I need a new power supply for my Black and Decker GS500 3.6V DC cordless grass clipper. I have not been able to find a Black and Decker replacement power supply. The original power supply label shows output 8Vac 250 mA.

I have a power supply with the unit label showing output AC 6V 300mA. The no-load measured output is AC 7V.

My AC power supply voltage is 25% lower than the design input to the charging circuit. I am not clear as to how that impacts the output of the charging circuit and its ability to charge the 3.6V DC battery.

Is it OK to use the lower voltage power supply? Will I damage the clipper charging circuit or the power supply? Will the charging circuit try to draw more amps than the capacity of the power supply?

This question was asked in Electrical Engineering and put on hold as off-topic.

Edit 6/8/2019 The charging circuit consists of the power supply, the battery pack with 3 Ni Cd batteries, and a 4-inch long blue device (see image). The black lead from the power supply is connected to the positive terminal of the battery pack. The negative terminal of the battery pack is connected to one end of the blue device. The other end of the blue device is connected to the black/white-stripe lead from the power supply.

enter image description here

  • EE has an issue where if something plugs into the wall, they don't think it applies to them. If you ask a question that's in any way related to a car it gets shipped over to Mechanics.SE.
    – JPhi1618
    Commented Jun 6, 2019 at 18:31
  • EE did point you towards an answer that was more than relevant... check out the comments , see electronics.stackexchange.com/q/442122/152903
    – Solar Mike
    Commented Jun 6, 2019 at 18:36

2 Answers 2


That may work, or it may fry your power adapter and/or tool depending on the design of the internal charging circuit. Your original charger supplied 2 Watts (8V * 0.25A), and your proposed replacement is only capable of delivering 1.8 W (6V * 0.3A). If your charger pulls more current to compensate for the lower voltage (which is likely, assuming it works at all), then most likely your power supply will either blow a fuse or overheat and melt.

One solution that might work for you if you're willing to do a little electronics work is a 9VAC adapter (which are cheap and common as dirt), plus a power resistor to drop that down to 8V. Here's one that can output a full 1A for less than $5. Since you know your device will be drawing about 250mA, and you want to drop 1V across the resistor, we can select a value with Ohm's law: 1V = 0.25A * R; R = 4 Ohms. That resistor will dissipate (0.25A)^2 * 4Ohm = 0.25W, so anything 1/2W or bigger will be fine. This would be one example of a suitable resistor.

EDIT: Another option, albeit slightly expensive, is an 8/16/24V Doorbell Transformer. It has the right voltage and more than enough current.

  • i would actually expect the power supply to run cooler on 6v than 8v, given the lower difference on the way to 3.6v. You CANNOT use a resistor like you suggest to regulate voltage. The voltage dropped by the resistor changes with load (see ohm's law), and the load is not constant on a battery charger. When the charger stops charging, the amps go down, and the resistor drops less voltage, thus feeding the circuit more than 8v. a diode (or 2) would be a better option, as the voltage drop on it isn't linear like an ohmic load, it drops ~0.6-0.8v depending on load, a much smaller range...
    – dandavis
    Commented Jun 6, 2019 at 20:29
  • @dandavis, you're thinking DC; this is AC so a diode won't work. And yes, the load isn't totally linear, and the voltage will float higher under lower load conditions, but that's true of the original power supply as well -- a small transformer like that will have enough internal resistance that it's probably above nominal voltage with no load applied. And 9V isn't so high above the nominal 8V that I would worry that it's going to fry anything.
    – Nate S.
    Commented Jun 6, 2019 at 20:40
  • And voltage doesn't create heat, current does. If the charging circuit draws more current to maintain the same power, then heat will rise, not fall. It doesn't matter that it's "closer" to 3.6V -- that conversion is internal to the tool, which won't matter if the power supply melts.
    – Nate S.
    Commented Jun 6, 2019 at 20:43
  • have you ever seen a "constant power" power supply, anywhere, outside of a SMU? You can use two diode in parallel, one reversed, to drop AC. I doubt the internals are based on an SMPS, so with linear supplies, the voltage drop from in-out, multiplied by the current, determines heat; 6 is closer to 3 than 8; less heat...
    – dandavis
    Commented Jun 6, 2019 at 20:44
  • SMU? I assume you mean a switched mode power supply? Because it's extremely likely that that's what's inside the charging circuit anyway, which is why I think it may behave in a constant-power way. But yes, you're right that a suitable arrangement of diodes could be made to work here as well.
    – Nate S.
    Commented Jun 6, 2019 at 20:49

I assumed the charger was a typical DC wall wart, but it appears that for some reason, they decided to use a real AC power supply. This means the charging circuit is complex and designed into the tool. Because of this, the wrong voltage may not work at all, but I still don't think a lower voltage would cause damage.

enter image description here

click to embiggen

Original answer retained below...

I'm simplifying some things here for the sake of brevity, but it should help.

Many "cordless tools" use very basic charging circuits. If they have Lithium batteries, the charging is more complex, but then they have removable batteries with specialty chargers, or they have the complex charging circuits integrated into them. What you plug in to the tool is just a "dumb" power supply.

I'll assume that a B&D grass clipper has a basic non-lithium battery. The way to charge these is to attach it to a power supply that is greater than the voltage of the fully-charged battery. Within reason, more voltage will charge faster and lower voltage will be slower. Since you have a 3.6v battery, anything over about 4v will charge it. Using a voltage higher than the supplied adapter will probably lead to overheating and damaging the battery. Their adapter is sized to give you a safe compromise between charging speed and safety/battery life. If you've notice, some "high speed" chargers for drills have a fan to cool off the battery while charging.

These simple tools normally monitor the voltage of the battery and internally disconnect the charger once the voltage shows the battery is "full".

So, all that is to say that the lower voltage adapter will probably work fine but may charge slower. It could charge much slower. There is a chance it won't work at all, but I think there is very little chance anything would get damaged. If the tool draws more power than the adapter is capable of, the adapter may get hot, but they are close enough in this case that I would check it after a few minutes, but not worry too much.

  • I think this might be true if the original power supply was DC, but since it was AC, that means that the AC/DC conversion and battery charging circuitry is inside the tool in this case.
    – Nate S.
    Commented Jun 6, 2019 at 19:40
  • @NateStrickland, I assumed he was just calling it an AC power supply. I haven't seen a cordless tool that comes with an actual AC-AC adapter. They are almost always a DC "wall wart" type thing.
    – JPhi1618
    Commented Jun 6, 2019 at 19:42
  • Now that I look at the actual charger it does have a strange connector and appears to actually output AC. Strange - shouldn't have assumed.
    – JPhi1618
    Commented Jun 6, 2019 at 19:46
  • 1
    It's true an AC wall wart would be less common, but OP said that it was labeled for 8VAC output. That's probably also why they are having a hard time sourcing a proper replacement.
    – Nate S.
    Commented Jun 6, 2019 at 19:46
  • most likely, a dc input would work. I'll bet that the first stop after the input jack (after the fuse, maybe) is a bridge rectifier to turn the AC into DC for the charging circuit to use, and you can feed a bridge DC and get the same result (smoother actually) as feeding it AC.
    – dandavis
    Commented Jun 6, 2019 at 20:32

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