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Old 15th December 2011, 07:15 PM   #11
benb is offline benb  United States
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Quote:
Originally Posted by Conrad Hoffman View Post
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.
To expand on that concerning construction: for any resistor or component that gets "warm," mount it elevated maybe 1/2 inch or more off the board. This lets air flow around all sides so it doesn't get as hot, and also the leads are longer, giving more distance for the heat to flow to the solder joint. Also, for several resistors that will get "warm," place them (in the PCB layout) with some minimum distance (1/2 inch, maybe?) in between them so likewise they have more air in between to keep them cool.

Looking at the link in the OP, the vertical case has two "lips" around the leads to elevate the body above the board, perhaps giving a little air flow between the body and the PCB. With wirewound resistors I'd like to know the inductance as well.

One last thing, the power rating only tells what the device will dissipate at the rated ambient temperature without breaking, it doesn't say the leads won't melt solder at that power! Tt can destroy other nearby components with its heat, but the manufacturer will say "yeah, but the resistor is still within tolerance, isn't it?"
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Old 15th December 2011, 07:43 PM   #12
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Originally Posted by eem2am View Post
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
Here's a quick simple reference, shows the area vs thermal resistance.

Heat Sink Temperature Calculator

Calculate the thermal drop from the derating "curve". Add to it the thermal resistance of two paralleled pc pads. If you have one square inch of pc pad, expect 43 degrees per watt for the pad area, added to the device thermal resistance...

Cheers, jn

Last edited by jneutron; 15th December 2011 at 08:04 PM.
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Old 15th December 2011, 08:17 PM   #13
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Default ran out of edit time...

Calculate the thermal drop from the derating "curve". Add to it the thermal resistance of two paralleled pc pads. If you have one square inch of pc pad, expect 43 degrees per watt for the pad area, added to the device thermal resistance...

Be very careful, I believe the derating they give is based on soldering at the plane of the standoffs. If you have longer leads, the lead thermal resistance will cause additional thermal drop.

The TJC for the unit is 5 watts gives 200 C rise, or 200/5 degrees c per watt, 40 degrees C per watt.

The thermal resistance of a copper lead is:

R = L/(A * K)

R is degrees per watt.
L is length in inches
A is area in square inches.
K is 10.2 for copper.

A 1 inch cube of copper has a thermal drop of 1/10.2, or .098 degrees C per watt.

A 32 mil diameter copper wire a tenth of an inch long:
A = 3.14*.016*.016 = .0008
R = .1/(.0008*10.2) = 12.2 degrees C per watt, or 122 degrees C per inch.

If you bend the leads and mount them flat, you'll use about .19 inches of wire, that's half the package thickness. That will be 12.2 times 1.9 = 23.18 degrees C per watt per lead, or 11.59 degrees C due to the leads...

So if you lay them flat and use 1 square inch of copper...

Tjc = 40 plus 11.6 plus 43, or 94.6 degrees C per watt.

If you have an ambient of 25C, you have 225/94.6 watts left...2.37 watts. and the board pads will rise 43 times 2.37, or 105 degrees C, to 130 C. Too high, as pointed out...but it certainly will NOT melt even tin/lead/silver eutectic at 179C.

ps..as a general rule, devices such as this are cooled by heat conduction through their leads. Note that the ceramic case is filled with an epoxy, and they do not specify the conduction coefficient of the epoxy..so trying to pull heat from the case sides is an unknown entity. There is a good possibility that they coated the resistor body with an RTV like coating to prevent the encapsulant from destroying the resistor when subjected to thermal cycling fatigue. That coating prevents heat conduction as well as eliminating shear forces on the ceramic.

Cheers, jn

Last edited by jneutron; 15th December 2011 at 08:35 PM.
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Old 15th December 2011, 08:56 PM   #14
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Quote:
Originally Posted by benb View Post
Looking at the link in the OP, the vertical case has two "lips" around the leads to elevate the body above the board, perhaps giving a little air flow between the body and the PCB.
The spacers are generally there so that the solder fillet will form properly, and so that the gap between can be cleaned after reflow.
Quote:
Originally Posted by benb View Post
One last thing, the power rating only tells what the device will dissipate at the rated ambient temperature without breaking, it doesn't say the leads won't melt solder at that power!
As I pointed out, the solder will not be in danger of melting. Board temps in the 130 C range will certainly not be good, and nearby comps can need derating as well. Tin silver is 232 C, lead tin 183C.
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Tt can destroy other nearby components with its heat...

Yes.

Cheers, jn
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