Calculating the AlephX

yldouright · 23rd January 2005, 04:05 PM

Netlist
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

Netlist · 23rd January 2005, 04:32 PM

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

yldouright · 23rd January 2005, 04:58 PM

Netlist
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.

Netlist · 23rd January 2005, 05:06 PM

Quote:

Originally posted by yldouright
we can discuss it line by line.

Shoot.

/Hugo

yldouright · 23rd January 2005, 06:23 PM

Netlist
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!

Netlist · 23rd January 2005, 07:19 PM

The spreadsheet is correct, indeed not 2A but 4A per fet.

/Hugo

yldouright · 23rd January 2005, 07:36 PM

Netlist
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?

Coulomb · 23rd January 2005, 07:40 PM

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.

Regards

Anthony

Netlist · 23rd January 2005, 07:40 PM

Sure, sometimes I read too fast.
Sorry if this was confusing you.

/Hugo

yldouright · 23rd January 2005, 07:45 PM

Coulomb
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

23rd January 2005, 07:40 PM	#8
Coulomb diyAudio Member Join Date: Jul 2002 Location: Ancaster, Ontario	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. Regards Anthony __________________ I like to stay current!

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23rd January 2005, 04:05 PM	#1
yldouright diyAudio Member Join Date: Aug 2002 Location: New York, NY	Netlist 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

23rd January 2005, 04:32 PM	#2
Netlist diyAudio Moderator Join Date: Jan 2003	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

23rd January 2005, 04:58 PM	#3
yldouright diyAudio Member Join Date: Aug 2002 Location: New York, NY	Netlist 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.

23rd January 2005, 06:23 PM	#5
yldouright diyAudio Member Join Date: Aug 2002 Location: New York, NY	Netlist 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!

23rd January 2005, 07:19 PM	#6
Netlist diyAudio Moderator Join Date: Jan 2003	The spreadsheet is correct, indeed not 2A but 4A per fet. /Hugo

23rd January 2005, 07:36 PM	#7
yldouright diyAudio Member Join Date: Aug 2002 Location: New York, NY	Netlist 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?

23rd January 2005, 07:45 PM	#10
yldouright diyAudio Member Join Date: Aug 2002 Location: New York, NY	Coulomb 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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