JNorton

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TopRudder,
Actually Multiplex makes a LN-5014 DC charger for $80.00 that chargers Lp just fine.
http://www2.towerhobbies.com/cgi-bin...&I=LXKSY1&P=ML

I'm still using my original 7aH lead acid gell. It is now 4 years old. My trips to the field are approximately 1/2 hour in length. My vehicle charging system puts out 14.0 volts after starting the car. It would be easy enough to put a 3 amp diode in series to limit the voltage to 13.3. I haven't an 1141 bulb or I would try it - I still might for curiosities sake. Basically all I am doing is to keep the gell cell topped off.

KISS

John

toprudder

Thanks for the link, I am looking for a decent field charger. However, the technical information in that link regarding charging does not mention anything about the charge method for lead-acid batteries. It almost reads like it is constant-current, which is a no-no for LA batteries. It should at minimum have a constant-voltage mode for LA.

JNorton

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http://www.multiplexusa.com/RC-comp/...rs/LN-5014.htm Top right of the page is a link to the instruction manual.

JNorton

My Feedback: (2)

TopRudder,
I've been looking for more information on the LN-5014 and came up short. The algorithum for LiPo is similar to that of charging Lb. I cannot conceive of a large corporation like Hitec / Multiplex failing to make a suitable charger. I do know that Red feels this is a good charger for the money but since I do not own one I cannot give any further information on it.

Perhaps this would give you more information.
http://www.rcuniverse.com/forum/m_3188884/tm.htm

John

iflynething

1.) Take off cap

2.) Put battery acid into mouth

3.) Swish until at field. The longer you live from the field the better.

4.) Once at field, put acid from mouth (you have been swishing, right) back into battery, and you're good to go...

5.) Wash mouth out with mouthwash (I'm going to say some Listerine or Scope)

6.) Repeat as necessary to get more charge throughout the day.



I hope this helps

~Michael~

toprudder

Thanks!

The manual is a little lacking in information. It still does not mention at all what the charging algorithm is for Pb batteries. It did mention that that nominal voltage must be entered, so that implies that it somehow limits the charge voltage. I still would like to know, but after reading the other message thread you pointed me too, I guess I will still keep looking, however I won't rule this one out.

I'm just a little leary about how it charges Pb batteries. I've had a real education on that during the last year with my camper. I just installed a new converter and battery in it, due to the inadequacies of the old converter.

AS-EE

JimTrainor, that 6 ohm of resistance for the battery internal is wrong. You more than likely just measured the current limiting resistance of the whole system. What you did was take 15.2 volts - 13.4 volts = 1.8 volts/ 300mA= 6 Ohms. That is the current limiting resistance AND NOT the battery's internal resistance.

JimTrainor

By gosh AS-EE, I think you've got it! You've mastered Ohm's law!
The "whole system" here is two wires, a fuse, the battery, the power panel's bananna jacks, and the meter.
I don't think the panel jacks, a wire, and fuse have 6 ohms of resistance. But then, I didn't think a lead acid battery could have a impedance of 6 ohms either (but there I could be wrong due to lack of detailed understanding of lead acid batteries... certainly 6 ohms would be astronomic for nicad - the only one I'm really familar with). Leading me to question my own measurement, or understanding of the battery (as I stated!).
Anyways, AS-EE has it all figured out, so I'll chime out. But I'll watch for AS-EE's own concrete numbers on the subject since I am interested in this.

AS-EE

I am way beyond the basic Ohm's law stuff. I am at the point where there are circuit networks where the resistors are not in parallel OR series and not even a combination of the two. Ohm's law alone will not help you in these networks. These types of networks must have special theorems applied such as Thevinin and Nortons, and my favorite being the Superpostion theorem. Also I am in AC networks which is many times more complex than DC networks. Finally, I am in Digital and semiconductor studies as well.


Anyways, I have a pictorial to show you how the current limiting resistor works:



You can see here that applying Kirchoff's voltage law around the loop will reveal that any voltage difference between the two batteries will ALWAYS be across the resistor. This voltage difference across the resistor will determine how much current will flow to the lower voltage potential E2.

JNorton

My Feedback: (2)

AS-EE
Actually the impedance of the battery is changing while it is charging. The voltage across the battery no load without the charger connected is 13.4 volts. The voltage across the battery connected to the charger is 15.2 volt and the current is 300 mA. Battery current in Lp batteries decreases as it becomes charged. This is a dynamic variable and cannot be explained as you are attempting to. Sorry.

See the graph - note the resting voltage.

John

EDIT-> http://www.homepower.com/files/battvoltandsoc.pdf

JimTrainor

Yeah, I get that.
I found a link yesterday with a paper that had charts of charge and discharge impedance for a 12 Ah lead acid battery.
AS_EE - keep up the good work at school. Seriously... models are good place to put some basic theory and practice to work.

toprudder

According to those numbers, and the graph, the battery was way past being fully charged.

JNorton

My Feedback: (2)

TopRudder,
Generic graph as I have no way of knowing what the actual battery used was. It is close enough to illustrate the point I was trying to make. Tough audience.
John

toprudder

John,

Sorry, I was not disputing the graph at all, in fact it reinforced my thinking that 15.2v was way too high of a charge voltage for the battery!

Like I mentioned earlier, I've had a real education lately in the care and feeding of lead-acid batteries. The converter in my camper (now) is a multi-stage solid state design. It starts at 14.4v (boost stage) which lasts for about 4 hours, then 13.6v (absorbtion stage) until the current drops below a certain point, then goes to 13.2v (storage/float stage). This is a safe way to quickly charge the battery without boiling the electrolyte excessively.

Now this is a regular flooded wet-cel, deep cycle type battery, but I suspect the chemistry is the same as the 7ah battery in question. The biggest difference is that 7ah battery is not vented. If it is held at 15.2v for a long period after being fully charged, then the electrolyte will boil and will increase the pressure inside the battery, perhaps enough to be dangerous. I've seen this type of battery swell and deform under this pressure.

I saved the PDF file you linked to, so I can use it as a reference. Thanks!!!

Also, I've read that the best way to determine the state of charge of a lead-acid battery (while charging or under load) is to measure the specific gravity in each cell. Not an option with a sealed battery!

Bob R.

AS-EE

Nah you think? I was just showing him that at the instant the battery is connected with a voltage 13.4 volts, that the difference between the two will be across the resistor. I could care less about what the voltage drops down to once the battery builds up more voltage potential. Do you want to know why? Because the voltage drop across the resistor will become less and less and therefore the current flowing to the battery will be less and less. This assuming of course that the charge voltage is not so high that it will damage the battery. Hence you make the charge voltage at a maxium of about 14.4 volts so that battery will charge up to that value.

When you design a charger you model a discharged battery for worst case scenario and select a resistor value for the maxium current you wish to have flowing to the discharged battery and afterwards you don't care because as I stated the voltage drop will become less as the voltage potential of the dead battery builds up.

I am pretty sure I did not have to tell Jimtrainor or any else that the voltage builds up on dead battery when it is charging. I have no idea what your trying to prove here Jnorton, but it's getting old real fast....

JNorton

My Feedback: (2)

AS-EE,
Where is the resistor that you show? There is none. I'll guarantee that if you open up the cheap 12 volt wall wart charger you won't find one. You can take a regulated lab quality supply and get the same answer - No resistor - it is the battery chemical composition itself. That was the point which apparently you missed. Jim Trainor put his charger on a battery and measured it open circuit voltage and then it voltage under charging. Once again there is no resistor, and lead acid batteries cannot explode when they reach zero volts because they never will reach zero volts. Both assertions you've made.
John

Two batteries - and a resistor? No there is one battery and it's internal impedance changing while it is being charged.

JimTrainor

What JNorton said!

To be perfectly clear: I measured the battery's open circuit voltage. Then I hooked up the charger and measured to voltage it output while charging the battery (again, just measure accross the battery terminal but with the charger attached.) I did that a few times, the numbers I posted were the final set, hence the battery had a "surface charge" when I measured it's open circuit voltage (hence was quite high).

AS-EE, you're clear that their is not an external resistor involved in this debate? We are, quite literally, talking about the battery's internal impedance. The Thevenin equivalent of the battery is a perfect voltage source, and a resistor. The physical reality is that there is just the battery - and it's not perfect!

The 6 ohms I computed (a number I still don't trust!) is the battery's internal impedance.

Take a look at this nicad battery spec sheet:

http://sanyo.wslogic.com/pdf/pdfs/CP-1300SCR.pdf

Part of the spec is the battery's internal impedance. We are talking about the same thing, but for a 7 Ah gel cell.

This thread has gone a haywire.
[sm=spinnyeyes.gif]

JNorton

My Feedback: (2)

AS-EE

From another post where you said I had no reasoning nor logical thinking. You sir are making comments on batteries you don't understand and applying simplistic reasoning thinking you're sharing revelations to the uneducated masses. The battery you are trying to model is a single node.

With that I'm going to retire from RCU for a while. Trying to answer questions and give good advise is work - sometimes your wrong.

John

JNorton

My Feedback: (2)

Only while charging. Discharge impedance is very low and a different animal. They are not the same thing as the battery reacts differently charging or discharging. With that I'm gone. See all you guys later.
John

JNorton

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I just had a good friend of mine die yesterday. We rode motorcycles, drank too much and chased women till we caught the ones that wanted to be caught. He was a diabetic. In the last 10 years he lost both legs, had a kidney transplant that failed last year. Last Friday he was told he was terminal. Yesterday I took off a half day from work, as a bunch of us were going to go to Jimmys at 4:30. I thought he'd like to hear the pipes on my Yamaha, then we could argue about whether his Harleys was louder. I never got the chance. He died at 2:30 while I was washing the bike.

AS-EE

That picture has a resistor that is EXTERNAL the symbol is a up there labeled V1, you must not know what the schematic symbol for a resistor is. THAT IS A CURRENT LIMITING RESISTOR I SHOWED UP THERE AND NOT A INTERNAL RESISTANCE OF A BATTERY. Please pay attention before you post nonsense.

IF YOU ARE SO CONFIDENT THAT THEIR IS 6 OHMS OF IMPEDENCE THEN HOOK UP AN ELECTRICAL LOAD THAT WILL PULL ONE AMP OF CURRENT AND YOU CACLULATE FOR ME THE VOLTAGE DROP ACROSS THAT INTERNAL IMPENDENCE OF 6 OHMS.

ALSO IF YOU ARE SO DAMN CONFIDENT IN THIS 6 OHM OF INTERNAL IMPENDENCE THEN HOOK UP A CAR BATTERY TO IT WHEN IT IS DEAD AND AND PRAY THAT YOU DID NOT RUIN YOUR GELL CELL BATTERY!!!!!!!

FINALLY, THE TRANSFORMER USED IN SEARS CHARGERS IS NOT YOUR AVERAGE TRANSFORMER AS I HAVE LEARNED FROM AN ENGINEER WITH A BSEE DEGREE ON ANOTHER TOPIC.

JimTrainor

AS-EE, you've gone off your rocker.
Signing off.

JimTrainor

That would explain it.
Now I'm going to have to put a load on the battery and measure the discharge impedance.

Sorry about your friend's passing.

AS-EE

I am not going off my rockers, but when I detect ignorance then I must correct it so you don't so something stupid as try to hurt yourself or give false information to others on this forum which will result in harm for them as well. I will not tolerate it one bit!!

AS-EE

THats not the topic I am talking about. I am looking at the aboves posts where you say no current limiting resistor is needed because of the 6 ohm of internal impendence. YOu are so wrong and this will be harmful because someone on here will believe you and hook up a car battery to a discharged gell cell battery because they will think that the 6 ohm of internal impendence will limit the current and this is wrong becuase it is not 6 ohms!! THe gell cell battery will be ruined instantly!!