RE: A Challenge

[Dan Strassberg <dan.strassberg@worldnet.att.net>]

At 02:15 AM 7/27/97 +0000, you wrote:
><<On Sat, 26 Jul 1997 23:13:25 +0000, Dan Strassberg
<dan.strassberg@worldnet.att.net> said:
>
>> don't know what the efficiencies of modern solid-state AM TX's run, but if
>> you assume around 75% (as I did), you come up with about 100,000 kWH/month*.
>
>A Harris DX-50 has a minimum efficiency (for a 1-kHz tone at 100%
>modulation) of 85%, according to the glossies I got from a nice
>engineer at WBBM.  (He also explained that 'BBM has a half-wave tower
>with a quarter-wave backup at the same site.  When there's an
>electrical storm in the area, they switch to the backup and ground the
>main antenna.)
>
Well, that makes my 75% number a bit low--but AMs are allowed to modulate
125% on positive peaks. So peak power out for a 50 kW AM is 2.25^2*50 kW =
over 250 kW. If the modulation is a symmetrical squarewave, the average
power out is 125 kW. 125/.85 = 147 kw. That would be the input power. I had
figured 100/.75 = 133.33 kw. so, even though I was a tad low on efficiency,
my result was still conservative because I didn't include the effects of
125% modulation on positive peakks. Now, admittedly, nobody modulates with
squarewaves (except during tests--well, maybe hard-rock stations do :-). And
my math is too rusty for me to redo the calculation using a sinewave for
modulation. But I think the 133-kw number that I used in my example turns
out reasonably close.

- -------------------------------
Dan Strassberg (Note: Address is CASE SENSITIVE!)
ALL _LOWER_ CASE!!!--> dan.strassberg@worldnet.att.net
(617) 558-4205; Fax (617) 928-4205

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