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Selecting proper MOSFET

Discussion in 'General Electronics Discussion' started by x2dz, May 25, 2011.

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  1. x2dz

    x2dz

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    May 25, 2011
    Hi everyone,

    I'm doing an automation project and am currently selecting the best (i.e. cheapest safe solution :) ) MOSFET for my application.

    The FETs will be mounted on a PCB and will be connected directly to an ATmega328 mC. The connection to the drain will be on the low-side ( meaning the FET will connect the load to GND/source).

    The largest load will be approx. [email protected] It will be an inductive load (a motor) and the FET protection has been thought of.


    Would a IRL540 logic level MOSFET be a good choice?

    Here's the datasheet ( http://www.bccomponents.nl/docs/91300/91300.pdf )

    This Power MOSFET is in a TO-220 package, has a R DS (on) = 0.077 Ohm with a V GS = 5V and according to my calculations should have no problems whatsoever handling the task it will be performing (and I'll more or less be wasting its potential on the sub Amp loads). I used P = I (of the load)^2 * R DS (on) .



    Thanks in advance :)


    p.s. The only thing that puzzles me is that there are two Power Dissipation figures mentioned in the sheet - one is approx. 50W (for the TO-220 package, in the beginning of the datasheet) and the other is 150W as a Maximum power dissipation rating for the FET. How can the package the FET is IN dissipate much less than the FET itself? What do these values represent?
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The 50W mentioned is in reference to the package, suggesting that it is typically used for dissipation up to this figure.

    The 150W dissipation is the absolute max and almost certainly refers to limits imposed by the structure of the MOSFET itself. Dissipation at this level for any period of time would be almost impossible to maintain without exceeding the maximum junction temperature (it would require an enormous heatsink).

    In your application, you are switching, so the greatest dissipation will be while the load is being switched, not when it is on (or clearly, off).

    And herein lies your problem., The gate will have a large capacitance, and the limited current available from the uC output will cause fairly long switching times.

    whether this is going to be a problem or not depends a great deal on how often the load is switched. The switching time is going to be very close to the total gate charge divided by the current available from the uC output.
     
  3. x2dz

    x2dz

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    May 25, 2011
    First off, thanks for your reply :)


    Second, about the power dissipation - what I understood from what you wrote is that virtually every MOSFET in a TO-220 package (absent a heatsink) will be able to dissipate no more than 50 Watts (given that the maximum dissipation of the FET itself is > 50W). Is this correct?

    Third. Basically I chose to use FETs because I don't like relays... Something to do with the clicking sound and the sparks on the inside just doesn't sit with me :) Meaning - they will be switching let's say (example figure) two times in a 10 minute interval. 4 minutes solid ON and 6 minutes solid OFF. That's all :)
    The mC is an ATmega328 - meaning every output port supplies a max of around 50mA of current @ 5VDC.

    All in all - do you think this MOSFET will have trouble performing the task I'm setting it up to do?

    Again, thanks for you reply!
     
  4. x2dz

    x2dz

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    May 25, 2011
    p.s.

    Although it doesn't matter to me how fast the FETs switch (on the order of ns) I did the calculation anyway - I get [1 829 ns] (100 ohm Gate resistor, V gs (threshold) = 1.5 V, 5V supply voltage, Qg = 64 nC).
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    No, what it means is that devices capable of dissipating 50W (or more) in this package will be able to be mated with a heatsink to allow dissipation up to about 50W with few problems. After you exceed 50W, things will start to get trickier.

    It's a rule of thumb. A bit like "A sedan is great for transporting up to 4 adults". Not every sedan can actually hold 4 adults. And you may be able to squeeze 5 or 6 into one, but it won't be comfortable. If you need to transport 8 people, you *might* be able to do it in a sedan, but things will be tricky.

    That sort of switching is very slow. It is unlikely that you'll see more heat than is generated by the current through the device when switched on.

    OK, let's assume 64nC and 50mA (or 50mC/s) The time required to switch the device on or off will be approximately ( (64x10^-9) / (50x10-3)) = 1.28us. Your calculation of 1.8us may be closer to the mark as I have not bothered considering a number of factors. (lets say it will switch in 2us)

    2us us actually quite slow, but considering the small load, the device isn't going to heat up much at all. -- if the load is 12V 4A, then the max dissipation is 48W for 2uS, which is about 96uJ, not enough to heat the device measurably (unless you do it many times per second). The dissipated power while the device is switched ON is about 1.2W. This is enough to make it quite warm without heatsink (the junction temperature would rise to about 80C above ambient), I would add a small heatsink -- something rated at between 10 and 20 degC per Watt would be fine -- this would limit the junction temperature to between 15 and 30 degC above ambient.

    You should have a resistor from the gate to ground to ensure that the mosfet stays OFF when there is no power to the circuit.

    I expect it will think it's on holiday :)

    It seems well over specced for the job. This is a good thing because it means it is unlikely to be a cause of failure even if a fault causes your motor to short and the mosfet has to survive until a fuse blows.
     
  6. x2dz

    x2dz

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    May 25, 2011
    Thanks for the reply :)

    You're saying that the 50W figure for the package is an indication of an easily achievable (via correct heatsinking) power dissipation. Got it. But what about the 150W for the MOSFET? What does this figure represent?


    How do you calculate how much energy (joules) would create a significant temperature rise?

    You get the 1.2W with this formula P = Id^2*R DS (on), right? And the 80 deg above ambient from the sheet (Max junction-to-ambient = 62 deg/W)? I'm asking this because I'd like to calculate how fast the FET would heat up to 80 above ambient and here's why - the 4A is basically the stall current for the motor I'll be using. Meaning I'd get to 4-5A in extreme situations and only for limited periods of time (I'll be monitoring the current with the mC and if it goes over a threshold this will lead to a stop). The normal/working current the motor will need would probably be closer to 2, 3A which gives on average about 0.48W of power... So if you could point me to a paper or give me the formula that computes the time needed to reach a certain temperature for certain power/temperature specs I'd very much appreciate it. :)

    The reason for this being that I have almost finished the PCB for the design and would like to keep it as small as possible and frankly I just don't have the room to put in a heatsink.

    Here it is if you'd like to take a look at it - http://imageshack.us/photo/my-images/833/routed4.png/

    I do have a resistor from gate to ground :) thanks for the heads-up.


    You say that but don't know that I originally thought of putting in ILR1004s in there - 130A monsters that would have gone wasted on this particular job :) It was just easier to use those since I know a supplier who has them in stock, but then I found out that the IRL540 is in stock as well so... :) Why waste the money?
     
    Last edited: May 26, 2011
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    It represents the highest dissipation that the junction can tolerate.

    You can probably imagine that a semiconductor does not heat up instantaneously. Whilst a certain amount of heatsinking may allow a constant dissipation of 50W, it will also allow dissipation of 100W at a 50% duty cycle (as long as the ON periods are not too long. However the maximum dissipation figure specified means that no matter how short the time, you should not exceed 150W dissipation.

    A Joule per second is a Watt. So if you are dissipating 10 joules over 2 seconds, the average dissipation is 5 Watts. The specs tell you the thermal resistance in degrees C per Watt. If that is 20 degrees C per watt, then the temperature rise could be as much as 100 deg C.

    Yep

    Yep, I estimated it based on that figure.

    For that you need to know the thermal mass of each part of the system.

    A simpler way is to average the power over some reasonable time period, like a second. The time period you choose needs to be short enough that the device will not achieve thermal equilibrium in that time.

    You could just suck it and see. The calculations show that at 4A continuous, the junction would not get anywhere near exceeding 175 degC. See how hot it gets. Even a small heatsink will help.

    But in general, the real answer is "You should have thought of that before you designed the PCB".

    If those 130A devices are in a TO220 package, they illustrate an interesting point. From memory the legs of a TO220 device will melt long before you get to 130A.
     
  8. x2dz

    x2dz

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    May 25, 2011
    Thanks for taking yet more time out of your day to help me out :)

    This time, just for fun, I'll start from the end of your post :D

    They are, in fact, TO220 devices - I was a bit puzzled, too, after seeing the figure, but... I guess this might be a bit like the Dissipation rating - e.g. maybe you could push 130A through the TO220 legs, but for say 2us ON,200us OFF :) I don't really know how it would be possible...

    Here's a link to the datasheet in question - http://www.irf.com/product-info/datasheets/data/irl1004.pdf - if you're curious you can take a look ;)

    I'll do some tests, for sure :) And I was thinking of reading up on heatsinking, anyway - maybe I'll find some formula which would enable me to calculate/simulate the temperature for a given scenario at a given point in time.

    I can, of course, always decide to alter the PCB design - it isn't carved in stone... I could probably squeeze in a heatsink or two for the largest loads... hm... I just looked at the design as I was typing and maybe there's a way I could attach a heatsink to one of the FETs and just leave it straight up (as opposed to bending it down and soldering a support to the PCB).


    I was hoping you'd know some formula to easily approximate the values, but thanks for trying anyway :) I'll do some digging and share what I find, if you're interested.

    Got it - no matter what you do you can never go over the maximum power dissipation written in the sheet.



    Do you happen to know the maximum safe temperature for a through-hole component mounted on a pcb (given that the TO-220's back will be bent up against the board)?

    Thanks again for all your help! I've now ordered the FETs and the fun stuff (testing) should begin soon :)
     
  9. x2dz

    x2dz

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    May 25, 2011
    Does someone have a clue as to what diodes can be used best as freewheeling diodes? (to block the back EMF from an inductive load when the power is removed)

    I have read different opinions on the matter and am currently trying to figure out what the best solution is - I have seen it done with standart 1N4007s, while some sources say it's better to use Schottky diodes... I guess both options work, but which is better?

    Also what should the current rating be for a diode used in such an application? - the diode mentioned above has its Imax=1A and Urm=1000V. This would not be enough to carry the back EMF of say a motor with 4-5A stall [email protected], would it? I'd need something rated @5Amps, but what should the rev voltage rating be?




    Thanks is advance!
     
    Last edited: May 27, 2011
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Given that your switching speed is quite low, A 1N4007 would probably work as good as anything.

    Normally fast diodes with low forward voltage drops are chosen so as to limit the exposure of the MOSFET to high voltage pulses.

    Lower speed diodes (rectifying diodes like the IN4007) have a significant capacitance and switch off slowly. This can mean that significant power is wasted in them if your MOSFET switches very rapidly. Slow switch on can expose the mosfet to high voltages.

    A diode with a low forward voltage drop is used so that current preferentially flows through the flyback diode rather than the intrinsic diode within the mosfet. The intrinsic diode generally has fairly poor characteristics and it is often best that it not be forward biased.
     
  11. x2dz

    x2dz

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    May 25, 2011
    Thanks for the reply! :)


    Wouldn't the 1N4007's If = 1A be a problem?

    After some reading up I reached the conclusion that the freewheeling diode should be (for voltages<100V):
    1) Preferably Shottky diode (because of lower forward voltage and quick recovery time).
    2) With about the same electrical (and most likely physical) characteristics as the MOSFET that it would be protecting.

    I went today to pick up my IRL540Ns and got an MBR1060 for testing purposes... Do you think its a better/safer way to go than a 1N4007? Its a Shottky 10A/60V diode in a TO-220 package.

    Here's the datasheet if you'd like to take a closer look. :) - http://www.fairchildsemi.com/ds/MB/MBR1045.pdf

    Thanks again for all your help - I think tonight will be the first series of stress tests for the poor little FET to see how well it fares in actuality. :)
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    That seems to be a good diode. It even has good leakage at elevated temperatures, something that Schottky diodes do not always have.
     
  13. x2dz

    x2dz

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    May 25, 2011
    Hi everyone,

    I've done some actual tests now and have concluded the following -
    [email protected] - almost no heating
    [email protected] - almost no heating
    [email protected] - slight heating, reaching about 30 degC in about a minute or so
    [email protected] - moderate heating
    [email protected] - FET getting a little hot to the touch in about half a minute
    [email protected] - the FET gets uncomfortably hot to the touch in about 10-20 seconds.

    Since I don't have one of those fancy IR thermometers at my disposal currently I'm stuck with the very unscientific "touch-and-feel" method of temperature approximation, but... All in all, I got some vital information based on actual tests rather than numbers :) I've decided that I'll include a heatsink just for the most heavy load-controlling FET (e.g. the one that will be controlling the motor). It turns out that heatsinking involves just as much (if not more) factors as picking the FET itself :) I've read a lot of info and would just like to put it out there so as to have someone read it and just say if I'm on the right track or not. Thanks in advance.

    p.s. I realize that I'd probably never have a failure even if I don't include a heatsink (or if I include the first I get my hands on), but I'd like to do the calculation just to be on the safe side :)



    I've calculated the following:

    - 150 degC as a maximum junction temperature (I've subtracted 25 from the 175 maximum given in the datasheet for a safety margin)

    - 50 degC as an ambient temperature (again with a safety margin)

    - P = I^2 * R ds (on) = 5^2 * 1 = 25W - 5A for the load (for margin) and 1 Ohm R ds (on) obtained from the "Normalized On-resistance vs. Temperature" graph for a value of about 25-30 degC Junction temperature. I read an article that suggested that when deciding on a heatsink you should look at the worst-case scenario. The wattage gets massively bigger when using values for R ds (on) gotten from this graph. Is this OK or is this over-protecting?

    - Maximum Junction-to-case = 1 degC/W (from datasheet)

    - Case-to-sink, flat, greased surface = 0.5 degC/W (again from datasheet)

    - [(150-50)/25]-[1+0.5] = 2.5 degC/W . This would be my worst-case-scenario - meaning anything < 2.5 degC/W would be OK for it :)

    Obviously this is way over the actual numbers I would ever need and I don't have a minority complex so I'll probably go with something more along the 10 degC/W line as Steve suggested. Again, the 5A (again exaggerated) is the stall current for the motor - meaning I don't think there will be a lot of times we'll be operating in this condition.


    I have two heatsinks in mind, but the problem is that the degC/W rating isn't given - just the dimensions. I've approximated their ratings based on this graph - http://robots.freehostia.com/Heatsinks/Graph1ex.gif

    - 100X70X15 - 105 cm^3

    - 100X55X18 - 99 cm^3

    By my accounts they should both be about 3-5 degC/W, but I'm a bit worried about just multiplying the values - I don't think this gives me the actual cm^3... Is this wrong?


    I couldn't find the formula that would give me the operating temperature for given parameters as Steve calculated - e.g. the "between 15 and 30 degC bit"...

    Could you possibly share how you reached that number?


    Thanks in advance to anyone who was able to get through all of that :D
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You've done a reasonably good job of calculating the heatsink required, except you made a few minor errors that mean you come out with a wrong result.

    That's really good because it dramatically illustrates the I^2 component of power!

    You first need to calibrate your finger (last post on that page)!

    Always worth doing. I would have done the calculation for 20degC ambient and no heatsink first. That would give you some idea of how much margin you have. Essentially Tj(max) - Tj(calculated) tells you how much warmer than 20degC ambient can get. However you never want to be running close to Tj(max). From memory you will roughly double the life of the component for each 10degC you can reduce its temperature.

    That's all fine

    Here's your mistake. That is the normalised on-resistance. You need to multiply the value you get from this graph by the value of on-resistance given in other graphs (typically calculated at 25degC or thereabouts)

    On Page 2 of the datasheet you will see specifications at Tj=25degC, and looking down you will the "Drain-Source On-state resistance" (Rds(on)). This is specified for both 4V and 5V Vgs. Let's use 4V, as it will be more conservative. This gives you a resistance of 0.1 ohms, when multiplied by the normalization value, you still have 0.1 ohms :)

    So the power is 2.5W, not 25W -- and if you continue from here you will find that your heatsink requirements are far less :)

    This is why I chose the Rds(on) for 4V Vgs rather than for 5V Vgs.

    yep

    Substituting the new figures...

    [(150-50)/2.5]-[1+0.5] = 38.5 degC/W

    A 38.5 degC/watt heatsink would be tiny.

    However... you really don't want the junction temperature to be so high. At the very least it will make the case so hot that you can burn yourself.

    I suggested a 10 to 20 degC per watt heatsink I think. This was based on the fact that the rating for a heatsink assumes it is mounted at a particular orientation in free air. This may not be the case. It is safe to assume that the heatsink will only give you half of the specified performance. So 10 to 20 becomes 20 to 40 in worst case practice.

    As you can see from the above, 10degC/Watt should allow you plenty of room for comfort even with the motor stalled.

    They're huge. what you want is something like this.

     
  15. x2dz

    x2dz

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    May 25, 2011
    Thanks for the quick response!

    I knew there had to be something wrong since the deviation from 0.077 Ohms (or 0.11) to about 1-2 (which I though originally to be Ohms but as you pointed out are indeed something like coefficients) seemed a little large.

    It turns out that I was a little off... On the order of 10 times :D :D :D

    I don't think I'll be able to get something like the heatsink you linked to, simply because its a little hard to get a little more exotic electronics stuff where I live :) I'll probably get to smallest one I can find and try how it works out :) When trying to "finger-measure" the temperature with a fitted heatsink, what should I be touching - the component itself or the heatsink? And where?

    Here are some pics of the heatsinks I can get -
    [​IMG]
    [​IMG]
    [​IMG]


    About the mounting of the heatsink - how would I go about that? Would I need plastic bolts, nuts and washers? What size (it's a TO220 package)? Would I need to buy thermal paste?

    p.s. About the washers - The girl at the electronics store has little clue about what she's selling other than - "Give me a resistor... OK", "Give me a capacitor... OK". So I assume it will be somewhat of a comical experience trying to explain to her what I want :D


    Thanks again for all your help :)
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The second one you show looks about right, although I'm making some assumptions about its size. Can you show it with a TO-220 device attached?

    It would be preferable to drill a new hole in the centre, but it's not essential (the heatsink looks like it was used for 2 TO-220 devices).

    My scale refers to touching the device, typically the epoxy part of the device. Understand that it's not exact :D

    hahah Give me a resistor. OK. Give me a capacitor. OK.

    You might try "Give me a TO220 mounting kit"

    As long as nothing is going to short out, you may as well not use an insulating washer. There is no guarantee that you can insulate those bolts from the heatsink anyway so you'd have to go plastic bolts (which would be part of the mounting kit).

    Yes, use heatsink compound. Don't get anything fancy. The entire aim is to have metal to metal contact with the compound filling the air gaps. Fancy "silver" heatsink compounds do very little other than separate fools and their money.

    The aim is to get a very thin layer of heatsink compound that fills microscopic gaps between the 2 metal surfaces. The amount you require is very small -- it's hard to apply too little, but it can be difficult to spread it if there is a very small amount. A good way to apply it is as follows:

    1) make sure that the device and the heatsink are flat and smooth -- they should sit flat against each other and there should be no visible gaps between them.

    2) place a smear of heatsink compound on one surface then place the device and the heatsink together.

    3) move the device on the heatsink in small circular motions to spread the heatsink compound. The circles should be small compared to the size of the device

    4) pull the device off the heatsink and examine both surfaces. You should see an even (almost transparent) coating with little peaks where the two surfaces were pulled apart. If some area looks thicker, or if the whole surface of the component is not covered, then put it back and do step 3 again. (if you need to repeat this more than a couple of times, wipe off the compound and try again).

    With practice you won't have to pull the device off to check -- it is actually a bad thing and to be avoided.

    Finally, place the bolt in place and tighten it.

    Next, fix the heatsink to the board, and finally solder in the leads.

    For best performance, the fins should be vertical so that convection can do its job correctly (with this heatsink it may be difficult unless you place the board on edge).
     
  17. x2dz

    x2dz

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    May 25, 2011
    Thanks for the reply!

    This would most likely trigger a response similar to - this


    Sadly I don't have any pics of the sinks with components attached - I'll most probably get the smallest one and see how things work out. I'm also thinking about borrowing an IR thermometer from a teacher from the university. I've seen it in his lab - he can run, but he can't hide :D

    Do you have an idea of a safe way to simulate the 4A-5A current since I don't have any device that consumes that much and works at 12V what I did the last time was I attached a Coax cable (the cable TV type) with a resistance of about 5 Ohms as a load switched by the FET. Is this OK (does it qualify as a load) or is its low resistivity effectively shorting my PSU?

    The PSU has Voltage and Current displays along with voltage and current limiting. It has internal electronic short protection as well as fuses (duh!). I was originally trying to get to 3-4Amps by connecting 2 12V dremels in parallel, but I just couldn't get it to more than 2.5Amps. With the cable @0.2-0.3V I can get about 4Amps and the supply gets a little warm to the touch (it's rated to about 5Amps so I guess this is normal)... Anyway I'm just asking what the easiest/cheapest way to simulate to load would be and whether the cable is in actuality the same as shorting the terminals (its about 2-3m long).

    Before I read the instruction and the steps for applying the paste I was just running it in my head and I ran it the exact same way :D Isn't that creepy :D

    About placing the component+heatsink on the PCB I was thinking of just soldering the legs of the FET in and just leaving it perpendicular to the PCB (which would be parallel to the ground). I guess it would just stay like this solely resting on the legs without a problem. Would you agree?

    Thanks again for your input - it has been invaluable.
    I'll be going to the shop tomorrow to get the heatsink :) and will be doing field testing shortly after :)
     
    Last edited: Jun 3, 2011
  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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  19. x2dz

    x2dz

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    May 25, 2011
    Hi again!

    I've now purchased the heatsink (smallest there was) and the thermal paste. Getting the components needed to insulate the two turned out to be harder than getting the radiator itself... But first thing's first...

    Where I'm at so far and what I've found (a picture is worth a thousand words...) -

    1) The heatsink didn't have a mounting hole drilled in so I had to rectify that :)
    [​IMG]

    2) The FET on top of the heatsink and the thermal paste
    [​IMG]

    3) The testing unit mounted on a breadboard (I know that breadboards aren't meant for high current applications, but its short duration so I think its OK... It does start to smell a little plasticky at 4Amps but... :D)
    [​IMG]
    (don't mind the MCU... It's not used ATM)

    4) Since I didn't have the right screw/nut I just put one in there to keep the sink roughly in place and splattered on some thermal paste to keep it cozy :D (I haven't cleaned up the residue on purpose because of the fragility of the thing as it is at the moment :D)
    [​IMG]

    5) The test rig in action - pushing [email protected] through the coax cable :D
    [​IMG]

    Sadly my PSU can't provide more juice than that (the transistors get hot at 4Amps in about 10-30 seconds even though there's a huge sink attatched to them)

    The good news - I guess the heatsink is way over the needed margin, because the FET doesn't even get warm when pushing the current :D I didn't think there'd be such a huge difference - from scorching hot in under 20 seconds to not even warm indefinetly :D So I guess mission accomplished :D

    I'll be going to another store later today to try and get the insulation stuff needed (and maybe see if they have somewhat smaller heatsinks). I'll report back with the finished thing :)

    Thanks for all your help, Steve!
     
    Last edited: Jun 10, 2011
  20. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    It's really hard to tel. I thought that heatsink was going to be MUCH larger.

    Looking at it, I'd say that it's probably around 7 degrees C per watt.

    Considering that the original package (not heatsinked) is probably around 65 degC/W, you're going from a rise in temperature above 80C (potentially) to one closer to 10C.

    It would be interesting to measure the temperature of the transistor both heatsinked and not and see what the difference is (and ambient - air - temp) so you can figure out the new thermal resistance.

    You can also do that before and after you fit an insulating washer to see what impact this has. It's probably of no consequence in this case, but getting a "feel" for these things is valuable. (also, measuring the voltage drop across the mosfet would be useful too. You can then determine the actual heat being dissipated and compare this to the results you got from the datasheet.

    You note that the transistors on your power supply get hot very quickly despite huge heatsinks. This gives you some indication of the efficiency of a mosfet. A relatively small heatsink can make the temperature rise almost not noticeable. (Of course those transistors are dropping a significant voltage - imagine what would happen to them without heatsinks!)
     
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