The Harbor Freight Tools 42294 is a cheap battery float charger. Plug it in and clip to your battery, and you're done. Except your battery is done too. Most of these chargers are set to around 14.4V which is a bit too high for float usage - it's great only for cyclic charging. At 14.4V it will ensure rapid charge and also loss of water from electrolyte if kept too long. This is problematic for AGM batteries.

Note that there is no current limiting on this charger other than the limit of the transformer and the 7805 regulator - and the robustness of the 7805s this isn't that bad. The 7805 will take the brunt of the short, and will get hot. Likely you will get most of the transformer's current limit as the upper bound and not hit the 7805 limit unless you're in a hot environment or hacked it as below.

Schematic of one incarnation of the tool.

I first saw the issue when I saw the web page at Desert Home. The concepts are indeed correct there, and like many other sites about this charger, they are mostly correct. However this is somewhat of a rebuttal to the design: Yes, HFT did make a mistake and designed the charger to leave the LED on even if the power goes out. I got a battery killed in this method and am kind of angry about it. However that was not the main purpose for R3-R5 and S9013 - these were put in to make sure you don't clip the battery on backwards! The LED will go out and the charger will be shut off if you clip the battery backwards. Nobody wants to come back with a battery reverse charged because thy hooked up the battery wrong. Unfortunately the power outage (or when the plug gets kicked out of its socket) situation will cause the battery to discharge through LED1, eventually leaving a ... dead battery to sulfate to death.

How do we fix this? Adding a diode will effectively kill the reverse protection detection - the LED will no longer turn off if the battery is hooked up wrong. Also keep in mind that the diode will drop about 0.7V, which may be just right for float charging anyway. If you never hook up backwards this might be the easiest solution; if reverse hookup is a problem...this would require some more surgery, which may be invasive. Haven't thought of an easy solution, yet.

Other questions?

1. How do we hack/modify this for 13.8V?

13.8V is the magic number for float charging (phase 3) of a nominal 12V lead acid battery where it's high enough to not let the plates sulfate and not cause water loss due to electrolysis. It also limits positive plate corrosion a bit. However Harbor Freight seems to have set it for 14.4V which is actually the best "cyclic" or "fast charge" voltage. Note that this is temperature dependent, but we'll have to ignore this as this device is too simple to incorporate it (or is it? I think we could add a 10KΩ thermistor in parallel to one of the resistors and we could get some limited temperature dependent control, which would be great in a hot and cold garage! However the thermistor would need to be place close to the battery and far from the hot, hot 7805).

The correct solution is to change R1/VR1 and/or R2/R2A/R2B. The suspicion is that VR1 was supposed to be a variable resistor to specifically allow tuning of the output voltage, but it was made to a regular resistor to cheapen the device. Increasing the resistance here will decrease the voltage output. Similar to a LM317 that has a characteristic voltage of 1.25V, the 7805 will try to maintain a voltage of 5V between OUT and GND. What you want to do is to tweak the resistance of R1 such that we get the right voltage across R2. Note that there will be current that needs to flow out of the GND pin so that needs to be accounted for so we need to add that in.

NOTE AND WARNING: A very likely issue with these resistor calculations is that using silver tolerance banded resistors will give up to a 10% error on the output voltage. Gold tolerance banded resistors will give up to a 5% error. Coupled with the inherent though low error in the 7805, you will likely have to tweak resistances a bit to get to the exact voltage you want. You may have to go with the more expensive metal film 1% resistors if you don't want to mess with a pot.

As designed, R2 is three parallel resistors: ( 6.8K || 1K || 1K ) = 466Ω. R1 is two resistors in series: 330+33=363 Ω. So the output will be whatever voltage that would satisfy the 363Ω resistor being equal to 5 volts - so that means 5V/363Ω amps goes through the R1 resistor, and that 5V/363Ω amps goes through the 466Ω resistor. We also need to add in current that comes from the GND pin of the 7805, about 5-8 mA. So that means that ((5/363 A + 6.5mA) * 466Ω) = 9.45 volts is dropped across R2, and the total voltage must be 9.45 + 5 = 14.45 volts, which is exactly what is seen from my unit.
Now if we want 13.8 volts we can work backwards. Subtract out the 5 volts we can't do anything with and get 8.8 volts that needs to be dropped by R2. Now we have a math solve for X problem here, we need to find a resistance for R1 that solves: 8.8 = 466 Ω * (5V / R1 + 6.5mA) - please mind units before messing up the calculation.
Rearranging the formula, we get R1=5/(8.8/466-6.5mA) and R1=404Ω. Since there's already a 330Ω resistor there we just need to have 404-330 = 74Ω. The next closest standard value is 75Ω, so all we need to do is replace the 33Ω resistor with a 75Ω, and we get our 13.8 volts.

2. What about for 6V batteries?

For a 6 volt supply, there are a few problems: first off, the 7805 will need a heat sink that wasn't needed when charging 12V batteries, so keep in mind. It will get about 3-4 watts hot and you don't want the 7805 to fry. Else the same calculation can be done: First off we need 6.9 volts to properly float charge a 6V lead acid battery. Now subtract the 5V of the 7805 and get 1.9 volts to drop across the bottom. Same sort of solving: R1=5V/(1.9V/466Ω-6.5mA) ... equals ...

-2KΩ...

Oh no! We calculated a negative resistance. Meaning it's impossible to get enough current flowing through R1 to get a 1.9V drop across R2. The quiescent current by itself will give 3 volts. So we need to modify R2 as well. To make the math easier, say we remove all three resistors R2/R2A/R2B and replace with a single 100Ω resistor. Then we can compute R2 = 5/(1.9/100Ω-0.0065) = 400Ω which should work. So the total change would be remove R2 R2A R2B, insert 100Ω resistor for R2. remove VR1, insert 70 ohm resistor (or replace both R1 and VR1 with two 200 ohm resistors) and we're done.

Keep in mind that the reverse protection circuit will not work at 6 volts, and you likely will need to reduce R5/R5A to get full current output.

I will need to get pictures of this thing once I crack this thing open... For now just refer to the Desert Home link above.