watts for a tv

Depends on the display technology, etc., but that's not at all unreasonable.

Bob M.

That sounds just a little low to me, but may be correct. My 17" CRT monitor consumes about 90 watts. My maybe 3 year old 20 inch CRT TV consumes 70 watts.

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Not implausible. Very energy efficient compared with some modern TVs.

Graham

Well my Sony E530 Multiscan monitor consumes allegedly 130W active but is 21" and does insane scan and pixel numbers. Plus makes LCDs look stupid.

Only running 1280x1024 @85Hz right now though.

Graham

Meanwhile the power brickette for my laptop eats 200W.

how can that be so much? Is the rating what it's actually consuming, or is that the maximum possible? If the CPU is the biggest power consumer, then that usage varies a lot, right? Or does the screen use more than the CPU?

(My Dell laptop's PS says 90 watts.)

*USUALLY*, the rating on a PS refers to how much total power (probably at multiple voltages) it can supply before something's magic smoke leaks out and ends all the fun.

If it's the usual "LCD with backlight" screen, I'd expect that operating the disk drive (spinning the platters, moving the heads, plus more power required to write) sucks more power than the screen, and quite possibly more than the CPU. Making stuff more substantial than electrons move is pretty energy-intensive. After the disk drive and CPU, audio output is another power-hog (once again back to the "making things move is expensive" concept) and any sort of wireless networking is, almost literally, throwing power out the window by the bucketful. (at least, anytime the network is "talking", it will be)

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All the time it's operative ? I bet my desktop doesn't exceed that.

Graham

No way. Laptop HDs are power misers these days. Invariable the CPU is one of the highest on the list probably after by the display but possibly ahead of it in those cases where fatheads HAVE to have the latest bleeding edge gizmo to write Word documents .

Graham

Keeping the platters spinning doesn't take much, they have really good bearings. The weight of the read arm is probably minuscule. Coding Horror has some measurements of the power consumption in a real laptop:

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I've never seen one over 90 watts. Unless it has a very fast charger for the battery?

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Everybody is ignorant, only on different subjects.

Keep reading. There are two sets of numbers. There is what it provides to the laptop (90W) and what the power supply itself consumes based on

125VAC.

In my case, my 65W Dell power supply uses 187.5W of AC power.

Not unless it's brick-sized or runs too hot to touch does it manage that. possibly you are confusing VA with Watts - they are not the same.

W= V * A

I know the supply voltage coming in, it's 125V.

explode.

Here are the numbers right from the supply:

AC: 100-240V 1.5A DC: 19.5V 3.34A

So either there's a different formula for AC wattage or what's printed on the supply is wrong.

I've always know Power = Voltage * Current

So 125V * 1.5A = 187.5W

19.5V * 3.34A = 65.13W

explode.

Perhaps the input is a maximum current draw, and the output is an average? You'd need to know the maximum input for rating circuit breakers etc. You'd neeed to know the avarege out to match it to the load. You would assume a decent switched mode supply could give a big boost when reqwuired (usually about double the rating?)

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explode.

W = V*A is only true for purely resistive loads, which your brick almost certainly isn't.

187.5 VA (now W). The current listed is likely the peak current, which is the rating needed for wires, switches, transformers, etc.

That would be the maximum power the PC could consume. Divide by .7-.8 or so and you get the maximum power drawn by the brick (from the wall).

But only if the voltage and current are in phase, which is not necessarily the case with AC. Hence the distinction between VA - which is what you get when you simply multiply the volts time the amps in AC - and watts, in which the calculation corrects for the possible phase difference. To get watts, which is the "real" part of volt-amps, the calculation is:

W = V * A * cos(p)

Where "p" is the phase difference between the voltage and current waveforms. Note that if there is no difference, cos(p) will be 1, and you get the simpler W = V * A.

I suspect, though, that what's really going on here is that you're confusing a peak or inrush current rating with the actual (typical) drawn during normal use.

Bob M.

unless A is a sine wave it's not that simple.