I’m working on a project, building a dummy battery for a piece of photography equipment, so I can power that device from an different, external battery.
The external battery I want to use to power the device, is Vmount battery that outputs 14.5v 95w
The device I want to power requires 11.1v 43.2 watts.
I’m just getting into hobby electronics and DIY electronics, so I’m already in over my head, and would appreciate some advice.
For some more information, I’m not sure what amperage the external battery outputs, or what amperage the target device requires. I suppose the best way to figure that out would be with a multimeter? test the external battery I want to use, and the removeable battery that came with the device I want to power?
And then when it comes to the power, I know the voltage needs to be close to 11.1v, as to not damage the device. But what would happen if the wattage and amperage was off?
In terms of the components I need, I know I need some kind of voltage regulator that can output 11.1v form my 14.8V coming in. But the only ones I can find that do that are all 20w. Outside of that, I don’t know what other components I would need. Do I need other components to control/regulate the wattage and amperage?
As for the battery that came with the device I want to power, it only has two contact pads. Negative and positive. So at least that makes my life easy.
Would love some help, even just understanding more. If this sounds like a dangerous, silly project for a beginner, let me know and I’ll abandon ship.
Welcome to the forum
Batteries can be tricky as there are a few factors to consider, we’ll try and get a few more specs from you and do the math to check that everything seems to be within safe limits and if it seems like a bad idea we should see that the numbers don’t line up.
To make sure we’re all on the same page (and for the benefit of anyone else who reads this thread later) I’m going to assume you’re familiar with Ohm’s Law, which we cover in a tutorial on our website.
Could you please send us a photo of any of the stickers on the batteries you have, or any spec sheets for them? We need to determine what the continuous current the battery can supply is, sometimes called a discharge rating.
Can you also double-check for me that the Watt ratings you have quoted are not Watt-Hour ratings? Watts are a measure of how much power we can draw from the battery, so we can calculate the continuous current rating from that. Watt-Hours are a measure of battery capacity, and tell us how much energy is in the battery in total, but not how quickly we can draw it out, so mixing up these two will give us big problems.
Finally, your load (the camera) will draw whatever current it needs to operate, you don’t need to regulate it, just ensure you have the headroom for whatever it will try and draw and that the voltage remains regulated to the correct level.
11.1V @ 43.2W is 3.9A. Your battery is capable of 95W and you require 43.2W so that is easily covered. What you need now is a buck converter to change the 14.5V down to 11.1V @ something like 5A would cover it fairly easily.
Be aware that although a buck converter may be rated at 5A READ THE FINE PRINT. At the relative high current of 4A a heatsink may be required. As a start search “buck converter” on Core web site.
Please take note of what Trent has to say regarding Watt/hrs. As he says this is very different from just Watts. I was taking what you stated as gospel and only considered Watts. This point only accentuates the need for ACCURATE reporting. It is the only way anyone is going to get any meaningful assistance.
Core have a device with a short lead time. SKU COM-18752. This is a buck boost converter with 8 - 36V input and 12V output with a max current of 6A. Considering 11.1V is LiPo’s version of a 12V battery and full charge voltage could be 12.6V this device might do. It is complete with heat sinking. You will have to look it up and make your own decision.