Understanding MOSFET power dissipation - Page 1
Forum section: electronics / beginners

author: danderson
author_rank: Contributor
timestamp: June 14, 2015, 11:34:57 pm
content: I'm building a configurable buck-boost power supply that needs to output, in the beefiest configuration, 16V@5A, aka 80W. I'm using a TI LM25118 as the controller IC. Per the datasheet equations, the components need to be sized for a peak inductor current of 18A.I'm now picking out the switcher FETs, and I'm not sure how to interpret the "power dissipation" figures in datasheets. For example,http://www.st.com/web/en/resource/technical/document/datasheet/CD00002071.pdfspecifies ratings:Drain-source voltage: 100VContinuous drain current at room temp: 26APower dissipation at room temp: 85WOn resistance: 60mOhmWhat I'm not sure about is how I determine if the power dissipation figure is sufficient for my needs. Ithinkthat the way I figure this out is by calculating the power dissipation through the FET's on-resistance, i.e. 18A * 0.06Ohm = 1.08W, which is very safe indeed... But that seems way off from the maximum rating, compared to the maximum current, so I'm worries that I'm getting it wrong.Is my method of calculating power dissipation correct? If not, what am I supposed to be doing?

author: BennVenn
author_rank: Regular Contributor
timestamp: June 14, 2015, 11:45:04 pm
content: There are a few very good design notes from the big manufacturers on calculating losses in the switch. Simplified it is a combination of RDS, gate drive dissipation and the time spent in the linear condution range. I'm sure the experts will jump in here with more info, but a quick google search will get you what you are after

author: rs20
author_rank: Super Contributor
timestamp: June 14, 2015, 11:48:04 pm
content: There are three things to check:-- Drain current: You've got 18A, which is less than 26A, so that's fine.-- Power dissipation: You've got 1.08W, which is less than 85W, so that's fine.-- Junction temperature: You need to take the thermal resistance from junction to ambient*, multiply it by 1.08W to get a figure in Kelvin, and add that to your max expected ambient temperature to find your peak junction temperature. If that's less than 125 degrees C or whatever the figure in the datasheet is, you're fine. If not, you need to re-think your heatsinking situation.The junction temperature test is almost always the one that catches people out, because they forget to calculate it. The power dissipation figure is for a crazy idealized situation where your transistor is bathed in nucleated freon.My instinct is that your design is fine, but do the calculations.* Or, if you're using a heatsink thermal resistance from junction to case plus the thermal resistance of the heatsink.EDIT: I concur with BennVenn, you need to figure out switching losses and add that into the power figure that you use for the junction temperature calculation.

author: rs20
author_rank: Super Contributor
timestamp: June 15, 2015, 12:10:19 am
content: Woah, woah, woah, you got your calculation wrong. This is what checking dimensional consistency is for:https://www.google.com.au/search?q=18A+*+0.06OhmThe answer? 1.08volts. P = I^2 * R; V = I * R. You're actually burning off more like 20W, which is going to need some serious heatsinking. All previous comments still apply w.r.t. calculating junction temperature.I think you need a different FET.Another thing to keep in mind is that your FET is only carrying 18A about 25% of the time, so you're going to be dissipating only 5W on average, but in 20W bursts. Figure 2 in the datasheet is there to help you figure out what the peak junction temperature will be in these circumstances. But, if you find that too complicated, you can just safely pretend the MOSFET is carrying 18A 100% of the time, and you'll end up with a design with some extra safety factor.Regardless, I think you need a different FET.

author: BennVenn
author_rank: Regular Contributor
timestamp: June 15, 2015, 12:23:10 am
content: 60mOhms is quite large for a modern MOSFET. An IRFB4110 has a 100v Drain-Source voltage at around 4mOhms. Bringing down rds dissipation to just under 2 watts. You'll likely loose a bit more than that with switching losses. And as rs20 says, this number is assuming 100% duty which you should never reach.

author: danderson
author_rank: Contributor
timestamp: June 15, 2015, 12:40:43 am
content: Thanks folks! 20W sounds much more in line with what I was expecting for such a beefy supply (woops, math is hard, clearly).The design is definitely getting some heatsinking, and possibly even some forced air for this particular configuration (I'm using the same layout for 5V@4A, 12V@3A, 22V@2A and 16V@5A, by swapping out various passives). So, there's definitely some thermal design that I need to do, and I'll be verifying performance with a thermal camera while under maximum load.And thanks for the tip regarding Rds(on), I'm new enough that I don't have good intuition as to what is high/low. Fortunately nothing is ordered yet, so it's easy to change.Thanks everyone for the wisdom!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 15, 2015, 03:25:17 am
content: Quote from: BennVenn on June 15, 2015, 12:23:10 amKnee-jerk reaction:Don't use the absolute smallest Rds(on).You *will* burn way more in switching losses!Such devices are only useful for battery management, low frequency rectification and stuff like that. They are hard to use at high frequencies and high efficiencies.As with all engineering, the question is not minimization, but optimization. Use only the smallest Rds(on) necessary for the application. When switching and conduction losses are about equal, you'll have it right.Also consider some series inductance (with snubber) in the supply. This is perhaps the second most common mistake.Tim

author: rs20
author_rank: Super Contributor
timestamp: June 15, 2015, 03:36:03 am
content: Quote from: AcHmed99 on June 15, 2015, 01:50:47 amSlightly pedantic, but this is quite often not a safe assumption -- what you're basically implying here is that k is always equal to delta on the "Thermal impedance" chart in the datasheet (e.g. figure 2 inthis one), which judging from the graph is certainly not always true! However, it iscloseto trueat high switching frequencies. E.g., at 30kHz, 20% duty cycle, you're looking at a thermal impedance 10% higher than you'd expect. Even so, that 10% is eating into your safety factors.If you can't be bothered reading the graph, you should work by peak current, not RMS current, to be safe. If you have to be wrong, always be wrong on the safe side.Quote from: T3sl4co1l on June 15, 2015, 03:25:17 amAgain, slightly pedantic, but itisactually minimization; but what you're minimizing is total losses (switching + conduction). But you're right, that'll tend to be when they're roughly equal. This is why I wish digikey had an interface to allow you to type in a custom formula to sort by!

author: BennVenn
author_rank: Regular Contributor
timestamp: June 15, 2015, 04:02:09 am
content: Quote from: T3sl4co1l on June 15, 2015, 03:25:17 amAhhh something I don't often consider. I'm often trying to push though a lot of current so I just assumed Rds losses would dominate. I'll definitely take this into account in my next desgn

author: rs20
author_rank: Super Contributor
timestamp: June 15, 2015, 01:16:09 pm
content: Quote from: AcHmed99 on June 15, 2015, 12:11:03 pmThe point that real-world waveforms aren't perfectly square is a good one - but the graph is still useable, and crucially, while it's not always accurate, at least it's inaccurate in the safe direction. And yes, when switching frequencies are high, you can often disregard the graph and just use RMS. I wasn't seriously suggesting using peak current (hence why the suggestion was attached to an "if") -- although if you have a design that works according to peak current, youdoknow that your design will definitely work, always. I'm just pointing out that the same can't be said for using RMS -- especially if someone blindly thinks that RMS will work when you're switching something at 60Hz or slower. I'm just encouraging people to understand what that curve represents.In particular, your comment that your measured case temperature are always correct suggests to me that you've completely missed the point of the graph in question, since that graph is all about avoiding short-term temperature peaks in the junction itself, which you can't possibly hope to observe at the case (let alone at the thermometer attached to the case). If the junction is operating at 10% duty cycle and rapidly oscillating between 50 and 170 degrees (i.e., way outside of safe region), the case might be at a safe-looking 60 degrees, but your junction is spending some of its time at 170, and getting damaged. So how long can my 10% duty cycle pulses be before my Tj peaks into a dangerous region? That's what the graph is for.