I have a laptop charger which has 19.5 V 3.33 A output. I want to use it for raspberry, which needs 5V and 2.5A and for monitor which needs 12V 3A. Is that charger strong enough to run the whole system? Is it possible to convert 19.5V 3.33A with a buck step down to 12V 5.5A which will be connected to monitor and then use an another step down to 5V 2.5A to feed the raspberry, or I have wrong idea about everything? Sorry I do not know much about electrical engineering. Thanks!

  • \$\begingroup\$ This is a bad idea. You'd be better off using power supplies designed for each output, or for suitable combined 5v/12v output. \$\endgroup\$ – Chris Stratton Dec 28 '18 at 20:08
  • \$\begingroup\$ Welcome to EE.SE! Keep in mind that "Is it possible ...?" is a yes/no question. In this case, the answer is "Yes". If you're asking us to design it for you, that would be too broad. You would need to specify something about the level of performance you expect as well as what constraints you have on the implementation. What is your specific question? \$\endgroup\$ – Dave Tweed Dec 28 '18 at 20:50
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    \$\begingroup\$ People who know nothing about EE nor basic Grade X11 physics should stick to buying solutions. \$\endgroup\$ – Tony Stewart EE75 Dec 28 '18 at 21:14
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    \$\begingroup\$ A rpi doesn't need 2.5 Amps tho. \$\endgroup\$ – Passerby Dec 28 '18 at 21:18
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    \$\begingroup\$ Tibor - Hi, "I have a laptop charger which has 19.5 V 3.33 A output" Just a warning that some laptop chargers sold on Ebay, Amazon, AliExpress etc. which claim to have a quoted current rating, don't really have that rating and can have poor quality or even unsafe internal construction. \$\endgroup\$ – SamGibson Dec 28 '18 at 22:04

Yes, it is possible

power of laptop charger is = 19.5 * 3.33 = 65 Watts

your loads power:

  1. raspberry pi = 5*2.5 = 12.5 watts

  2. Display = 12 * 3 = 36 watts

    total power = 48.5 watts

Total power < power of charger,

you can use it. it will work

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    \$\begingroup\$ Conversation of energy is not violated, sure. But actually implementing it will require some effort and will probably not turn out economical. \$\endgroup\$ – Chris Stratton Dec 28 '18 at 20:48
  • \$\begingroup\$ DC-DC converters are pretty cheap. If you can convert 19.5 V to 12 V with 80% efficiency, and then 12 V to 5 V with 80% efficiency, the numbers work out almost exactly. You get 4.33A @ 12V, the monitor takes 3A, and the remaining 1.33A converts to 5V @ 2.55A. Of course, if you convert directly from 19.5V to 5V, you get almost 3.2A to play with. \$\endgroup\$ – Dave Tweed Dec 28 '18 at 21:00
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    \$\begingroup\$ Uh, 12v at 5.5a is 66 watts. \$\endgroup\$ – Hot Licks Dec 29 '18 at 1:08
  • \$\begingroup\$ @HotLicks: Uh, Satish corrected the mistake in the question of thinking that the input to the RPi stepdown converter would be the same current as the output. \$\endgroup\$ – Ben Voigt Dec 29 '18 at 3:43

The answer to your header question is NO!

Input 19.5x3.33 = 64.935 Watts.
Output 12x5.5 = 66 Watts.

So your new output power is higher then your input power which is against the law of conservation of energy.

But according to your question details your equipment requires only 48.5 Watts. That is do-able but only if your buck step down converter (or converters) has/have at least 75% efficiency.

But as Chris Stratton already said: it is better to use supplies which are dedicated for the circuits. You will be spending time and effort on something special, whilst there are cheap, ready made solutions. The idiom "penny wise, pound foolish" comes to mind.

  • \$\begingroup\$ sounds like an interesting project Tibor is trying, and he'll probably learn a lot. "it is better to use supplies which are dedicated for the circuits" assumes he isn't doing this to "just" learn more for fun \$\endgroup\$ – Blundell Dec 28 '18 at 21:37
  • \$\begingroup\$ there are cheap ready made switcher modules on every online marketplace. so long as you de-rate them 30% or so on current they work well. \$\endgroup\$ – Jasen Dec 29 '18 at 8:00

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