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I am trying to build a power supply for my LCD monitor.

It consumes 19.5V and 1.7 amperes.

I am using a 12 volt 16 ampere lead acid battery. I have used a step up boost converter to increase the voltage to 19.5V, but I am not able to decrease the current. It is still 16 amperes only.

If I use this power supply with 19.5 volt and 16 amperes, will it damage my LCD monitor?

Davide Andrea
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    Don't worry, law of conservation of energy has decreased your amp rating from 16 when you stepped up the voltage. Besides your lcd monitor will consume only the current it needs. – Mitu Raj Nov 06 '21 at 17:34
  • `It consumes 19.5V and 1.7 amperes.` ... that is somewhat incorrect ... this is more accurate ... `It requires 19.5 V and can draw up to 1.7 A` – jsotola Nov 06 '21 at 18:30
  • think about this: the power supply that you are replacing is probably able to source 1.7 A ... the monitor probably draws only a few mA in standby mode ... why does the monitor not burn up when in standby mode? – jsotola Nov 06 '21 at 18:37

1 Answers1

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Golden rule:

Voltage is pushed, current is pulled

That's a poetic though slightly imprecise way of stating that the load (the monitor) sets the current, not the source (the battery). At a given voltage, the load decides how much current it wants, and that is all that it takes.

The fact that the source could provide more current to some other load that wants more current, has no effect on this particular load. Therefore, no, the monitor will not be damaged.

Davide Andrea
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