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Old 19th November 2009, 07:38 PM   #5981
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Quote:
Originally Posted by Tea-Bag View Post
45 watts of sand-cast wirewounds BUNDLED together
As in bundled so that they touch eachother, like two smacking teabags ?
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Old 19th November 2009, 07:48 PM   #5982
Tea-Bag is offline Tea-Bag  United States
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and duct-taped to my neck to ease the pain from arching my neck
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Old 19th November 2009, 07:54 PM   #5983
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Quote:
Originally Posted by jacco vermeulen View Post
As in bundled so that they touch eachother, like two smacking teabags ?
eww
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Old 19th November 2009, 08:14 PM   #5984
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I'm a tea-seholic, mea maxima culpa.

I bought these square tea-bags in GB last spring, attached to eachother as a string in the box.
If i put two side-attached bags in a pot they stick like folding leaves and it takes long for the tea to reach prime time, i have to put them in one by one.

Power resistors are like little heatsinks, they need plenty clearance to cool properly.
Bundle them together and their temperature will skyrocket, despite having plenty power reserve.
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Old 19th November 2009, 09:44 PM   #5985
udailey is offline udailey  United States
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An amp that will drive, if I recall, "2 Ohms without burping" probably is not having trouble with speakers or cables. I wish I knew that answer but my first thing to eliminate as a problem would be the speakers/cables.
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Old 20th November 2009, 12:55 AM   #5986
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Depending on wether Bluewater has wired or not the limiters.
In the second case, an intermittent short circuit could be lethal.
I am thinking about a thin strand of copper at one end of the cable for instance.
He said he replaced the channel which burnt again.
So, i immagine the cause is external.
...Better to ask Bluewater if he uses limiters.
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Old 20th November 2009, 07:27 AM   #5987
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Default It is a little off topic isn't it :)

Quote:
Originally Posted by Tea-Bag View Post
I was hoping for some help there.

24 Volt supply, with 16 volts quiescent across 16 ohms.

I=V/R so that's 16/16 which is 1 amp.
Watts = I*V so that is 1*16 which is 16 watts.

That's under no signal conditions.
Worst case is the FET turning on and putting 24 volts less the drain source saturation voltage.
Lets go worst case and say 24 volts.
Using the above that's 36 watts.

The amp is AC coupled at the input however so that's not going to happen.

You need to choose a resistor (data sheets !) that will dissipate around 20 watts (preferably a non inductive component too) in the conditions you require ie free space, or attached to a heatsink where you have calculated the temperature rise etc and know it is within the operating limits of the resistor.

And my somewhat "unkind" last comment.

If we take the power out (headphones) as say 20mw and the power input of the amp as 24 watts, then powerout/powerin * 100 gives an efficiency of 0.083%

Draw your own conclusions on that one

Last edited by Mooly; 20th November 2009 at 07:52 AM.
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Old 20th November 2009, 09:16 AM   #5988
Beftus is offline Beftus  Netherlands
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Quote:
Originally Posted by Mooly View Post
If we take the power out (headphones) as say 20mw and the power input of the amp as 24 watts, then powerout/powerin * 100 gives an efficiency of 0.083%

Draw your own conclusions on that one
As a long as it sounds divine, who cares about the lowish efficiency?
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Old 20th November 2009, 09:48 AM   #5989
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The planet.
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Old 20th November 2009, 09:49 AM   #5990
massimo is online now massimo  Italy
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As a long as it sounds divine, who cares about the lowish efficiency?
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