# X

Discussion in 'Power Electronics' started by John29, Nov 1, 2016.

1. ### John29 Thread Starter New Member

Oct 13, 2016
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Hi
I have an rfid reader (http://media.digikey.com/pdf/Data Sheets/ThingMagic PDF's/MERCURYDevKit.pdf) with these specifications(as seen in datasheet): • In: 90-264 V, 0.4 A, 47-63 Hz • Out: +9 V @ 1.4 A • Max total output power: 12.6 W I would like to create an external battery module instead of connecting the reader to a wall. I would like to have some precisons on how to do it, which battery to choose and how long would the battery last ( I would like it's the run-time to be at least 12h)?
Thank you very much

2. ### AlbertHall AAC Fanatic!

Jun 4, 2014
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1.4A for 12 hours is 16.8Ah @9V. That's a pretty big battery. The standard 18650 lithium cells are 2.6 Ah @3.7V so you would need around 16 of them for your purpose (assuming no losses along the way).

3. ### John29 Thread Starter New Member

Oct 13, 2016
18
0

HOw to you know that 1.4A for 12 hours is 16.8Ah @9V ? Why would I need 16 batteries like that?

4. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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Your first step is to measure how much current the reader takes when in use, that will dictate which battery to use, chances are it probably takes 500mA to 1amp, so your looking at a 9V sla or lithium battery.

5. ### crutschow Expert

Mar 14, 2008
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The required battery capacity equals the current times the time or 1.4A*12h = 16.8Ah.
Then you divide this value by the battery capacity to determine how many batteries you need.

But as Dd noted, you need to measure the actual current used, since it is likely significantly less than the rating of the power supply.

6. ### John29 Thread Starter New Member

Oct 13, 2016
18
0
How many 9v sla batteries would I need for my objective? If I use lithium battery, do I need to create some 9v feeding voltage with them? I am asking that because I am afraid that my rfid reader will burn if he has 9V in it's input and the batteries give 30 V output for exemple.

7. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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You can use a 12v sla battery, and a Lm2596 buck pcb to drop the voltage to 9v.

8. ### crutschow Expert

Mar 14, 2008
22,223
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The batteries would be put in parallel and/or series to get the required 9V.
But to determine how large a battery you need you first need to measure the current drawn by the RFD reader.
Do you know how to do that?

9. ### John29 Thread Starter New Member

Oct 13, 2016
18
0
I can do it with an ampere-meter. I connect the black cable to the ground and the red to the rfid reader output (in the antenna connection), right? And after for exemple if I have the value of 500 mA, I should use that value instead of 1.4 A for the calculation 1.4A*12h = 16.8Ah., right?

10. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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yes, i would operate the rfid with a valid keyfob to see if the current increases, and use that value.

11. ### crutschow Expert

Mar 14, 2008
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Since the operation of the keyfob would typically be a very low duty-cycle, I don't think that needs to be included in the battery capacity calculations.

12. ### crutschow Expert

Mar 14, 2008
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Wrong.
You break the connection between the power supply output and the RFID reader power input, and connect the ammeter in series from the power supply output (red meter lead) to the RFID power input (black meter lead).

13. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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The rfid reader will activate a relay upon a valid keyfob to open the door, which will increase the current...

14. ### crutschow Expert

Mar 14, 2008
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I understand that.
But I think the duty-cycle for that will be low enough that it won't have a significant effect on battery life, which is the subject of this thread.