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
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
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.
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.
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.
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.
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.
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.
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.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.
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?"
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
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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
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
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The spacers are generally there so that the solder fillet will form properly, and so that the gap between can be cleaned after reflow.
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.
Yes.
Cheers, jn
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