I confirm... IRFB4110 Mosfet CAN BLOW !!!

dirty_d said:
The7, its likely that the current wont be able to reach the max, depending on how many pwm cycles it takes the motor current to go from 0A to maxA, if it reaches maxA in 1 cycle the MCU will lower the PWM cycle the next cycle. it likely works like this:

1. read throttle voltage
2. read battery current
3. if battery_current < 75A, PWM_duty = throttle_voltage / 5.0V; else if battery_current >= 75A, PWM_duty = PWM_duty * (75A / battery_current)
4. repeat

its probably much different than that but you get the idea, if the motor current ever reaches the max it will only take 1 pwm cycle for the MCU to lower the duty cycle to bring it down.

You have present an algorithm with very fast response.
 
yes but its far from a perfect algorithm, consider if you have the throttle maxed, with a high load, the current will go over 75A, so the algorithm cuts back the pwm duty cycle, now its under 75A, so the next pwm cycle will have 100% duty cycle again. it will keep jumping from 100% back to whatever % is calculated to bring it down to 75A, likely the actual motor current will stay near 75A because the inductance keeps it from changing fast enough, but still its nasty. maybe a simple ramp up would fix that.
 
doc could you run through your reverse polarity protection?
I know you posted how to do with a diode and fuse, but you have two diodes used to protect, wondering how have set them up.
or if have posted somewhere already
 
to drbass:
just wondering about the 3k resistor using to bullet proof board, assuming thats dropping voltage down for low voltage circuits?
Would that mean ( using I=V/R) say for 48volt: I=48/3,000=16mA ( enough to run the board?)
at 86volt would be 28mA
v2 crystalyte boards i have are using 91ohmn gives at 48v (48/91) = 530mA
at 86volt ( 86/91) = 950mA, checked another brand board and gives about 500mA also ( its a 24v board).
 
The7 said:
eP said:
So for Doc's case (limit=75A based on PWM) your assumption is wrong !!!

I have to add f_PWM must be high enough to keep motor current ripple as low as possible to avoid motor's core saturation.
Hi eProf.

Thanks for pointing out these two vital points.

I would like to seek your opinion on the following two cases.

Case 1
If the rider tries to FULL throttle at stall, the response time of the current limit may not be fast enough to reduce the 100% duty cycle to a lower value.
During that short period of 100% duty cycle, the motor current could have reached a very high value (very close to the max current as if the motor has NO inductance) because the frequency of the PWM will have no effect in 100% duty cycle.

Case 2
The frequecy of the PWM will always in effect if the rider is partial throttle at stall. The "steady state" PWM duty cycle of this case is same as that of case (1).

The steady state PWM duty cycle is not 38% any more. However, it could be computed with all the known values of R, L and f_PWM by iteration. I would guess it to be 40% to 60%.

Your stupid ePupil.

ad 1&2 if the limit activation is substantially delayed (for hundred miliseconds for example) than you cannot do calculations based on that limit.

However you should be aware motor's rpm cannot drop immediately to zero, so even at that case ( of immediate limit absent) is highly probable that limit was reached long time before (again 100ms for example ) the stall occured.

So before you begin your calculations you have to answer yourself: was the limit activated or the limit was absent for substantial amount of time .
But to do that you have to know how much time is delay and what was rpm deceleration (rpm per milisecond ).

You cannot mix assumptions/calculations the limit is present (activated) and the duty is 100% (for the sake of substantial delay = limit is temporary absent) because it is confusing.
 
drbass:
tried 3k and then 1.5k as replacement for big resistor in v2 controller (not tried analog), resistance too high controller wont run, I could well be missing something on this. Measured the current v2 low volt circuits use is 30mA no throttle, 40mA motor running ( approx)
 
Have no idea how google dumped me here. But I gotta add my 2c .
In my time as a Lift Technician I have come across almost every type of drive and motor system imagneable so theres a few things I need to point out.

PWM Current Limit Issue:
Basic Answer, was the controller current limiting?. Definitly. With bigger 3 Phase AC motors even at 0RPM they will still be running a PWM of many KHz.. however the PWM never moves much less than 80% or so. cos the high frequency PWM (a square wave too... lots of back EMF in those!, Remember, we're not working with nice pretty sine waves here) at say 80% PWM you have massively increased the reactiance of the motor, even bringing it up to be many tens, even hundreds of ohms basicly its a LR Filter and in this situation this would keep the average current down in the motor (The motor would NEVER see 180 amps because of the iron induced reactiance)

2: Does the current limit have time to kick in?..... 3 seconds! thats eons when you happen to me a uC running at many MHz.. unless the code is written really poorly(I'm assuming the current limiting here is intagral to the controller, not some aftermarket addon, cos that'd make things totaly different)

Calculations regarding IRFB4110 Mosfets
Right... everyone seems to have missed reading the datasheet, ya have to read between the lines.. and you have to keep in mind semiconductors are quantum mechanical devices... and as such don't quite work the same way when you look at them hard. First off, the current rating of them is not even close to 180A. its 75A Max @25 degrees C Junction temprature... course the junction wasnt at 25.. the high current pulses warm it up a bit... this inccreases the Rds(on) at also does something funny, it increases the Vds... which in turn leads to more resistive losses inside the unit.. which leads to more heat.. higher Rds Higher Vds(they're not linerly related... this is not a resistor. its a semiconductor V=IR does not apply)



So... My theory:
The current limit was either poorly coded, or nonexistent in the controller (Say by using a CA to do the current limit). caussing a very high current to flow, basicly toasting the FETS... easy :D
 
cheetor said:
Have no idea how google dumped me here. But I gotta add my 2c .
In my time as a Lift Technician I have come across almost every type of drive and motor system imagneable so theres a few things I need to point out.

PWM Current Limit Issue:
Basic Answer, was the controller current limiting?. Definitly. With bigger 3 Phase AC motors even at 0RPM they will still be running a PWM of many KHz.. however the PWM never moves much less than 80% or so. cos the high frequency PWM (a square wave too... lots of back EMF in those!, Remember, we're not working with nice pretty sine waves here) at say 80% PWM you have massively increased the reactiance of the motor, even bringing it up to be many tens, even hundreds of ohms basicly its a LR Filter and in this situation this would keep the average current down in the motor (The motor would NEVER see 180 amps because of the iron induced reactiance)

2: Does the current limit have time to kick in?..... 3 seconds! thats eons when you happen to me a uC running at many MHz.. unless the code is written really poorly(I'm assuming the current limiting here is intagral to the controller, not some aftermarket addon, cos that'd make things totaly different)

Calculations regarding IRFB4110 Mosfets
Right... everyone seems to have missed reading the datasheet, ya have to read between the lines.. and you have to keep in mind semiconductors are quantum mechanical devices... and as such don't quite work the same way when you look at them hard. First off, the current rating of them is not even close to 180A. its 75A Max @25 degrees C Junction temprature... course the junction wasnt at 25.. the high current pulses warm it up a bit... this inccreases the Rds(on) at also does something funny, it increases the Vds... which in turn leads to more resistive losses inside the unit.. which leads to more heat.. higher Rds Higher Vds(they're not linerly related... this is not a resistor. its a semiconductor V=IR does not apply)



So... My theory:
The current limit was either poorly coded, or nonexistent in the controller (Say by using a CA to do the current limit). caussing a very high current to flow, basicly toasting the FETS... easy :D

Methinks this is a bit like teaching granny to suck eggs.

No offence intended, but coming here with a first post that plucks an antique thread from the grave and then proceeds to "educate" us all, as if we're all a bunch of ignorant fools, on the how and why of the obvious is probably not the best way to make friends and influence people. In fact it makes you look like a bit of a tosser.............

FWIW, there are a lot of us here who can give you chapter and verse on the way controllers work, the challenges of remaining within the complex SOA of power FETs, the even bigger challenges of dealing with lead and trace inductance and the pitfalls of using inadequate commutation capacitors on the controller bus. Some also know a fair bit about FET gate drive requirements, too, as well as the complex and unequal dynamic current sharing that can occur with large banks of paralleled FETs in packages with poor internal thermal performance.

This antique controller that Doc was playing with all those years ago when he blew those FETs had some known problems. It had a fairly low max commutation frequency for one thing, and switched the FETs rather slowly for another. It was superseded a long time ago now, although it's fair to say that the current crop of Chinese controllers don't excel at providing good gate drive for the large paralleled arrays of TO220 FETs they insist on using.

What, exactly, was the point of unearthing a thread from 2008 like this? I can only see one point, to give yourself an opportunity to make a first post that was an excuse to show off your limited understanding of how the things that we've been working with for years work.
 
4110 is 180A rated but only at 25'C

Then what would the amp rating be at 50'C ? I run my 4110 12 fet 90-100A peak with 70+ volts.
 
cheetor said:
Calculations regarding IRFB4110 Mosfets
Right... everyone seems to have missed reading the datasheet, ya have to read between the lines.. and you have to keep in mind semiconductors are quantum mechanical devices... and as such don't quite work the same way when you look at them hard. First off, the current rating of them is not even close to 180A. its 75A Max @25 degrees C Junction temprature... course the junction wasnt at 25.. the high current pulses warm it up a bit... this inccreases the Rds(on) at also does something funny, it increases the Vds... which in turn leads to more resistive losses inside the unit.. which leads to more heat.. higher Rds Higher Vds(they're not linerly related... this is not a resistor. its a semiconductor V=IR does not apply)
D
Your comment about folks misreading the datasheet ends up a little closer to home than I think you wanted it to. And no reading between the lines is necessary. :)

The theoretical max continuous drain current rating for the IRFB4110PbF is not 75A, it's 120A at 25C (bond-wire limited). And that's 25C case temperature, not junction temperature.

And a FET is absolutely a slave to Ohms law!! Vds, Rds(on) and Id all have the same relationship as V, R and I do in a resistor. Can that Rds(on) value change over time? Of course. But, the same thing happens with every resistor out there. All resistors have a temperature coefficient that causes the resistance to change as it is heated or cooled. You can still calculate the drain current in a FET at any time knowing Vds and Rds(on). The relationship is completely linear between these three parameters.

Of course, in a FET the relationship between Vgs, temperature, Rds(on), and drain current can get very complex to calculate. But you can get quite close checking the curves in the datasheet. None of this has nothing to do with Ohm's Law though.
 
TMaster said:
4110 is 180A rated but only at 25'C

Then what would the amp rating be at 50'C ? I run my 4110 12 fet 90-100A peak with 70+ volts.
The theoretical max continuous current rating is still 120A at a FET case temperature of 50C. In fact, it's 120A all the way up to over 105C case temp. But, this only applies if you have a cooling setup that can keep the case temperature at that level and never exceed the 175C junction temperature rating!!! In fact, you can go up to 670A when pulsing current, but you're still limited by the junction temperature rating of the FET. And always remember, the controller's case temperature is much, much lower than the junction temperature of the FETs. Even the temperature of the body or the metal tab of a FET is a lot lower than its junction temp.

Another problem is that, depending on how good a gate driver the controller uses (i.e., how fast the FET is turned on/off) and how well the FETs are cooled, you can blow a FET at wayyyyyy below it's datasheet current ratings. And this doesn't even begin to take into account all the other ways a FET can be blown. :)

[Edit] Changed "175V" to "175C"
 
IRFB 4110 FETs, like any other FET (or semiconductor device, come to that) have a multitude of interactive limits that have to be taken into account. The spec sheet isn't that easy to decode without knowing the way that different operating conditions interact.

Working through the data for this device (from the IR datasheet, attached):

The maximum current that the silicon on the device can take is 180A, but if you look closely you will see that there is a small footnote, 1, linked to that parameter that states that there is a package limit of 75A, so the true maximum current is theoretically 75A, not 180A.

Next you need to look at two other device parameters, Rdson (the 'on' resistance of the FET) and the thermal resistance, Rthjc and Rthcs.

Rdson gives you a way to calculate the power, in the form of heat, that the junction will generate when the FET is turned on and conducting current. Let's assume that the current flowing through the FET is the package limit of 75A and the junction temperature is 25 deg C. At the typical value of Rdson (0.0037 ohms) the power being lost in the silicon will be 75² x 0.0037 = 20.8W.

Knowing the initial power that the junction has to dissipate, we can now work through the temperature rise that it will produce. The two figures of Rthjc and Rthcs can be added to give the total thermal resistance from the junction to the heatsink. A good heatsink (bigger and heavier than the ones typically used on controllers) might have a thermal resistance of maybe 2 deg C per watt. This would give us a total thermal resistance of around 0.402 + 0.5 + 2 = 2.902 deg C per watt.

So, for our 20.8W the junction temperature will increase to 20.8 x 2.902 = 60.36 deg above it's initial temperature, making the junction temperature around 85 deg C.

It doesn't end there though, as if you look at the way that Rdson varies with temperature (fig 4 on the data sheet), then you'll find that it will be around 1.5 times the value at 85 deg C that it was at 25 deg C. This means that the heat being generated at the junction is now 75² x 0.0037 x 1.5 = 31.2W. With 31.2W of heat being created the junction temperature rises further to 31.2 x 2.902 = 90.5 deg C above the initial temperature, or about 115 deg C.

As you can probably guess by now, this process is running away. The more heat at the FET junction, the higher the Rdson and the greater the heat generated. Unless something is done to reduce the current the device will exceed it's junction temperature limit quite quickly and then fail.

The intelligent way to stop this happening, if you have to use FETs with such poor package thermal limitations, is to have a means of rolling back the current limit with FET temperature. This would provide a measure of protection to the controller from conditions, like sustained hill climbing, where the controller may be running hard on the current limit for long periods. This is hard to do with these small package FETs though, because the case temperature will always lag the junction temperature by a fair bit, due to the poor package thermal resistance. There is a time factor in this, too. Short bursts of high phase current may be fine, as they won't generate a sustained temperature rise in the FETs. The FETs can take junction temperatures up to 175 deg C for moderate periods of time, but obviously their performance at that high junction temperature is very much poorer (Rdson would be around 2.5 times greater, for example, meaning much higher controller losses).

Thankfully, the collective knowledge on here is far greater than it was a few years ago when Doc started making mega-controllers from Chinese parts that we didn't wholly understand at the time!

Jeremy












View attachment 232-20788-IRFB4110PbF.pdf
 
Hi Jeremy,
The IRFB4110PbF datasheet has been updated from your 11/05 version. The rating is now 120A, not 75A. :)

See the attached datasheet from 04/08:
http://www.irf.com/product-info/datasheets/data/irfb4110pbf.pdf
 
Thanks, John. The problem we have is that, as far as I can tell, we have no way of knowing whether we have the newer, better bond wire, packages or the older ones with the 75A limit. I guess if we buy from IR directly, or an approved source, then we may be reasonably assured, but do you know of a way to identify the newer from the older devices?

As far as the rest of the points made go, the 120A limit just makes the thermal case more severe. Substituting 120A for the FET current for the 75A I used in that example above, just makes the temperature rise greater and the junction temperature limit is reached more quickly . 120A gives 53.3 watts at 25 deg C (junction temp), which would push the junction temp to around 155 deg C above ambient (maybe 180 deg actual) which is over the junc temp limit from the off. Even with a hypothetically perfect heatsink (one with 0 deg C per watt thermal resistance) then the junction temperature at 120A will be in runaway, leading to over-temp failure, so the true continuous current limit will be well below that figure (53.3W gives 48 deg C rise from 25 deg C with a perfect heatsink, pushing junction temperature up to 73 deg C, where Rdson will have increased to around 0.005 ohms, increasing junction loss to around 72W, pushing temperature up to around 90 deg C, where Rdson will have increased further pushing junction temp up over 100 deg C, where Rdson increases further, etc, etc).

For all practical purposes, with a real-world heatsink, then I doubt you can push more than around 40 to 50A though a 4110 reliably. You may get away with high peak currents for short periods, but the controller would always be skating on the edge of failure if anything got a bit hot.

Jeremy
 
Jeremy, I'm not sure there actually are two versions (120A and 75A bond-wire limit) of the IRFB4110PbF.
Datasheets are often updated without any changes to the device itself as additional testing is done and more data is gathered from customers, etc. The initial specs can be set to a lower, more conservative, level to keep customers from screaming that the device can't meet its specs and then the specs updated to the actual capabilities of the device at a later date. But, I don't know if that applies here. Just that it happens.

IMHO though, from any even remotely reasonable source for that FET, you'll be sure to get one that meets the April 2008 specs. If anyone is buying from e-bay, or other sources like that, I think the least of their problems is a possible 75A bond-wire limit. :)

I agree with your limits. 20A is typically just fine with a decent heatsink and 40A-50A is a reasonable max.

<begin personal rant towards no one in particular>
But, these numbers are essentially useless since we don't know how effective the cooling is. A 4110 can easily burn out at 2W. I realize that many times we want a rough guess at the capabilities of a particular FET, but IMHO that's caused a lot more trouble than the estimate was ever worth.
<end personal rant>

But, it's been a tough working weekend and perhaps I'm just feeling cranky. :D
 
Jeremy Harris said:
Thanks, John. The problem we have is that, as far as I can tell, we have no way of knowing whether we have the newer, better bond wire, packages or the older ones with the 75A limit. I guess if we buy from IR directly, or an approved source, then we may be reasonably assured, but do you know of a way to identify the newer from the older devices?

As far as the rest of the points made go, the 120A limit just makes the thermal case more severe. Substituting 120A for the FET current for the 75A I used in that example above, just makes the temperature rise greater and the junction temperature limit is reached more quickly . 120A gives 53.3 watts at 25 deg C (junction temp), which would push the junction temp to around 155 deg C above ambient (maybe 180 deg actual) which is over the junc temp limit from the off. Even with a hypothetically perfect heatsink (one with 0 deg C per watt thermal resistance) then the junction temperature at 120A will be in runaway, leading to over-temp failure, so the true continuous current limit will be well below that figure (53.3W gives 48 deg C rise from 25 deg C with a perfect heatsink, pushing junction temperature up to 73 deg C, where Rdson will have increased to around 0.005 ohms, increasing junction loss to around 72W, pushing temperature up to around 90 deg C, where Rdson will have increased further pushing junction temp up over 100 deg C, where Rdson increases further, etc, etc).

For all practical purposes, with a real-world heatsink, then I doubt you can push more than around 40 to 50A though a 4110 reliably. You may get away with high peak currents for short periods, but the controller would always be skating on the edge of failure if anything got a bit hot.

Jeremy


You're off a bit Jeremy. You're calculating as it the tab was isolated on the 4110. It's not, so unless you're making a brushed motor controller (or have 3 seperate heatsinks), you've gotta have a silicone pad or kapton film layer between the FET and the heatsink.

http://parts.digikey.sg/1/1/139393-heatpad-220-009-sp400-7403-09fr-51.html

Even in the situation with a perfect heatsink that never increases in temp, realistic continous current is substantially lower due to the large decrease in thermal conductivity from the needed insulator material.

Fortunately the first part of the slope involved in thermal run-a-way is a nice gentle grade, as we inherently know the parts work fine at >80-100amp bursts during the acceleration stage, then cool while cruseing along at a current level below the thermal run-a-way current level, and our battery capacity tends not to be much over 2kw-hrs, so it tends to survive OK. It's when nuts like myself go for those 50mph extended hill climbs that makes for excellent demonstrations of the impending FET thermal-run-a-way.
 
Generally with paralleling N FETs you should not assume a "times N" increase in current capability. Without doing some serious device matching and balancing, SQRT(N) is a much better assumption.
 
liveforphysics said:
You're off a bit Jeremy. You're calculating as it the tab was isolated on the 4110. It's not, so unless you're making a brushed motor controller (or have 3 seperate heatsinks), you've gotta have a silicone pad or kapton film layer between the FET and the heatsink.

http://parts.digikey.sg/1/1/139393-heatpad-220-009-sp400-7403-09fr-51.html

Even in the situation with a perfect heatsink that never increases in temp, realistic continous current is substantially lower due to the large decrease in thermal conductivity from the needed insulator material.

Fortunately the first part of the slope involved in thermal run-a-way is a nice gentle grade, as we inherently know the parts work fine at >80-100amp bursts during the acceleration stage, then cool while cruseing along at a current level below the thermal run-a-way current level, and our battery capacity tends not to be much over 2kw-hrs, so it tends to survive OK. It's when nuts like myself go for those 50mph extended hill climbs that makes for excellent demonstrations of the impending FET thermal-run-a-way.

I just included the probable thermal pad/mica washer/kapton tape thermal resistance in the heatsink figure, but you're right, I was probably still pretty optimistic. My guess is that the typical controllers, with TO220 FETs mounted to spreader bars using crappy silicone pads have a total junction to air thermal resistance of around 5 or 6 deg C per watt.

Jeremy
 
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