Battery based tank circuit (tested) - Page 10

nzKAOSnz · 10-05-2004, 03:22 PM

Quote: Originally Posted by MatrixPC

You can buy a Sproggy MK3.5 as a kit or just a PCB. With a kit you still need to buy the main Maxxim or Linear Technology and National Semi (for 12V) chips.
Oh, changing car oil isn't that hard. For the past 15yrs I only brought my car to dealer for changing oil 5 times because they offer free car wash with any service.
I can change the oil in my Matrix within 15 min but about an hr on the Supra because the oil filter location on the Supra is freaking hard to reach. It stuck between the side of engine block and the Turbo charger.

You should try to change the plugs on a subaru legacy :S Now i knw why they use plugs that are good for 100000ks

Arathranar · 10-16-2004, 03:52 PM

Quote: Originally Posted by Ricky327

Bedside that, the limiting resistor was put there mainly to limit the current being drained by the starter motor. Why not use a diode to isolate? because I didnt want any voltage drop on charging...this way the backup battery get the full charging voltage from the alternator.

But the resistor will cause exactly the problem you are trying to avoid, surely? 0.9A will cause a 13.5V drop. 0.1A will cause a 1.5V drop. A typical diode drop is equivalent to a changing current of only 0.04A. A diode seems far preferable since it prevents the crank taking any current from the tank battery.

I was intending to use the 12V accessory line to power the tank battery so that when the car is off, there is no way for the PC to drain the main battery. An alternative (may even be required if the switched accessory circuit is not up to it) is to use the acc line to switch a relay hooked up to the unswitched 12V line. That is the same as suggested earlier in this thread.

Ricky327 · 10-16-2004, 09:32 PM

Quote: Originally Posted by Arathranar

But the resistor will cause exactly the problem you are trying to avoid, surely?

Consider the worst case, the starter motor have ZERO resistance.

Now how much current can a starter motor possibly drain out of that tank battery via the limiting resistor?

Ricky327 · 10-16-2004, 09:48 PM

Through a resistor it is able to supply the full alternator voltage into the tank battery.

Through a diode it is always 0.6-0.7V less than the alternator voltage that goes into the tank battery. This way it can never get charged up fully.

The graph shows how much charging voltage the battery requires, depending on the temperature. As you can see even pumping the alternator voltage directly into that battery may not be enough to charge it properly.

Based on 20degC the 12V battery need around 13.5V for standby use, while for cyclic use it need 14.4V ATLEAST.

Now subract the 0.6V drop by the diode from the 13.4V generated by the alternator, we end up we only 12.8V going into that tank battery. We are now well below the charging spec of the battery...12.8V is not even mentioned in the graph for the 12V battery.

Can you see where the problem is?

Ideally the voltage should be stepped-up to whatever the battery need and as much current is pumped into it without exceeding any limits...MCU controlled perhaps. But then how far do we wanna go for the sake of charging a simple SLA battery?

Arathranar · 10-16-2004, 10:36 PM

Quote: Originally Posted by Ricky327

Through a resistor it is able to supply the full alternator voltage into the tank battery.

That's not true. That was my point. There will be a voltage drop across the resistor. Even with a very low charging current, the drop across the resistor will be more than that across a diode. So you probably end up with a lower charging voltage than with a diode or an incredibly slow change. Or possibly even both.

My planned tank circuit will run off the switched 12V and only employ a single diode. That way the crank cannot affect the tank battery output voltage at all. Since the main battery voltage drops to 12V when the ignition is switched off, that's important to avoid the PC getting too low a voltage whilst it's shutting down.

Ricky327 · 10-16-2004, 11:13 PM

A fully discharged battery will draw in alot of current but as it charge up it will come close to ZERO. How can the resistor drop voltages when the battery is not even draining any current?

Sure the voltage drop is alot but it will go down to zero as the battery charge up.

For Float Standby Applications
Charged at 2.275 volts per cell continuos. Battery will
seek its own current level and float fully charged.
However, users should be aware that when charging
from fully discharged, the battery can draw an initial
charge current of approxiamately 2cA. Care should
therefore be taken to ensure that this initial charge
current (if ungoverned) is within the output capability
of the equipment. Final charge current at 2.275 volts
per cell is typically between 0.0005cA to 0.004cA.

From YUASA datasheet, look above it says :

Final charge current at 2.275 volts
per cell is typically between 0.0005cA to 0.004cA.

So for the 0.8AH battery we have the final charging current of 0.8 x 0.0005 = 0.004A

The voltage drop is therefore only 0.004 x 68 = 0.028V that is no where near the 0.6V that can be dropped by the diode you proposed.

The idea you propose had been discussed long before I started this thread...its is an old idea. And I guess you already worked out why I modified it for my own need.

If the simple diode work for you then go ahead. If there is such thing as ZERO drop diode or I can somehow increase the alternator voltage up to 15V then that idea is perfect...but then YUASA recommend that the charging current should be limited, now we are back to the resistor idea again.

Ricky327 · 10-16-2004, 11:33 PM

Just to add, I have measured it before and there is hardly any drop on a fully charged battery.

Just in case you wanna know how much maximum current can be drawn out of the tank battery...

(Assuming 12.5V out of the tank battery)

It is 12.5/68 = 0.18A which is not exactly alot, thats just equivalent to 2 small relays being powered by the battery for 2-3 seconds...we are taking about exteme case here.

If we assume the starter motor drop the main battery down to 10V while the tank is trying to supply the starter motor at 12.5V via the 68ohms

now that will be 12.5-10/68 = 0.037A

That figure can almost be seen as negligible...and that figure is more realistic. Can we say thats alot to drain out of the tank for 2-3 seconds? no I dont think so.

Arathranar · 10-16-2004, 11:45 PM

Quote: Originally Posted by Ricky327

So for the 0.8AH battery we have the final charging current of 0.8 x 0.0005 = 0.004A

The voltage drop is therefore only 0.004 x 68 = 0.028V that is no where near the 0.6V that can be dropped by the diode you proposed.

I'm not sure I follow the calculation above, but it seems that's when the battery is already fully charged, which isn't very interesting. Long before it gets there the current will be higher and the voltage drop across the resistor will be much higher, significantly slowing down the battery change.

Ricky327 · 10-17-2004, 12:22 AM

I'm not sure I follow the calculation above, but it seems that's when the battery is already fully charged, which isn't very interesting.

Thats exactly right, the calculation is for a fully charged battery. The point of it is to prove that the drop goes down to ZERO as the battery charged up and it is capable of supplying the full voltage from the alternator. Not fully charging the battery is a very bad idea anyway. On the diode it will never be able to supply all the voltage available from the alternator...ever.

Long before it gets there the current will be higher and the voltage drop across the resistor will be much higher, significantly slowing down the battery change.

Again that is correct, the charging current should be limited as recommended by the datasheet. Sure the charging may be slow but why would you let the battery goes completely flat? Thats the whole reason why a good charger have the abilty to increase the voltage up to whatever it take just to make sure the full current is being pumped into that battery until it is fully charged up.

Even if the full 13.4V is directly connected to the tank it will still limit and find its own charging current, from the datasheet >>> Battery will seek its own current level and float fully charged

Imagine you connected the tank battery directly to the alternator and you have an internal short (fault) in that battery? Imagine the current drain and what will happen to that small battery installed under the passenger seat

Heck I dont even wanna think about it.

Yup I agree with you that the charging is slow, but it solve the other problems Im having.

Maybe your alternator is pumping 15V then you are all fine with the single diode design.

I know mine only output about 13.4V which is the standard for most. I have tried the single diode on my setup before and the battery never get charged up properly, it was only getting charged to maybe 40%...hence the reason for this new setup.

Up to you really which setup that work best for you, Im not saying my setup will work for everyone it just that I found some few problem I have encountered with the traditional setup.

Ricky327 · 10-17-2004, 01:01 AM

Heres the float charge graph, just incase you havent seen it before :

The 68 ohms resistor value on a 0.8Ah battery gives about 0.25CA the charging current as recommended by the company.

As you can see it take about 12hours to charge up a battery from 100% flat with the setup above...thats fairly standard on most charger. Although the charge voltage should be 13.65V but we can only have 13.4V from the alternator so Ill expect the charging time to be alittle bit more or maybe the battery will never get charged up to 100%.

The point is even the whole 13.4V connected to the tank battery may not be enough to charge up properly. So what chance does a 13.4-0.6V (minus diode drop) charging the battery to 100%?

I know you can increase the current to whatever it need to get the battery charged up in like 4 hours by using the diode directly...but why? and risk of having the other problems, like no protections and not being able to charge fully therefore destroying the battery. Isnt 12 hours enough since they are the standard rate for most chargers? some charger may even take 24 hours.

Arathranar · 10-17-2004, 01:02 PM

Quote: Originally Posted by Ricky327

Imagine you connected the tank battery directly to the alternator and you have an internal short (fault) in that battery? Imagine the current drain and what will happen to that small battery installed under the passenger seat

A fuse will also handle catastrophic failure like that and is a preferable solutio in my mind. I'd like to know I had a problem rather than allowing permanent controlled current drain from my main battery.

Quote: Originally Posted by Ricky327

Yup I agree with you that the charging is slow, but it solve the other problems Im having.

I'm still trying to understand those charts, but it seems to me that the resistor will increase the charging time by an order of magnitude or more from those charts. Whilst the diode solution will slow it down only a small amount but limit the max charge. Since max charge is *only* interesting if you need 100% to survive the crank (in which case the circuit is a failure since the slow charging means you'll have to be driving a long time before it can survive a 2nd crank), that makes the diode solution far preferable for all scenarios.

I'll be using a 5.1Ah since it wasn't much more expensive than a 0.8Ah battery ($10 vs $3), I have space for it and I'll be driving a 200W dc-dc psu with an AMD mAtx board. Hence I was plenty of juice. The 0.8Ah would probably work but there was no point going that route.

What exactly is CA? Is cell-amps? I.e. the current needed per cell in the battery? How do you know how many cells there are in a battery?

Arathranar · 10-17-2004, 01:07 PM

Quote: Originally Posted by Ricky327

The 68 ohms resistor value on a 0.8Ah battery gives about 0.25CA the charging current as recommended by the company.

Ok, I'm really confused here. 0.25A will give a 17V drop across the 68ohm resistor. How does that work? As far as I can see, the resistor guarantees you don't get anywhere near the charging current the battery can take.

Quote: Originally Posted by Ricky327

I know you can increase the current to whatever it need to get the battery charged up in like 4 hours by using the diode directly...but why? and risk of having the other problems, like no protections and not being able to charge fully therefore destroying the battery. Isnt 12 hours enough since they are the standard rate for most chargers? some charger may even take 24 hours.

I think this is the crux of our different viewpoints. As far as I'm aware, lead-acid batteries don't get destroyed by lack of full charging (unlike Ni-Cd batteries). So I don't care if I never fully charge my battery - as long as it's sufficient to survive crank. I do care if it takes so long to charge that it ends up failing at that job after a series of short drives.

Ricky327 · 10-17-2004, 02:16 PM

Heres the proper charger for an SLA battery :

http://www.maplin.co.uk/Module.aspx?...dID=&doy=17m10

I do have this one but only used it on the bigger battery I had, a 2.8Ah.

Look what it say its maximum charging current is 350mA a constant voltage charger. I havent opened it to see whats inside but Ill assume its just a constant voltage with current limit set to 350mA and this is designed for an SLA battery right upto 9Ah...mine is only 0.8Ah and Im pumping 200mA.

As I type this, Im charging the 2.8Ah battery I got. The battery is flat and guess how much current the charger is pumping? not surprisingly its 380mA...should be 350mA accoring to the spec.

Read this thread :
http://206.131.241.58/ubb/ultimatebb...c&f=1&t=002162

And tell me how much current I should really pump into that 0.8Ah without blowing it. How much are you planning to pump into your battery?

There probably 1,000 more site that will tell to charge at 1/10 of the capacity if it was to be done safely.

If I want to do this properly and charge at a much faster rate then Ill probably use a PIC or a dedicated controller, together with temp sensor, voltage sensor, current sensor, timer, step up the voltage and so on.

Really you should just try your setup and see if it work good for you. There are so many way to charge up the battery, take a pick and let us know whats the outcome.

Ricky327 · 10-17-2004, 02:31 PM

So I don't care if I never fully charge my battery - as long as it's sufficient to survive crank.

http://www.buchmann.ca/Chap10-page5.asp

Have a read about sulfation due to undercharging.

Or try this, leave your battery flat for a while and see what happen.

the slow charging means you'll have to be driving a long time before it can survive a 2nd crank

So you are saying you crank your engine for 30 mins? that the tank battery gone flat? and can no longer survive the next crank?

if you need to crank that long then you better worry about how you gonna take the car home than worry about surviving the crank

but it seems to me that the resistor will increase the charging time by an order of magnitude or more from those charts. Whilst the diode solution will slow it down only a small amount but limit the max charge.

If you have the chance measure the charging current flowing into the battery when connected through a diode...see if its much higher than what the graph says. I havnt done this so I wouldnt know.

Arathranar · 10-17-2004, 02:59 PM

The full quote was:
Since max charge is *only* interesting if you need 100% to survive the crank (in which case the circuit is a failure since the slow charging means you'll have to be driving a long time before it can survive a 2nd crank)

Your response took one phrase totally out of context and tried to mock it.

If you don't need 100% (which must be the case since the circuit wouldn't be consistently effective otherwise), then getting the max charging voltage is not interesting since your circuit is not so heavily dependent on having a fully charge tank battery in order to maintain the voltage into the auto pc. Many people on this forum have used the diode only solution with great success. So clearly having a fully charge battery is not a requirement.

I'll read up on that thread about undercharging.