Need Legal Advice On Ohm's Law

[Carl Eyman]

I have a HF 14.4 V drill almost 4 years old. Last week I could no longer charge the batteries. Investigation turned up that the power cube (transformer/rectifier) was only putting out 10.5 volts. Since radio shack didn't carry 18v 500 ma cubes I found one on e-bay for $!! including freight. Troube is it is 1.1 amp. Now here's where I need Ohm's law advice. My reasoning was that if it purt out 18v to the same batteries that the old one did, the current would be the same, right? What happened was the led came on indicating charging as it should but went out again in a short time (maybe 20 minutes) The batteries are both seemingly fully charged. The voltage is up and the drill works well. The charger does get warm. The no-load voltage on the cube is about 21 or 22v, though marked 18.

I suppose I'll fry my batteries, but they must be near the end of their natural lives anyway. Anybody got any words of advice?

[Chris Fite]

The important value is the voltage. It needs to be right to avoid damage to the battery. The amperage is simply the total capacity of the charger. If it is too low, it cannot keep up with the demand.

[Effie Lever]

Ohm’s law may not work here because your batteries present a more complex resistance to the charger then a simple resistor. In general, a higher voltage charger will try to force higher charging current through the batteries and like you suspect, it could fry them.<O:p</O:p

Effie.

[Randy Meijer]

Chargers are designed to charge at a specific voltage "AND" rate(amperage). To much or too little and you have potential problems. In general your best bet is to buy an identical replacement from the original manufacturer. If you're satisfied with the drill, buying a new one may be the most economical alternative. I have found that for modestly priced drills, buying a pair of replacement batteries or getting the batteries rebuilt, costs almost as much as buying a complete new drill!!!

[Lee Schierer]

I think you need to get the same amperage and voltage from your charging device. Your old charger produced .009 watts of power. (P=IE) Your new charger is producing .0198 watts of power more than double. Unless your batteries can disipate this much power (heat) you could have some well done batteries.

With that said, if you can't find any other charging source, then the batteries aren't much use dead. I would go with the 1.1 mA charger and not get upset if the batteries fail shortly. You will have gotten more life out of them than you would have without the charger. If they last a couple of years you are well ahead of the replacement cost.

[Carl Eyman]

Lee: Your advice is right on, I think. This wasn't posted as much as a need for advice as a conundrum concerning two power sources both nominally 18v producing much different results. The drill, 2 battteries, and charger only cost me $65 four years ago. If this $10 charger extends the life of the system for a few months, It will be worth it.

[Jon Toebbe]

As others have noted: you're more than doubling the applied power to the battery. This is not a good thing. Best case scenario: dramatically shorter battery life. Worst case scenario: nickel-cadmium shrapnel. These things can explode. A much more likely outcome is that you'll start an electrical fire (still a pretty bad case).

Be safe.

[Randy Meijer]

.....Your old charger produced .009 watts of power. (P=IE) Your new charger is producing .0198 watts of power more than double. Unless your batteries can disipate this much power (heat) you could have some well done batteries..... I would go with the 1.1 mA charger.....If they last a couple of years you are well ahead of the replacement cost.

1. !8 volts @ 500 ma is 9 watts, not 0.009 watts.
2. !8 volts @ 1.1 amp is 19.8 watts, not 0.0198 watts.
3. The new charger is a 1.1 amp charger, not a 1.1 ma charger.
4. If you consistantly overcharge your batteries, they won't last for 2 more years and you could have other serious ramifications as mentioned above.

Having said that, one must realize that there are two parts to most charging system. The first is the "cube" that plugs into the wall and supplies the electricity. The second is the device that monitors/regulates the charging, contains the indicator lights and acts as the charging stand. Regulators can be very simple or very complicated. Some simply pass through the "cube" voltage and display a light indicating that charging is occurring. Others regulate and sense charging voltage and amperage and adjust(reduce) the charging current as the batteries come up to full charge. Some even have a maintainance mode where the batteries receive a trickle charge if left in the charger continuously. That being the case, you may or may mot be damaging your batteries. If there is a regulator in your stand and it is still functioning properly, then you should not have any problems at all. On the other hand, if you have no regulator, then your batteries will not be lasting too much longer. If the batteries get warm when charging that is OK. If they are getting hot, you may have a problem. When you measured the 10.5 v supply voltage was that the voltage out of the cube or out of the "stand?"

[Bill Fields]

HF stuff varies in quality. I have many 18v batteries and chargers and they seem to work OK.

Waiting for them to offer an 18v jigsaw compatible w/ their battery charging suite.

HOWEVER--this is an opinion that some Creekers will dispute--buy from low-end, cut-rate HF, and you take your chances.

Example: Could not resist a $165 price for thier copy of a 12" compound sliding dual miter saw, for $165. Looks OK--parts broken upon reciept.

Everthing seem to be with tolerances

Bill Fields

[John Hart]

Carl,

I think you should be looking at Capacitive Reactance rather than Resistance in this case and note that the instantaneous current draw at initial "On" time is at maximum. As the charge in the battery is increased, the current draw goes down. This is a little more complex than straight Ohm's Law as Effie pointed out, and also, as Randy said, not necessarily a function of your power source, but more a function of the base's ability to regulate that charge. I think the fact that your charger got hot, indicates that it is delivering a lot of current and MAYBE exceeding it's ability to properly regulate the current. The fact that you acheived a "quick charge" indicates that the batteries were able to handle the extra current that was being delivered. The possibility of eventual destruction of the batteries due to these conditions really lies in the specifications of the charger, the batteries, and their tolerances. If you were worried about it, you could always cut the delivered amperage to the charger in half (to 500mA) by placing a high wattage resistor across the power coming out of your transformer cube. This would require more calculation and, again, the specifications of the charger would need to be known.

[Lee Schierer]

1. !8 volts @ 500 ma is 9 watts, not 0.009 watts.
2. !8 volts @ 1.1 amp is 19.8 watts, not 0.0198 watts.
3. The new charger is a 1.1 amp charger, not a 1.1 ma charger.



Yep, I screwed up the decimal point, should have used my slide rule. :D