Electrical – Is it OK to use an AC power supply with output voltage lower than specified on the original power supply to recharge a cordless tool

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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

Best Answer

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.