Thread: Battery based tank circuit (tested)

Ricky,
In your original post (schematic), shouldn't your top diode (from main battery/alternator) be on the OTHER SIDE (left side) of the current limit resistor? If not, won't current begin flowing from your backup battery to the main battery during cranking, limited only by the resistor? This would effectively drain the backup battery if the main battery drops significantly during cranking, and may pull the backup battery's voltage down with it. The current drain is not the issue, but pulling down it's voltage may be enough to reset our pc.

Perhaps I'm not understing this correctly........

Ok, I haven't read the whole thread, but I'm using a sproggy at the moment: It's a modified 2.5 version, and it does not survive the crank (my cars is a diesel tho).

Infact, the engine needs to be running in order for the hard disk to even spin up (which makes it a pain to power the carputa in the house). I very much like robiewp's idea... I think I'm gonna use it since I have a backup battery and I'm sure I've got a spare relay somewhere.

I understand that Opus can survive a crank with voltage drop to 8 or 9 volt,how do they do it?anybody have a schematic or something?

My Sproggy 2.8+ does survive a crank, long cranks in ten degree F weather.

There is alot of information on this board about the various Sproggy designs if you The PSU is named for an early member of this forum who designed it in the first place. Mastero currently markets a variant or two. I think his 2.8+ version is the best - that's why I have 2 It does have some hard to get parts, though. And there is some question over whether they will work properly with EPIA M motherboards (it appears to only be a startup problem, some kind of check the MB is doing,) although several have reported success with big caps on the input. If I had one, I'd try just shorting out pin 14 (has anyone tried this?)

Mine have had no problems surviving cold starts in several vehicles. It is important to have a fat wire directly to your battery - my current setup has an 8 AWG, although that is surely overkill.

In your original post (schematic), shouldn't your top diode (from main battery/alternator) be on the OTHER SIDE (left side) of the current limit resistor? If not, won't current begin flowing from your backup battery to the main battery during cranking, limited only by the resistor?

Hehe, I knew someone will mention this.

As I said on the original post, the current that can be drained out of the backup battery by the starter motor is limited by the 68Ohms resistor. I have tested this in the worst possible case by shorting out the connection where the main battery goes...of course whithout the main battery. The terminals are shorted as if a realy heavy drain is made by the starter motor, the maximum current that get drain out of the backup in this extreme condition is = 12.5/68 = 0.18A, assumingg the voltage across the backup is 12.5v. Of course this extreme drain will never happen as starter motor only drop the main battery by 10.1V in my case. So in normal crank (12.5-10.1) / 68 = 0.035A is drained out of the backup. This current is nothing compared to 2.7A drained by my laptop.

The resistor was placed there deliberately so that the backup do get the full charging voltage from the main battery. As I said before, original tank circuit will never charge your backup fully unless your alternator put out more than 14V to compesate for the voltage lost caused by the diode.

I hope that clearer now

I shall post the update soon how long the battery last with a current drain of 2.7A

EDIT :

A 0.035A additional drain out of the backup battery for 2-3 second during crank is almost nothing really.

Just for curiosities sake, how big a battery are we talking about? Voltage, Ampherage and anything else I need to know. there are alot of batteries for sale on eBay...maybe my next project?

The battery on test is the smallest I can find...thats 12V 0.7Ah as labelled on the diagram. This should last a few minutes at 2.7Ah drain.

If you want a heavier load then use something like 2Ah or more, of course the limiting resistor is also lowered and will cause more drain on your back up when the starter kick in. But then this higher drain is compensated by the bigger Ah rating of your battery.

Bear in mind, the pupose of this circuit is to survive a crank and not to extend the running time of your PC. So even a batttery that can only last for 5mins is far more than enough as you only need to hold the voltage for 2-3 seconds during crank.

or a "tank circuit" could be used for extended carputer use. the same designs should work

Ricky,

I'm going to give your circuit a try. I'll let you know how it comes out. Here's what I plan to implement . Let me know if you see any major flaws in the approach.

thx

MikeH,

I dont know much about the radioshack DC-DC converter, The one I used is the targus one and can work right down to 11V before cutting out. Your setup look fine, yes I know some poeple here might say its complicated by I guess I can understand why you did it that way.

But ill recommend you use a higher Ah rating battery and recalculate the resistor value. Use a diode with the lowest voltage drop you can find with the right current rating of course.

I know 0.32V drop do exist :

http://uk.farnell.com/jsp/search/sea...0000~026006000

Take a pick, I know the lowest voltage drop are kinda expensive. So you have to decide on this one.

I have done a test on the backup battery being fully charged, it did last about 3mins at 2.7A drain. I dont know your total drain but Ill expect it to be more thats why I recomend you use a bigger battery.

Good luck and let us know what happen