15A current limit?

How can yo know this?

Are you confusing house wiring codes for current with this DC application?

7 volts?

NO, the limiting device will need to dissipate 12 volts at 15 amps until the voltage of the battery being charged rises.

FETs have no advantage over bipolars in "typical voltage drop". With 2 power bipolars, some resistors, and a heat sink this can be done.

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  |  __O    Thomas C. Sefranek   WA1RHP@ARRL.net
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Are you trying to charge the aux battery from the main battery or connect both batteries to the one charger? Both are possible but require different circuits.

The normal reason for connecting two batteries in parallel like this is "jump starting" a car that has a flat battery (jump start = UK English slang). In that case you need wires that can handle much more than 15A and more like 150A.

If you are planning on using croc clips to connect the aux battery then they WILL be connected together one day (=15V) or if you connect a duff battery that has a short circuit it's also 15V.

If you connect the battery the wrong way around it could be 15+12=27V or if you have the two batteries connected and the aux is reversed it's 24V.

If the aux battery is in a car and you crank the engine it's 15V with some noise spikes.

No it's more like 225W (or 405W if battery reverse connected)

I know this because the aux battery will never be discharched below 11V and the vehicles charging system will never be great that 15V, so 4V potential difference

I used a table to automotive wiring current capacities.

SEVEN? why 7?

The device will have 4V potential across it... not 12 (Vce=4)

I thought fets had ~0 ohms ON resistance, whereas bipolars suffer a diode drop.

This is in a vehicle. The only time the batteries will be connected together is when engine is running and alternator is charging. When engine is running, I want to limit current to 15A to aux batt.

NO, I will NOT be jump starting using this aux battery.

The whole thing will be protected with a 20A fuse, just in case something goes awry.

If I drive the vehicle into a tree at 70mph, bad things will happen to... but since I don't plan on doing either, and theres a 20A fuse in line, I feel pretty safe.

In addition to the 20A fuse, the whole circuit will be sitting behind an NO-relay, that will only be closed when ignition is ON ( open while off and while starting)

P=IV?? 4V*15A=60W, no?

Yes that's right. I guess we thought you should allow for the case when the aux battery is completly flat or sort circuit, in which case it's....

15 x 15 = 225W

I was also allowing for reverse connection (27V)

27 x 15 = 405W

Now that you have explained the situation further I guess you safely assume

60W.

Try a variation/cut down version of this design...

Yes, it's always difficult to decide what to put in the original post. Two much detail, and nobody will read it.. To little, and everyone makes assumptions..

Any.. I still don't know how to build the current limit device... Any ideas?

CWatters wrote:

Also this app note...

Actually that's for much lower currents but it might give you the idea.

5V.

O.K.... If you say so.

application?

I WISH FETs had 0 Ohms... They are now appearing with low on resistances in the milliohms. I have measured saturation voltages in the MICROvolts in small signal bipolars. Yes, MICROvolts. Anyway, power Bipolars can get down to .3 volts drop in saturation. If the FET had a 1 ohm saturation that would be 15 VOLTS at 15 amps. But... they are better these days.

In any case, why do you care? Even with a bipolar burden of 1 volt, you still have to drop your 4 volts somewhere. So what burden voltage will you allow the circuit to use with your FET solution? ?The 12 to 11 volt path, leaving 1 volt for the circuit regulating 15 amps? It can be done, but it involves a current sense amplifier.

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  |  __O    Thomas C. Sefranek   WA1RHP@ARRL.net
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That's NOT a reason, it's a belief. Why do you believe the .6 volts is VERY important? Why do you believe .6 volts? (Most power bipolars can saturate at .3 volts or less!)

If you want to use an FET, you will encounter the 3-5 VOLT gate drive requirement. An N-Channel low side driver would make sense, as you have 12 volts to work with. Conversely, a P-Channel high side driver.

SO again, you need to sense the current in order to limit the current. (You have not said WHY you want to limit the current...) Resistively, it makes sense to keep the burden low, as it is voltage loss similar to your belief about Bipolars. Immagine a resistor in series which will give you 1 millivolt per amp. You need an op-amp to amplify that low voltage up to the gate drive voltage range.

But, using either an FET or a bipolar transistor you will encounter a voltage drop in the device. (A power FET could have a .05 ohm RDSon) [many power FETS have higher RDSons!) that times 15 amps still gets you back to bipolar voltage drops. .75 volts!

It sounds like your trying to design a battery charging a battery, (A 12 volt battery charging a 12 volt battery is poor!) this is BEST done with a DC-DC converter.

If they really are both 12 volt batteries, use a relay and forget the current limit.

E-Mail me if you would like more details on building your solid state low dropout FET relay.

Tom

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  |  __O    Thomas C. Sefranek   WA1RHP@ARRL.net
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What made you decide on a 15 amp limit? 19ga wire will carry 30 amps without overheating and considerably more for a short time (such as at the beginning of the charging cycle. I don't really think you will need any elaborate current control unless your auxilary battery is very large.

Where can I find data in the current carrying capacity of wiring for automotice applications. Note, this wire will be running in body panel channels, wiring harnesses, etc... so 'free air' current capacity numbers will not be applicable IMHO.

- jim

p.s. The 15 Amp limit is the charge rate at which I want to recharge the batteries, per manufacturers recommendati> What made you decide on a 15 amp limit? 19ga wire will carry 30

18ga copper wire is 5.5 milliohms per foot, so you would lose about 0.8 volts at 15 amps on a 10 foot run, and the wire would dissipate 12 watts which is a little high. If you go to 12ga, the resistance is only 1.6 milliohms per foot and voltage loss is .24 volts over 10 feet, and the heat generated is 3.6 watts which is probably nothing to worry about. If you don't want to lose the .24 volts, go to 10 ga wire which is only 1 milliohm per foot.

-Bill

It sounds like you are trying to build a 15A max trickle charger.

Your maximum voltage drop is 4V, and you want to limit the current to 15A.

Try this:

You tune it for the particular current you want by tuning the voltage out of the trimmer to be the same as the voltage across the current sense resistor R1. For this design, at 15A, thats 150mV above ground. Thats a 6.2V zener, I think. Using a different zener means different resistor values in that area.

The p-mosfet shown won't work, because it will die given the current (its model was handy in my drafting program.) You'll have to find a power mosfet that will support the power and current ratings you need. Also, the current sense resistor, R1, should be at least 5W rating. You can find current sense resistors at mouser or digikey that have this kind of rating.

The current rating for the mosfet you need is something like 25A (since you are using a 20A fuse.) Power rating = 25A x max voltage differential between batteries. If your battery has a short, that would be 15A x 15V = 225W, so watch out; even with current limiting, you can still fry the mosfet and set your RV on fire.

You also will need a big heatsink.

Regards, Bob Monsen

A crude & simple method is to run the current thru a light bulb. A headlight takes about 4 amps, so 4 in parallel will limit the curent to

16 amps at zero batt voltage. The voltage drop will get very low at trickle currents.

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free men own guns - slaves don't
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Usually if you are going to charge a second battery in an RV or towed vehicle you would want a diode isolator such as this one...

This way you won't need a solenoid and there would be no voltage drop from the diode isolator.....There are a bunch of similar solutions if you do a google search..... Hope this helps.........My 2 cents on the subject....Ross