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 | |----||--|
| |---[shunt]---| D2
|
|
- -- GND ---------------------------------- GND

(Copy & paste to notepad if that does not view properly).

D1 would be 300A rating, D2 30A.

So, looking at RS, I have a choice of "Schottky Barrier Rectifiers",
"Rectifier diodes" and "Standard recovery diodes". The main difference being
that the Schottkys have a lower voltage drop, about 0.70v vs 1.4v of the
rectifier ones. The Schottkys are also higher frequency rated.

For my application I'm passing DC through these, not AC. Is there a reason
*not* to use the Schottky diode, as the lower voltage drop would be
advantageous.
Differences in physical package type etc are irrelevant for my application.

Anything else I should consider?

Thanks in advance,

Alan.

In article ,
"Alan" writes:
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.

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).

--
Andrew Gabriel

Alan wrote:

So, looking at RS, I have a choice of "Schottky Barrier Rectifiers",
"Rectifier diodes" and "Standard recovery diodes". The main difference
being that the Schottkys have a lower voltage drop, about 0.70v vs 1.4v of
the rectifier ones. The Schottkys are also higher frequency rated.

For my application I'm passing DC through these, not AC. Is there a reason
*not* to use the Schottky diode, as the lower voltage drop would be
advantageous.
Differences in physical package type etc are irrelevant for my
application.


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:

http://www.chipcatalog.com/IR/300CNQ035.htm

I 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.

Anything else I should consider?

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

On Sun, 5 Feb 2006 18:16:46 -0000, Alan wrote:

Anything else I should consider?

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.

--
Nigel M

Alan wrote:
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 | |----||--|
| |---[shunt]---| D2
|
|
- -- GND ---------------------------------- GND

(Copy & paste to notepad if that does not view properly).

D1 would be 300A rating, D2 30A.

So, looking at RS, I have a choice of "Schottky Barrier Rectifiers",
"Rectifier diodes" and "Standard recovery diodes". The main difference being
that the Schottkys have a lower voltage drop, about 0.70v vs 1.4v of the
rectifier ones. The Schottkys are also higher frequency rated.

For my application I'm passing DC through these, not AC. Is there a reason
*not* to use the Schottky diode, as the lower voltage drop would be
advantageous.
Differences in physical package type etc are irrelevant for my application.

Anything else I should consider?

Thanks in advance,

Alan.


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

On Sun, 5 Feb 2006 18:16:46 -0000, "Alan"
wrote:

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

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.

wrote:
Alan wrote:

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

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.

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

"Alan" wrote in message
...
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

snip

Should never connect batteries in parallel (without protection), just choose
a bigger battery in the first place.

If a cell in one should fail due to say a internal short (quite possible in
lead acid), the other 3 will dump their load through the failed battery with
quite interestingly devastating consequences. I have seen the exploded
remains of a largish (using 100Ah batteries) UPS where some clever spark
added a second battery to improve its capacity and after a month or two one
of the batteries ruptured, exploding the UPS case and spraying suphuric acid
gel everywhere. Bit of a mess, including all the 415V switch gear and
cabling it "ate" away.

On Mon, 6 Feb 2006 16:12:30 -0000, Ian_m wrote:

Should never connect batteries in parallel (without protection), just choose
a bigger battery in the first place.

********.

--
Nigel M

In article , says...

Should never connect batteries in parallel (without protection), just choose
a bigger battery in the first place.

If a cell in one should fail due to say a internal short (quite possible in
lead acid), the other 3 will dump their load through the failed battery with
quite interestingly devastating consequences. I have seen the exploded
remains of a largish (using 100Ah batteries) UPS where some clever spark
added a second battery to improve its capacity and after a month or two one
of the batteries ruptured, exploding the UPS case and spraying suphuric acid
gel everywhere. Bit of a mess, including all the 415V switch gear and
cabling it "ate" away.




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?

In article ,
Gary Cavie wrote:
Should never connect batteries in parallel (without protection), just
choose a bigger battery in the first place.

If a cell in one should fail due to say a internal short (quite
possible in lead acid), the other 3 will dump their load through the
failed battery with quite interestingly devastating consequences. I
have seen the exploded remains of a largish (using 100Ah batteries)
UPS where some clever spark added a second battery to improve its
capacity and after a month or two one of the batteries ruptured,
exploding the UPS case and spraying suphuric acid gel everywhere. Bit
of a mess, including all the 415V switch gear and cabling it "ate"
away.

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?

It is total ******** 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.

--
*A picture may be worth a thousand words, but it uses up a thousand times more memory.

Dave Plowman London SW
To e-mail, change noise into sound.

On Mon, 6 Feb 2006 22:46:43 -0000, Gary Cavie
wrote:

just how often does it cause an accident?

Talk to some off-roaders who use electric winches.

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

Dave Plowman (News) wrote:
In article ,
Gary Cavie wrote:
Should never connect batteries in parallel (without protection), just
choose a bigger battery in the first place.

If a cell in one should fail due to say a internal short (quite
possible in lead acid), the other 3 will dump their load through the
failed battery with quite interestingly devastating consequences. I
have seen the exploded remains of a largish (using 100Ah batteries)
UPS where some clever spark added a second battery to improve its
capacity and after a month or two one of the batteries ruptured,
exploding the UPS case and spraying suphuric acid gel everywhere. Bit
of a mess, including all the 415V switch gear and cabling it "ate"
away.

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?

It is total ******** 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.

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.

Dave Plowman (News) wrote:

It is total ******** 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.

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.


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.

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

In article ,
The Natural Philosopher wrote:
It is total ******** 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.

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.

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.

--
*I don't have a license to kill, but I do have a learner's permit.

Dave Plowman London SW
To e-mail, change noise into sound.

On Tue, 07 Feb 2006 00:24:11 +0000, Andy Dingley
wrote:

On Mon, 6 Feb 2006 22:46:43 -0000, Gary Cavie
wrote:

just how often does it cause an accident?

Talk to some off-roaders who use electric winches.

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

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

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

cheers,
Pete.

In article .com,
wrote:
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.

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

--
*No husband has ever been shot while doing the dishes *

Dave Plowman London SW
To e-mail, change noise into sound.

wrote:

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.

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?

--
Andy

Have a look here

http://www.rcbatteryclinic.com/parallel.html

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

On Tue, 07 Feb 2006 11:14:11 +0000, Pete C wrote:

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

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 ?

On Tue, 07 Feb 2006 13:46:44 +0000, Andy Wade
wrote:

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.)?

Yes. There's not even much difference to a single plate in a jar, if
the surface of the plate is damaged in patches. Remember that lead acid
plates are the same metal for both electrodes - it's just a transient
surface coating (mostly lead peroxides) that makes them different. If
you cause this coating to vary across a plate's surface, then all sorts
of currents can flow.

However you obviously won't get this effect between plates in adjacent
cells. This is just one reason why when lead acids fail, they kill a
single cell quickly in favour of the others.

Andy Wade wrote:
wrote:

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.

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.)

Good point.

There's series resistance, which is low but not nothing, so the wiring
wil take at least some of the heat, there's limited energy in the
discharging battery, and thermal capacity in the overheated battery.

With a single cell failure there will be more heat dissipation to
neighbouring cells than would occur with a whole battery.

I dont know the outcome, I've had batteries go ape but that was
regulator failure.


NT

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

snip

Thanks for the help guys.

Looking at my options:

I can't put the diodes at the meters - as the voltage drop across the diode
would mean that the meters would not display anything - everything would go
via the shunt. After all, the voltage across the meters will only be a few
mv, even at full current.

Contactor with op-amp: - a possibility, however a contactor big enough will
cost more that I'm willing to pay!

Schottky diode - I can get a 400A diode for about £50 - so not too bad. I'll
fuse the output of the unit under this (using a 355A fork-lift fuse and
holder) so it will never see over-current.
Most high useage will be a few seconds at a time jumping a car, and the only
continuous load will be from running the 3kw inverter - which draws upto
250A. As long as the diode is on a suitable heatsink we should be OK. I can
live with the 0.6v voltage drop - you'd usually loose this over "average"
jump leads anyway - and in my case I've over-engineerd the leads somewhat
that the voltage drop would be minimal.

Changing zero point of existing ammeter - nice idea, but it would mean a
home-made scale and bugger it up somewhat. If I can find a matching
centre-zero ammeter from the supplier and get it to do this, I'll consider
it.

I'll report back on what I end up doing and the results.

Alan.

Alan wrote:

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

Thanks for the help guys.

Looking at my options:

I can't put the diodes at the meters - as the voltage drop across the diode
would mean that the meters would not display anything - everything would go
via the shunt. After all, the voltage across the meters will only be a few
mv, even at full current.

Contactor with op-amp: - a possibility, however a contactor big enough will
cost more that I'm willing to pay!

you not got 10p in your pocket? Copper washers plus a solenoid is your
cheapest option. Thats how I did it the one time I needed to switch
100s of A.


Schottky diode - I can get a 400A diode for about £50 - so not too bad.

kinell


I'll
fuse the output of the unit under this (using a 355A fork-lift fuse and
holder) so it will never see over-current.

complete and total non sequitor. Circuits like this tend not to
survive.


Most high useage will be a few seconds at a time jumping a car, and the only
continuous load will be from running the 3kw inverter - which draws upto
250A. As long as the diode is on a suitable heatsink we should be OK.

it'll die sooner or later.


NT

wrote:

fuse the output of the unit under this (using a 355A fork-lift fuse and
holder) so it will never see over-current.

complete and total non sequitor. Circuits like this tend not to
survive.

Quite so. Also check the voltage drop across the fuse, which will not
necessarily be negligible. (And it could all be done with an op-amp and
two 1N4148s in any case.)

--
Andy