jlkonn

My Feedback: (1)

John,
Man!
My head is starting to hurt already![]
I'm going to have to hook the meter up both ways to measure volts and then current!
[X(]
You are making me work too hard!
Gotta go take the daughter's two golden for a walk.
I will keep you posted this afternoon what happens with the pot.
Thanks!
JLK

toprudder

I suspect the pot is to adjust the voltage. Sometimes a fixed voltage regulator is made into a variable voltage regulator by putting a pot in the feedback circuit. If I remember the pinout of the 7805, the left leg is input, the middle leg is reference (usually 0v or ground) and the right leg is the output. If the pot is used to adjust the voltage, the middle leg of the 7805 will be tied to the wiper of the pot, one leg of the pot will be tied to 0v, and the other leg of the pot will be tied to the output of the 7805 (probably through a resistor).

Is this charger intended to be used only for lead-acid batteries? The reason I ask is that if it is intended for Nicads, the regulator is probably wired to act as a constant-current source. Not good for lead-acid batteries. It would be wired differently if that were the case.

I have a regulated, variable power supply that I built when I first got into electronics. I set it to 13.6v and connect it to my 12v battery to charge it. The regulator that I used (317) has current limiting built into it, something like 1.5 amps. If the battery is really low, it will start at 1.5 amps but by the time the battery is charged the current will be less than 100ma.

John, you are right, the range is small and a good meter is needed. I tend to be too frugal sometimes. It was hard for me to throw away an old analog meter that was 30 years old, even though a few ranges did not work any more. []

Bob R.

jlkonn

My Feedback: (1)

Bob,
The face of the charger states it is for lead acid and sealed 12 volt batteries.
It was made by L R Taylor, a company that made a lot of RC charging equipment.
I too hate to throw things away.
That's why I'm messing with this charger even though I just bought a maintainer that works great!
I have my Dad's old RCA volt/ohm meter somewhere.
It has a huge analog display.
I need to find that thing.
I think he built it back in the '60s.
JLK

JNorton

My Feedback: (2)

It's not that bad. I would suspect as does TopRudder that it is a voltage adjustment. Not necessarily. You want to measure the chargers output on the battery. This charger could have been designed to charge the battery and then switch to a float charge. You'll need to read it with the battery connected.

John

AS-EE

Here is two of the three types of meters you asked for Jnorton:

0-1.5 Ammeter



0-15 Voltmeter



Meter is from radioshack at:

[link=http://www.radioshack.com/product.asp?catalog_name=CTLG&category_name=CTLG_011_008_002_000&product_id=22-410]www.radioshack.com/product.asp?catalog_name=CTLG&category_name=CTLG_011_008_002_000&product_id=22-410[/link]

AS-EE

If you read the voltage at the output of the charger without the battery connected then no wonder it read 15.68 volts. Digital Voltmeters have very high imepedence when measuring voltage and therefore you will not pull a lot of current to sufficiently drop voltage across the current limiting elements inside the charger. It's not until you place that battery on the charger and then measure the voltage.

AS-EE

Here is a way to charge your small lead acid batteries on your way to the field or to keep it maintained:



I don't recommend going any higher than about 950mA since many small gell cells can only handle a maximum initital current charge of 1 amp. Oh and I did not get very specific on the diode part about the discharge. When I said discharged I mean the gell cell having greater voltage potential if the car battery were to go to a dead voltage and thus the diode will prevent the gell cell from discharging into the car battery.

jlkonn

My Feedback: (1)

John,
Yes, the charger was designed to reduce the current.
It has an ammeter that reads from -0- to 500ma.
As the condition of the battery approached full charge the ma's would drop to nearly -0-.
JLK

JNorton

My Feedback: (2)

Try this. Meter is 0 to 1 mA. Internal resistance is 50 ohms to make it simple. For a full scale deflection then you must have 50 mV. Current shunts are across the meter and must carry the full load. You must ask yourself if your answer is resonable. Your 1.5 amp shunt has a resistance of 11.33 ohms. Mine is a 149 mA shunt has a resistance of .3355 ohms. The 1350 mA shunt (1 mA meter + 149ma shunt) has a resistance of .0370 ohms. This reflects the 50 mV it takes to drive the 1 mA meter full scale.

Also shown is the switching arrangment to allow one meter to be used as a 0-15 volt, a 0-150mA current and a 0-1.5 A meter.

Ain't this fun.
John

JNorton

My Feedback: (2)

Super. Let us know how it works.

jlkonn

My Feedback: (1)

OK,
The pot adjusts current.
I now believe the charger is working correctly although out of calibration.
What I am going to do is plug my 7A field battery into the maintainer and be sure it is "topped off".
Then I am going to connect the L R Taylor and adjust the pot until the ammeter on it is deep into the lower side of the green.
I am going to then use it to charge the field battery the next time it needs it and see how it goes.
JLK

JNorton

My Feedback: (2)

Kool. Keep in mind to check the voltage across the battery after it says it is charged. Hopefully it will drop to 13.4 or less then everything would be golden as you could leave it connected to the battery.

John

JNorton

My Feedback: (2)

Ok let's review this circuit.
Incoming voltage variations 13.8 to 14.4
Battery charge state 11.9 fully discharged - 14.4 charging.
Vf forward voltage drop of the 1N5408 = 1.2 volts @ 3 amps
If = 3 amps
Vrm = 1000 volts

You should always do an analysis for all variations to make sure the circuit will work under all conditions.

So if I have 13.8 volt input and the battery is 11.9 volts there is a 1.9 volt difference
14.4 - 11.9 = 2.5 volt difference

The diode will always have a voltage drop from .7 to 1.2 voltage depending upon the amperage drawn.

So if we take the worse case 13.8 volt input - and a flat battery of 11.9 volts less .7volts from the diode drop there will be 1.2 volts across the dropping resistor which will limit the current available to the battery to I = E/R 1.2/2= 600 mA

At the high 2.5 - .7 = 1.8 volt or 900 mA. At 900 mA the diode drop might be higher than .7 but I don't have the graphs in front of me.

So the design should work fine. There is the however the variation in charging current that under some applications could bite you.

Good job,
John

JNorton

My Feedback: (2)

Ah heck I'm tired. I forgot to check voltage at the battery! I should not have used a dead battery voltage 11.9 volts to calculate the drop because you need at least 13.5 volts for the battery to charge. At 13.8 volts which is the lowest I've seen a car alternator if you subtract the diode drop the battery won't charge or it will be right on the edge. If you changed the diode to a Schottky type which has a .3 volt drop you'd be okay. I'd eliminate the isolating diode because if the car battery discharges where you would need it to function you've got bigger problems! Like how to get home.

John
EDIT-> Repeat after me a Lp battery is not a linear device!! It won't accept a charge unless the voltage is higher than 13.5 better at 13.8 and should be above 14.2 volts.

JNorton

My Feedback: (2)

AS-EE,
In the light of a new morning it occurred to me that perhaps you'll take the attempted analysis of your circuit as a personal attack on you. I assure you it was not. I had envisioned us setting around a table critiquing circuits. My buddy Jimmy's visitation is tonight with the funeral Saturday. I had so much fun designing the metering circuit that when I couldn't sleep I looked at yours. A controls engineer today spends most of his time at a PC designing logic to be downloaded into a PLC, not with math designing descrete components. It was fun and showed just how rusty I am.
John

AS-EE

In class I am in the process of designing a frequency counter using MOD 16 counters and a reference oscillator with a 555 timer.

JNorton

My Feedback: (2)

I'm sure you'll have fun with it. If I ever win the lottery and no longer need to work I'm going to turn into a professional student. Being in school is a great time in your life. Enjoy it.
John

speedster 1919

My Feedback: (3)

If you two boys are done JOISTING now. I have one other ? I am using the converter now in cig plug. With the van shut off the charge is stopped. If I stop for a while and make a customer call ,Do I have to remove the Hangar 9 charger leads hooked to 12v gel cell. Will it discharge the battery hooked up and no charging. I don't know if a diode is in the wall wart or 12v inverter. My cig lighter plug is only active with key on run or acessery.

JNorton

My Feedback: (2)

Use a DVM meter and find out. Put it on the current selection move the positive lead to current and coonect the battery using the meter like you would a piece of wire to the charger.

JNorton

My Feedback: (2)

You have another possible problem if you think about it. Modern chargers are very sophisticated. Turning the power on and off while they are in the middle of a cycle is a sure way to get stuff screwed up as is taking the load on and off before the charging cycle is finished. Think about it.

By the way the word is joust.

John

speedster 1919

My Feedback: (3)

Yes your right , I didn't spell check. The I is next to the U on the keyboard..............

JNorton

My Feedback: (2)

If you don't have DVM to check for leakage from you charger to the battery pack when the charger is unplugged use a #44 or a #46 pilot bulb from Radio Shack. Use it in place of the positive lead to the battery. If you have any leakage current present that could drain the battery it will glow.

EDIT-> What you are looking for is not a white light but just the hint that the filament of the bulb is glowing orange.