# How long will the batteries last when hooked up to this light?

The light being used is here: https://www.amazon.com/Germicidal-254nm-U-shape-150mm-Purification/dp/B01CV2W12E

Lamp Current: 5.0mA Wattage: 8 watts

It will be hooked up to 2 AA batteries in series The batteries are 1.5 V and 3000 mAh

How long will the batteries last?

My first assumption is to do this: 𝐵𝑎𝑡𝑡𝑒𝑟𝑦 𝐿𝑖𝑓𝑒= 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦/(𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝐷𝑟𝑎𝑤)=(3000 𝑚𝐴ℎ)/(5 𝑚𝐴)=600 ℎ𝑟𝑠=36,000 𝑚𝑖𝑛

But that doesn't use the wattage? Is this correct?

• The amazon page says that they have a Starting Voltage of 450 V (they are discharge lamps), and there is an inverter between them and your batteries. Amazon description does not say what is the input voltage of the inverter and if it is compatible with your 2 AA batteries: check this, in a comment it is said 12V input, but I couldn0t see it. After triggering the lamp (see initial discharging voltage of 900 V), the inverter stabilizes between 4 and 9 mA feeding the lamp, and this shall be highlighted in your question and used as ref. – andrea Sep 24 '17 at 8:20

First of all, you're ignoring that the inverter requires 12 volts. Two AA batteries in series is only 3 volts.

Second, the 5mA is the lamp current at high voltage. At 12 volts, the current will be higher. Using the 8 watts, at 12 volts, it would be at least 0.667 amps (probably a little higher due to inefficiencies in the inverter). Let's assume 1 amp.

If you use 8 AA in series to get 12 volts, at 3000mAh, you're looking at 3 hours, probably less as the voltage output drops as the batteries discharge, its unknown exactly how low it can go before the inverter shuts down.

Your formula for estimating battery life is corrected. To apply it correctly you need to use the battery discharge current not the lamp current.

Between the batteries and the lamps of your product is a DC-DC converter that steps up the voltage from 3V to Vlamp. 8W is likely the lamp load power.

Assume the DC-DC converter is 100% efficient. The power drawn from the batteries will also be 8W.

$$P = IV$$

$$I_{batt} = \dfrac{8}{3} = 2.67 \text{A}$$

Using a discharge current of 2.67A yeilds a battery life of a little over an hour.