battery drainage compustar + drone

I have a 94 MR2 and last year I had a CompuStar FT6000AS and Drone Mobile installed by a local shop. The vehicle is not a daily driver. Prior to the install, the car would start fine even after 3-4 weeks of sitting. After, this became 3-4 days. Thinking I have a bad battery, I got a new one with very high cold crank numbers; same problem. I had the shop run tests and they said everything is normal; the car is only drawing minimal power; I believe the numbers were in the 0.08-0.09 range; I assume that's amps. After continuing to deal with this issue, I took the car in to a mechanic who right away said it's most likely the alarm. I told him the story; he tested and said there's huge draw, varying from 3.5V to 9.5V. He unplugged my alarm brain and all has been good since then. Is this normal for the drone to use so much power? I would appreciate some feedback as at this point I'm not sure how to proceed. I spent all this money and have no alarm!

The "normal" draw for that vehicle should be around 0.04 the Comp. should add around 0.02 and the drone about 0.03 = 0.09 so right on the ball park there, how good is your battery and yes, I'd expect with lack of use a week to 10 days without the drone, add the drone and the figure of 3-4 days is spot on! Just re-read your post, the mech is talking out of his back side! How can anything on that car draw less than 12Volts? (Actually 12.6).

And those voltage figures, it doesn't draw volts it draws amps, i.e. current.

FWIW, I agree. Voltage is not a "power draw". Maybe he means Watts - ie, 3.5W to 9.5W (the latter equivalent to a 10W dome light) implying about 300mA to 800mA = 0.3 to 0.8A. That's about 10x more than what your shop reckoned it drew (~90mA) and Howard reckons it should draw. I had an HU that drew ~125mA (because BOTH its constant +12V & IGN/ACC +12V were connected to constant/battery +12V) and that caused a noticeable battery voltage drop after ~4 days (a 40AH battery). Note that high (C)CA has no effect - only the AH matters wrt to capacity. (Tho being pedantic, high CA & AGM batteries will discharge quicker than "plain" lead-acids, but such differences are negligible except with high current discharges.) I'd assume a few options: - disconnect the drains. (Probably undesirable.) - start the car and run for ~5 minutes every few days. (IMO also undesirable when at home.) - a trickle charger. (When at home. Ensure NOT a cheap unregulated charger (even if 1A or 100mA!) - it must not rise above (say) 14.4V when the battery is (near) fully charged. Best is a smart charger that reduces to float voltage (~13.4V) when the battery is full.) - a low voltage cutout. That may be the best choice in case you have no charger or it fails. Solar panels & regulators/chargers are another option. (Never a "raw" solar panel - they can exceed 14.4V etc - even if an appropriate (say) 1 to 2 Watt panel, tho 5W etc are more common.)

Thanks for all the info and advice guys. I was sure it's amps but then the mechanic showed me on his device and it was in volts. He tested it by disconnecting the ground from battery's negative. I've changed the battery once a year thinking it's been the problem. The current battery is one year old; I had it tested and it's fine. The draw without alarm is 0.03; with compustar and drone it's at 0.09. Exactly as you said. I live in a highrise so solar panel is out and so is battery charger since there's no outlet. Would starting and running the car for 5 minutes every 2-3 days solve the issue? I don't think the battery would be charged enough. Sometimes I boost the car, leave it on for 10 minutes and turn it back off. Half hour later it wouldn't turn on again. The low voltage cutout; this sounds like the best option. So the alarm would get completely disabled?

Running/idling the motor for 5 minutes every 2-3 days is bad for the motor and it's debatable if the battery will recharge back to full in that time frame. The best solutions, without disconnecting anything, is to either store the vehicle connected to a float charger or before the battery goes dead get out and drive the car for at least 15 minutes at normal road/highway speed. Regardless of how little the draw, cars with electronics that run 24/7 don't do good in long-term storage - the draw always runs down the battery faster than the battery would self-discharge. Also a draw of 3.5V to 9.5V isn't as bad as your mechanic makes it's out to be but you will probably want to have someone else do your electrical/electronic work in the future.

Well as I said, 5 minutes should be enough to replace most lost charge [ POST EDIT - see below!! ], but that depends on your alternator, and I'm so used to considering lost cranking charge as opposed to whatever flattening drain you have - ie, cranking won't flatten the battery anywhere near enough to prevent an immediate off & 2nd & 3rd crank etc whereas your flattening is preventing a crank. (And for lost crank, I usually say 5-10 minutes is enough to replace 90-95% of the cranking charge.) Now I'd assume the various experts also checked your charging voltage? It should be 14.2-14.4V, maybe higher after first starting. If it's lower, or if it's lower at idle, recharge takes longer. And if it's 13.8V or lower, it is not charging the battery properly and it may be a weaker than normal battery. (Greater than the olde 13.8V 'standard' charging voltage - ie, 14.2-14.4V - is required to rejuvenate old style batteries (namely to reverse sulfation that occurs with non-full batteries) and is generally required for newer the technology batteries that incorporate Calcium. It'd be nice to know your full battery voltage after a good run and after its 'surface charge' has dissipated. It might be 13.5V or higher with surface charge but once that has dissipated it should be ~12.7V (normally 12.67V). A general rule for surface charge is it may take up to 24 hours to fully dissipate on its own, or maybe a few up to 15 minutes with headlights or hibeams on. Let's assume you have a 60AH battery and it's 50% discharged - ie, using simple conservative rules = 12.7V fully charged voltage minus 5 x 0.1V = 12.2V. That's using the generally conservative but simple "0.1V per 10% of battery capacity" rule. In practice it's usually more - maybe 1.3V per 10% - so 50% discharged may be 12.0V; not 12.2V. Incidentally, whilst deep cycle batteries can be discharged 50%, the recommendation for crankers is no more than 20% discharge - ie, ~12.4V (such voltages being measured Open Circuit aka no load, but 100mA should be close enough to no load). Assume your alternator supplies 6A for recharge (6A being 10% of 60AH which is a common maximum recharge current specification, tho in practice that is usually exceeded in typical vehicles - at least for 10's of seconds to about a minute after starting). So 50% of a 60AH battery means we have to replace 30AH which means 5 hours at 6A. Assume ~30% battery inefficiency => ~8 hours (Who's the idiot that reckoned 5 minutes?) Or 5 hours at 8A. Etc. That seems excessive... Let's tackle it the other way... Assume 100mA (instead of 90mA). That's 24 x 0.1 = 2.4AH per day, Hence 3 x 2.4 = 7.2AH over 3 days. That's 7.2/60 = 12% discharge assuming a 60AH battery. The 20% recommended limit occurs after (60 x .2 = 12)/.1A = 120 hours = 5 days with a drain of 100mA for a 60AH battery. The calcs above are VERY rubbery. You may have a different capacity battery and that effects how long you can discharge it (at 90mA or 100mA etc). And if your xAH battery is weak or old, it may only be 80% of that capacity. And you may charge at far more than 6A. F.ex my former 40AH AGM cranking battery after minutes of unsuccessful cranking etc would initially recharge at 45A (about 5x more than its specified max recharge current of 20% of AH - ie, 8A) but that would drop to 10A within a minute. And even if the alternator can spare 100A, the battery may only recharge at 3A. The recharge current is V/R (from Ohm's Law V=IR hence I = V/R) where V is the difference between the charging voltage and battery internal voltage (eg, 14.2V & 12.2V => 2V) and R is the battery's internal resistance that increases with state of discharge, and age (and temperature etc). At best for wet FULL lead acid batteries of ~60AH, R may be 15mR (milli-Ohm) => I = 2V/0.015 = 133A - not that it will be a mere 15mR when 50% discharged, and not that the alternator can supply 133A... Sorry there's so much of my thinking aloud above, but it may prove useful. And it certainly blows my (pathetic) "5 minutes" statement out of the water!      The LVCO (low voltage cut-out, aka battery protector etc) would disconnect the offending load - ie, the alarm. Though I often recommend the ~$20 MW728 battery protector with 10A switching capacity, I won't in this case because it draws 10mA when merely sitting there monitoring, and more when its relay is switched on. I don't recall its "on" draw, but it is probably in excess of 50mA which is significant compared to your load. IE - if it's 90mA, your battery reserve time is halved (approximately). The was an AUD$22 kit from Oatley electronics but I think that has been discontinued (Oatley kit K227 12-24V Dual Battery Controller). It has a standby consumption of 50uA, or 500uA with status LED, and uses a latching relay with an 80A contact rating. Hence it's a mere 50uA or 500uA (ie 200 or 20 times less than the OFF MW728) except when changing relay state - ie, off to on or on to off. There would be low current circuits - maybe using MOSFETs instead of relays - but I'm unsure of commercially supplied units or kits. (Are you keen to DIY build an electronic circuit? Or hack some commercially available offering?) A key consideration is the hysteresis - ie, LVCOs or battery protectors as with voltage sensing "battery isolators" have a disconnect (load) voltage and a reconnect voltage. When a load is removed from a battery, the battery self recovers and increases its terminal voltage. You'd probably want the alarm/load to NOT reconnect unless the voltage exceeds (say) 12.6, ie the battery has been reasonably recharged. (I had a circuit for the MW728 to be isolated once it doodie the load off, tho that involved a relay. But the same principle could be used on others, and a MOSFET used instead of a relay.) FYI - I have written about a long unused but fully charged 40AH AGM (identical to the AGM I mentioned above) that caused "hunting" - the MW728 with cut-in voltage of 12.5V and cut-out of 11.8V when powering a 4A cooler would switch in & out over 30 second cycles. That was due to a "weak" battery and it was solved one the AGM has a 'proper' charge off the alternator - ie, a high current blast that blew away the cobwebs. Yep, alternators are often better than battery charges (because they can supply heaps of Amps). And yet again I recall on of my gurus saying "batteries are more of an art than a science".

oldspark wrote: Well as I said, 5 minutes should be enough to replace most lost charge [ POST EDIT - see below!! ]..... And yet again I recall on of my gurus saying "batteries are more of an art than a science".

You wouldn't happen to have a different username on a model and era specific motorcycle forum would you. If not, I've seen your brother. None of that short story addresses why 5-minutes of idling is bad for an internal combustion motor but anyway I'll revise my recommendation to meet oldspark's criteria. Using a 4-stage battery maintainer is the only way to rectify the battery drain issue in a car that needs to be constantly protected with electronic security measures but isn't daily driven.

10 days have gone by since I disconnected the alarm and all good so far; car started today just fine. I guess I'll try removing the Drone Mobile and see how long the battery lasts with just the Compustar.