I bought myself one of those mister fans from Home Depot that runs on the batteries for their drills. I thought it would be nice to be able to hook up a 12 volt battery or plug in power supply to also provide power for it, so I I soldered a boost converter with a barrel jack to the terminals where the batteries connect and set the boost converter to 18 volts. This works great, the fan doesn't know the difference. I can plug it in or I can use the battery. The issue I ran into is that when I leave an 18 volt battery plugged into the fan, when I come back it is completely drained.

Partially I'm curious why this is happening. I haven't looked at the IC driving the boost converter, but I would assume that it wouldn't be powered up unless there was power on the input side.

I was wondering if anyone had suggestions on additions I could add to the circuit to keep it from draining backwards into the boost converter (I'm assuming that's what's happening, as it did not drain the battery before I added the modification)

Here is a link to the same type of boost converter I am using.
10 Pieces DC-DC Boost Converter 2A Adjustable Module Voltage Regulator Board Input Voltage 2 Volt-24 Volt to 5 Volt-28 Volt Output Voltage (Without USB Connector) https://www.amazon.com/dp/B089Y7NDCR/ref=cm_sw_r_apan_i_Q2VKMBCHQ829G8FF7X97

 

It sounds like you have two power sources connected to the fan. I'm not clear on where the switches, if any, are located. You may want to isolate the two power sources from each other with diodes that allow current in one direction only. In this way the fan will take power from the source with the higher voltage. Something like this:



I'm showing Schottky diodes because they have a lower voltage drop than normal diodes.

 

okay, that was my plan. I am just a little surprised that it could flow backwards considering every diagram I've seen of a boost converter has a diode and a capacitor between the + and - lines as seen below. I suppose the capacitor will have some leakage current, but to drain a 2 ah battery in a few days seemed like a crazy amount of leakage for a capacitor. Any thoughts on where the current could be going?

 

show us how you connected the boost converter to the fan.

Bob

 

The boost converter will be reading the output voltage so it can regulate the amount of boost. So, there's some drain through the voltage divider.
https://www.olimex.com/Products/Breadboarding/BB-PWR-3608/resources/MT3608.pdf
https://www.electrothinks.com/2021/08/MT3608-2a-dc-to-dc-step-up-power-boost-converter-module.html
And a cheap Schottky diode might have more leakage in reverse bias than a silicon rectifier.

 

okay, that was my plan. I am just a little surprised that it could flow backwards considering every diagram I've seen of a boost converter has a diode and a capacitor between the + and - lines as seen below. I suppose the capacitor will have some leakage current, but to drain a 2 ah battery in a few days seemed like a crazy amount of leakage for a capacitor. Any thoughts on where the current could be going?

View attachment 268375

A boost converter will also have a feedback loop resistors in parallel with the capacitor which will act as a load

 

Attached above is the picture of how I had it wired. I measured 260 uA of current flowing into the boost converter. I guess that could be the voltage divider. It seems crazy that 260 uA would deplete that battery in 1-3 weeks. My math says it should be more like 8 months. Of course the batteries aren't starting fresh.

I tried putting the diode in series with the +vout but the boost boards appeared to be morally opposed to such a setup I had 3 burn out their ICs in quick succession after adding the diode. I don't know if I can really blame this on the diode but I never had the issue before adding the diode. I may just accept the lower fan speeds from 12v.