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Old 15th December 2011, 10:12 AM   #1
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Default Continuous power dissipation of resistor?

Hello,

The MVM5 family of wirewound resistors is rated at 5W.....however, the temperature de-rating chart does not say how much PCB copper area the resistor has to be soldered to in order to be able to achieve that power rating continuously.

..do you know how much PCB copper area is required for the MVM5?


http://www.seielect.com/Catalog/SEI-VM_MVM_LVM_WVM.pdf
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Old 15th December 2011, 11:20 AM   #2
marce is offline marce  United Kingdom
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With it being a leaded device, and even if attached to a plane it will have thermal releif to allow soldering, and it is mouted in a way that the heat dose not travel down the lead, the PCB copper will not make much difference. We would de-rate these types by 50%. There are better thermaly efficient packages that allow the use of a heatsink or heat spreader. Or some more exotic SMD parts, but these can be hard to solder effectively without a reflow oven.
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Old 15th December 2011, 11:22 AM   #3
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The de-rating chart shows % of rated power vs. AMBIENT temperature.

You should not need to care about device case temperature, copper size, etc., as long as the air temperature around the device is acceptable for the % of rated power dissipated?

EDIT: ...and what marce said

Last edited by discrete; 15th December 2011 at 11:24 AM.
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Old 15th December 2011, 11:28 AM   #4
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Frankly, if you're running close to de-rating limits, and you're not limited by corporate penny-pinching design rules, then I'd just get a 10W jobbie.
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Old 15th December 2011, 03:47 PM   #5
AndrewT is offline AndrewT  Scotland
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Yes,
50% de-rating is a good rule of thumb for much in electronics.
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Old 15th December 2011, 04:00 PM   #6
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Running anything at 100% design limits is asking for trouble.

Common sense says keep things cool.

As Andrew T says, derate by at least 50%.
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Old 15th December 2011, 04:06 PM   #7
AndrewT is offline AndrewT  Scotland
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In Civil Engineering we work to 100% all the time.
But there is a caveat.
The design rules already include a Factor of Safety (FoS). This FoS varies depending on the operational and constructional conditions.

Woh betide any Engineer that uses 101% and it subsequently falls down.

In Electronics and Elecrtronic components, I have no idea what FoS the manufacturers build in. I have no idea how that FoS might vary depending on operational conditions.

All I have are the ratings stated on the product and/or in the datasheet.
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Old 15th December 2011, 04:26 PM   #8
Bill_P is offline Bill_P  United States
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The data sheet shows conflicting information. In the electrical specifications chart it lists full power at 70C. The power derating curve makes it look more like 55C for full power. Other manufacturers publish a family of derating curves that vary depending on the wattage rating of the resistor. Here there is only one curve for all wattages, a bit unusual and maybe not very accurate.

Taking the information from the power derating graph, at 55C and full power the resistor temperature will be 250C. That's extremely hot and will cause the PC board to discolor in short order. The slope of the graph is about 40C/Watt and the PC board can take 105C at most for the most common board materials. Starting from an ambient of 25C a temperature rise of 80C is the limit. That equates to 2 Watts dissipation in the resistor.

The copper on the PC board will not significantly aid in resistor cooling. Air flow would be helpful and the mounting position for the resistor should be well ventilated.
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Old 15th December 2011, 04:28 PM   #9
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by Bill_P View Post
.........That equates to 2 Watts dissipation in the resistor.
There's a fluke and a half !
Almost = 50% de-rating.
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Old 15th December 2011, 04:49 PM   #10
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FWIW, I've run sub-miniature MOX resistors near their max power rating, soldered to long fine traces, and they fell off the board! The redesign increased the trace size well beyond what was needed current-wise, and cured the melting solder problem. Significant dissipation does occur through the leads.

On a related note, I've seen early failure of electrolytic capacitors where one trace ran directly to a hot power resistor. The heat gets conducted right into the body of the capacitor and kills it.

You need to manage your heat transfer with as much diligence as the electrical functions of the circuit.
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