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This question already has an answer here:

So, I just blew the fuse in my 17" monitor using a universal adapter with the settings

Input: 100-240V 1.5A(1.5A) 50/60hz

Output: 15/16/18.5/19.5/20/22/24V (max of 70w)

I want a clarification of sorts on exactly how to compute Wattage on a device such as a monitor using a universal adapter with the above setting\selection. On the back of the monitor, I have rating 12V 4.16A.

What I typically do to calculate power is to multiply the Voltage and Current numbers

 **12 * 4.16 => 49.92W**

Now to set the Adapter for the monitor, I computed that by setting the adapter to 22W and assuming the amperage to be 1.5A as written on the back of the adapter, I could get about

 **I computed that 24V x 1.5 A => 36Watts** 

Given that it is lower than what I computed as the wattage needed for the monitor, I decided to try it anyways. But, as I said, I heard a pop sound and some smoke came out of my monitor. I would appreciate some help in understanding my error here so I can be more careful next time when using a Universal adapter such as the one I used

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marked as duplicate by Ignacio Vazquez-Abrams, Olin Lathrop, PeterJ, Daniel Grillo, Dave Tweed Aug 4 '15 at 15:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ The monitor is rated at 12V, so you applied 24V?!? That is a recipe for BOOM. Also the 1.5A is the maximum "Input" current of the PSU (that it draws from the mains supply), not the output current rating. \$\endgroup\$ – Tom Carpenter Aug 2 '15 at 18:44
  • \$\begingroup\$ I am not sure I follow. It is rated 12 V however, total wattage was higher than what was supplied by the adapter. I have never had to be specific on voltage and current instead I have always used the formulae P=VI when computing compatibility before adapter and device. Is this a wrong approach? \$\endgroup\$ – user272671 Aug 2 '15 at 19:29
  • \$\begingroup\$ About the output current, how do I get that from what I was given on the Universal adapter? \$\endgroup\$ – user272671 Aug 2 '15 at 19:30
  • \$\begingroup\$ The current should either be either (a) in a datasheet for the supply somewhere, (b) printed as a graph vs voltage on the supply, or (c) assumed to be Pout/Vout (so for the 24V setting that would be 70/24=2.91A). \$\endgroup\$ – Tom Carpenter Aug 2 '15 at 19:43
  • \$\begingroup\$ Power is not what you should be looking at initially when determining suitability. If something is rated at 12V, you should not feed it 24V (unless it specifically says you can) - aside from magnitude, it is no different from having a device rated for 110V only and feeding in 240V, something will go bang. \$\endgroup\$ – Tom Carpenter Aug 2 '15 at 19:45
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All the prior comments are useful.
Read them carefully.
This says the same things as they have already said but in an extended form.

Summary:

  • Fuse may be blown.
  • Worse may have happened.
  • Match voltages.
  • Ensure power supply current rating is high enough.

First you need to match voltages.
Voltage is the "driving force".
The device will take what current it needs at rated voltage.
If you apply too much voltage the device will be "ver pressured".
Your source must be able to supply the current that is needed at rated voltage BUT the voltage must be correct.
Too much voltage may destroy from voltage alone or from a current "spurt" stored in a capacitor in the supply. Too little voltage will generally not allow enough current to flow, will not operate circuits that need a certain voltage and will sometimes destroy.

ASSUME for this purpose

  • That the load is resistive - that is OK for this purpose.

  • And understand (or accept) Ohms law. Which can be written
    I = V/R.
    R= V/I
    V = IR.

Monitor looks like R = V/I = 12V/4.16A =~ 3 Ohms.
This is the "effective" resistance and will vary as the device does various things during operation.
This is the LOWEST R it will be and max current.
It will usually have a higher effective R and lower current in operation.
BUT when it needs 4A it needs 4A - if your adaptor is unable to supply 4A then it can NEVER supply what the monitor sometimes needs.

Next IF you applied 24 V to the 3 Ohm load
then I = V/R = 24/3 = 8A.
You are TRYING to make it take 8A.
The adaptor will not supply more than 1.5A steadily but may have an output capacitor that will supply an 8A pulse (or more).
IF you are lucky you may have blown a fuse when the 24V cap charged the input cap on the monitor.
If you are unlucky you destroyed a component not rated to take 24V.

Wattage: Wattage is "what happens" - it is not usually something you set. ie
You apply V.
The device draws I at a given V
You calculate resultant wattage.
While you CAN get constant wattage sources they are very unusual and specialist.
FWIW: Watts = V x I = V^2/R = I^2 x R
These are all the same formula rewritten using the Ohms law equivalences above.

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