Charge 12v lead-acid battery via pulse width modulation

I need to charge the 12 volt engine start battery in my sailboat from the 12 volt house battery in the boat. The house battery is the one connected to the alternator. I can't just parallel the two batteries for charging, since the house battery is a long cable run away from the alternator, and when a heavy charge is going into the house battery(50 amps or so), the voltage that the start battery would see could be as high as 15.5 volts. I only need to supply about 3 amps when the engine is running, just enough to run the fuel pump and ignition.

I can use a linear regulator, of course. But I thought perhaps I could use a pulse width modulator driving a switch (FET ot BJT) to apply the

15 volts to the battery in pulses, with the average voltage being around 13.6 volts. (Sort of a switching regulator without the inductor).

If I had the battery home with me, I could test the idea. But it is

100 miles away.

Any battery experts out there who would care to comment?

Thanks Larry Bradley VE3CRX Remove "removeme" from my e-mail address for direct mail Ottawa, Canada

(use the e-mail address above to send directly to me)

Reply to
Larry Bradley
Loading thread data ...

Use a real buck regulator with an inductor, and avoid nasty peak currents in your pass device.

Question: If the "house" battery is so far from the alternator, are you using remote sensing?

...Jim Thompson

--
| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
 Click to see the full signature
Reply to
Jim Thompson

Yes, using remote sensing, so the regulator sees the actual house battery voltage.

The peak current was what I was afraid of. I'm no switching regulator guru, although I've read quite a bit. Probably the best way to go. Any suggestions as to where to get an inductor suitable for the 5 amps or so I need?

Larry Bradley VE3CRX Remove "removeme" from my e-mail address for direct mail Ottawa, Canada

(use the e-mail address above to send directly to me)

Reply to
Larry Bradley

I've been known to just take any old filament transformer (E-I), rewind to suit, then stuff paper shims until it doesn't saturate ;-)

...Jim Thompson

--
| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
 Click to see the full signature
Reply to
Jim Thompson

Jim Thompson wrote in news: snipped-for-privacy@4ax.com:

why would the voltage to the house battery be 15.5V? Isn't that TOO HIGH?

ISTR that the charge voltage shouldn't be more than ~14v.

And why not just run a string of Schottky diodes to drop the charging V to the start battery?

--
Jim Yanik
jyanik
 Click to see the full signature
Reply to
Jim Yanik

Lead acid cells have a big TC. See my patents.

Ugh :-(

...Jim Thompson

--
| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
 Click to see the full signature
Reply to
Jim Thompson

Somebody at the boat store could probably give a better answer than anyone here; _somebody_ with a boat must have had a similar experience.

(or try to find a boating or RV newsgroup.)

Good Luck! Rich

Reply to
Rich Grise

I'm a little confused, Larry.

First, the long cable run conveys no information. What is the wire size? I assume you're not on an aircraft carrier so the run length is maybe 20 feet round trip?

Restoring the energy from starting the engine would probably not be affected by the cable run provided the wire size is not ridiculously small (e.g., 18awg or smaller).

Providing the 3 amps for ignition and fuel pump could actually be handled by 18awg wire.

It seems the only real issue is that 15.5 volts is likely to lead to gassing in the engine start battery unless the temperature is really cold, in which case you're in the wrong cruising waters.

But 15.5 volts is really on the high side for lead acid, AGM, or gel batteries as I recall. You certainly should not see these batteries drawing 50 amps with a terminal voltage of 15.5 volts. Are you seeing that?

So my suggestion is use the proper regulator for your house battery and it may solve your start battery charging problem. Ideally, both batteries are the same type. If your house battery is an AGM or gel type, then you will not damage your lead acid starting battery with the setup you already have, provided the house battery regulator limits charging voltage to whatever your AGM or gel battery requires.

Chuck

Reply to
Chuck

Those who tout familiarity with lead-acid batteries would do well to read my patents before deciding what are realistic voltages.

...Jim Thompson

--
| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
 Click to see the full signature
Reply to
Jim Thompson

I always get caught up like this - I ask a specific question, but don't provide enough background, so I end up sending peple off track. Sorry about that.

The alternator, of course, is on the engine. The house battery is 25 milliohms of resistance away (32 feet round trip of wire, plus various connections to switches, terminal posts, etc). This is all unchangeable, unfortunately.

The alternator regulator is a 3-stage regulator, and senses the battery voltage at the battery terminals, so the battery never gets more than 14.4 volts (the nominal gassing voltage). But 50 amps and 25 milliohms results in about 15.6 volts at the alternator. Another poster asked about using diodes to drop the voltage - fine when the charge rate is 50 amps, but when it drops down to 5 amps and charging at 13.5 volts, the start battery doesn't get charged at all.

The start battery is connected by heavy gauge wire (via a switch) to the alternator. It starts the engine, of course, but also provides power for the electric fuel pump and the electronic ignition - about 3 amps. The brief load of starting the engine doesn't discharge the battery much, so there is no real need to charge the start battery with a large current. Being able to provide 3 to 5 amps is plenty. And I have a separate 115 volt charger for when I'm in the marina.

So the problem is "How to limit the voltage from the alternator, which can be as high as 15.6 volts, and as low as 13.5 or so, to around 13.2 volts for the start battery?"

Three alternatives came to mind. a) a linear regulator. No problem, I can easily do this. Built lots of them before. b) a switching regulator. Never done this before, but I know what is involved. Getting and inductor might be a pain. c) Just use pulse width modulation (a simple FET or BJT switch) to drive the battery. The switch when on would put 15.6 volts to the battery for a short time. But the average voltage would be 13.2 (or whatever). A previous poster suggested that this could result in the current during the on time being quite large.

There is a 4th alternative. You can buy a device that does exactly this - it will charge the start battery from the house battery - I presume it is just a DC-DC converter - maybe a SEPIC arangement to enable it to handle input voltages both above and below the output voltage. But they are expensive, and I like to brew up my own stuff at any rate.

Thanks for the interest and ideas.

Larry

Larry Bradley VE3CRX Remove "removeme" from my e-mail address for direct mail Ottawa, Canada

(use the e-mail address above to send directly to me)

Reply to
Larry Bradley

Ahh. Thanks for the elaboration, Larry. It all makes sense now.

You might use a zener in conjunction with an automotive relay (30A or so), or for possibly greater efficiency, a zener with a power mosfet. Or a zener, pot, mosfet and relay. With a handful of parts, virtually no losses, and no switching hash, you can set the trip voltage so the starting battery is connected to the alternator only when the alternator voltage is below the level you set.

Once your house battery soaks up some charge, the voltage drop (across the 25 mohms) will decrease and alternator output will be at an acceptable level for the starting battery. I think I'd be surprised if you were sustaining a 50A charge rate for much more than an hour.

If you decide to go with the pwm approach, you might look into the TL494 chip driving a power mosfet.

Good luck

Chuck

Reply to
Chuck

Is it physically feasible to clip a pair of #16 or #14's to the house battery itself, (16 o4 14 because it has to go all the way back to the motor) and let the start battery just float-charge? Lead-acid is lead-acid, you know; they all have the same voltage, and when the battery is fully charged, it will draw negligible current.

Have Fun! Rich

Reply to
Rich Grise

And miss out on all the fun?

It's much better to build it yourself.

--
http://www.wescottdesign.com
Reply to
Tim Wescott

Chuck:

I actually have somthing like that now. A few years ago I built my own amp-hour meter using a PIC 18F452 - it measures both house and start battery voltage and current.

It is programmed to actuate a solenoid relay to connect house and start batteries together when the engine is running (it monitors the ignition switch).

Last year I added a large diode in series with the relay contacts - it drops the voltage by about .8 volts to the start battery. When the house battery charging current gets down to a reasonable level, the program bypasses the diode via a small relay to apply the full voltage.

It's a kudge, so I thought the regulator trick might be "neater".

50 amp charge is about 1 hour, then tapers off.

Larry

Chuck wrote:

Larry Bradley VE3CRX Remove "removeme" from my e-mail address for direct mail Ottawa, Canada

(use the e-mail address above to send directly to me)

Reply to
Larry Bradley

Larry Bradley VE3CRX Remove "removeme" from my e-mail address for direct mail Ottawa, Canada

(use the e-mail address above to send directly to me)

Reply to
Larry Bradley

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.