Doc EXTREME 36 FETS controller

[CamLight]

The water measured waterflow was: 235ml per minutes
The ambiant temp of the water at input was 20.9 degree C
The water at the output of the heat sink tube was 24.3 degree C
The resistor temperature close to the heat sink was at 34.4 degree Celsius
The dissipated power was exactly 50Watts

CAMLIGHT HERE ARE THE DATA YOU ASKED FOR:
34.4 degree C at the resistor.. the closer as i was able to put that sensor and firmly pressed against the resistor and aluminum heat sink.
with 235ml per minutes

Doc

Thanks Doc!

I'll use your numbers...
Assuming 20.9C ambient, 34.4C heat sink temp, and 50W power, we get the following:
34.4C - 20.9C = 13.5C temperature rise.
13.5C / 50W = 0.27C/W heat-sink-to-ambient thermal resistance for the setup you described.

Using tostino's numbers (adjust as necessary for whatever power dissipation levels are finally determined to be the proper numbers to use)...
- 36 FETs with 311W to be dissipated.
- The better Sil-Pad's at high pressure (over 100PSI for the TO-220 mounting pressure) = 1.75C/W case-to-sink thermal resistance. This can shoot up to over 6.0C/W for the lousier performing pads at lower pressure!!!!!
- IRFB4110 junction-to-sink thermal resistance - 1.11C/W.

Soooooo, here goes...
- Total junction-to-sink thermal resistance for each FET = 1.11C/W + 1.75C/W = 2.86C/W.
- Equivalent junction-to-sink thermal resistance for 36 FETs = 2.86C/W / 36 FETs = 0.0794C/W.
- Junction-to-ambient thermal resistance for 36 FETs on that bar with the above-mentioned cooling setup = 0.0794C/W + 0.27C/W = 0.349C/W
- Temperature rise for the FETs = 311W * 0.349 = approximately 109C junction temperature rise, above ambient, for the FETs.

This assumes.....
- All 36 FETs are on that one bar.
- Each FET is dissipating the same amount of power. This ain't going to happen. But, you can recalculate as necessary if the power levels for any individual FET is higher or lower (e.g., for high-side and low-side FETs).
- The water stays at the same temperature all the way down the bar. This ain't going to happen either. Let's assume that it goes up 10C at full power (maybe?), then the FETs operating where the water exits the block will be operating 10C higher than the ones near to where the water enters the block.

Assuming your ambient (water) temperature is 21C - 31C over the length of the block, your FETs will be operating at 130C - 140C if dissipating 311W.

That's not a problem at room temperature but at higher ambient (water) temperatures you'll start bumping up against the maximum recommended junction temperature for reliable, long term operation, 145C. If you want to exceed that temperature, you've got another 30C before you reach the abs. max of 175C.

 

[CamLight]

Here are the measured temp at different time after i disconnected the power to the resistor but i left the waterflow unchanged

34.8 at 0sec
33.2 at 15 sec
30.2 at 30 sec
27.7 at 45sec
26.1 at 60 sec


Any data to calculate with that?

Doc

As you mentioned, this can help you determine the thermal inertia of the system. To use this number, you'll need to know the duty-cycle for the controller and how it changes over time in order to adjust the temperature rise/fall of the system as you ride. But, the numbers will change for every ride you take and you'll go nuts trying to figure out the FET temperatures this way. IMHO, it's best to use the dissipated power numbers for an average ride and add temperature protection OR use the worst case numbers (highest power dissipation at the highest ambient temperature) and design for that without needing over-temperature protection.

 

[tostino]

Tostino, Does your spreadsheet calculate heat from an entire 36 fets controller?

If so, assuming that i use 150A continuous battery current , that's 375A of phase amp according to the 2.5x factor of the current multiplication effect with the pwm. previoulsy discussed on past threads http://www.endless-sphere.com/forum...p=289898&hilit=current+multiplication#p289898

with 375A of phase amp that's 2739 watts at 100% duty cycle .. unless at 100% throttle, current multiplication factor fall to 1.0 ??

Ahh now I rememeber! The simplest way to calculate it is this: phase current = battery current / duty cycle.

It's not a fixed ratio like you thought.

My spreadsheet calculates the losses for one phase if I remember right... But it should be pretty accurate for the whole controller since a single phase will always be "on", right? I have it so you would put in the resistance of each bank of 6 fets. So each bank of 6 fets should see that heating for 1/3rd of the time

 

[tostino]

Hrmm, so from your calculations Cam, we will need MUCH better heatsink to ambient numbers than Doc is seeing right now.

This can be achieved by better water flow, and also more surface area for the water to contact the copper tubing. I am gonna use more runs of smaller tubing to achieve this. You may have the same amount of water flow through a few smaller tubes as a single larger one, but the amount of surface area the water is able to contact is greatly increased. I believe this will have a huge impact on cooling effect.

Either way, I can't see any possible way that this is worse than air cooling, and yet people with air cooled 18 fets are pushing 100 amps from the battery no problem. Hell my old 12 fet controller pushes 100 amps battery side and is limited to 165 phase amps. It is still working perfectly fine.

More water flow seems to be the answer we're all looking for.

 

[Doctorbass]

The waterflow i used for the test was just FOR TEST and i used lower waterflow to get bigger delta Temp to try getting more accuracy fo rthe calculations.

I can easy use 10X better waterflow in the same tube.

Also we must remember that this aluminum bar i used was ONLY to calculate one side of teh controller... in the final projetc, i will use two watercooled aluminum bar ( one for each side)

Plus, i will alsi use the heat sink o fthe aluminum box to contribute to dissipate the heat.. so i will have both aircooling and watercooling.

I also think that if we was able to draw 100Amp ( up to 250A phase current) on a single 18 fet controller without any waterflow

We will be probably be able to draw 200 amp on a 36 fet by combining watercooling and aircooling as well.

Sice we are not 100% sure, that's why i also plan on being able to replace the mosfet easy if something blow.


Big thanks John for your time and all calculations and explanations!

Doc
Doc

 

[karma]

could have a heat-pipe as a backup :wink:

im using it on my small 9 fet :wink:

 

[tostino]

Oh gosh, I can't believe I forgot about having two different heatsinks for each controller. That means at the current crappy flow rate, you need to cut in half the c/w rise for the calculations.

That means that 0.0794C/W + 0.135C/W = 0.2144C/W
311w * 0.2144C/W = 66.6 C junction rise over ambient at the same heat output That sounds a whole lot better than 109 C rise at that same crappy water flow to me!

We now need to know the difference in C/W for different water flow rates so we can try figuring out the relationship between the two.

 

[CamLight]

My pleasure Doc! Good luck with the build.

tostino,
Good point about the flow rate vs. thermal resistance numbers. I'm kind of leaning towards a 10X flow rate increase only improving the heat sink's resistance numbers a bit though. Not much surface area to pull heat from. But, I could be very wrong about that.

Doc,
Could you increase the flow rate to the value you'd use in the final configuration and then see how much power it took to raise the heat sink temperature (in the same place, next to the resistor) to the same value as before, 34.4C?

 

[texaspyro]

And remember that you have to extract the heat from and recirculate that water... we still need the numbers with the water/glycol/beer recirculating.

Instead of a normal heat sink/dinky copper tube, how about using some aluminum square tube (like http://cgi.ebay.com/2-square-x-1-4-w-6063-aluminum-tube-x-36-long-/130340697367?pt=LH_DefaultDomain_0&hash=item1e58e93517#ht_500wt_722). Cut to length, weld some plates with hose fittings on the ends. Drill and blind tap for FETs. You now have a huge aluminum thermal mass and plus a jug load of water to keep things cool.

 

[CamLight]

Oh gosh, I can't believe I forgot about having two different heatsinks for each controller. That means at the current crappy flow rate, you need to cut in half the c/w rise for the calculations.

That means that 0.0794C/W + 0.135C/W = 0.2144C/W
311w * 0.2144C/W = 66.6 C junction rise over ambient at the same heat output That sounds a whole lot better than 109 C rise at that same crappy water flow to me!

We now need to know the difference in C/W for different water flow rates so we can try figuring out the relationship between the two.

The temp rise is as you calculated but the numbers to use are different as the heat sink resistance is still the same but the FET resistance and power levels are different.

Let's roll the numbers....
- Assuming two heat sinks, each 0.27C/W sink-to-ambient, in setup used by Doc previously.
- 311W / 2 sinks = 155.5W each sink
- 36 FETs / 2 sinks = 18 FETs
- Heat sink-to-ambient resistance is the same = 0.27C/W
- FET w/SilPad junction-to-sink resistance is the same = 2.86C/W
- Equivalent junction-to-sink resistance for 18 FETs is now 2.86C/W / 18 = 0.159C/W
- Total junction-to-ambient resistance is now = 0.27C/W + 0.159C/W = 0.429C/W
- Temperature rise from ambient is now = 0.429C/W x 155.5W = 66.5C

So, your temp rise number was right, but I wanted to list the calculations for others who might want to run through the numbers for different variable values.

 

[ZapPat]

- Equivalent junction-to-sink resistance for 18 FETs is now 2.86C/W / 18 = 0.159C/W

It seems to me that there are only 6 FETs in parallel sharing a thermal path at any time on the PWM'd phase, not 18. So that number would be 3 times higher, about 0.477oC/W per FET group, using those *good* silpads (isn't that a contradiction in terms?). So that changes the other dissipation numbers a fair bit...

Total junction-to-ambient resistance is now = 0.27oC/W + 0.477oC/W = 0.747oC/W
Temperature rise from ambient is now = 0.747oC/W x 155.5W = +116oC


I have also looked at tostino's spreadsheet a bit and I think there might be some revisions to be made, but I will have to decrypt the compacted formulas to see what's going on. That high side loss formula is more complex and there is no explanation of the constants used (1 000, 10 000, 1 000 000), so maybe I'll just try and make it up from scratch and compare the results. Tostino - does this spreadsheet take into account the dissipation of the phase that is being held low along with the one that's being PWM'd? This is an extra burden on the low side heatsink, specially at low duty cycles with high currents since you have an extra group of low side FETs conducting the phase current all the time in addition to the PWM'd phase losses.


Anyways, I can't wait to see a video of Doc actually using this beast with a motor that can take all that phase current! And Doc please try and find something better than those silpads unless your 6 FET heatspreader's surface is really wonky, you'll be able to push that +116oC junction temp over ambient down to almost half by doing just this! Use those belleville washers and some medium strength thread locker if you're really worried about screws loosening. You're setup will also be pushing the FETs down in the right place I believe (over the die), so you'll be able to use much more screw pressure without worries this time.

Good luck Doc, and make us another good video to enjoy!
Pat

 

[tostino]

Zap, I had Rhitee05 help me with the calculations last year. You can see the discussion here: http://endless-sphere.com/forums/viewtopic.php?f=2&t=19655&start=30

I honestly don't fully understand all the mechanics of what is going on, and would really like to learn more.

 

[tostino]

Okay, it slowed down at work so I had time to look at the spreadsheet again.
The high side calc is: =($C2^2*$D2*$A2) + IF($A2 = 100%,+(1/2*$B2*$C2*((1/1000000) + (1/1000000))*(1000)),+(1/2*$B2*$C2*((1/1000000) + (1/1000000))*10000))

What i'm doing here is: ($C2^2*$D2*$A2) calculate phase resistance * duty cycle
What i then do here is: IF($A2 = 100%, check if duty cycle = 100 or not
If duty cycle = 100%: +(1/2*$B2*$C2*((1/1000000) + (1/1000000))*(1000)) I keep the switching losses in the calculation at a lower frequency (1000 hz) to simulate the commutation frequency of the motor
If duty cycle != 100%: +(1/2*$B2*$C2*((1/1000000) + (1/1000000))*10000)) I calculate the switching losses at 10,000 hz (note, the (1/1000000) part is meant to be 1us turn on, and turn off time)


On the low side we have: =1.3*$C2* (1-$A2)+ (C2^2 * D2)
What is happening here is: 1.3*$C2* (1-$A2) calculate the diode losses
Then here: + (C2^2 * D2) I add in the phase resistance losses

 

[CamLight]

- Equivalent junction-to-sink resistance for 18 FETs is now 2.86C/W / 18 = 0.159C/W

It seems to me that there are only 6 FETs in parallel sharing a thermal path at any time on the PWM'd phase, not 18. So that number would be 3 times higher, about 0.477oC/W per FET group, using those *good* silpads (isn't that a contradiction in terms?). So that changes the other dissipation numbers a fair bit...

Total junction-to-ambient resistance is now = 0.27oC/W + 0.477oC/W = 0.747oC/W
Temperature rise from ambient is now = 0.747oC/W x 155.5W = +116oC

Arrgghhh....phase currents always screw me up.
Thanks for the correction!
We're back up to very, very toasty with those FETs now.

And Doc please try and find something better than those silpads unless your 6 FET heatspreader's surface is really wonky, you'll be able to push that +116oC junction temp over ambient down to almost half by doing just this! Use those belleville washers and some medium strength thread locker if you're really worried about screws loosening. You're setup will also be pushing the FETs down in the right place I believe (over the die), so you'll be able to use much more screw pressure without worries this time.

I agree! Those SilPads are just awful heat conductors.
Might be required in this case since the surfaces are hand-prepared and Doc is using relatively thin metal (the copper) for the heat spreader? Only takes a mil or two from perfectly smooth and/or perfectly flat to ruin the thermal coupling to a FET. Doc does have a nice mounting setup for the FETs, I agree.

 

[Doctorbass]

Again, a BIG thank to all of you that try to get me in the right way for those thermal dissipatin calculations!

I plan on soldering each 6 fets group together on a thin 20mil thick copper sheet to better share the current between those and also to share the heat.

That should also help increasing the thermal transfer fron he mosfets to the silpad sice the total area for all 6 fets will change by around 2.5X the surface.

Actually, what is the most complicated is soldering these 6 group of 6 fets to the copper sheets WITHOUT damaging them and also keeping them well aligned to let each 18 pins to still be aligned to the 18 holes of the pcb!!

Its easy to pre-apply solder to the copper, than pre apply solder to the fets.. but melting them together and keep all 6 fets aligned... IS A REAL PAIN !!!!

I think i will make a jig...

I already tried with couples of old SCR i had that are now useless ( t0-220) and that was really difficult!

I must hold all 6 together, press them to the copper sheed while the solder melt ... to not overheat them... and cool them fast and pray that all them was not too stressed...

Doc

 

[whatever]

just a note for anyone in australia or nz, the fisher paykel washing machines use a water cooled aluminium tube that is designed specially to fit t0-220 fets with clips to hold them in place, the fets are quite spread out plenty of room for a big controller use, also commonly can find the washing machines on cleanup days etc

 

[Doctorbass]

just a note for anyone in australia or nz, the fisher paykel washing machines use a water cooled aluminium tube that is designed specially to fit t0-220 fets with clips to hold them in place, the fets are quite spread out plenty of room for a big controller use, also commonly can find the washing machines on cleanup days etc

That's interesting since new washing mechine are using brushless motor. I guess that they are using the water for washing as coolant... it's pretty brilliant.

Have you any pictures of their controller board?

I found these info about those:

The air tube from the tub pressure dome connects to a nipple on the upper right-hand side of the module. The module has a female port that connects with the water inlet valve nipple making a water-tight connection. Water is run into a heat sink tube inside the control board module to keep it cool. It’s the only water-cooled control board I’ve ever seen in the appliance world!

http://fixitnow.com/wp/2004/10/30/a...ation-fisher-paykel-dc-drive-washing-machine/

And other doc i found:

http://www.cnczone.com/forums/stepper_motors_drives/67757-fisher_paykel_smartdrive_ac_stepper.html





Doc

 

[rhitee05]

Doc,

Using a heat spreader on the backside of the FETs is a really good idea, but I think that 20 mils is probably too thin to do very much good. I suspect something like 1/8" or 1/4" copper bar would perform much better. I guess you'd really need some FEA simulations to know for sure!

You'd probably have much better luck soldering the FETs onto the bar using reflow soldering. It's not too uncommon for hobbyists to use a toaster oven to approximate a reflow cycle and they seem to have good luck with it. Even with a crude cycle it would probably be less stressful on the FETs than the very high point heating you'd get with an iron. You'd just need to put down some solder paste, set the FETs into place, and bake - much easier to get the FETs positioned correctly this way, and you'd probably get more even contact on the backside, too.

If you were really brave, you might even be able to do this using a torch to heat the backside of the copper (moving quickly to try and keep the heat even). Not sure I'd try that with your 20 mil copper, though, it would probably have to be the thicker bar for that to work.

 

[karma]

doc have you picked out a pump if so what model?

i should have this one Friday to dismantle.

http://www.tigerdirect.ca/applications/SearchTools/item-details.asp?EdpNo=5267687&CatId=1871

should be interesting to see what the pump looks like. and the price is right

found a link showing the innards i can work with this

http://www.frostytech.com/articleview.cfm?articleID=2564

cheers

 

[Arlo1]

Hows this going doc?

 

[tostino]

Karma,
That pump is the same one I have in my PC, because I didn't feel like putting out for a real watercooling setup. It really doesn't seem high end at all, and there are much better options available i'm sure. It is meant for cooling a single processor. Thats it. It has low waterflow, and the radiator is actually pretty small for our use I think.

 

[Doctorbass]

Hows this going doc?

Well.. I'm little busy at work these days so when i comeback to home i try puting some of my time on that. I did not touched that since the last watercooling test video i did.


Spring is coming and i'll need to finish all the project for teh begining of summer... THERE IS ALOT TO DO!!

Doc

 

[Doctorbass]

Doc,

Using a heat spreader on the backside of the FETs is a really good idea, but I think that 20 mils is probably too thin to do very much good. I suspect something like 1/8" or 1/4" copper bar would perform much better. I guess you'd really need some FEA simulations to know for sure!

You'd probably have much better luck soldering the FETs onto the bar using reflow soldering. It's not too uncommon for hobbyists to use a toaster oven to approximate a reflow cycle and they seem to have good luck with it. Even with a crude cycle it would probably be less stressful on the FETs than the very high point heating you'd get with an iron. You'd just need to put down some solder paste, set the FETs into place, and bake - much easier to get the FETs positioned correctly this way, and you'd probably get more even contact on the backside, too.

If you were really brave, you might even be able to do this using a torch to heat the backside of the copper (moving quickly to try and keep the heat even). Not sure I'd try that with your 20 mil copper, though, it would probably have to be the thicker bar for that to work.

Eric, Yes, I know that the copper is thinny.. but i dont really have the choice since i dont have more thickness available on the actual setup.. dont forget that i already have a 1/4" thick aluminum "L" Bar and copper tube under that copper sheet to help thermal conductibility.

The copper sheet is just to better transmit heat from the fets to the silpad.

If i would use thicker copper sheet, than i would need to redesing everything to make it fit.

Putting just copper as heatsink would be perfect.. but for the high side.. that would be a pain since i would need 3 seperate heat sink ( one per phase) and watercooling that woudl be a pain too...... plus the need of replacement of a fet would be apain too sice all the other fet would unsolder once i heat up the bar..

 

[karma]

Karma,
That pump is the same one I have in my PC, because I didn't feel like putting out for a real watercooling setup. It really doesn't seem high end at all, and there are much better options available i'm sure. It is meant for cooling a single processor. Thats it. It has low waterflow, and the radiator is actually pretty small for our use I think.

after seeing the pic's of the inside yes it would be. perfect for me. i was just wondering what doc picked out. :wink:

 

[ZapPat]

I plan on soldering each 6 fets group together on a thin 20mil thick copper sheet to better share the current between those and also to share the heat.
That should also help increasing the thermal transfer from the mosfets to the silpad since the total area for all 6 fets will change by around 2.5X the surface.

20mils seems a bit thin for a heatspreader, Doc, but I can understand that you are already having enough problems soldering the FETs to them as it is! You do have a very good point about your heatspreader helping out your silpads, but you probably won't see a 2.5X Rth improvement despite the 2.5X larger surface area because of the heatspreader's thinness. Maybe someone knows the thermal math behind that kind of stuff and could help? *EDIT*: I see Eric and Doc have already discussed this point, so I guess we'll see how the 20mil does.

Thanks for the spreadsheet explanations BTW, tostino! Eric (Rhite05) knows his stuff for sure, so I'll review that thread before working on the spreadsheet. I'll probably cut the calculations up into separate parts and describe each of them (kind of like you did above), so that anyone wanting to refine or modify it will be able to easily. One thing I did figure out is that the phase being held low (the non-PWM'd one) *is* accounted for in the spreadsheet by the [I_ph^2 * R_fet] term on the low side calculation, written as ($C2^2 * $D2). One thing to take note of about the spreadsheet's low side loss number is that there are actually two sources each with their own separate thermal paths and dissipations (low side diode losses on the PWM'd phase and then resistive FET losses on the non-PWM'd phase), since each loss term represents a separate group of low side FETs. That means we actually need to analyze three distinct Rth's and their associated dissipation numbers, rather than only two. These are:
- Losses on the high side FET group of the PWM'd phase resulting from both conduction and switching events
- Losses on the low side FET group of the PWM'd phase resulting from the FET's passive (off-state) diode conduction
- Losses on a second low side FET group of a non-PWM'd phase resulting from the FET's On-state resistive conduction

Pat