Dear Bill,
The figure 3817.5 Watts is simply the Anode Supply
consumption at full bore of 2545 Volts times 1.5 amperes.
This is from a "black box" model where the entire amplifier is just seen
as a 1,691 ohm resistor.
Of course, the amplifier will not be cranking at full bore, key-down, A0A
for any time. I just wanted to design in some reliability.
Thanks,
Hal
On Sun, 08 Jan 2006 16:46:54 -0800 Bill Turner <dezrat1242@ispwest.com>
writes:
>
> ORIGINAL MESSAGE:
>
> At 07:36 AM 1/8/2006, HAROLD B MANDEL wrote:
> >The power supply is rated at 2545 VDC at 1.5 amperes.
> >
> >Hence, the power supply runs 3817.5 Watts through
> >the amplifier.
>
>
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>
> This part of your question puzzles me. I see you are in the US, so
> you will be running max output of 1500 watts. Most amplifiers are
> about 60-66% efficient, so the input will be around 2500 watts.
> Where
> does the figure of 3817.5 watts come from? At no time should you be
>
> putting that through the amp or lots of smoke will escape.
>
> With 2545 VDC to the amp, the current should be a bit less than one
>
> amp when tuned up. Since your meter's resistance is .33 ohms, a
> single diode across it will begin to conduct at about 1.5 amps. I
> would use it that way, knowing the reading will never go above 1.5
> amps. If you put two diodes in series, they will begin to conduct
> around 3 amps. Since the meter is rated for two amps anyway, no harm
>
> is likely to result. Meters are damaged by a massive overload, not
> by
> 50%. Your choice.
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