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31st January 2004, 08:21 PM  #21 
diyAudio Member
Join Date: Sep 2002
Location: North of Boston

total bias
The question I was trying to get answered. Is does the total bias = the draw out of the transformer?
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MikeW 
1st February 2004, 09:53 AM  #22 
diyAudio Member
Join Date: Aug 2001
Location: Ingolstadt Germany

yes
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een ooievaar is geen konijn want zijn oren zijn te klein! 
14th January 2005, 12:47 PM  #23 
diyAudio Member

wuffwaff
I'm with MikeW on this one. I put in the following: Voltage=20V Bias=8A AC Gain=50% Mosfets=4 Peak Current=8A I don't understand why the dissipation per FET is 80W when the dissipation is 20V x 2A (ie:8A/4). Is your calculator assuming that the bias is only burdened by 2 of the four FET's in this the case above? 
14th January 2005, 01:03 PM  #24 
diyAudio Member
Join Date: Aug 2001
Location: Ingolstadt Germany

Hi,
the voltage is symmetrical so total voltage from plus to minus would be 40V. 40Vx8A=320 Watt is 80W/fet After rereading your question I can also give you the following answer: the current is not devided by 4 but by 2, so 4A is flowing in each half. It flows through the active current source into the output stage. This means 4A per fet with each fet only seing half the supply voltage: P=4x20=80 watt William
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14th January 2005, 01:13 PM  #25 
diyAudio Member

wuffwaff
Thanks for your reply. I have been trying to reconcile what PSUD2 outputs to your spreadsheet so if I am to understand you correctly, the FET will dissipate double the voltage rail times the amps going through the FET, correct? I had been told that the 50% AC gain reduces the dissipation back down to the bias times the rail. I just read your edit: By this new information, it would appear that each FET will now dissipate 160W since this new model indicates 40V times 4A. Something is wrong here. When you say that each FET only sees half the voltage, do you mean to say that is only sees +/10V of the 20V rail or only the positive or negative side of the swing? 
14th January 2005, 02:02 PM  #26 
Warp Engineer
On Holiday

ok lets see....
Voltage=20V Bias=8A AC Gain=50% Mosfets=4 Peak Current=8A each side of the AX will have 4 amps running thru it and each of the 2 Fets / side will have 20V across it... 20x4 = 80Watts.
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 Dan 
14th January 2005, 02:50 PM  #27 
diyAudio Member

AudioFreak
Thanks for sounding in but your reply doesn't address how the alledged 80W are produced. Isn't the voltage swing 40V as wuffwaff suggests? If the amps in that case are 4 per FET then why isn't the dissipation 160W? How does the AC current gain factor into the expected power dissipation or isn't it a factor at all? I should also point out that many of the members whom I have interviewed that have built the AlephX experience junction and heatsink temperatures that suggest the dissipation is usually a litte more than half of what the wuffwaff spreadsheet and you claim in your post and I haven't even started to address the disparities between the power numbers indicated in the wuffwaff spreadsheet graph with those presented by PSUD2. Your curt and definitive reply indicates that all this must be very elementary to you but perhaps you can address my questions more completely in the future? 
14th January 2005, 03:03 PM  #28 
diyAudio Member
Join Date: Sep 2003
Location: Upstate NY

40 v is the maximum potential swing, but we are talking about no signal conditions (worst case for class A amp heat dissipation)
Since you have 4 amps per side, and we'll assume that your absolute DC offset is near 0, each transistor sees 20 volts. Since 4 amps flows through each transistor, you have 80 watts per transistor. Look at the junction temperature that produces with a real heatsink. Yikes! 
14th January 2005, 04:12 PM  #29 
diyAudio Member

BobEllis
I am assuming that you built the AlephX and like many others, relied on the wuffwaff spreadsheet to model it. Did you measure the junction and heatsink temperatures? If so, how did they compare to the numbers produced by the spreadsheet? 
14th January 2005, 04:50 PM  #30  
Warp Engineer
On Holiday

Quote:
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