# Simulate a battery with an AC wall transformer [closed]

Entry level EE has battery question. I use a heating element (vaporizer) that takes a 1.5V AA rechargeable 2000 mAh battery. It comes with an AC wall adapter that is OUTPUT 12V DC @2000mA so 24 WATTs right? That passes through some circuitry that has a variable knob so you can control the heat (which I assume current) and simulate the battery. It's much more powerful than the actual batteries are though. I'm sick of using rechargeables.

Anyway, the power adapter burnt out (just the circuit side, not the transformer) and I don't want to spend 80 bucks on another one. I'd like to simply plug another wall adapter directly into the device. I don't need the variable setting, whatever power it comes out as (as long as it's comparable to the battery it should be fine).

Can I use a wall transformer with 1.5V at ~16 AMPS and get the same result? Do they make such adapters at like Radio Shack? If I use higher voltage will the device still work as long as we don't exceed the 24 WATTS that the original power adapter was using as input? It is a small incandescent light and a heating element, nothing more.

Opinions?

• 16 A at 1.5 V won't come standard, but you can wind one yourself if you like if you have access to a transformer of suitable VA rating. – winny Aug 14 '18 at 6:57

## 1 Answer

It is necessary to match the voltage output of the existing device. With more information on exactly what the load is (Resistor value/effective ratings, same for bulb) it will be easier in any case to make sure that your replacement doesn't have the capability to cook the vaporizer. That said, you are correct that you should not exceed the intended wattage, and that is likely enough information. If you can use a multimeter to measure the open circuit voltage of the fully charged battery, if that is >12V, it would help to know what it is.

Depending on the load, the output current will likely be determined by Ohm's Law:

$$I = E / R$$

Where $I$ is current, in (A)mps, $E$ is voltage, in (V)olts, and $R$ is resistance, in Ohms($\Omega$)

We also have Watt's Law for calculating Power($P$), in (W)atts:

$$P = I \times E$$

to calculate power use at a given voltage and current. We can combine the two formulas if you wish to calculate power directly from voltage and resistance:

$$P=E^2/R$$

The resistance of your load (assuming it is purely resistive and disregarding the resistance of an incandescent light bulb increasing as it comes on) stays the same, and to control the power of the load, you need to be able to vary the voltage across it, from the maximum voltage down to whatever the minimum voltage is that it's worth turning it on at. If the load draws 2A at 12V, it likely equates to 6 ohms, although there will be a lower resistance before the light bulb turns on, and the resistance of the resistor will likely increase at least marginally as the element heats up.

As I read this, your adapter is probably a 0-12V DC 2A power supply. It's not clear whether it is current or voltage controlled. Pictures of the board and product info/labels would be useful. Your options likely amount to:

Finding a replacement product (this will necessitate figuring out what exactly you have)

Building a replacement (Same as above)

Repairing the one you have (may be possible without even knowing what it is or does.