Mofset rating question - Page 1
Forum section: electronics / beginners

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 12, 2016, 11:04:12 pm
content: This seems like a really basic thing, but I just want to make sure. Say I have a mosfet rated at 60v and 50a, does it really mean it can carry 50 amps, at up to 60v? Or is there a gotcha with that? I imagine the wattage rating plays a role too, as you don't want to exceed it's max power dissapation, but that can be solved with proper cooling. I guess if using it linearly then you want to pay more attention to that as well but for simplicity sake let's just assume using it as a switch.Idealy you would not design something to use the full rating of a fet, but just wanted to make sure the actual capability is not much lower due to perhaps another stat I need to look at.Just as an example I'm looking at this one here:https://www.fairchildsemi.com/datasheets/RF/RFP50N06.pdfMaking a basic SMPS as a little project with some spare and bought parts so just want to get a general idea of rating. My diodes will probably be the bottleneck as they're rated at a few amps only. Was thinking I may have to use an array of mosfets but one is probably going to be plenty.

author: uncle_bob
author_rank: Unknown
timestamp: July 12, 2016, 11:33:38 pm
content: Quote from: Red Squirrel on July 12, 2016, 11:04:12 pmHiIf you put 50A into a gizmo at 60V you will get 50 x 60 = 3,000W of power in that gizmo. I suppose that somewhere there is a device that is designed for liquid nitrogen cooling and for the price of a new BMW it will handle 3KW on a continuous basis. For the rest of us, you get products that have a power handling rating of say, 131W at a junction temperature of 25C (second page of the data sheet). If you look at the third page, it will tell you about "single pulse" ratings of the device.So simply put .. no. The data sheet gives you a pretty good idea of what the device actually can do.Bob

author: helius
author_rank: Super Contributor
timestamp: July 13, 2016, 12:03:30 am
content: The power dissipated in a device according to P=IV is proportional to the currentthroughthe device, and the voltagedroppedacross it. This voltage drop follows Ohm's Law V=IR. The resistance of a FET channel is not constant, but changes as the gate turns on and off. When the gate is off, the resistance is near infinite. When it is on, it is very low, only a few milliohms. So, when the FET is off, its resistance is extremely high, and no current flows. The drain to source voltage Vdss increases to whatever voltage is in the external circuit. You could view the power consumed as P=V^2/R, with near infinite R there is no power. When the FET is on, its resistance is tiny, and so the voltage drop is also very low. The channel passes effectively as much current as the circuit can supply, and it dissipates power as P=I^2R. R is very low but not insignificant, so as I^2 becomes 40,000 you're going to get hot...Normally the entire available supply voltage AND current is not dissipated in the FET because you are using it to switch a load that consumes power. There are cheaper ways to make a resistive heater.tl;dr: The voltage rating of 60V is when the FET isOFF. The current rating of 50A is when it isON.Both states consume little power compared to the linear region, where the IV product is higher.

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 13, 2016, 12:17:24 am
content: Ok so long story short, you can't go by that rating alone. What is the typical rule of thumb then? Like, can that mosfet handle say, 5 amps at 12 volts? What's the easiest way to know without getting into complex math formulas and deep theory? Do I in fact have to go by the watt rating as well? That particular mosfet has a rating of 131w, does it mean that the volts and amps times can't exceed this? Or is that just how much it can handle as far as thermal dissipation? I don't imagine the entirety of the power is going to be dissipated into the fet, only a very small amount. The rest will be dissipated into whatever the load is.

author: uncle_bob
author_rank: Unknown
timestamp: July 13, 2016, 12:21:40 am
content: Quote from: Red Squirrel on July 13, 2016, 12:17:24 amHiThe power the device will dissipate is dependent on the device's ratings *and* the size of the heatsink you put it on. A reasonable rule of thumb is that any time you go over a watt on something like this device, it's worth looking at the heatsink and derating information. So current running *and* voltage across the device *and* volts times amps > 1 ... check the ratings.Bob

author: Signal32
author_rank: Frequent Contributor
timestamp: July 13, 2016, 12:22:26 am
content: As helius explained, for your purpose where you are using the MOSFET as a switch there is no need to worry. You should only care about the 50A limit. If you're going to usit as a linear device, such as voltage regulation, constant current etc then you would need to worry about the voltage across it. In your case if you fully turn it on to 0.022R then even at 50A the voltage across it would be 1.1W for a total power dissipation of 55W which is under it's rating.

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 13, 2016, 12:45:44 am
content: Quote from: Signal32 on July 13, 2016, 12:22:26 amOh ok yeah that's what I was asking, if I use it as a switch, and not linearly. The ratings just seem high for their size, so I figured maybe there was a catch. It seems to me the leads alone are not thick enough to actually carry 50a but I guess that's also it's absolute max rating, so in a typical device you'd probably have multiple of these to share the load and not actually load them at their max.

author: Audioguru
author_rank: Super Contributor
timestamp: July 13, 2016, 12:56:16 am
content: You correctly said that its maximum power dissipation is 131W. But its datasheet shows the graph of only when its case is 25 degrees C or less. How will you cool it, its case gets very hot when it is dissipating power.

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 13, 2016, 01:22:23 am
content: I imagine something like this would have a HUGE heat sink block and tons of air flow, if it was actually at it's max.In my current case I'm only looking at pulling like 4-5 amps at very maximum, but I just got curious when I saw that the current rating is so high for something that is physically that small. Ex: for 60 amps you typically need 6 awg wire which is quite thick compared to the leads.

author: uncle_bob
author_rank: Unknown
timestamp: July 13, 2016, 01:23:22 am
content: Quote from: Red Squirrel on July 13, 2016, 12:45:44 amHiAs a switch, it depends very much on the gate voltage and the effective resistance between drain and source. If it was (say) 0.01 ohms, you have I^2 *R to deal with. At 50A, I^2 is 2500. Even at 0.01 ohms you are into a 25 W heat level on the device in the "on" state. Depending on your gate voltage you could be much lower, but you could also be higher.Bob

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 01:38:58 am
content: Quote from: Red Squirrel on July 13, 2016, 01:22:23 amThe point is, you can never get to the continuous MAX (power rating) in practice AND have its temperature <= 25 deg C. (Unless you cheat and unrealistically use liquid Nitrogen or something).Because real life heatsinks will be well above ambient temperature (i.e. > 25 deg C), under heavy loads, > 100 Watts.Hence the MAX allowable dissipation DROPS as the heatsink temperature rises. So the 131 Watts, is NOT POSSIBLE (with exceptions, such as short pulses, ignoring datasheet specs, and other methods) in real life (unless you CHEAT and use liquid Nitrogen or something).Also the continuous MAX rating, would probably give you a SHORTENED component lifetime (MTBF).

author: uncle_bob
author_rank: Unknown
timestamp: July 13, 2016, 01:51:24 am
content: Quote from: MK14 on July 13, 2016, 01:38:58 amHiWell .... never say never ....If you have a gizmo that runs over a wide temperature range (say -55C to +85C) and it needs to self heat:The max heat demand will be at -55. If the heatsink faces out (as it probably does), it will indeed be mighty cold, even with a goodly amount of power into it. A 0.5 C/W heatsink (not to crazy) will "only" rise 70C with 140W into it. It still will be below (barely) 25C in this case. You may indeed say that I contrived the numbers so it works that way (I did). You can also say that devices like this don't exist ... they do (OCXO's with regulated supplies). It would be very unusual for one to dump 140W into a heatsink. Something around 3 to 5W would be a bit more realistic. In that case, 0.5C/W is a bit off as well.... If you dig into a fancy microwave system, you might find something in there that works the same way and *does* dump 140W into a heatsink.Bob

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 13, 2016, 02:09:30 am
content: Yeah was speaking more in theory as well, I know heat has to be taken into account and you'd have to derate to some extent. That's where the more complex theory would come in.So basically you can in fact go by the max amperage rating but have to take the resistance into account as well to find out how many watts it will dissapate and derate based on operating temp.I imagine that idealy you probably want to design so that you don't use more than say, half of the rating in normal conditions, so that it can somewhat handle abnormal conditions too. For something like a big PSU or any high power stuff, probably would want a thermal cut off anyway.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 02:30:28 am
content: Quote from: Red Squirrel on July 13, 2016, 02:09:30 amI agree with that.If you want it to be very reliable and/or last a long time and/or tend to work first time, keeping the temperatures reasonable, is a good idea.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 02:42:01 am
content: Quote from: uncle_bob on July 13, 2016, 01:51:24 amYes and No.Technically and in practice, that would work, just as you describe.But if the Electronic design engineers want the device to be very reliable and/or last a very long time. They still will probably want to keep below (with a suitable safety margin), the MAXIMUM specification(s).Also (speaking from experience), going even a little (e.g. >= 10 deg C), above the maximum junction temperature can (and will, I've done it MANY times), VERY rapidly destroy the device. E.g. stuff melts inside the device.

author: KL27x
author_rank: Super Contributor
timestamp: July 13, 2016, 02:43:11 am
content: QuoteThere is a catch.If you're using as a switch with a slow switching speed:60V at 50A = total load resistance of 1.2R. The on resistance of the FET is 0.022ohms. So at this load, the FET is going to drop 1.8% of the voltage/heat. Or about 54 watts.According to the datasheet, it can dissipate a theoretical of 131 watts. Per the data, the temp differential to achieve this would be about 151C. The max temp rating on the part is 150C. So if the ambient temp around the device were -1C, then it would be possible to achieve this wattage dissipation with passive cooling.To sustain 54 watts, the differential must be around 62 degree. So if ambient is 30C, you ought to be able to sustain 50A at 60V with IC temps of 92C. Not quite the temp of boiling water. This is hotter than I would run a FET continuously. But depending on your duty cycle and period, it might be fine. I would treat these heat dissipation figures are what the package could theoretically dissipate, if it is the weakest link, provided good heatsink. Without heatsinking, you can't come close to even that. Temp will probably exceed 150C at 50A load at 60V. But again, with a reasonable duty cycle and period, it is rated to "sustain" this 50A just fine for maybe a couple seconds at a time, even with no heatsink. It can handle the current, if you can get rid of the heat within the restrictions of the package.These are just ballpark figures, AFAIC, and needs real world testing if you decide to push things to the limit. Without a heat sink, I wouldn't try to dissipate much more than 5 or 10 watts thru a 150C rated TO-220, continuously. So maybe derate the part by about 10-20% of the max in this example.Switching losses are another story. Obviously when you switch FET on/off, at every point in between on and off you are going to momentarily dissipate much more heat thru the FET. The faster your frequency, the tighter you need to drive the gate. And the more you will have to derate the part.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 03:05:25 am
content: Quote from: KL27x on July 13, 2016, 02:43:11 amThe datasheet says:QuoteSo taking the 10 Watts you give above ...10 (watts dissipation) x 62 (deg C, see datasheet, Ambient cooling resistance) = 620 deg Cso we would need about -600 deg C ambient. (Or -450 deg C, with a much higher junction temp).Since absolute zero is around -273.15 deg C I am interested in knowing where the ambient temperature is that low (-600 deg C) ?(Above is a Joke/sarcasm, sorry).tl;drMaybe a watt or even a couple of watts is a more realistic limit. For a non-heatsinked part, and no/little air movement.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 03:13:10 am
content: Quote from: blueskull on July 13, 2016, 03:09:24 amMy understanding is they were talking about NO heatsink (and I assumed no PCB copper cooling), and NO forced air cooling.If I have misunderstood, SORRY.

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 13, 2016, 03:13:26 am
content: Ah right makes sense that there would be more losses when switching, as for a very short time it is operating in it's linear region and the resistance is higher then. I imagine in super high efficient applications there are ways to drive the "on" faster/harder using capacitors or something.I think the hotter it gets the higher the resistance is too right? So you could get thermal run away if thermal dissipation is not handled right. Just look at video cards and the MASSIVE heatsinks they put on there. Those things are dissapating like 500 watts in a relatively small area, compared to say a 500w heater.If I was to drive 50a through a fet I'd be looking at a pretty big heat sink I imagine.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 03:17:36 am
content: Quote from: Red Squirrel on July 13, 2016, 03:13:26 amMosfet drivers, are a good way (but potentially expensive) of rapidly switching them, as they tend to have significant gate capacitance, to overcome.

author: KL27x
author_rank: Super Contributor
timestamp: July 13, 2016, 03:19:39 am
content: There's another gotcha in figure 2. At 150C case temp, max current is only 20A. So no. You can't just plug in a T0-220 into a PCB and make it switch 50A at 60V.

author: KL27x
author_rank: Super Contributor
timestamp: July 13, 2016, 03:22:33 am
content: QuoteYeah, I figured 10W as throw it out there maximum number to make a point, that no, you can't start to think you can dissipate 54W out of a TO-220 without heatsinking. Unless it's rated to operate at 1000C, lol.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 03:25:46 am
content: Quote from: KL27x on July 13, 2016, 03:22:33 amI agree, and fully understand.Sorry about my earlier post.I know what you mean.Forum posting is a pain in the neck. It is so easy to post late at night (or something) and accidentally make a comment which is a bit technically WRONG. Then find you have been torn into, the next day, when you check the forums.

author: KL27x
author_rank: Super Contributor
timestamp: July 13, 2016, 03:29:29 am
content: I didn't want to be wrong on the low side, then it makes my argument weaker. As it is, it is more rock solid with your nitpicking. The previous answers of yes you can pass 50A at 60V, as long as you aren't switching fast are plain wrong, then. Agreed?At ambient of 30C and perfect passive cooling/heatsinking, it looks like you will top out somewhere south of 40A at 60V, max theoretical sustained fully on state, at temps approaching 90C at the device.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 03:38:42 am
content: Quote from: blueskull on July 13, 2016, 03:15:53 amWe seem to be talking about slightly different things.When I said NO heatsinking, I meant NO heatsinking AND NO heatsinking by PCB copper either, which is unrealistic, I know, as usually there will be some.With decent PCB copper (or even (PCB) aluminium substrate) etc, you can get quite good results.Off hand, I'm NOT sure about radiated heat cooling of the device ?I probably assumed it was included in the free air ambient thermal resistance, but I'm NOT SURE about that (as regards datasheets) ?I presume it is defined in the small print of the datasheet, and/or described in the test circuits, often towards the end of datasheets.

author: KL27x
author_rank: Super Contributor
timestamp: July 13, 2016, 03:44:25 am
content: This is a good question, which would be practically useful to have an answer, then.Out of a TO-220 package. Soldered to a 1/2 oz board. Stuck say flat against board with just a copper footprint slightly larger than the back of the IC for additional heatsink. How many watts can an average TO-220 dissipate at ambient of 30C, while maintaining maximum temp of package of 85C? On average. This is kind where I was throwing out 5-10W... And that only considering this FET is rated for 150C max operating temps.In real world, I find even 1-2W too much for such TO-220 package device as LT78XX regulators, without heatsink, for longterm reliability.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 03:49:03 am
content: Quote from: KL27x on July 13, 2016, 03:44:25 amYour calculations make sense!After having a quick read/skim through the following ...http://www.ti.com/lit/an/spra953b/spra953b.pdfI've found out how technically complicated and hard, doing even the "free air", calculations are.It seems that the datasheet figure, already DOES include the PCB and other real life physical elements of the situation.Example of the complexity, involved.The angle of the thing, changes the air flow and hence the results (approximately).

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 04:03:13 am
content: Quote from: blueskull on July 13, 2016, 03:48:35 amThanks for the reply.That's useful to know, because unlike through hole circuits, modern surface mounted parts are NOT so easy to heatsink, like they use to be.Small components seem best for good radiation (cooling), as bigger components lose lots of surface area, relatively speaking.QuoteI've seen that method used, quite a lot. Usually with mosfets or power transistors (BJT's).

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 04:16:45 am
content: One tip, if you DON'T already know this, or it might have been mentioned (possibly by me) in this thread, already.Mosfets tend to have significant gate capacitance, so if you need decent switching speeds (to minimise switching losses etc). Then you will probably need mosfet drivers, which can be a bit pricey.This makes it much easier/faster to drive it from an MCU's port pins. E.g. PWM.

author: KL27x
author_rank: Super Contributor
timestamp: July 13, 2016, 06:13:59 am
content: https://www.eevblog.com/2016/07/06/eevblog-895-bec-pro-model-airplane-regulator-testing/somewhat relevant, at around 20 minutes, anyway.

author: MK14
author_rank: Super Contributor
timestamp: July 13, 2016, 06:40:33 am
content: Quote from: KL27x on July 13, 2016, 06:13:59 amThat video is a very good find, and has lots of relevance.It helps show the realities of how difficult it can be to cool electronics down, without fan assistance. Also it shows how things misbehave when overheating etc.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 13, 2016, 10:19:04 am
content: Absolute maximum ratings are single points in the V,I grid. They aren't simultaneous.Check the SOA (Safe Operating Area) for how much it can dissipate, for different time durations (single shot). You will note the right edge is defined by Vmax, the top edge is defined by Rds(on) and Imax, and a slope (note the log-log grid, so these are curves of constant power*) between those sides, which tells how much power dissipation you can get (assuming ideal cooling).*If you see a steeper slope, that means the voltage is high enough in that region, so that single spots within the die will overheat and cause failure. This is known as second breakdown. Historically, MOSFETs had been able to dissipate up to their thermal limits, but modern high-density switching MOSFETs are now nearly as prone to 2nd breakdown as BJTs are. (BJTs, in turn, can be designed to avoid this effect, and you will find audio power amplifier types that claim "free from 2nd breakdown" as a result.)If you need to dissipate high power, at high voltages, linear type MOSFETs are preferred. They are somewhat more expensive than the regular (switching) kind, but that's the price you pay for active power dissipation!Note that, if the SOA doesn't provide a DC curve, it's probably because it's not recommended. Note also that, even if it does, that's assuming ideal cooling. You must look up the RthJC(max) parameter, and add your heatsink, and thermal interface, thermal resistances to that. This new RthJC is used to calculate the practical power dissipation limit. Typically, you'll be strained to dissipate more than 50W from a TO-220, or 100W from a TO-247, or 150W from a TO-264, or...Tim

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 14, 2016, 01:32:51 am
content: On subject of mosfets I just realized something. On most if all buck converter topoligies, the "switch" (fet) is on the high side. But fets and transistor normally go on the low side! So how does this work? I also read you need to drive a fet with MORE voltage than the input, though from playing around with them that does not seem to to be the case, you only need a few volts depending on the fet. Am I missing something here? Maybe I missread something. Or do buck converters usually use P channel? From what I'm finding most fets used are N channel though.If I switch on the low side, it means the high side is always going to be at the same potential as the input, making it impossible to measure the voltage, as it will always read as being same as input.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 14, 2016, 01:42:47 am
content: Gate drive voltage is relative to source. Maybe you're measuring it "wrong". Maybe you need a bootstrap driver instead (change your perspective).Tim

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 14, 2016, 02:02:16 am
content: I'm hoping to keep it simple. It seems to work if I use it low side though... so I must have read wrong that the drive voltage has to be higher than input.But only issue I put it on the low side of the circuit, that means the positive stays "on" right through from input to output, so I can't properly measure it at the other end to decide if I need to increase or decrease pwm cycle. I hope that makes sense.I want this circuit done soon, time goes by too fast, I had all year to work on this and next thing you know that time has come. It's for a portable solar system that I want to bring camping. Worse case scenario I'll forget measuring the solar side, and just put the charge controller back. Idea is I want a display of solar side volts/amps and load side volts/amps but the charge controller's grounds arn't at the same potential and arn't isolated so it made it impossible to measure, then decided I'll just make my own basic controller but think this is more complicated than I expected. All buck converter topologies I see drive on the high side, but as far as I know you can't do that with a fet or transistor. At least not easily.Edit: I think I'm starting to see where the bootstrap would come in now, I think what I had read is that the voltage has to be higher if you want to drive on the high side, and that's what you need a boostrap circuit for. Is that correct? So say you are switching 20 volts, the bootstrap circuit would give 23ish volts to gate? Then that lets you put the fet on the high side.

author: batteksystem
author_rank: Regular Contributor
timestamp: July 14, 2016, 02:13:19 am
content: Quote from: Red Squirrel on July 13, 2016, 03:13:26 amAs the discussion is going on mainly on the current side, I think it would be important to remind you to allow margin on both voltage and current rating. Never use a 60V device with a 60V application. Some slight over-voltage will breakdown the MOSFET real quick.

author: Red Squirrel
author_rank: Super Contributor
timestamp: July 14, 2016, 03:04:22 am
content: Quote from: batteksystem on July 14, 2016, 02:13:19 amYeah makes sense. My question was more based on in theory, ex: it won't just blow up the second it hits that amount.I imagine there are rules of thumb for how you derate components. Personally as a noob I would just play it safe and over engineer everything then check thermal values while testing it.

author: uncle_bob
author_rank: Unknown
timestamp: July 14, 2016, 03:11:16 pm
content: Quote from: Red Squirrel on July 14, 2016, 03:04:22 amHiOk, since you keep bringing up rules of thumb:1) Never use a part at more than half the power rating unless you really know what you are doing.2) The power rating *always* must be fully calculated for your heatsink and ambient conditions.3) Never take shortcuts when it comes to power ratings.That's pretty much the full list.Bob