Heat Calculations of Electronic Components

[fifty60]

I have a 60W power supply. The spec sheet says that it is 83% efficient, draws 1.7A max on the input at 100VAC.

1.7A * 100 = 170VA
170VA * (1 - .83) = 28.9VA

170 - 28.9 = 141.1 VA

The power supply is only 60W. I would expect that after doing the above calculations I would end up with 60 instead of 141.1.

So, out of the original 170VA, how much is lost as heat? Do I need more than the efficiency to calculate this?

Is 28.9VA being lost as heat, or is 110VA being lost as heat (170-60)?

 

[gar]

151027-0839 EDT

In an AC circuit average voltage times average current is not necessarily a measure of real average power. An RMS measurement is a type of average measurement.

However, it is true that instantaneous voltage times instantaneous current averaged is equal to average real power over the time of averaging.

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[fifty60]

Interesting. So the 100V is definitely the RMS value, but you think the 1.7A may not be?

1.7/Sqrt(2) = 1.2

1.2 * 100 = 120 VA

120 * (1 - .83) = 20VA

Seems to be getting closer, but still doesn't make sense. There are still 40VA not accounted for...

 

[gar]

151027-0907 EDT

I did not say that. I expect that both may be RMS. You do not understand what instantaneous, average reading calibrated to RMS, and RMS measurements are.

Separately I made measurements on two power supplies I have. One was a computer switching supply, PF = 0.62 . The other a transformer, rectifier, and capacitor input filter, PF = 0.92 .

.

 

[Smart $]

Gar hits on a key point about power factor. Your calculation does not account for it... i.e. just VA is not considered actual input power.

 

[GoldDigger]

Gar hits on a key point about power factor. Your calculation does not account for it... i.e. just VA is not considered actual input power.

Especially when you are calculating efficiency.

 

[Jraef]

Your question was about HEAT, not amps.

So aside from all of that, if you are using the power supply to power up devices inside of a control panel, ALL of the power consumed from it is actually going to be heat in that panel. Not all 60W, because presumably your PS is larger than your load. But if you look at the load of your devices being powered by the PS, all of that energy (watts) is being expressed inside of that box, plus whatever losses the PS has as a percentage of that.

So for example lets say you are powering up a bunch of relays, an electronic device of some sort and maybe a few pilot lights. If you add up the power consumption of all of those devices and it comes to 39W, then you assume a 13% overhead (losses in the PS), you have 45W of heat being put into the enclosure, worst case (meaning all of the relays are on, all the pilot lights are lit, the electronics are drawing full power). So for calculating for heat gain inside of the box, i.e. for sizing a fan or an AC unit, you use that entire value. Safer (easier) bet is to just use the 60W.

 

[gar]

151027-1543 EDT

fifty60:

Many of the questions you ask on this forum are reasonable to ask, but you should do some research and study on your own before presenting the questions here. You need to put some of your own time into trying to figure out answers. This means think logically and probe the question from more than one direction. Try to figure out what is the correct question or questions to ask.

In this case you should have at least done some measurements on the power supply. Using a small inexpensive instrument called the "Kill-A-Watt EZ" measure the input current, voltage, power, VA, and power factor under various load and supply voltage conditions.

We don't know at what operating point the various values you provided were obtained. Was maximum efficiency that at full ouput load or some lesser value.

Assuming for some of your data. Efficiency is for full DC load, 60 W. Efficiency is the total overall efficiency at full DC load. Then
Input power = 60/0.83 = 72.3 W. Thus, heat loss is 12.3 W.

Again assumptions --- 100 V AC RMS is the lowest rated input voltage, this is a switching supply, and at full load and minimum input voltage the maximum full load input current is 1.7 A. This equals an input VA of 170.

The calculated power factor is 72/170 = 0.42 . Can you experimentally verify any of these values. The 12.3 W is the hardest to verify. It is relatively easy to verify whether their claim of 83% efficiency is valid.

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[fifty60]

Understood. Jraef, Gar. Thanks for the input information. It has been a while since I have thought about efficiency and PF, that is my fault. I can figure out the rest with the direction you have provided. Thanks!

 

[steve66]

I would ignore the input current, since you don't know the power factor.

It puts out 60 watts, and its 83% efficient. So 60/.83 = 72.3 watts.

So about 12 watts would be lost as heat.

 

[GoldDigger]

I would ignore the input current, since you don't know the power factor.

It puts out 60 watts, and its 83% efficient. So 60/.83 = 72.3 watts.

So about 12 watts would be lost as heat.

"So about 12 watts would be lost as heat"... in the power supply. The rest will be lost as heat in the loads and if they are in the same box they will heat that box, as Jraef pointed out.

 

[Carultch]

Interesting. So the 100V is definitely the RMS value, but you think the 1.7A may not be?

1.7/Sqrt(2) = 1.2

1.2 * 100 = 120 VA

120 * (1 - .83) = 20VA

Seems to be getting closer, but still doesn't make sense. There are still 40VA not accounted for...

That RMS rule of sqrt(2) is ONLY TRUE for linear loads with no energy storage components. In otherwords, pure resistors. And usually, nominal voltages and currents already refer to the RMS value drawn from the outlet.

There are other reasons why power would be different than volts*amps, and that is where power factor comes in to play.

 

[LMAO]

I have a 60W power supply. The spec sheet says that it is 83% efficient, draws 1.7A max on the input at 100VAC.

1.7A * 100 = 170VA
170VA * (1 - .83) = 28.9VA

170 - 28.9 = 141.1 VA

The power supply is only 60W. I would expect that after doing the above calculations I would end up with 60 instead of 141.1.

So, out of the original 170VA, how much is lost as heat? Do I need more than the efficiency to calculate this?

Is 28.9VA being lost as heat, or is 110VA being lost as heat (170-60)?

I don't get it? you said it is 60W so its max current draw should be 60W/100V=600mA not 1.7A...?

 

[GoldDigger]

Do you not understand the concept of power factor and reactive power?

 

[Smart $]

Do you not understand the concept of power factor and reactive power?

I think the "not" part is plainly evident. :blink: