Schottky diode maximum temperature - Page 1
Forum section: electronics / projects

author: shadewind
author_rank: Frequent Contributor
timestamp: October 19, 2011, 10:31:01 pm
content: I'm measuring the temperature of a Schottky diode on an LED driver I'm making and I'm trying to figure out what the maximum temperature it can deal with is.It's this one:http://www.vishay.com/docs/94119/10mq100n.pdfI'm looking at Fig 4 on page 3. My average forward current is 0.31 A and my duty cycle is 0.4 so if I'm interpreting the graph correctly, I should be good up to about 145 degrees C. Am I doing this right? It sounds a bit high! Of course, I'm nowhere near these temperatures but it's not exactly cold either.I suddenly realized I really don't have any clue how hot the "hot" components usually get in power designs...

author: sub
author_rank: Regular Contributor
timestamp: October 19, 2011, 11:07:27 pm
content: The spec lists a maximum junction temperature of 150 degC, so this sounds alright, though a bit close. Remember that the thermal resistance between junction and case will cause any measurements that you make to be low unless you take it into account.Looking at the diagram, you'll note that the caption says current vs. maximum allowable temperature rather than the other way around---the idea I suppose being that you would pick your maximum temperature and then use that to figure out the current limit, rather than the other way around. Since it says absolute temperature rather than temperature above ambient, I doubt it will provide a useful way to estimate the junction temperature from the current.

author: shadewind
author_rank: Frequent Contributor
timestamp: October 19, 2011, 11:18:19 pm
content: But it doesn't say "Allowable Junction Temperature", it says "Allowable Case Temperature". Also, since it says "Allowable Case Temperature" and not "Allowable Current", I'm assuming that you check to see what your current is and then figure out the maximum allowable temperature.

author: IanB
author_rank: Super Contributor
timestamp: October 19, 2011, 11:53:44 pm
content: The maximum junction temperature is 150 C, so with any given current there must be sufficient temperature difference between junction and outer case/leads to conduct away the generated heat. The graph is saying that at a DC current of 0.4 A (for example) a case temperature of ~146 C will provide a sufficient temperature difference to maintain the junction at or below 150 C. If the case temperature is any higher the junction temperature will exceed the maximum of 150 C.Of course this graph represents the absolute limit and you don't want to design anywhere close to the limit if you can help it.

author: shadewind
author_rank: Frequent Contributor
timestamp: October 20, 2011, 12:13:49 am
content: Quote from: IanB on October 19, 2011, 11:53:44 pmNo of course not. Measuring with a thermocouple right at the device shows the temperature to be ~ 90 C. I'm gonna check in the MOSFET as well.Another thing which seems to be a bit worrying is that the inductor gets pretty hot as well. It's one of these:http://www.epcos.com/inf/30/db/ind_2008/b82477p4.pdfMeasuring at the top of the inductor gives me about 80 C. But I'm wondering a bit about the data sheet. On page 4, it says "Rated temperature" at 85 C but at page 2, it says "Temperature range up to 150 °C".

author: IanB
author_rank: Super Contributor
timestamp: October 20, 2011, 12:22:27 am
content: Quote from: shadewind on October 20, 2011, 12:13:49 amI doubt the thermocouple is telling you the actual surface temperature of the device, unless you use some kind of thermal compound and insulation to attach the thermocouple to the device surface. The device could be hotter than your measurement. An IR thermometer is a better way of measuring surface temperatures.The rated temperature of the inductor is the ambient temperature at which it can safely handle the rated current. If the ambient temperature is higher you must de-rate the inductor according to the graph at the foot of page 6.

author: shadewind
author_rank: Frequent Contributor
timestamp: October 20, 2011, 01:22:26 am
content: I see. Do you have any ballpark estimate on how big the difference could be? I know there are lots of factors in play here of, course.I seriously doubt that this heat could be coming from the resistance in the inductor itself so I'm guessing it's coming from the FET and diode since they are all attached to the same net which also acts as a heatsink. Maybe I should have added thermal reliefs for the inductor?I'm assuming the heat from the FET is mainly from switching losses since the RDS(on) of 0.045 ? shouldn't be enough together with 0.5 A average current to cause any significant heat. The datasheet of the TPS40211 (the controller) suggested a resistor between the gate drive pin and the gate but maybe I should remove that to increase the switching speed.

author: IanB
author_rank: Super Contributor
timestamp: October 20, 2011, 02:39:02 am
content: Quote from: shadewind on October 20, 2011, 01:22:26 amThat's an interesting question. I just did a quick test with a big power resistor that I heated up by putting it across a power supply. I measured the resistor surface temperature with an IR thermometer and by placing a thermocouple against the surface. When the IR thermometer was reading 135 C the thermocouple settled out at a steady reading of 110 C. Both thermocouple and IR thermometer read within 1 degree of each other when measuring a control temperature where cooling effects on the thermocouple are eliminated.So I think you might assume the surface temperature is from 10 to 30 degrees C higher than the thermocouple measurement, depending on the air temperature surrounding the thermocouple. If you insert the thermocouple into an enclosed space that has had a long time to reach equilibrium then the thermocouple reading will be closer than if measuring an exposed surface in the open air.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: October 20, 2011, 04:06:03 am
content: Those IR thermometers are subject to the emissivity of the surface. The thermocouple needs grease and good contact, but also needs very fine leads to avoid distorting the reading. Heck, it's a wonder we can measure the temperature of anything! I've also had problems where PCB traces transfer a surprising amount of heat from one part to another. I didn't note the core of the inductor, but core loss data can be hard to interpret. I've had cores heat up under conditions I'd never have expected it. I wouldn't expect an air core to heat up except under extreme conditions.

author: Simon
author_rank: Unknown
timestamp: October 20, 2011, 05:36:01 am
content: Quote from: Conrad Hoffman on October 20, 2011, 04:06:03 amthat would indeed take some doing

author: shadewind
author_rank: Frequent Contributor
timestamp: October 20, 2011, 11:39:25 am
content: Quote from: Conrad Hoffman on October 20, 2011, 04:06:03 amThat's a ferrite core I believe. Are you saying that the heat might actually be from the inductor itself?

author: shadewind
author_rank: Frequent Contributor
timestamp: October 20, 2011, 02:28:46 pm
content: It seems pretty clear to me that it's the FET which generates most of the heat. When changed the 15 ? resistor at the gate and replaced it with a 0 ? resistor, the switching time halved (72 ns down to about 34 ns if I remember correctly) and the temperature measured by the thermocouple (which isn't the actual case temp, I know, but it's a reference at least) dropped significantly and it now measures around ~70 C at the FET, ~60 C at the diode and when I put the thermocouple under the inductor (you can get it under there), it gives me about ~ 60 C as well. At the top of the inductor, I get about 55 C.This is what the layout looks like, by the way:I'm bit worried that maybe I should have added thermal reliefs to L1 but all I kept thinking about during design was the fact that a lot of current will potentially pass through L1 and read somewhere that it can be a good thing for the inductor to take the heat off of the FET and diode. Would there be any point in adding thermal reliefs to L1, would it make a difference? Would it be a good thing?

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: October 20, 2011, 02:37:42 pm
content: All the different ferrite types have different losses for different frequencies and power levels. If you're good with a data sheet and smarter than I am, you should be able to calculate the temperature rise. There's also skin effect on the wire to take into account. IMO, they can all take substantial heat, though some types more than others, so if there's margin on the inductor and you need to cool associated parts sure, route some heat through the traces. Low impedance,minimum loop areaand short distances are so essential for switcher designs, I'd avoid any thermal reliefs in the critical loop. IMO, if you're getting 60 degree C readings max, the design is likely good.

author: shadewind
author_rank: Frequent Contributor
timestamp: October 20, 2011, 06:42:15 pm
content: Quote from: Conrad Hoffman on October 20, 2011, 02:37:42 pmBut that's measured with the thermocouple as well, you realize that? But anyway, you're saying that these are values one would expect for something like this? The switching frequency is rather high (800 kHz) so I guess the losses are too.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: October 20, 2011, 08:11:04 pm
content: Yi! 800 kHz is fast. Switchers at that frequency and higher are a bit of an art, one I'm not that skilled in. I've had problems at 100 kHz! Any slow edges will increase dissipation by a lot, as will the wrong core material or non-optimum design of the inductor; you need wire losses about equal to core losses for best efficiency. If you measure the efficiency, simple power in vs power out, what do you get? Since the power difference will be dissipated in the supply, does it have the surface area to do it?

author: shadewind
author_rank: Frequent Contributor
timestamp: October 20, 2011, 08:42:24 pm
content: Quote from: Conrad Hoffman on October 20, 2011, 08:11:04 pmYou can see the board above. This big area to which Q1, L1 and D1 are attached is the main heatsink area which extends to the back as well with an identical shape. I thinking about reducing the switching frequency to half of the current value which should high enough to not warrant larger components but reduce the power dissipation.I think I'll measure the efficiency and draw a graph vs. input voltage as it is now and get back to you.

author: shadewind
author_rank: Frequent Contributor
timestamp: October 20, 2011, 10:22:53 pm
content: There we go. Efficiency is about 90% which to me sounds pretty good considering everything.Here are the measurements:http://dl.dropbox.com/u/2462319/leddriver.pdf

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: October 21, 2011, 12:04:54 am
content: 90+% efficiency is super. Sure, people do better, but it takes a lot of development. Just make sure the heat is getting out where it needs to and you should be ready to light LEDs.

author: shadewind
author_rank: Frequent Contributor
timestamp: October 21, 2011, 05:36:21 am
content: I believe I can afford to drop the switching frequency to half anyway and halve my switching losses and thus increase the efficiency even further. I cannot do anything about the diode power dissipation since that's Iout * Vf (it doesn't seem to get much better than 680 mV at these reverse voltages).I believe that I can even keep the inductance since the ripple current doesn't have much bearing on the output voltage ripple in a boost converter, only the input ripple and that's still better that what's in the reference design.