How long can an ordinary 50 ohm 1206 resistor survive having 30 volts across it? How about 60 volts?
- posted
5 years ago
-- John Larkin Highland Technology, Inc lunatic fringe electronics
How long can an ordinary 50 ohm 1206 resistor survive having 30 volts across it? How about 60 volts?
-- John Larkin Highland Technology, Inc lunatic fringe electronics
P = E^2/R = 30^2/50 = 18 watts DC or CW power. Rated power handling varies with the type of 1206 resistor. The highest power I could find in the list for 1206 footprint is 2 watts as series PCAN (aluminum nitride): I have no clue how long it will last at a 9x overload. Extrapolating from the PCAN1206 graph on the data sheet, the chip temp will rise to a surface temp of about 500C at 18 watts.
-- Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558
Hummm.... 1.1414 times longer than a 25 ohm resistor? :)
I suspect that the answer will be under a millisecond. I was just wondering if anybody has any better numbers, like a typical destructive number of joules.
My fault condition will be 60 volt pulses at a few MHz, RMS voltage about 30. I can probably ignore the 60 volt part, since the pulses last under 100 ns, so it's the RMS that will pop the resistor.
I can do a simple circuit that detects the overload and shuts off the pulse source. I can do that in maybe 50 or 100 us.
Ordinary 1206 thickfilms are good for a watt if the end-caps are soldered to good copper pours. That's irrelevant for a short pulse overload. There are suggestions that thinfilms are better for pulse overload.
-- John Larkin Highland Technology, Inc lunatic fringe electronics
The first-order handwave is that a 1206 thickfilm can absorb some number of millijoules in a short pulse, independent of resistance. Wild guess is 20 mJ. For a short pulse, the heat doesn't have time to diffuse very far out of the cermet resistance element. A 1206 can easily stand 1 watt for 1 second, which is a full joule.
I suppose I'll have to blow some up.
-- John Larkin Highland Technology, Inc lunatic fringe electronics
Oh, you wanted a guess(tm).
1206 weighs about 10 grams / 1000 pcs Specific heat of alumina = 800 Joules/Kg*Temp The Al Nitride chip is rated to survive to 155C. Ambient temp at 25CSpecific_Heat = (Watts * Seconds) / (kg * dTemp) J/Kg.K
800 = (18watts * sec) / (0.01Kg * (155-25))Seconds = 800 * 0.01 * 130 / 18 = 57 seconds
It will probably be longer as there will be some radiation loss and conduction loss through the PCB.
At 60 volts, the power dissipated will be 4 times as high. The relationship is linear, so the resitor will blow in 1/4th the time.
-- Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558
Oh, you wanted a guess(tm).
1206 weighs about 10 grams / 1000 pcs Specific heat of alumina = 800 Joules/Kg*Temp The Al Nitride chip is rated to survive to 155C. Ambient temp at 25CSpecific_Heat = (Watts * Seconds) / (kg * dTemp) J/Kg.K
800 = (18watts * sec) / (0.01Kg * (155-25))Seconds = 800 * 0.01 * 130 / 18 = 57 seconds
It will probably be longer as there will be some radiation loss and conduction loss through the PCB.
At 60 volts, the power dissipated will be 4 times as high. The relationship is linear, so the resitor will blow in 1/4th the time.
-- Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 ================================================== Does "1206 weighs about 10 grams / 1000 pcs" mean each resistor weighs 10/1000 grams? If so your answer should be 57 msec, right? If not, I'll go quietly (:-)), just tell me what it does mean.
I think that needs to be divided by 1000.
AlN is a really good heat conductor. Ordinary resistors are alumina, not so good. But for short pulses, the resistance film is almost adiabatic - the heat doesn't have time to diffuse - so the joule capacity goes down.
-- John Larkin Highland Technology, Inc lunatic fringe electronics
Yes.
2nd page (the pages are not numbered) is a chart with the mass (weight) per 1000 pieces.
Oops. Somehow, I always manage to screw up even the simplest arithmetic. I forgot to divide the 10 grams by 1000. Therefore, you're correct. It should be 57 msec. Corrected version:
1206 weighs about 10 grams / 1000 pcs Specific heat of alumina = 800 Joules/Kg*Temp The Al Nitride chip is rated to survive to 155C. Ambient temp at 25CSpecific_Heat = (Watts * Seconds) / (kg * dTemp) J/Kg.K
800 = (18watts * sec) / (0.00001Kg * (155-25))Seconds = 800 * 0.00001 * 130 / 18 = 57 milliseconds
It means that I shouldn't eat dinner while grinding numbers. Thanks for spotting my latest math error. (Grumble...)
-- Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558
But first orders being what they are, it misses the truth of the matter.
A better observation is that most resistors have a P ~ t^(-1/2) curve, though the exponent ranges from 1/3 to 1/2.
This reflects the _diffusion_ of heat into the body, a process that cannot be expressed as poles (first order and etc.). Which also means the energy is not constant at small scales, it varies too (as sqrt(t) or so).
Most resistors also give an endurance of 5x rating for 5 seconds, though for a chip resistor that will be much smaller, like 2.5x, or 1-2 seconds.
If your resistor doesn't have a starting point for basing this curve on, don't use it. There are plenty of pulse rated parts to choose from...
Tim
-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website: https://www.seventransistorlabs.com/
My appologies. I screwed up. My guess(tm) should have been 57 msec.
shows that the specific heat of AlN is 780 to 820 Joules/Kg.T. shows that the specific heat of Al203 is 451 to 955 Joules/Kg.T. I found some other sources using 800 J/Kg.T so I used that for both AlN and Alumina. Metallic aluminum is about 900 J/Kg.T. Not much variation in specific heat capacity among the various aluminum compounds.
What is different is the thermal conductivity. From the above URL's: AlN = 60 - 177 W/m.K Al2O3 = 12 - 38.5 W/m.k
Thermal conductivity only helps if you have a good thermal conduction path to a suitable radiator (or evaporator). The Vishay data sheet for the PCAN1206 resistors at were tested using: "Thermal imaging and load life testing was conducted mounting one device to a 1.6" x 3.7" test card with 3.5 mil copper plating on both surfaces. Thermal vias on 50 mil centers were utilized for heat transfer between surfaces of the test card." which means the test resistor was getting plenty of help removing the heat by the PCB. If so, your 1 msec guess is closer to the mark than my 57 msec guess(tm). It also means that an AlN and Alumina resistors will both blow up at about the same time if they don't have any additional heat conduction and radiation paths.
-- Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558
Mount it on the underside of the PCB and it will drop off when the solder melts.
Cheers
-- Clive
We're talking large fast-ish pulses. I don't think the heat will diffuse far enough in 50ms to melt the solder and even if it did, the surface tension of liquid solder is more than enough to prevent a 1206 resistor from dropping off.
Jeroen Belleman
and even if it weren't, dropping off would be a very slow process relative to the timescales of interest. And solder melting isn't the same as the R dying etc.
NT
Depends on the orientation of the PCB. If mounted vertically the resistor will float off its pads and drop off within a few seconds.
It's a shame more companies don't give pulse dissipation information. One exception: Vishay's CRCW-HP "Pulse Proof High-Power Thick-Film Chip Resistor". This part has a double-sided printed element, and has curves for single pulse and continuous pulses. The 1206 curve shows 4 ms for 20 watts, for n
Oh fun, are you going to give an answer too? So everything has ~3J/cc*K (volumetric heat capacity) (.12*.06*.02 * 2.54^3) ~2.3 E-3 cc, say a 200K rise?
2.3E-3*200/3 ~ 150 mJ. (Assuming no math errors) then at 18 watts it's about 8 m sec.?George H.
3.2 seconds.
The Panasonic ERJ T08 looks like it can handle 1.5 as much power as the Vishay part at 4ms, or 20W for 10ms.
-- Thanks, - Win
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