Electronics advise - Diode choice help needed

[Alan]

Hello,

I'm trying to source a suitable large current (300A) diode for an application I have.

Background: I have a unit I made consisting of 4x 85ah leisure batteries wired in parallel mounted in a trolley, with a 3kw inverter / charger. The unit has a

12v output to use for jump-starting cars, via a (suitibley heavy-duty) plug/socket arrangement to croc clips rated at 300A continuous. Cabling is 40mm2 low-voltage flexible type for all the "heavy side". The inverter has a mains input to allow it to work in UPS mode and to charge the batteries when mains is present. I also use an external heavy-duty leisure battery charger to charge the "unit", which can deliver upto 25A continuously. This connects via a plug/socket.

The unit has 3 panel meters for battery voltage (0-30v scale), battery output current (0-300A scale) and charge current (0-30A scale). The two ammeters are fed via shunts.

The charge meter is connected via the external charger socket so displays charging current when this is connected.

So far so good, all works as designed and provides enough stored power to run the inverter for ages or start virtually any 12v vechile.

Problem: If I use the mains input on the inverter (which charges the batteries), or once a car has started from this unit (and it's alternator has started working), I get a negative current displayed on the main battery output ammeter, because the charge is going to the batteries via the output path. This is doing the ammeter no good, but secondly I have no idea *how much* current is going in.

Solution: I plan to add a large (say 300A rating) diode to the "output current path", and a smaller one (say about 30A) in the opposite direction, each directing the current to the output or charge ammeter as required.

ASCI art circuit diagram:

• ------------------(A)-----| | | o/p meter | B | |----|>|-- a |---[shunt]---| D1 | t | | t | | e |-----(A)-----| |---- Load / charge input r | charge meter| | y | |----|

[Andrew Gabriel]

Change the zero point of the 300A meter so it reads -30A -- 0A -- +270A (or with appropriate shunt modification or addition of series resistor,

-30A -- 0A -- +300A). Then you don't need the diodes or the charging meter. This is what I did when I built a similar (but lower current) unit some

30 years ago (and I've still got it).

[Tim S]

Without commenting on the overall design, as I haven't had time to digest it yet, so to directly answer this query:

Use Schottky - no disadvantages and one very important advantage that you've half alluded to: the forward voltage drop is half that for a simple silicon rectifier. That's half the heat to get rid off.

Here's a part that might be suitable:

have no idea where you'd buy it.

Now, at 300A continuous, you need to get rid of 210W of heat (as opposed to

420W)! That'll need a moderately efficient heatsink, maybe with a fan.

If however, as the application is jump starting, you'll presumably have a low duty cycle, so if you can bolt the diode to a lump of metal with enough thermal mass to absorb the energy output for the maximum single shot run time. Unless you're starting a row of knackered cars in sequence, I would have thought your max run time at 300A would be a couple of minutes and even then, not at 100% duty cycle, then a long rest period, plenty for it to cool down.

I'll have another read of the whole thing after tea if that's OK - I'll be back if I think of anything.

Tim

[Nigel Molesworth]

Why not put the diode between the meter and the shunt? Ten you only need a small one, and you won't get a voltage drop.

[meow2222]

I cant say I followed the setup details fully, but I think I see the problem ok. The diode will not solve it, it would cause further problems. 1.5v @ 300A is huge P_diss for any diode. The 1.5v drop would stop your batteries charging other batteries, and eat a good 10% of your output. Also there is as far as I can tell, no 300A current limit on the battery's output, so your diode wouldnt survive anyway.

I was going to sugest a centre zero meter, but Andrews suggestion sounds better in this case.

If you actually did need a diode, I would have suggested an opamp to detect direction of current flow plus a relay. Then youve got a 0v drop slow diode.

NT

[Andy Dingley]

No one ever needs 300A diodes. Redesign it so that your high current path is through a contactor, switfched in as necessary (quite possibly shorting out a lower current diode).

Schottky is good and has no drawbacks, except that they used to be expensive. Now nearly all of the high current diodes are Schottky, just to reduce the dissipated power in them.

[meow2222]

also forgot to mention the huge currents and correspondingly low Rs, plus the nature of some of the kit youre using, means inductive kickbacks may occur. These will kill diodes, especially schottkys with their inferior V_r.

NT

[Ian_m]

[Nigel Molesworth]

Bollocks.

[Gary Cavie]

I have heard this said many times, but have done it myself several times without any problems. My old Hiace had two batteries paralleled up straight from the factory, so I'm not sure just how big a problem it is. Obviously, if you happen to be unlucky, it is a problem, but just how often does it cause an accident?

[Dave Plowman (News)]

It is total bollocks with lead acid batteries - but not with Ni-Cads. And I've never known a lead acid cell to fail short circuit - it's the reverse. And even if it did, another battery connected in parallel wouldn't deliver enough current to blow things up. Unless several cells failed short circuit at once.

[Andy Dingley]

Talk to some off-roaders who use electric winches.

If you parallel batteries, and you hammer the batteries, you get wiring fires.

[The Natural Philosopher]

Both Nicads and lead acids can be discharged in parallel with no special circuitry..and lead acids can be charged in parallel - the voltage/charge curve equalizes the charge..Nicads can be trickle charged in parallel but not delta peak charged.,

In all cases the voltage/charge-discharge curves are such that no especial protection is required - no diodes - nothing.

[meow2222]

Overheat the battery by drawing too much current for too long and the pates buckle. They can also short from deposits accumulating at the bottom.

Lets see, with an offload V of 13.2v, one shorted cell means a 5 cell battery seeing 13.2v, 20% above the 2.2v is being applied per cell. To get that in terms of familiar numbers, this is equivalent to a 6 cell battery receiving 13.2+20% = 15.84v. I would think this would fairly seriously overheat it. 6v batteries paralleled would be much worse, 24v batteries would probably sit there happily all day.

NT

[Dave Plowman (News)]

IIRC, parallel connection with Ni-Cads isn't recommended unless being used quickly as they tend to self discharge faster. Or something like that. Of course if you need a higher than normal power pack using commonly available cells you might have no real option.

[Pete C]

Is it the wiring between batts, or to the winch? ;)))

I suppose plates can buckle and short if the battery is abused.

cheers, Pete.

[Dave Plowman (News)]

Depends on how long it takes for the batteries to equalise.

[Andy Wade]

All these nasty effects can take place _within_ a cell, leading to the healthy part of the cell feeding current to the damaged part, and internal heating. Practical cells consist of many pairs of plates in parallel, so is there any difference between plate pairs in parallel in one jar of electrolyte and plate pairs in parallel in more than one jar of electrolyte (assuming equal temperature, etc.)? (That's a genuine question, BTW, I don't know the answer.)

Another interesting question is when paralleling batteries should one (if possible) strap all equipotential points, i.e.

• + | | ----- ----- | | | | --- --- --- --- - - - - | | |_____| | | | | --- --- --- --- - - - - | | |_____| | | | | --- --- --- --- - - - - | | | | ----- ----- | | - - Like this /or/ like this?

[The Natural Philosopher]

Have a look here

is a lot of useful info on using and not abusing cells here.

[Andy Dingley]

Between batteries - if they didn't use either multiple contactors, or a Schottky diode.

You have two batteries connected by a high-current unfused cable. You then abuse the batteries and deep cycle at least one of them. Of the dozen cells in there, sooner or later one will fail - and the way of such things is that a cell fails, not that they all gradually sag in unison.

Now with a 2V potential difference on the ends of a chunky piece of copper, how can you _not_ have a problem ?