Mosfet heat dissipation gut check - Page 1
Forum section: electronics / beginners

author: swat3d
author_rank: Contributor
timestamp: September 10, 2014, 12:42:10 am
content: Hi,I'm a self taught electronics hobbyist. I know there are lots of articles and posts about mosfet a power consumption and heat dissipation, I've been through a lot of them. I think I got it but I'd like a gut check from you guys.I'm building the electronic load from Dave's videos.Here is the mosfet I'm using:http://www.nxp.com/documents/data_sheet/PSMN022-30PL.pdfI already had a heatsink lying around, it's a 3.4C/W.The max Tj for the mosfet is 175C, to keep it safe I've chosen 125C.For the ambient temperature I've chosen 30C.The Rtjb junction to base thermal resistance is 3.6C/W.I'm assuming a base to heatsink thermal resistance of 0.5C/W for the thermal compound.I've done as follows to calculate the max power I can dissipate.P= (150C - 30C) / (3.6C/W + 0.5C/W + 3.4C/W) = 16WThe temperature of the heatsink should be 16W * 3.4C/W = 54.4C ? Or 84.4C ?Now the Ron resistance at a junction temperature of 150C is 0.035Ohm.The max Vds voltage drain to source of the mosfet is 30v so I picked 25v.To calculate the current that can flow through the mosfet at 25v with a thermal resistance of 0.0035Ohm that will take the junction temperature to 150C I do the following.I = sqrt(16W / 0.036Ohm ) = 21ADoes this make any sense?If I wanted to find a suitable heatsink instead:Total thermal resistance = (150C - 30C) / 16W = 7.5C/WHeatsink thermal resistance = 7.5C/W - 3.6C/W - 0.5C/W = 3.4C/WAny feedback would be more than welcome, for some reason I'm suspecting the numbers I'm getting in terms how much current/voltage can run through the mosfet look kind of high to me.Thanks a lot I'm advance !

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 10, 2014, 01:34:28 am
content: Quote from: swat3d on September 10, 2014, 12:42:10 amFair enough, that'll give you headroom -- about 50C worth, which means you could practically stick it in an oven and guarantee it still working! Or, let the heatsink clog up with fuzz in a really dirty environment or something.Quote54.4C is the temp rise. At 30C ambient, that's 84.4C.Not nearly touchable, but quite safe by silicon standards....Where did 125C go? The 125C limit would be 12.7W.QuoteUmm..0.0035 is a typo obviously, but ohms aren't thermal resistance. What are you trying to calculate, and what variables do you need to solve it?You mention Vds, but don't go anywhere with it. To finish that thought, you'd have no more than 16W / 25V = 0.64A. Note that this is Vds alone, not including other voltage drops (current sense resistor; ballast resistor maybe?), so the terminal voltage of your load may be a little higher.Rds(on) + Rshunt + any other R in the circuit, is what limits V/I at the lowest end of the operating range. Due to Rds(on) alone, you'd have to drop:V = sqrt(P * R) = 0.75VI = sqrt(P / R) = V/R = 21.4Aassuming the other resistances in the circuit can handle that much current. The total voltage drop, obviously, has to include those resistances (V = I*R, add up all the V drops).QuoteThis is the correct value, for the transistor alone anyway. It didn't sound like that's what you were going to calculate... but maybe your hands "got ahead of your mind" as sometimes happens to us all?..More importantly, the amount of current your circuit can handle varies depending on how much voltage it has to drop.The relation:I = P/Vis an hyperbolic curve, so you have this curve of V vs. I, and anywhere below that curve you're safe. But it's really nonlinear (division), so it's hard to calculate just willy-nilly with analog circuitry. If you want to implement hard limits, analog wouldn't be the best way -- you might implement it with software and some ADC/DAC action, which would be perfect if you want it programmable anyway.You could also take the angle that, while it could do 20A at the lowest voltages, who's ever going to use that much, anyway? Let's just limit it to 0.64A across the board, so we never have to worry about weird cases. Or, since that's not very much, you could have selectable ranges, so maybe it only does 0.64A at the full 25V range, but it'll allow up to 3.2A on the 5V range, and so on. Each range would be a rectangle, with the corner touching that 'max power' curve I mentioned above.QuoteYes, this is the correct way to calculate back. Note again this is for T_J = 150C, not the 125 you chose at the very beginning.Tim

author: swat3d
author_rank: Contributor
timestamp: September 10, 2014, 05:46:46 am
content: Hi T3sl4co1l,Thanks a lot for the reply, I really appreciate you taking he time!The target junction temperature I was going for was 150C and not 125C, typo.I guess I got lost into the details, I started trying to calculate what heatsink I would need to run it at 25V, 1.5A and I didn't even consider I may not needed a heatsink. First time working with mosfets.Then I started playing with the formulas to better understand the relationship between V,R,I,P.I guess it makes sense that 54.4C is above ambient since (16W * 3.6C/W) + 84.4C would be 142C, close to the 150C target junction temperature. Is that right?So after calculating how much power I could dissipate with that given heatsink, I was trying to find out how much current makes 16W given the 0.036Ohm Ron resistance.It turned out to be the 21A.Now this is what I can't get my head around. Assuming here is no shunt or anything else dropping voltage in the circuit, current is shorted to ground.The 21A figure, how does that tie into the voltage? I guess it's as you mention V = sqr(P * R) = 0.75V.So in order to have 21A of current flowing through 0.036Ohm there has to be a voltage of 0.75V.When I set my PSU to 10V and current limit to .5A, the voltage goes down to 0v and the current limit led turns on. I guess is trying to supply more current than it can, therefore drops the voltage to produce more current? Why doesn't it just leave the voltage at 10V and only delivers the 500mA? That would be 5W. Isn't that what current limiting should do? It doesn't matter how much current the circuit is asking for, if it's set to 500mA it should only deliver that?0.5A * 0.5A * 0.036Ohm = 9mW0.009W * 3.4C/W = 0.03C above ambient for the heatsink(0.009W * 3.6C/W) + 0.03C = 0.062C above ambient for the junction.So starting again from scratch.Specs25V at 1.5A = 37.5W(150C - 30C) / 37.5W = 3.2C/W3.2C/W - 3.6C/W - 0.5C/W = -0.9C/W.It would be impossible which I guess I makes sense since the datasheet specifies the max power dissipation is 41W at 25C ambient and 175C junction.(175C - 25C) / 41W = 3.6C/W which is the junction to case thermal resistance, in which case you would need an ideal heatsink with a coefficient of 0C/W.So the specs gotta be lower than that.Given that I already have the heatsink and. Calculated the max power dissipation at 150C and 30C ambient is 16W, I know I want to pull 1.25A at least, 16W/1.25A=12.8VThe shunt resistor is same as Dave, 10 10Ohm resistors in parallel.1.25A / 1Ohm = 1.25V1.25A * 1.25A * 1Ohm = 1.56W / 10 = 156mW. The power rating for the resistors should then be 1/4W.Then the power dissipated by the mosfet will get lower after subtracting the 1.25V drop across the shunt.Does that make sense?I still don't understand how to tie power, voltage, current and Ron resistance. The calculations I've done above don't take the 36mOhm resistance into consideration.When you do P = V * I that's the power, but not the power dissipated i assume. Power dissipated needs to take into account the resistance, right? With a resistance of 0Ohm there is no power dissipation.@ModifiedI think I got it. So for a resistance os 0.036Ohm in order for it to dissipate 16W, 21A need to flow through it at a voltage of 0.76V.If I run it from a higher voltage, let's say the 12.8V found earlier it could only pass 1.25A and it will produce those same 16W.That's why you mentioned before having different currents at different voltage ranges or limiting it across the board.By limiting it to 1.25A, at lets say 5V it would be 6.25W.6.25W * 3.4C/W would set the heatsink to 21.25C above ambient.6.25W * (3.4C/W + .5C/W + 3.6C/W) = 46.8C above ambient for the die junction.Is that right?Thanks a lot in advance.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 10, 2014, 08:00:37 am
content: Quote from: swat3d on September 10, 2014, 05:46:46 amYes, give or take rounding I guess.QuoteIt can't deliver 10V AND 0.5A because the power supply would have to somehow reach out and change the load resistance to 10V / 0.5A = 20 ohms. It's not smart enough to do that... and frankly, neither am I!The current drawn from a voltage source is determined by the load resistance. Or alternately, the voltage developed by a current source is. If it's a current- and voltage-regulating type supply, it will choose whichever option yields less power (so it doesn't get overloaded).In other words, if you plot the points of (V, I) that the power supply can deliver (its load line), you will find it traces half of a square on the graph. For load resistances greater than 20 ohms, it will be a straight vertical line segment (constant V from 0-0.5A), and for lesser, it will be a straight horizontal line (constant I from 0-10V).Note: such a bench supply is NOT an ideal Thevenin voltage source (or Norton current source). Though it has a well-defined open circuit voltage (10V) and short-circuit current (0.5A), the equivalent output resistance is NOT 10V / 0.5A = 20 ohms. It is not a linear function of V(I), but a piecewise linear function.(Which is, in turn, a crude approximation of the real case. If you measure closely, you will see the voltage isn't constant along its segment, but it varies by some millivolts; likewise the current slope will not be perfectly orthogonal. The crossover region between the two segments will also not be an ideal sharp corner, but probably rolls off, following a curve, going gently from one to the other. The origin of this curve is probably related to some exponential function, such as a diode-OR function. And if you look *really* closely, you may see drift, noise, hysteresis and all other sorts of nasty effects.)What the supply is doing is, for currents less than 0.5A, it is intended to approximate a voltage source, which really means it has an equivalent output resistance significantly less than the operating range (i.e., << 20 ohms). Real regulators are in the milliohm range, so that, over a span of amperes, the voltage only changes by some millivolts -- a pretty stable voltage for most purposes. But if you try to draw more current, the voltage ceases to be ideal, and the approximation breaks down rather dramatically.Anyway, back to dissipation;QuoteCorrect. So if you want to dissipate more power, not only do you need a larger heatsink, but you need a better transistor, or more of them.QuoteA transistor is a poor approximation of a resistor to begin with (Rds(on) is only nearly true when the MOSFET is fully saturated), and you are purposely using it in the non-resistive region for this project -- instead, there are many ways you could equivalently describe a transistor in this state. A voltage-controlled resistor would be one poor description, but suffice it to say: the effective load your power supply sees will NOT be Rds(on), but something much larger than that.The only relevance Rds(on) holds in this circuit, is how low the voltage can go while maintaining the requested load current. Since you said the shunt resistance is 1 ohm, it clearly doesn't do you much good if Rds(on) is significantly smaller than 1 ohm -- if it were 0.1 ohm instead of 0.035 ohm, you can only go down to 1.1V instead of 1.035V -- but who cares, they're both pretty damn close to 1V, and pretty low in any case (would you ever really need it to work at less than 2, or even 3V?).Tim

author: swat3d
author_rank: Contributor
timestamp: September 11, 2014, 01:57:23 am
content: Thanks a lot guys for the answers I really appreciate it.I'm getting my head around it reading your explanations.So of course, when I set my PSU to 10V .5A that's the limit, it won't go above 10V or above .5A as you said but if I short it which I'm almost doing but connecting it directly to the drain, the source to ground and 5V at the gate, there is only whatever Ron resistance the mosfet has for that given current/temperature/Vgate. The PSU can't magically modify the load to add resistance and maintain the 10V.I was understanding it wrong I thought that the PSU would actually generate that voltage.So the maximum specs given for the mosfet, 30V, 30A, 41W mean, you can have a combination of V/A that will produce 41W through the mosfet staying within the voltage and current limit.Which means if you want max current 30A then 41W/30A = 1.36V.If you want max voltage 30V then 41W/30V = 1.36A.To produce the max current you'd need a resistance of 1.36V/30A = 0.045Ohm.To produce the max voltage 30V/1.36A = 22Ohm.I would say, wait a minute, the min resistance needed for the max current is higher than the min resistance do specified in the datasheet so the mosfet should actually be able to run more current than that. The min resistance in the datasheet is 0.019Ohm for a Vgs = 10V and Id = 5A. But after looking at Figure 13 I can see that Ron is not linear and it depends on multiple factors, T, I, Vgs, etc. at a Vgs = 10V and Id = 5A matches that of the table, the higher the current the higher the Ron and at 30A is not even specified so it could be that at 30A is actually 0.045Ohm.Or does it rely on external components limiting the current to not exceed hose values?Am I on the right track here?Now, those are absolute max values for a Tj=175C and Tb=25C of course you want to reduce those to a safe zone.So I guess the best approach is to decide the specs first as Dave usually says. It should be a total power that's below the max 41W of course.Let's say 20W for instance. For those 20W we need to decide the max current and from there figure out the voltage or the other way around. I'm gonna go for a current of 1A to keep it simple.20W/1A = 20VI will be sensing the current with 10 10Ohm resistors so:Vdrop = 1A*1Ohm = 1VP = 1V*1A = 1W/10 = .1W per resistorThe mosfet therefore will see 19V, it will have to dissipate 19W.To keep a die temperature of 150C at 35C ambient that's a raise of 115C.115C/19W = 6C/W total thermal resistance6C/W - 3.6C/W - .5C/W = 1.9C/W heatsink needed.To verify the results,19W * 1.9C/W = 36.1C , 71.1C ambient for heatsink(19W * .5C/W) + 36.1C = 45.6C, 80.6C ambient for thermal compound(19W * 3.6C/W) + 45.6C = 114C , 149C die ambient give or take decimals here and there.Those numbers are for maximum current and maximum voltage, anything below that will result in less power therefore will be completely safe.I hope I got it right this time!!Thanks!

author: swat3d
author_rank: Contributor
timestamp: September 12, 2014, 01:41:02 am
content: Hey, just to close this topic, I tested it today on the breadboard and the numbers are spot on.Thanks a lot for the help!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 12, 2014, 06:29:12 am
content: Quote from: swat3d on September 11, 2014, 01:57:23 amYup! If you turn it on hard, it'll either draw a whole lot of current, or yank the voltage down. Whatever the circuit around it behaves as (current limited, or constant voltage, or some resistors, or whatever).With some resistance 'below' the FET, current flow causes source voltage to rise which reduces Vgs, which eventually will throttle it down and put it in the linear (constant current) region. If you further control gate voltage to regulate output current, you get an even better one.And, your numbers look right.Tim