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Old 23rd January 2005, 04:05 PM   #1
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The spreadsheet for the 8A entry described above indicates that each of the 4 FETS will burden 4A each! This is the same for both the 1.0 and 1.2 version of the spreadsheet. When you say use a little more to stay in the SOA is that to mean double the amps/power dissipation? That seems peculiar to me
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Old 23rd January 2005, 04:32 PM   #2
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I meant:
Use a couple more (mosfets) to stay in the SOA.
Of course you'll always need multiples of four in the AlephX.
4,8,12,16 and so on.
The SOA would be something like max 30W per device so for 8A/22V supply you need 12 mosfets, for 10A/22V this would be 16.
Please do take a look at the spreadsheet. Many cells have a small red mark in the top right corner. Just moving the mouse into these cells will reveal a lot more info as well.

/Hugo
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Old 23rd January 2005, 04:58 PM   #3
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I am quite familiar with the spreadsheet and have it open with the example posted above. The little red dots do not explain this 'peculiarity' in the calculations. Perhaps you should revisit it so that you can refresh your memory and we can discuss it line by line.
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Old 23rd January 2005, 05:06 PM   #4
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Quote:
Originally posted by yldouright
we can discuss it line by line.
Shoot.

/Hugo
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Old 23rd January 2005, 06:23 PM   #5
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Excellent. I will be using the 1.2 version of the spreadsheet for our analysis. Earlier in this thread we agreed that cell B13 indicates the total current that must be burdened by the FETs so that an 8A entry with 4 FETs as per the original hifizen boards will result in 2A through each FET and require 250mΩ source resistors. Plug in those numbers and you will see that the source resistor value is half and the current values double!
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Old 23rd January 2005, 07:19 PM   #6
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The spreadsheet is correct, indeed not 2A but 4A per fet.

/Hugo
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Old 23rd January 2005, 07:36 PM   #7
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I'm sorry but to clarify, are you reversing your previous statement when you stated that the 8A total bias will result in 2A per FET? If the spreadsheet is correct, how does 4x4=8?
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Old 23rd January 2005, 07:40 PM   #8
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Am I completely wrong, I thought only two Fets see the load as the other two are current source??? Thus 4A being the load current.

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Old 23rd January 2005, 07:40 PM   #9
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Sure, sometimes I read too fast.
Sorry if this was confusing you.

/Hugo
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Old 23rd January 2005, 07:45 PM   #10
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Nice of you to joing the party. The spreadsheet clearly states "total bias for one channel being one monoblock". One monoblock includes both current source and gain FETs so clearly this is an unfortunate description. It is amazing to me how many people accept what is written for gospel
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