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I have purchased a 300W Modified Sine Wave Inverter. I am also connecting a 12V 18A/hr SLA battery to it.

I am using it to power an RFID reader. The input power characteristics of the power block are: 100-250V, 50-60Hz, 0.5A.

The output characteristics are: +48V DC, 0.4A. Please note the power block connects to the RFID reader via an ethernet cable only. Hence I cannot just connect a DC supply directly to the RFID reader.

How would I calculate how long the battery will last before I need to charge it, do I use the input or output characteristics of the power block? (Is there an equation I could also use).

Also, I would also like to connect my laptop charger to the inverter while the inverter is powering the RFID reader. The input characteristics for the laptop charger is: 100-240V, 50-60Hz, 1.2A

Output characteristics are for the laptop charger are 19V DC, 2.73A.

Will it be okay to connect the laptop charger and the RFID reader at the same time, or will this exceed the limitations of the battery and the inverter?

I know how to calculate the duration of a battery given a DC to DC setup. But know I am going from DC to AC to DC again. Do I use the input characteristics of the power supply blocks i.e. the currents or the output characteristics of the power blocks?

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    \$\begingroup\$ this is totally overkill. Your RFID reader's "block" is DC supply with a voltage that's probably lower than 12V. A simple buck converter would have powered it directly from the battery with a fraction of the cost and effort of your inverter. (and a way, way higher efficiency) \$\endgroup\$
    – Marcus Müller
    Dec 19, 2019 at 21:09
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    \$\begingroup\$ Other than that, this question has really been asked a hundred times here. \$\endgroup\$
    – Marcus Müller
    Dec 19, 2019 at 21:10
  • \$\begingroup\$ and that's really just one of very many questions. \$\endgroup\$
    – Marcus Müller
    Dec 19, 2019 at 21:10
  • \$\begingroup\$ @MarcusMüller firstly a simple buck converter will not work due to the RFID reader setup. It can only be powered from the block which only accepts an AC source. Indeed I wished it could have been powered from a buck converter. Secondly no it does not answer my question. \$\endgroup\$
    – JoeyB
    Dec 19, 2019 at 21:13
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    \$\begingroup\$ This is no answer, and I'm not trying to belittle you. Power is power. It's clear where it has to come from, and it's clear how much you need of that and where, so I really don't understand what you're asking here, sorry. \$\endgroup\$
    – Marcus Müller
    Dec 21, 2019 at 15:55

1 Answer 1

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The values you provide are not sufficient to determine the runtime of the RFID reader, either with or without the laptop. The power brick output specification means that the power brick is able to supply up to 0.4 Amps at 48V, which is a power of around 20W. This implies that the power consumption of the RFID is less than 20W. We have no information, how much below 20W it is. It might be 3W as well as 17W.

The power supply input specification is even less helpful for the question at hand. The power supply is able to cope with input voltages between 100 and 250 Volts, and it promises to not blow a 0.5A (slow-blow) fuse. As 0.5A at 100V is alrwady 50W and 0.5A at 250V is 125W, I am positive that the supply is never going to take that much real power, although at 100V it might consum 50VA apparent power. Typical actual consumptions of a 20W switch-mode supply are up to 2W if the output is not loaded and up to 30W if the output is loaded it the maximum capacity.

A 300W inverter might very well have an idle power consumption of 10W, and a full-load consumption of 420W. If the load to that inverter is 30W, the consumption of that converter could be around 50W from the battery

Your battery provides a total energy amount of 12V * 18Ah = 216Wh. The estimated runtime of the RFID reader alone powered through the 300W inverter and the 20W power brick is thus likely between 20 hours (most energy lost in the converters, total consumption around 10W) and 4 hours (the RFID device utilizes the 20W supply all the time at maximum permissible load, total consumption from the battery around 50W).

The only way to provide better numbers is to actually measure actual power consumption instead of relying on the specs of the power brick.

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