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Old 11th June 2008, 06:41 PM   #51
djQUAN is offline djQUAN  Philippines
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I rechecked the computations again as MPP should be a low loss core but it was highest? also, I found graphs for B versus freq losses so I added some more except for hi-flux as I couldn't find any...

1pc Material 60 T150
area: 0.887cm2
volume: 8.31cm3
mag lenght: 9.38cm
turns: 44
inductance: 125uH

B=666.188378 gauss @ 80V
B=866.044892 gauss @ 104V
H= 235.779957 Oe @ 80V
H= 306.513944 Oe @ 104V
graph based, 80V = 11.634w
graph based, 104V = 19.94w

2pcs Hi Flux T106 stacked
total area: 1.318cm2
total volume: 8.56cm3
mag lenght: 6.49cm
turns: 22
inductance: 177uH

B=896.675gauss @ 80V
B=1165.678gauss @ 104V
H=170.386441 Oe @ 80V
H=221.502373 Oe @ 104V
graph based, 80V = ??
graph based, 104V = ??

2pcs MPP T130 stacked 66nH/t2
total area: 1.396cm2
total volume: 11.56cm3
mag lenght: 8.28cm
turns: 37
inductance: 177uH

B=503.369gauss @ 80V
B=654.379gauss @ 104V
H=224.609 Oe @ 80V
H=291.99 Oe @ 104V
graph based, 80V = 2.08w
graph based, 104V = 3.70w

looks like I can use the MPP cores that are coming here.
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Old 11th June 2008, 07:16 PM   #52
luka is offline luka  Slovenia
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Hi

I wish I could help, but can't
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Old 12th June 2008, 01:19 AM   #53
Eva is offline Eva  Spain
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All your calculations seem right...


2pcs Hi Flux T106 stacked
total area: 1.318cm2
total volume: 8.56cm3
mag lenght: 6.49cm
turns: 22
inductance: 177uH

B=896.675gauss @ 80V
B=1165.678gauss @ 104V
H=170.386441 Oe @ 80V
H=221.502373 Oe @ 104V
graph based, 80V = approx. 9W (outside table)
graph based, 104V = approx. 18W (outside table)


This is assuming a 147u Hi-Flux core (Arnold HF-106147-2) which is the one that results in approx. 177uH inductance with 22 turns and 2 cores stacked.

----------------------------------------------------

2pcs MPP T130 stacked 66nH/t2
total area: 1.396cm2
total volume: 11.56cm3
mag lenght: 8.28cm
turns: 37
inductance: 177uH

B=503.369gauss @ 80V
B=654.379gauss @ 104V
H=224.609 Oe @ 80V
H=291.99 Oe @ 104V
graph based, 80V = 1.4w (for 60u Arnold A-071065-2)
graph based, 104V = 2.3w (for 60u Arnold A-071065-2)


In other words, cool

----------------------------------------------------

The problem that you are going to have with 60u MPP is that inductance drops to 50% at H=100 Oe and to 20% with H=200 Oe, so the amplifier may not like very low impedance loads (2 ohms). 26u is more suitable for that purpose because inductance drops to 70% at H=100 Oe and to 50% at H=200 Oe. Anyway, your 60u arrangement is ok for 4ohm operation. 50% inductance drop at full peak output current is ok.

Congratulations for understanding all the calculations, particularly H which is very important for sizing the output inductor of any class D amplifier to handle the maximum output current imposed by supply rails and rated load.

I don't know "60" material either, but an output inductor should never idle near 90șC. This is a very bad design practice. Iron powder cores suffer from thermal aging, they progressively lose their permeability and the hotter they operate the faster it happens. A badly designed inductor can become useless within a few months or years.
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Old 12th June 2008, 06:00 AM   #54
djQUAN is offline djQUAN  Philippines
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oh no!!!

I need that amp to be able to run at 2ohm full power. the online sources you posted doesn't seem to have the 26u MPP cores. any idea where I could get some? the 60u ones are from the ebay page you posted a while ago...

I'll do a bit more computation if I could lower H value to less than 200 since the amp originally has 125uH inductance and I increasd it since I thought the increased inductance could take care of the drop at high currents.

edit: with 30 turns, we get about H = 180 at 80V and about H = 240 at 104V so we are still above but a little lower.

core losses will definitely be higher but probably not burning hot.
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Old 12th June 2008, 12:51 PM   #55
Eva is offline Eva  Spain
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The original inductor was probably not able to handle 2 ohms either. The chance to see an output inductor failing is almost non existent unless there is something wrong with the design.

20% of 177uH is still 35uH which isn't too bad at 77Khz when the output approaches the rails. Using many turns always helps because you end up with higher inductance despite the higher drop due to stronger H. This is good as long as the control loop remains stable across the resulting inductance range, which is something that you should test. Resonance of the first LC section would rise from 3800Hz to 8500Hz when inductance drops from 177uH to 35uH. What are the LC values in the second section?

Which switching MOSFETs and how many are employed in the class D section and in the PSU? How much current is required to trigger the protection? Is the circuit really sized to handle 40A-50A peak?? I have serious doubts. Those fashion class D amplifiers are not that good.

Anyway, consider finding some free space to add one or two stacked core(s) near the first pair and wired in series. The hot (switching) end should be connected to the first pair (short wire to avoid EMI).
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Old 12th June 2008, 03:23 PM   #56
djQUAN is offline djQUAN  Philippines
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first LC section (originally) is 125uH and 10uF. the next one is an unlabeled drum core inductor and 1.8uF film cap with a 2R + 10uF zobel.

main fets are a two pairs of IRFP264N. I don't know the overcurrent limit.

PSU uses 4 pairs of IRF3205, 2" diameter main trafo, 4X 820uF 160V per rail, rectifiers are F30D40D.

I'm suspecting that the original inductor failed because the windings rubbed and shorted or that it got hot enough that the varnish melted off and shorted to each other.

how do I test the stability of the control loop when everything is installed? does it have to be loaded? pushing full power? what do I look for? I have a sine/square wave signal gen so I could use that. no high power dummy loads though.

here are some latest pictures of the amp. it still has the T106-125perm hi-flux cores installed.

Click the image to open in full size.

Click the image to open in full size.

Click the image to open in full size.
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Old 12th June 2008, 05:47 PM   #57
Eva is offline Eva  Spain
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If the loop is stable, you will get clean sinewave output (no 500Hz-10Khz ringing) and clean clipping with and without load.

This amplifier was never designed to reliably drive 2 ohms, take a look at the datasheet of IRFP264N, max allowed drain current is 44A at 25șC and 31A at 100șC. Also, 40A output current is reflected back to the primary as 66A per IRF3205, which is a bit high (not to mention 50A which results in 83A per IRF3205).

The amplifier probably blew because it was driving hard a 2 ohm load, and it will probably blow again in the same conditions.

Couldn't you measure the pot inductor? I doubt this one can handle 40A either, it's very likely to saturate at 25A or less given the size.

The current limit has probably a lot to do with those four grey resistors.
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Old 12th June 2008, 06:13 PM   #58
djQUAN is offline djQUAN  Philippines
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the amp may not work with full power but will probably be fine with music?

I think power loss from battery voltage sag/cable loss/SMPS output sag are also factors that prevent this amp from blowing up when driven hard into 2ohms. just note that the 80V rails are unloaded so they will definitely be lower when driving a speaker.....

the four resistors are indeed for the current limit sense but I have no idea what the values are since the color codes don't make any sense. also, I don't have any means to measure very low resistances.

IRFP264 is 31A at 100degC that is per fet right? this has four. two per rail and is running half bridge so it should be within limits.

I could take out the inductor and measure it but it would be a lot of work. I'll try to take it out and post later.....

edit: the inductor is glued in pretty tight so I can't remove it without fear of breaking it.
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Old 12th June 2008, 06:52 PM   #59
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Ithink this amp is 2R stabil, coz speaker impedance goes up a lot, maybe they did count on that

Use knife and cut in the middle of glue, if it is not hard as epoxy
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Old 12th June 2008, 07:20 PM   #60
djQUAN is offline djQUAN  Philippines
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there are a lot of things in the way to get a knife under the core. believe me, I have to break it to take it off. not to mention burning the PCB tracks off to remove the thick copper wires soldered to the board.
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