Did I kill my battery charger?

I have a 6 amp Schumacher battery charger I bought new in the late 1980’s. I’ve used it perhaps a total of 30-40 times over the years, and it’s never had any problems before.

The other day, I hooked up the charger to a battery, and pegged the meter. I quickly double checked the connections, and realized that, perhaps for the first time in my life, I had connected the cables the wrong way and didn’t notice before turning it on. When I connected them correctly, the meter read about 5 amps and slowly started to go downwards, which is normal behavior for this charger. I use an automatic setting, where the charger is supposed to gradually go to zero as the battery gets charged up.

When I checked the battery several hours later, the meter was showing 4 amps. I could hear a little bubbling in the battery, which is supposed to mean that the battery is fully charged. Normally, the charger would have gone to zero by then, since the battery was fully charged several months ago and hadn’t been used. I was only topping it up.

When I connected the charger to the same battery the next day, the meter showed 4 amps, but I could hear the battery bubbling, so I disconnected it.

My guess is that something in the charger is now damaged, but I’m not 100% sure. I think the charger is old enough that it doesn’t have the reverse-polarity protection of more recent models.

If you hear bubbling, likely you are overcharging the battery and may have damaged it. I would start by checking the acid levels and add water if needed. (Use distilled if local water is even a little hard. Note: spring water is NOT distilled and likely is worse for your battery than tap water.

After getting the fluid levels right I would bring it into an auto parts store and have them test it.

Good Luck

You probably popped a diode or a polarized capacitor in the charger. Time for a new charger…

As the charge on the battery increases, the internal resistance of the battery increases and the amount of current automatically decreases (I think Ohm’s law has something to say about this, although Congress may be in the process of repealing it). The fact that the internal resistance increases causes the charging rate from the battery charger to automatically taper off.

Since you “pegged” the meter, you may have just damaged the meter and it reads high. You might put some sort of load across the charger leads and check the voltage reading with a volt meter set on the DC range. On the other hand, since it is a 6 amp charger and you probably have gotten your money’s worth over the years, you may want to just buy a new charger.

I think Tardis’s comment about a bad diode may be correct. Getting a new charger would be a good thing since the newer ones do have better features.

I also very muched liked Triedaq’s comment about Ohm’s Law. There could be changes coming to this old and dependable law if Congress has its way with it. Way to go Triedaq.

I’m sorry, but this is wrong. The internal resistance of the battery will decrease slightly as the battery reaches full charge.

The reason that charging current drops off is still Ohm’s Law, but the resistor is in a different place and it doesn’t change. The battery charger supplies a particular voltage. For purposes of this example, let’s say it’s 14 volts. There is also some resistance in the charger (and the cables and the battery) Let’s say that resistance is 1.5 ohms. Let’s also assume that the battery voltage is 11 volts before charging.

Connect the battery, and the charger is 14 volts, the battery is 11 volts, so there is a 3 volt drop across the 1.5 ohm resistance. This causes 6 amps of current flow into the battery. After charging a while, the battery voltage is now 13 volts. The drop across the (still) 1.5 ohm resistor is now 1 volt, so the current is now 0.667 amps, just over 1/10 of what it started at. Eventually, the battery reaches 14 volts and the current becomes zero.

Of course, it is a little more complicated than that, and the current never really falls all the way to zero, but you get the picture.

If the charger produces 14 volts and the battery can only muster 11 volts, the difference in potential is 3 volts. When the battery’s output becomes 13 volts, the difference in potential is 1 volt. If the charger’s voltage remains constant at 14 volts. then there is more opposing force when the battery is at 13 volts than when it is at 11 volts. Since
I(current) = E(voltage)/R(resistance) and the voltage from the charger remains constant, then I = 14/R and R must increase for I to decrease. This opposing force of the battery is the resistance encountered and the lower the opposing force, the higher the current flow. It seems to me that to the charger, the resistance to current flow is increasing as the battery reaches full charge.

What kind of a battery are you charging?? Perhaps the battery has a shorted cell and is now a 10V battery…The charger will continue to charge at a high rate, overcharging the 5 good cells. That may be what you hear bubbling…Those old Schumacher chargers were designed with reverse connections taken into consideration. They are pretty tough…

I’m sorry, but you are quite wrong here. Of course, it’s not like I’ve designed battery chargers as part of my job, oh wait, I have…

These are pretty tough little units, I have two of them and I prefer them over the newer “smart” chargers. They don’t get much simpler than these chargers. Do you have a DVM? With the load (battery) connected, measure the DC voltage and check the ripple on the AC scale to see if you toasted one of the rectifying diodes.

Depending on the type and size of battery you were charging, you may want to see if it was the battery that got damaged. Measure the unloaded, terminal voltage of the battery and post back. Also, details on the type and capacity of the battery in question…

then there is more opposing force when the battery is at 13 volts than when it is at 11 volts.

Keep in mind, you’re charging a capacitor, not a pure resistance. Tardis’ simple explanation does a good job of conveying the essence of it…

If you take a discharged capacitor and put an ohmmeter across the the terminals of the capacitor, the ohmmeter will begin swinging up indicating a higher resistance as the charge on the capacitor increases. The 1.5 volt battery of the ohmmeter is charging the capacitor and as the capacitor becomes charged, less current flows from the battery of the ohm meter and the meter indicates a higher resistance. In the case of the storage battery, I probably used the wrong term “internal resistance”. As the charge increases on the battery and its voltage increases, the difference in potential between the charger at 14 volts and the voltage of the battery decreases. so less current flows.

I think that the problem here is that you are looking at a battery as if it were a resistor. It is not. A battery is a voltage source is series (not parallel) with a very small resistance. Perhaps you need to read up on this and do some research. Batteries and capacitors do not work the way you think they do. You can’t measure the internal resistance of a battery or a capacitor with an ohm meter, because ohm meters don’t work correctly when a voltage source is involved. To measure the internal resistance of a battery, you measure the battery voltage with no load applied. Then you apply a known current load and measure the voltage while the battery is driving that load. The difference between the no-load voltage and the loaded voltage, along with the amount of current the load was drawing allows you to calculate the internal resistance.

I just double checked the battery with an analog voltmeter with a 10v and a 50v range and nothing in between. (So I had to use the 50v scale.) The battery showed only 11v, and a small charged UPS battery shows 14v, so your bad battery theory must be right. It’s good I charged up the UPS battery and checked it today. (Earlier it was in an unknown state of charge, and the Schumacher meter read very high when I connected it.) For final confirmation today, I also connected the charger to that UPS battery, and the meter read zero.

So it looks like the charger is fine, and I need to lug that spare battery to the recycler. Could that brief reversed connection have killed a cell in the battery?

I guess that almost anything is possible, but I wouldn’t expect it to. How old was the battery? Maybe it went bad on its own.

Two possibilities come to mind; one, the battery sulfated during storage and the reverse connection was a coincidence or B. the reverse connection warped one of the cell plates and it is now shorted. The latter is why I asked about the battery size/capacity. Told you they were tough little units!

Yes, you are correct. I got out my physics book last night and worked it through. The internal resistance of a battery does decrease as it approaches full charge and in applying Ohm’s law in this case one has to look at the differences in the voltage between the charger and the battery. As this voltage difference decreases, as you pointed out, the current decreases. The resistance does change in the direction you stated, but this is negligible in the effect on the current. The perfect battery, then, would have zero resistance, and produce an infinite amount of current. I guess it has been too many years since I studied this in physics.

The battery is a maintenance free 12v car battery with about 500 CCA. It was used for 4-5 years in Minnesota in a car always parked outdoors, so it got a workout in the winter. The last time I topped up the battery (with the same charger) was 3-4 months ago, when the charger’s meter went from a few amps to zero in maybe an hour or two, maybe even less than an hour.

I don’t have a digital voltmeter. Since the charger now seems OK (see my previous post), I’m not in a hurry to replace the so-so analog voltmeter I’ve been using for many years.

What should the AC ripple be if the charger is working OK?