okay, how do you recover an auto-saved reply....? GRRRR!