Thank you for such a well explained reply. I haven't got one of these modules with me now but I will try and find the gate driver part number.It would probably make it even worse...
A common problem with cheap devices using large power FETS is the circuit not having enough gate drive current, so the power device does not switch as fast as it could - and it's when it is part on that the power dissipation is very high.
The existing device has a typical gate capacitance of 7.1nF and maximum almost 9.5nF
The IRFP4568 are rated 10.5nF typical. Having double that is going to slow the switching far more & mean the FETs waste a lot of power.
ps. The module is apparently a "DPX800S"
What input & output currents & voltages is it running at?
I use it at various voltages but mostly 12V, 36V and 52V on the input.Remember the output limit is 10A; you can only [theoretically] achieve 800W at over 80V output.
At less than 80V, the wattage max is 10A x o/p volts, by the makers spec.
What input and output voltages are you using?
Once I have the module with me I will post the pn for the gate driver. Do you know of any mosfet I can get that is better than the 055N15A, maybe I could use a single fet which has a lower resistance rather than using two?Also of concern is gate drive R typically used to suppress parasitic L
transients, eg. its Pdiss. Make sure that R has sufficient power handling
capability.
Regards, Dana.
Bi-polar transistors do need resistors to balance their current, but MOSFETs have a positive temperature coefficient for their on-resistance so will inherently tend to share the load current when used as switches, no source resistor needed.I don't use fets/mosfets, but i guess it's the same for bi-polar transisters. So yes, you will need low resistance resistors on each source pin.
So on a BMS which has multiple mosfets I don't need to try and place the cable in the centre of them? I always thought this would help spread the load but from what you are saying they do it themselves due to their positive temperature coefficient?Bi-polar transistors do need resistors to balance their current, but MOSFETs have a positive temperature coefficient for their on-resistance so will inherently tend to share the load current when used as switches, no source resistor needed.
Thanks. I'll grab some temperature data from the stock setup then buy one of the mosfets and re measure. I used to think the main specification on mosfets was the rdson but it makes sense that the gate capacitance is extremely important. Most of the things I have that use mosfets are all power converting. Edit: I just checked and TK56A12N1 is in a TO-220 package? Will the heat transfer negate the gains or will it still be worth it? I assume there is nothing in a TO-247 package that is available at the moment. I can source the TK56A12N1.The dissipation due to "on" resistance at 30A (eg. 15A average at 50% duty cycle) is near enough half a watt, half the time - so a quarter watt; near enough irrelevant.
All the heat is due to dissipation while it is changing between fully off and fully on.
Something like one of these may actually work better, as it's got reasonable current and voltage ratings, but much lower gate capacitance, so should switch faster:
I am guessing this is referring to the crutschow post? Ballasting = resistors? Are you saying that because the mosfets are generally within close proximity of each other on the PCB it is still a good idea to use ballasting? What is ballasting?If relying on Rdson matching of paralled MOSFETs w/o ballasting I would be
concerned with this -
View attachment 137793
Looks like a significant current non sharing in the making.
And typically since multiple MOSFETS typically in the same thermal environment
they all experience similar heat rise, mitigating some of the beneficial effect of
rising Rdson with T. Not to lose full insight this is all non linear and starting at different
points on the Rdson vs T curve for each device I would still want to swamp out that variation
with ballasting unless I match at device test.
Regards, Dana.
Ballasting in this context is to insure MOSFETs that are paralleled, when on,
one of them do not hog most of the current thereby overloading it. So one
adds R in series to mitigate device to device Rdson variation and Vth varia-
tion where one turns on before the others. crutschow is correct, there is a
+ tempco to Rdson such that when MOSFET has more current it heats up and
its Rdson mitigating some of the current rise effect, eg. negative feedback.
But one has to do analysis to see if that effect insures no device is ever over
its ratings due to all changes such as initial Rdson tolerance, board layout
(slight R changes in traces), etc.. In fact in RF work this becomes very important
for RF Power MOSFETs because of all the AC parasitic involved resulting in RMS
power differences between them. Note also Rdson variation with T is non
linear, complicating analysis even more.
So at high currents using low Rdson MOSFETs one adds R in series to lessen the
effect due to simple load sharing differences. The sim I showed shows this problem,
but not a complete analysis as the device specs only show a typical and max Rdson
value, no minimum, so the worst case is not handled, eg. one MOSFET at minimum,
one at maximum, for the analysis.
Simply think of the problem as a simple voltage divider effect with load and the
paralled R's, and accounting for all tolerances.
Lastly note ballasting at high currents typically done with trace length versus
discrete R's.
Regards, Dana.
PS : By the way many high power MOSFETs actually have internal ballasting due to
their architecture. To insure no hot spots developed on die. You can google tha
He's vastly overcomplicating things for you - just stick a pair of identical FET's in parallel, with heatsinks etc. and see how hot they get, compared to a single FET. Bear in mind, you're still dissipating the same amount of heat, just spread across two FET's and heatsinks.So it's best to over spec the mosfet and use resistors or design the trace to accompany the mosfets. How do you work out the value of R or is it just a generic value?
To be honest I like to learn anyway. What you're saying makes things extremely easy though as sharing the heat load is the goal I am trying to achieve and measuring the amount of heat is directly linked to the resistance or performance/efficiency when they are in parallel. I think I will post the pn for the gate driver just to make sure as I don't want to buy another batch of mosfets as mousers shipping is ridiculous. Rjenkinsgbs suggestion of the TK56A12N1 seems like a good idea to be honest. Is it possible to deliver more current into the gate to achieve the same effect, I know it would be easier to change the mosfet but I wondered if it is possible to do that with a different gate driver? Also would using 2x TK56A12N1 be silly or overkill?He's vastly overcomplicating things for you - just stick a pair of identical FET's in parallel, with heatsinks etc. and see how hot they get, compared to a single FET. Bear in mind, you're still dissipating the same amount of heat, just spread across two FET's and heatsinks.
To be honest I like to learn anyway. What you're saying makes things extremely easy though as sharing the heat load is the goal I am trying to achieve and measuring the amount of heat is directly linked to the resistance or performance/efficiency when they are in parallel. I think I will post the pn for the gate driver just to make sure as I don't want to buy another batch of mosfets as mousers shipping is ridiculous. Rjenkinsgbs suggestion of the TK56A12N1 seems like a good idea to be honest. Is it possible to deliver more current into the gate to achieve the same effect, I know it would be easier to change the mosfet but I wondered if it is possible to do that with a different gate driver? Also would using 2x TK56A12N1 be silly or overkill?
I can't. These modules are already really fragile and shorting them blows nearly everything else on the PCB leaving them uneconomical or too difficult to repair. Unfortunately though I do not know enough about electronics to be able to make a better decision other than a guesstimate. This is why I ask in places like here because you guys are clearly way more experienced in this subject. Any advise is always appreciated!Yes you can always throw parts into a design and burn them up, thats a positive learning
experience. Or take reasonable well established principles and protect in advance as the
posted ap notes show from the experts. Good engineering generally does the latter. Of
course you can always check the "identical" box when you order parts. let me know who
will ship to that precise spec.....
Regards, Dana.
I will post the pn for the gate driver and then it should hopefully be clear if swapping out the 055N15A for something with a lower gate capacitance will help. From what the other guy said it doesn't sound like it can be anything else as 1/4W certainly isn't what it is turning into heat. Hopefully he is right and it makes a noticeable difference. I wonder why they didn't figure that out and use a fet with a lower gate capacitance in the first place though.As I see it, spreading the load across two transistors and two heatsinks can only help. 'If' the FET isn't switching fast enough, then increasing gate current will help - but it all depends on how 'slow' it might be, and if the difference is great enough.
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