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#5981 |
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diyAudio Member
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As in bundled so that they touch eachother, like two smacking teabags ?
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Looks like Sponge Bob has killed another thread. |
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#5982 |
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not politcally affiliated
diyAudio Member
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and duct-taped to my neck to ease the pain from arching my neck
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#5983 |
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diyAudio Member
Join Date: Mar 2004
Location: ohio
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#5984 |
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diyAudio Member
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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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Looks like Sponge Bob has killed another thread. Last edited by jacco vermeulen; 19th November 2009 at 08:16 PM. |
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#5985 |
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Passive Aggressive
diyAudio Member
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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.
Uriah
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You can purchase LDRs anytime to build a standard LDR attenuator or to build my new LDR Attenuator "A Lighter Note". Email me. diyldr@gmail.com |
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#5986 |
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diyAudio Member
Join Date: May 2008
Location: On the moon.
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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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#5987 |
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diyAudio Member
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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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#5988 |
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diyAudio Member
Join Date: Nov 2008
Location: Netherlands
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#5989 |
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diyAudio Member
Join Date: May 2008
Location: On the moon.
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The planet.
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#5990 |
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diyAudio Member
Join Date: Dec 2002
Location: Milan, Italy
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