toprudder

Guys,

We can argue about it all day long, but the battery does what it does, regardless of how we try to explain it.

The observation that it has 6 ohms internal impedance is just that -- an observation. It apparently has 6 ohms internal impedance. The two voltages (13.4 and 15.2) were both measured at the battery terminal, and has nothing to do with any external circuit resistance.

The 13.4v was an artificial surface charge, not the true resting voltage. The true resting voltage of a fully charged lead-acid battery is approx 12.6v.

A lead-acid battery can't be modeled as a perfect voltage source with a resistor -- there is chemistry involved.

The apparent internal impedance changes depending on the charge state of the battery, and the charge current. It will also appear MUCH different depending on whether it is being charged or whether it is under load. Again, there is chemistry involved.

Lets go flying![8D]

Bob R.

JimTrainor

Indeed chemistry.... I guess the "equivalent discharge impedance" I calculated is the energy going into reversing the chemical reactions that occured to discharge it plus some electrical impedance.

What do you think AS-EE, shall we through in a big capacitor to model the energy being absorbed by the battery as it charges?

Good thing AS-EE isn't studying CE or this could get really messy.

Trying to ignore this... but failing.

AS-EE

I did not model the battery with the internal resistor. I was trying to show to you that if you measured 15 volts on the output of the charger with nothing connected to it then that is the open circuit voltage. When you applied the battery the voltage was at what the battery was. This is becuase there is an external impendence source inside your charger that has the voltage difference of the two different voltage sources across it. Your charger, if it is the good kind probaly uses a transistor circuitry setup to limit the current. If its a sears charger then transformer is specially designed to drop in voltage when current is pulled from it. If the transformer on the other hand was not designed like that then you will need a way to limit the current to the battery or you will have a huge surge of current flowing to the battery which may be more charge current then it was designed for and thus will damage it.


Now do you understand what I was trying to tell you???

He measured the output voltage of the charger to be at 15.2 volts and not across the battery. (battery was not connected). Then he connected the battery and measured the value of 13.4 so where did the extra voltage go? It was dropped inside the charger and not in the battery...

Now do you see the light?

JimTrainor

AS-EE you still don't get it.

Actually, that's not what he did (err... i did...).

The charger was *connected* to the battery when the 15.2 was measured! I was measuring the voltage that the charger generated in order to push current into the battery. That's the whole point of the measurement! The goal was to determine the difference between the battery's resting voltage, and the charge voltage, subtract those, then measure the charge current, then divide to get the a restistance value. Which came out to be 6 ohms... doohhh!

The internal workings of the charger are irrelevant here. They may be interesting, but not relevant to the point I was making.

AS-EE, lets review what I did, as reported in post #12:

1) disconnect everything from battery and measure voltage across terminals. Result = 13.4 volts
2) connect charger to battery and measure voltage across the terminals. Result = 15.2 volts.
3) connect charger with the meter in series and measure current. Result = 300 ma.

To summarize: The two voltages (13.4 and 15.2) were both measured at the battery terminal, and has nothing to do with any external circuit resistance.

AS-EE

Thank you for clearing this up.

AS-EE

Now the next question is how fast did it go from 13.4 to 15.2 volts? I need the duration of time.

With my custom made charger I went from 12.71 to 13.3 on the battery. I don't think it is the resistance in the battery, but a chemistry process. We need a CE on here.

JimTrainor

The charger's output jumped higher than the battery's open circuit voltage faster than I could got the meter on it after pluging the charger in... mostly. That final bit of settling (say last .2 volts) did take a few seconds. But I don't know what to attribute that too. Maybe the battery, maybe the charger, maybe both.... dunno.

AS-EE

You know this instant change only occurs when the battery is fully charged and when we put this fully charged battery on the charger, that causes it to jump up real fast, but at the discharged voltage of about 11.8 I found that it stayed at the level for a minute or so then it gradually goes up more voltage in a non-linear fashion. Do you have the charging graphs for your battery?

JimTrainor

Nope. Nothing so fancy!
Just me and my meter.

JNorton

My Feedback: (2)

AS-EE
Really? What false information have I posted?

jlkonn

My Feedback: (1)

JNorton,
Here is the pix.
JLK

JNorton

My Feedback: (2)

AS-EE,
You disparage my talents and abilities. You say I cannot tell what a current limiting resistor is.

Here is a photograph of a +/- 0 to 15 volt tracking regulated power supply @ 1.5 amps. There is also a 5 volt 1 amp regulated supply. The transformer was wound with an extra set of windings for the 5 volt power supply. It also has a current meter that is switchable 0-1.5 A or 0 to 150 mA f.s. reading. It was built before you were born in 1972. I've been doing this a long time. The only thing I have done with you is question some of your comments that indicate your ignorance. The fact of which you blithely ignore claiming it is me that is ignorant is somewhat damning. When you can design and build your own power supply and make your own meter shunts I might be inclined to listen to you.

John

JNorton

My Feedback: (2)

JLK In a previous PM to me asked about an old maintenance charger.
He said his Digital meter is 20 years old. "My 7A sealed cell battery shows 13.6v with nothing attached to the
battery. The charger shows 16.6v without being attached to the battery. It was over 14v when attached to the battery."

Fully charged Lp batteries show between 12.8 and 13.2 volts depending upon how long they been setting - your meter is reading 13.6 which indicates to me it's probably a half volt off. If you can find an accurate meter and the charger reads between 13.0 and 13.4 volts it will act as a float charger. My feeling is that your charger is working. Radio Shack sells some inexpensive digital meters.

Just my two cents,
John

PS What are the numbers on the TO-220 case (three lead plastic flat object) on the wall? If they are LM317 or a 7812 or 7813 the are regulators. Could also be a series pass transistor.

JNorton

My Feedback: (2)

AS-EE
Since this is the second time you've called me ignorant. Since we all stand in awe of your technical prowness why don't you design a 0-15 volt meter then switch it to a 0-150 mA and a 0 - 1.5 amp current meter. Please show the schematic diagram and the resistor and shunt values. After all it's just Ohm's law and you've told everybody your way beyond that.
John

AS-EE

That was not towards you, Jnorton. That was towards Jimtrainor until I realized the we had a miscommunication and I found out how he measure the 15 volts. I was under the impression that he measured the voltage with no battery connected, and well to find out he hooked up a fully charged battery that is at 13.4 volts, but then suddenly it jumped to 15.2 volts and this is what he meant. If you look up a few a posts you will see that I thanked him for clearing up this miscommunication that we had.

JNorton

My Feedback: (2)

I beg to differ it was in response to my trying to point out to you that a battery is a single point node. You used a two battery model with a dropping resistor to indicate charging impedance. A battery is a chemical device and as such does not react in a linear fashion. That the impedance relected by charging cannot be modeled as a pair of batteries and a dropping resistor. I also pointed out that the battery resistance in the discharge mode is totally different and used a graph to illustrate. For which you called me ignorant.

What you use for a model can be mathematically sound. Everything correct by it's application can be wrong. You keep defending the correctness of the model. A 12 volt lead acid battery cannot get below 9 volts. Chemically it is impossible for it to do so. Yet you stated that at zero volts the battery would explode. A zero ohm load will draw tremendous current would but a battery won't, do you see the difference? The battery will not even accept a charge at 9 volts. It doesn't at like a resistive load. Do you understand?

What I proposed and use is to connect the field battery to the car electrical system through a 1 amp fuse. My battery after a day of flying is never below 12.7 volts the fuse doesn't blow and for the short amount of time 1/2 hour the battery doesn't over heat.

Thank you for your time.

John

jlkonn

My Feedback: (1)

John,
The numbers on the 3 lead dealy are as follows:
904
LM340T5
7805
What does that make it?
Is it attached to the back of the aluminum face plate for heat sink effect?
I'll drop by Radio Shack today and see if I can pick up a digital multimeter.
I'll let you know how I do then.
Thanks!
JLK

jlkonn

My Feedback: (1)

John,
What is a "tracking regulated power supply"?
What is it used for?
Thanks!
JLK

AS-EE

Here is what I said earlier since you did not read the above posts:


I THOUGHT JIMTRAINOR MEASURED THE 15.2 VOLTS WITHOUT THE BATTERY CONNECTED. HOWEVER HE DID NOT SO I WAS ARGUING ABOUT THE WRONG THING THE WHOLE TIME.








HERE IS WHAT HE REALLY DID AND WHAT I THANKED HIM FOR THE MISCOMMUNICATION I HAD WITH HIM:



THANK YOU FOR YOUR TIME MISTER!!

JNorton

My Feedback: (2)

It's not worth the time.

toprudder

Off the top of my head, the 7805 number indicates a positive 5 v regulator. I believe the LM340T5 indicates the same thing.

Yes, it is attached to the faceplate for heat sink effect. There is probably an insulator under the tab, but may not be necessary. There should be some silicone heat sink compound on there as well (usually white in color).

Before you buy another meter, I would double check to see if your old one is bad (I doubt it is). Let the 12v battery set for 24 hours without being connected to anything. (you could speed this process up some by putting a small load on it for 15 minutes). Check it with the meter again and see if the voltage isn't less than 13v, probably close to 12.6 or 12.7.

If you have just taken the battery off charge, you will measure a surface charge that is higher than the resting voltage. That is probably what you saw.

JNorton

My Feedback: (2)

Only reason I suggested a new meter was that the one he has is 20 years old and was used as a combination tachometer.
Because the range for a float charge is so small to check it an accurate meter is a needed.

jlkonn

My Feedback: (1)

Just back from Radio Shack.
I am the proud owner of a new digital multimeter.
John was right...the old meter is off.
The new meter reads 12.92 volts on the battery.
15.68 volts on the charger without it being connected to the battery.
Too high right?
What's that potiometer do on my charger in the picture?
JLK

JNorton

My Feedback: (2)

A tracking regulator supplies equal but opposite voltage. So if you needed +5 volts and - 5 volts with a common ground you can turn one knob and both sides of the power supply "track". Used mainly for OP amps.

JNorton

My Feedback: (2)

Turn it and see. Could set the current allowed or the voltage. Hook it up to a battery, hook the new meter across it and find out. Meters are really helpful.

John