Outer Limits 4x4 Board

i went down to buy a treacle charger from kmart yesterday, picked up the cheapest one i could find expecting that it would just be a slow charger. Had a look on the back and it recommended a larger one for the battery i want to charge (700 CCA overlander) they didnt have the massive one on the shelf, but the next size down was about $90 and i dont really want to spend that much. Are they just trying to upsell them, or do i really need a big one.

The honey ones at the least taste alot better...

Buy this.

20A max current capacity, 13.8V regulated for float charging. I won't use my K Mart special again.

Cheers,

Scott

hillbilly wrote:The honey ones at the least taste alot better...

NJ SWB wrote:Buy this.

20A max current capacity, 13.8V regulated for float charging. I won't use my K Mart special again.

Cheers,

Scott

can i have your kmart special seeing as your not using it.

seriously, cheers for that, i didnt even think of just getting a power supply. will it work the same as a charger, im not real good at electronics so i try to steer clear of it all.

mickyd555 wrote:

NJ SWB wrote:Buy this.

20A max current capacity, 13.8V regulated for float charging. I won't use my K Mart special again.

Cheers,

Scott

can i have your kmart special seeing as your not using it.

seriously, cheers for that, i didnt even think of just getting a power supply. will it work the same as a charger, im not real good at electronics so i try to steer clear of it all.

It should work better than a charger. I've not seen a K Mart style charger deliver much more than 3 amps for more than a few seconds. But they typically deliver too much current for too long - if you leave them connected too long they will overcharge the battery and kill it.

This power supply is regulated at 13.8V (and mine is regulated well), the same as your car charging system should regulate at. You can't overcharge your battery (unless the power supply fails.)

Hook it up, walk away for as long as you like. Time to fully charged will depend on how "flat" the battery is, but a few hours should add enough charge to start your truck.

Cheers,

Scott

thanks Scott

Its actually a battery i scored off my neighbour. He just quit working at century-yuassa and gave me his camping battery. I had a look and its a overlander 700 it reads 11.2 V at the moment, but i wanted to charge it up a bit so i can keep it stored untill i put it in the truck (i have to put my duel battery charger in first.)maybe i should just hook it up temporarily.........

That power supply won't be too happy charging a fully discharged battery.

Battery chargers are designed to limit the voltage output if the current drain is too high; whereas that supply will try and pump out a regulated 13.8V up to 20Amps max. A flat battery will draw more than 20 Amps and most likely trip the in built protection circuitry (which it hopefully has?!) causing the supply to shut down.

Something to investigate before purchasing one.

Good point. I don't know about this one, but other SMPS I've worked with will allow their voltage to "sag" to whatever they can cope with, up to a point - then the overload/short circuit protection shuts them down.

If the voltage sags the output should regulate quite effectively (lower charge voltage, so the battery won't try to draw as much.)

Alternatively, a low value/high wattage resistor in line will help limit current for a dead flat battery. Say 0.5 ohm/10W IN SERIES with the battery - NOT across the battery/SMPS. One conductor in a 30m extension cord could be suitable for such a purpose. Or use two conductors - one in the +ve lead, another in the -ve.

OK - perhaps this is getting a little too complicated for the electrically challenged. Who would buy a current limiting add-on for a power supply, if I build them?

Scott

Put the crap battery in your truck to recharge it
Then use the charger to maintain the charge

Make sure you keep the battery off the ground when stored
Best kept on a plank of wood sitting on two bricks

cheers
Wayne

NJ SWB wrote:..... Say 0.5 ohm/10W IN SERIES with the battery - NOT across the battery/SMPS. One conductor in a 30m extension cord could be suitable for such a purpose. Or use two conductors - one in the +ve lead, another in the -ve.

You'd need a lot higher Wattage than 10 Watts!

As an example, say the battery is sitting at 6 Volts - the resistor would have to drop 7.8Volts at 20 Amps. The power rating for the resistor would need to be greater than 7.8 x 20 = 156 Watts!!!

My advice- buy a proper battery charger.... the risk of a fire is too great.

SuperCheep have some good heavy duty Aussie made ones that are fully automatic.

murcod wrote:

NJ SWB wrote:..... Say 0.5 ohm/10W IN SERIES with the battery - NOT across the battery/SMPS. One conductor in a 30m extension cord could be suitable for such a purpose. Or use two conductors - one in the +ve lead, another in the -ve.

You'd need a lot higher Wattage than 10 Watts!

As an example, say the battery is sitting at 6 Volts - the resistor would have to drop 7.8Volts at 20 Amps. The power rating for the resistor would need to be greater than 7.8 x 20 = 156 Watts!!!

????

Will the battery voltage still be at 6 volts while it's charging? Only if it's completely shagged and beyond redemption. You can't recharge a lead acid battery at 1 V per cell. You need more than 2V per cell before you can start to charge the battery, so terminal voltage shouldn't stay below 12V for long. We're talking seconds, or less.

I haven't seen a 10W resistor which will start a fire before it fails open circuit - what are you on?

Now, rethink your example. 0.5 ohms at 20A? You're dropping 10V from 13.8, which leaves 3.8V across your 6V battery.

More likely, after who knows what start conditions, we'll quickly move to a terminal voltage of 12V plus, with an indeterminate current. A 0.5 ohm resistor in line, with a voltage drop of less than 1.8V. Current should be less than 4A, and power less than 8W.

But I could be wrong. Please educate me.

Scott

Now, rethink your example. 0.5 ohms at 20A? You're dropping 10V from 13.8, which leaves 3.8V across your 6V battery

.

My example wasn't the best, but you get the idea. Even in your example Scott, 10V x 20 Amps is 200 Watts! Yes, it shouldn't stay at that level for long- but what if the battery has a couple of shorted cells?? Even a short surge at 10x the power rating of the resistor is very risky.

A resistor will quite easily start a fire if it isn't mounted on a proper heatsink or is left on something combustible. I'm an electronics tech and have seen it happen often enough! Fibreglass PCBs with holes burnt in them from resistors burning out is a fairly regular thing to see (and that's in circuits that supposedly have the proper rating resistors fitted.)

In my books it's not worth the risk. Buy a proper battery charger.

murcod wrote:

Now, rethink your example. 0.5 ohms at 20A? You're dropping 10V from 13.8, which leaves 3.8V across your 6V battery

.

My example wasn't the best, but you get the idea. Even in your example Scott, 10V x 20 Amps is 200 Watts! Yes, it shouldn't stay at that level for long- but what if the battery has a couple of shorted cells?? Even a short surge at 10x the power rating of the resistor is very risky.

If you're dropping 10V across the resistor the battery is severely damaged - 4 out of 6 cells would be damaged. If the battery is this severely damaged the resistor will get hot almost instantaneously, and can be disconnected before any damage is done. It's not something which will normally appear only after hours on charge.

murcod wrote:In my books it's not worth the risk. Buy a proper battery charger.

The cheaper "proper" battery chargers (certainly anything less than $100) cannot be left to float charge - if you don't disconnect at the right time you're damaging the battery. In my experience, with a degree in electronics, it's not easy to determine "the right time" to disconnect. With the power supply, who cares?

The chances of a problem with a dud battery on the power supply, vs overcharging any battery on a sub-$100 "battery charger"?

I maintain the power supply is better value for money.

YMMV

Scott

Let me put it another way then.

Your 10W 0.5 Ohm resistor (when flowing 20 Amps) will only be able to handle a maximum 0.5V dropped across it before it decides to have a melt down. ie. the battery would have to be sitting above 13.3V before the resistor was within it's design parameters. Of course those figures will all depend on the internal resistance of the battery and how much charge current it will accept.

Running components at their maximum rating is never a good idea. Guaranteed the leads would desolder themselves at the temps it would be operating at!

Anyway, we agree to disagree! Perhaps a 130Watt headlamp globe would be a smarter proposition if a load is desired to be placed in line?

Back to the original question :

Are they just trying to upsell them, or do i really need a big one.

It should still charge the battery, however it will take a very long time. I've got a 4 Amp RMS charger and that's slow charging 4 cylinder petrol engine batteries (ie~400CCA). If the battery is dead flat it usually takes around 12hrs before it reaches the float charge level.

while you two were playing "im smarter than you"

i put the dual battery charger in and all is good

nah, thanks for the help guys good tech.

mickyd555 wrote:while you two were playing "im smarter than you"

I just don't want to see someone burn their house down!

Glad to hear you've got it safely sorted .

murcod wrote:Let me put it another way then.

Your 10W 0.5 Ohm resistor (when flowing 20 Amps) will only be able to handle a maximum 0.5V dropped across it before it decides to have a melt down. ie. the battery would have to be sitting above 13.3V before the resistor was within it's design parameters.

Now you're confusing the hell out of me. 20A through a 0.5 ohm resistor would drop 10V - so you'll only have 3.8V across the battery.

Ohms Law takes precedence here. You seem to be focussing on the 10W and ignoring the 0.5 ohms.

A 10W load carrying 20A can only drop 0.5V - yes. The resistance of this load would be 0.025 ohm. I'm talking about a 0.5ohm resistor, which can drop 2.2V before it exceeds it's 10W rating.

I don't think you're modelling the battery properly in your calculations. 13.8V less 2.2V leaves 11.6V - that's not charging. To be charging (unless it's stuffed) a 12V battery will have a terminal voltage above 12V - the resistor shouldn't be dropping more than 1.8V, which is within it's rating. You won't get the full 20A, but current limiting was the object of the exercise. Once the battery has some charge (have NFI how to determine when) it shouldn't draw 20A, so the resistor can be removed again.

Either way, I think your idea of a headlight has nailed it - perfect for this application. As current drops the resistance will drop too.

And I will also agree that this setup is probably less than ideal for anybody who may be electrically challenged.

Is this fun for you too?

Scott

murcod wrote:

mickyd555 wrote:while you two were playing "im smarter than you"

I just don't want to see someone burn their house down!

Glad to hear you've got it safely sorted .

I do - I bought marshmallows!

Well done!

NJ SWB wrote:Now you're confusing the hell out of me. 20A through a 0.5 ohm resistor would drop 10V - so you'll only have 3.8V across the battery.

Ohms Law takes precedence here. You seem to be focussing on the 10W and ignoring the 0.5 ohms.

A 10W load carrying 20A can only drop 0.5V - yes. The resistance of this load would be 0.025 ohm. I'm talking about a 0.5ohm resistor, which can drop 2.2V before it exceeds it's 10W rating.

What I was trying to explain (probably very poorly!) was a 10 Watt resistor, forgetting it's resistive value, flowing 20 Amps can only drop 0.5V before it's power rating is exceeded.
ie. using P=VI , V= P/I
V=10/20=0.5V

So, if the 0.5 Ohm 10W resistor is dropping greater than 0.5V @ 20 Amps then it's power rating is being exceeded.

As far as how much load a charging battery is going to draw, well it will vary greatly depending on the type of battery and it's charge level. Something like an Odyssey battery can have as little as 0.002 Ohms internal resistance and fully discharged could draw a huge amount of current initially when charging.

I personally have discharged a battery (accidentally) way below 10V and have successfully charged it, plus after charging it held charge for over a month. I lost count how many times I did that in my previous car, it was often so flat that the interior light would be barely on..... Not a good practice, but the battery was still going strong after five years when I sold it.

It's hard to say how much current a flat battery could potentially draw from that supply??? I'm just suggesting to play it safe, high power resistors do get VERY hot.

It would also be interesting to establish if that supply has a voltage foldback function or if it actually cuts out when 20 Amps is exceeded?

NJ SWB wrote:To be charging (unless it's stuffed) a 12V battery will have a terminal voltage above 12V - the resistor shouldn't be dropping more than 1.8V, which is within it's rating. You won't get the full 20A, but current limiting was the object of the exercise.

murcod wrote:As far as how much load a charging battery is going to draw, well it will vary greatly depending on the type of battery and it's charge level. Something like an Odyssey battery can have as little as 0.002 Ohms internal resistance and fully discharged could draw a huge amount of current initially when charging.

Yes. Which is why I suggested an external resistor, to limit the current flow. Add 0.5 Ohms to the 0.002 Ohms internal resistance and calculate your current flow.

Scott

Yes, but I'm still a bit dubious about potentially how much voltage difference there could be between the battery and power supply output. That's the big factor in how much power the resistor will have to dissipate, and where such a circuit could come unstuck.

I notice Super Cheap have got a dual power supply / charger for sale for $199.

-Charges then maintains battery in peak condition
-Short circuit and polarity protection
-DC power supply output 3.5V -> 15V
-current up to 10 A

It's got both current and volt meters (only analogue) and decent banana binding posts for the power supply output. The usual alligator clips are used for battery charging.

murcod wrote:Yes, but I'm still a bit dubious about potentially how much voltage difference there could be between the battery and power supply output. That's the big factor in how much power the resistor will have to dissipate, and where such a circuit could come unstuck.

For the battery to be charging the terminal voltage must be greater than 2V per cell (more than 2.15, from memory) in order to reverse the normal "discharge" reaction.

You've got me curious now - I want to test this, but I don't have any spare batteries to play with. I took two to the dump just before I moved. But I might be able to hook up enough spare headlights to pull more than 20A, to test the overload performance.

The Super Cheap unit sounds good, but I'm too cheap to spend that much. I bought the Jaycar unit to run my car fridge inside - Autofridge wanted over $200 for a 5A unit.

Cheers,

Scott

I'm tempted to buy one when they've got one of their 20% off sales. The power supply side would be useful to have, although I don't know if the output is regulated once set?

So many questions - ain't it always the way?

The "current up to 10A" could be a clue. Is it 10A at 3.5V / 35VA limit? This would be barely more than 2A at 15V.

If the variable output isn't regulated it wouldn't be terribly useful. I can't think of too many uses for it...

Let us know once you buy one.

Scott

It's 12V / 10Amps - that's a definite. Even if it wasn't regulated it would still be useful for running 12V equipment.

I tried Googling the part number but didn't have any success (PCS10A)

I bought one of the Super Cheap combined charger / power supplies yesterday- 20% off meant $159 reduced from $199.

It appears very well made (in China....) with decent sized alligator clamps and a nice case. While I haven't tested out the amount of current output and regulation of the power supply, I can tell you it is a proper three stage charger.

It will initially bulk charge and supply a constant 10 amps to the battery until the battery reaches 14.7V. It will then keep the voltage at 14.7V for up to 2 hrs with the charge current gradually reducing. The third stage is when it switches to float charging mode and keeps the output at 13.7V (meaning you can leave it connected permanently to a battery - it says it will switch itself off too?)

It's got an internal cooling fan that appears to only come on when needed. The adjustment potentiometer for the power supply side of things is a ten turn type which will make accurate voltage settings a breeze. The panel meter is also about as accurate as you can get (ie. 14.7V on my multimeter looks about right on the analogue meter readout... Hmmm future mod with a built in digital LCD meter readout....)

The instructions also state it has in built over current protection that will be activated when the current reaches "15-20% over" the maximum of 10 Amps.

It has decent full sized banana plug binding posts on the front panel for the charger/ power supply outputs (same outputs are used for both with a switch to control which mode it is in.)

For the money it appears like a good buy. BTW it comes with a 12mth warranty.

http://www.projecta.com.au/catalogue/cid/3/asset_id/15

i use 2 of these on my landcruiser. One for each battery. They work great. My car starts better than ever.

My Landcruiser is a second car so it doesn't get driven much. I got them cause i went through my first set of batteries in 12 months.