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Choosing power supply, how to get the voltage and current ratings?

I'd like to power my new LG M2450D monitor with a basicXL BXL-NBT-U02 universal AC-DC power adapter. This PSU is rated 90W max and supports 15-24V output voltage. The original monitor PSU (PA-1650-68) was fixed at 19V and 3.32A.

Now, doing the math:

90W / 19V = 4.74A

Of course this is an ideal value, what is "safety margin" i should assume for the real max amperage?


marked as duplicate by Leon Heller, Kortuk Sep 22 '12 at 22:21

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  • \$\begingroup\$ Sounds OK. Note that some power supplies will not make max power at max voltage (Murphy saw Dave Tweed's sensible sounding suggestion coming and decided to play games with it). BUT it sounds suitable. \$\endgroup\$ – Russell McMahon Sep 22 '12 at 18:31
  • \$\begingroup\$ @RussellMcMahon: If you're going to throw out a comment like that, you're going to have to provide an example ... because I can't think of one. \$\endgroup\$ – Dave Tweed Sep 22 '12 at 20:20
  • \$\begingroup\$ @DaveTweed - it would be hard for me to give an example except by eg photographing one when I next see one. I've seen a few that say basically that. I MAY have been casting the net wider than stricytly correct in that they may be eg 12V in and variabke out that do this. I'll keep an eye out. \$\endgroup\$ – Russell McMahon Sep 22 '12 at 21:16

To be conservative, you should assume that the maximum power rating of the universal adapter occurs at the maximum output voltage, and that all lower output voltages are limited to the same current.

In this case, 90W/24V = 3.75A, so you should be good to go with the output set at 19V with this amount of current.

  • \$\begingroup\$ OPs replacement choice appears to be a good fit from a spec standpoint. \$\endgroup\$ – Adam Lawrence Sep 22 '12 at 18:11
  • \$\begingroup\$ This is a nearly constant and repeating question and we attempted to make a main question for this very common question, I do think it would be valuable for you to add your information to that post @daveTweed. :) \$\endgroup\$ – Kortuk Sep 22 '12 at 22:21
  • 1
    \$\begingroup\$ @Kortuk: I take your point, but I'd actually diagree that this is an exact duplicate. Fundamentally, this is a question about (conservatively) interpreting ambiguous specifications; it just happens to occur in the context of selecting a power supply. If I were to add my answer to Olin's question, I'd have to add an explanation about why I was making my point. \$\endgroup\$ – Dave Tweed Sep 22 '12 at 22:54
  • \$\begingroup\$ @DaveTweed It seems to be another how can I buy a PS for my consumer device, you could add another answer about this situation, really up to you. \$\endgroup\$ – Kortuk Sep 22 '12 at 23:01
  • \$\begingroup\$ @Kortuk - This was slightly different in main intent from most others. He understands Wattage abd Watts = V x I. He is asking how close he can get to the nominal spec. eg if He knows Padvertised and has V_target, how much greater does Padverised / Vtarget = Ioutput_notional have to be that I target to be safe. eg to make a new example if he has a 20V, 4A target and a Padvertised = 100 Watt supply then Inotional at V target = P/V = m100W /20V = 5A. As I target is 4A, he is asking if it is safe to assume that the 25% extra current from 4A to 5A is enough in oractice to be safe. ... \$\endgroup\$ – Russell McMahon Sep 23 '12 at 3:33

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