# watts the deal

so, i have this ac adapter:
http://i1190.photobucket.com/albums/z449/m75214/122811195907.jpg
it says it's about 1.5 amps. so, why can't i say 1.5A * 120V = 180W
and say it's 180 watt?
it says on it that it's 90 watts though, which is confusing me.
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wrote:

I doubt very seriously that this thing is dissipating 90 watts making 90 watts. It would get so hot you couldn't touch it. I agree the output is 90w but I bet the real input is some combination that ends up being about 100w. It is a "wide mouth" switching supply so it can tolerate a huge difference in input voltages while making that 90w at 19v. I bet the 1.5a is the peak amps, and it averages down throughout the cycle.
Have you ever looked at current on one of these things on a scope? It is a very ugly wave form.
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On 12/28/2011 09:11 PM, bob wrote:

Hello, and while I haven't seen a schematic of the Dell PA-1900 AC adapter it probably uses a switching (flyback)type design. Under load these adapters often become very warm during normal operation and I would expect energy conversion efficiency on the order of about 75-85% depending on input AC voltage level and DC load current. I would treat the AC and DC current values of 1.5 A and 4.62 A, respectively, as maximum (not to exceed) values. Your calculation (corresponding to an efficiency of 50%) assumes that we have these maximum values under design load (computer and/or battery charging) conditions which might not be the case. Sincerely,
--
J. B. Wood e-mail: arl snipped-for-privacy@hotmail.com

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The 180 watts is the input power. Multiply the output... 19.5v * 4.62A = 90.09 watts
That means the transformer uses 180 watts to produce 90 watts making it about 50% efficient.
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"Rich." wrote in message

The 180 watts is the input power. Multiply the output... 19.5v * 4.62A = 90.09 watts
That means the transformer uses 180 watts to produce 90 watts making it about 50% efficient. -----------------------------------dk---------------------------- It could be a bit more efficient than that- input Volts times Amps isn't necessarily real power- there is a power factor involved due to inductance, capacitance or nonlinearity causing harmonics. This Power factor will be less than 1. Pout = Vin*Iin *pf* efficiency
Don Kelly cross out to reply
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So, if I pay 10 cents a kilowatt hour, is this thing costing 1.8 cents an hour?
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"bob" wrote in message

So, if I pay 10 cents a kilowatt hour, is this thing costing 1.8 cents an hour?
------------------------------------------------dk ----------------------- No. Krw answered this- correctly. Think of a right triangle with the hypotenuse being V*I (ac only) and the base being the real power- what you pay for-, and the opposite side being the "reactive power" or volt-amperes-reactive (VARs) which represents energy that is shuttled back and forth between load and source with an average of 0. Residential consumers don't pay for this directly (it increases losses a bit). The power factor is the cosine of the angle between the V*I and the real power.period This is the classical approach but- -in recent years , with more switching supplies, rectifiers etc, there are harmonics which affect the power factor in a messy way.
Don Kelly cross out to reply
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No, you pay for real power. You don't pay for the imaginary part. The power factor tells you how much is real/imaginary but that information isn't here. All you know is the current and voltage at the input, not it's PF.
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wrote:

The device has too many unknowns, measure the AC current at the input at your full rated load and use those numbers.
Tom
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"Tom Biasi" wrote in message

The device has too many unknowns, measure the AC current at the input at your full rated load and use those numbers.
Tom That won't work. In general it doesn't work for AC--something that was discovered over 100 years ago. That's when the power factor idea came into being. Don Kelly cross out to reply
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wrote:

I know Don but since he has so little info and is concerned about operating costs the power factor most likely is insignificant in this case.
Regards, Tom
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"Tom Biasi" wrote in message

I know Don but since he has so little info and is concerned about operating costs the power factor most likely is insignificant in this case.
Regards, Tom
I believe that newer wall warts are in the range of 70% efficiency, older ones are less than this. So with 90 watts output a newer unit would draw about 130 watts at unity pf and that is what you pay for. Note that for 180VA input at 70 % efficiency, the pf is then about 0.7 My point was that the cited 180VA input for 90 watts out doesn't mean 180 watts input.
You are probably correct in that the difference between 1.8 cents/hour and 1.3 cents per hour may not be a big deal to the user. You are also right in that there is little information to go on.
Have a good 2012 Don Kelly cross out to reply
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No, it's not insignificant. VA(max)0, P(output)�, that indicates that the PF could be .5. It can't be known just from the data given, though.
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On Sat, 31 Dec 2011 18:08:22 -0500, " snipped-for-privacy@att.bizzzzzzzzzzzz"

Agreed, but I don't have confidence in those numbers, they could have underrated the published output. Since I don't know anymore and don't have the unit in front of me I'll have to pass on further comment. Regards, Tom
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I *highly* doubt it. However, looking at it again, they are probably listing the over-current device's rating. That's the number I've always used on these labels.

Unit? Are we talking about the same thing? http://i1190.photobucket.com/albums/z449/m75214/122811195907.jpg
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On Sat, 31 Dec 2011 19:05:40 -0500, " snipped-for-privacy@att.bizzzzzzzzzzzz"

What would you like me to call it?
Tom
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Tom Biasi Inscribed thus:

Note that the PSU, a DELL PA10, is rated from 100 to 240 volts input !
--
Best Regards:
Baron.
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bob wrote:

Its good for 90 Watts out (19.5V * 4.62A) given some inefficiency, it will need somewhat higher input power. But not 180 Watts. That thing would be glowing red hot if it consumed 90 Watts internally.
Some of the excess input current is to allow for poor power factor. So its not real power consumed by internal losses.
Also, what you have there is probably a switching power supply. So the 1.5A represents the maximum draw at 100V, not 120V. Which makes the device's input rating 150 Volt-Amps (not 180 and not Watts).
--
Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com
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