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Old 5th December 2006, 06:08 AM   #1011
PWatts is offline PWatts  South Africa
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Flodstroem,

As with all bias issues it's always confusing how people refer to it. With 5A I meant in total from the supply i.e. both rails. At 2.5A per rail, it gives 5A total, or 620mA x 8 output devices.
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Old 5th December 2006, 12:39 PM   #1012
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Hi Pwatts,
if you want to refer to bias your way then spell it out each time to avoid confusion.
This is the method used in the spreadsheet but it is quite unconventional.
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Old 5th December 2006, 01:51 PM   #1013
PWatts is offline PWatts  South Africa
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OK OK fine.. but it should be clear to figure out anyway since the total bias is after all the total bias per transistor added together.. and knowing it's 620mA each it's not rocket science.

From a heatsink perspective it makes sense to refer to the total bias since that is what the sink needs to dissipate, but from the amp side the per rail value is more sensible.

Besides, anyone building the amp should at least know the basics of how to calculate reasonable values for 100W into 8ohms or it's likely to blow up anyway
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Old 5th December 2006, 02:08 PM   #1014
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Mmm, like this one that even a year later is good for the sensation of substitute embarrassment ?

Quote:
I put the pcb back to the psu, I did put on my amplifier and the pcb started to burn. On the side where the three mosfets were wich get hot, the connectionstripes burned through.
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Old 5th December 2006, 10:33 PM   #1015
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Once again, anyone that is a late-comer that would like a couple of boards just needs to dial into the WIKI and replace my name with thiers.
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Old 5th December 2006, 11:47 PM   #1016
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Quote:
Originally posted by PWatts
From a heatsink perspective it makes sense to refer to the total bias since that is what the sink needs to dissipate, but from the amp side the per rail value is more sensible.

Heatsink dissipates power, not current. Bias current flows from +Ve to -Ve, and it is regulated at one single point (Vbe multiplier). It makes much more sense to me to say that my amp runs 2A idle current at +/- 50V (that is 200 watts of power to be turned out into heat) rather than say I've got 4A idle. I mean, if you measure current anywhere in the circuit, there is no one single point where your ammeter will show 4A, is it ? Same current flows through both branches of output - you can't add them. But you can add voltage, 'cause between +V and -V you do have 2 times 50 volts. The voltmeter will measure 100V between rails. I think it is a bit more than just semantics.

Bratislav
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Old 6th December 2006, 06:21 AM   #1017
PWatts is offline PWatts  South Africa
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Jeez OK I should've stuck to the conventional format I suppose.

Heatsinks do of course disssipate power - but with the voltage a fixed parameter (well hopefully with a proper transformer anyway), current increase will increase power in direct proportion so there's nothing wrong with that statement.
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Old 6th December 2006, 02:48 PM   #1018
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Default don't feel bad...

...there is no single answer that will suffice for all people and all possibilities...and it's clear your answer will have answered some peoples unspoken questions...

Power in the heatsink is a big deal for this amp, giving people more info may help them understand what's proportional to what and thereby gain better insight into their particular build. This is unambiguously a good thing...

...and this is key; except for exact clones, there is a good chance no two of these amps are the same, thus each builder/designer must understand the heat issues, they need to make adjustments and end up with a stable, reliable amp.

I expect we will see the same questions many more times before we are done, and answering them different ways, from different perspectives is needed to give everyone the best chance of understanding...

Stuart
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Old 6th December 2006, 11:49 PM   #1019
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Power in the heatsink is a big deal for this amp
Based on the original mkII bias current, what would be the minimum C/Watt heatsink ?
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Old 7th December 2006, 12:29 AM   #1020
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Default thats easy...

2.9 amps, 57 volt rails == 330 watts per channel, with 30c temp rise...you need heatsinking to the tune of 0.091c/watt, per channel.

To give (computer geeks) an idea of what that translates to: A stock P4 cooler (with fan) supplied by intel is slightly better than 0.5c/w, so minimally you'd need 6 of them per channel, with their fans...perhaps 8 might be better, one per transistor...of course 8 fans would be kind of loud...

Unfortunately CPU heatsinks are designed to use fans, they are much less efficient without them...

http://www.overclockers.com/articles373/socketA.asp

Stuart
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