Power Supply Instead of Battery

Off Topic
1

Hi All!
I have a handheld RFID scanner that I want to replace the battery with a power supply. I can easily spec the PS to the voltage of the battery, but I am not sure how to spec the current of the power supply. I can’t find specs on the battery, or the device. I need to figure out how much the device draws at max. I could get the max battery current (voltage / internal resistance), but I expect this is much higher than the device actually needs.
CSL CS101 Handheld RFID Reader - EasyBadges It is basically this unit.

I have reached out to the manufacturer, but honestly, I don’t expect them to help me out.
Has anyone else tackled this kind of problem in the past?
Open to any suggestions or guidance!
Thanks!
Nate

2

It’s a 1400 mAh battery, and the continuous usage battery life listed in 1.5 hours

so 1400 mA/H * 1.5 H = 2100 mA, so 2.1 Amps. I would say if you are anywhere greater that 2.5 you should be good.

What’s the current listed for the battery charger? I would think you could match that and be pretty good.

3 Likes
3

Good call! Charges at 25VDC @ 2.5A
Thank you!

4

Well, the voltage for the charger is different than the output voltage of the battery, so 19.8V X 2.5A = 50 watts. the battery charger is 25V X 2.5A = 62.5 watts. (I was thinking that the voltage would be equal, but it’s not). So I would maybe make sure the watts are higher than the battery charger. So at 19.8 V the for 62.5 watts, the amperage is 3.1 amps.

You can’t really go too large on amperage, anything over what you need is fine. I don’t know what the costs differences are for the different wattages of power supplies, but if the cost is negligible from 50 to 100 watts, go with the bigger one.

1 Like
5

Homer Simpson Cartoon GIF

4 Likes
6

The charger handles two batteries at once. So, I am guessing 12v batteries with a nominal voltage of about 15v.

7

image
image838×292 39.1 KB

I thought I saw 19.8 but it’s 14.8. (and that might be dependent on which manual I found, you should look at the actual manual you have or the battery)

so 62.5 / 14.8 = 4.22 amps

but yeah, if the charger does 2 at a time, you should be able to cut the wattage roughly in half.

The earlier comment still stand though, excess wattage won’t hurt you if the voltage is spec’d correctly.

1 Like
8

Thank you! We are going to print up an insert to convert the battery to barrel jack plug. :crossed_fingers:

9

1 Like
10

Can you believe the manufacturer suggested that it is not recommended to replace the ($100+) battery with a DC power supply? Can you think of any reason a battery would be recommended instead of a power supply? I always thought a power supply provided more stable voltage and current.

11

It is a detriment to their bottom line, you will no longer have to buy 100+ dollar battery from them :rofl:

1 Like
12

A battery would be a cleaner power source. Less voltage ripples, spikes, etc… But if they designed their power circuitry properly, that shouldn’t matter unless you’re using a noisy $2 AliExpress special…

3 Likes
13

It’s probably about designing for the power source too. I wonder what kind of circuitry is in the battery. There is going to be some kind of control system, and I am curious if you don’t have that if it will still work.

1 Like