[AMRadio] AM Audio


Bob Bruhns bbruhns at erols.com
Sat Aug 20 10:51:01 EDT 2005


Yes, that will work.  But remember, you want equal
voltages across the resistor and the inductor.
Surprisingly, this will not be 50% of the applied
voltage, but it will be about 71% of the applied
voltage.  This adds up to about 1.414 times the applied
voltage!  You should confirm this by measuring both
voltages directly.

This happens because the impedance of the inductor is
reactive, and the current is equal, so the phases of
the voltages differ by 90 degrees.  This phase shift
causes a partial cancellation when the voltages are
added, so 71 + 71 = 100 instead of 142.  It's
interesting to observe.

  Bacon, WA3WDR


----- Original Message ----- 
From: "Geoff" <w5omr at satx.rr.com>
To: "Discussion of AM Radio" <amradio at mailman.qth.net>
Sent: Saturday, August 20, 2005 10:30 AM
Subject: [AMRadio] AM Audio


> Well, I completed the task yesterday of installing a
Reactor for the
> 250TH rig.
> (Push-pull Class C 250TH link-coupled final,
modulated by push-pull
> Class B 250TH modulator)
>
http://w5omr.shacknet.nu:81/~w5omr/hamstuff/AM-Stuff/Titanic/modulator/new_titanic.jpg
>
> Six inductors in series, 5 of which are known values
> 10Hy @ 500mA
> 10Hy @ 500mA
> 9.9Hy @ 500mA
> 12Hy @ 400mA
> 16Hy @ 450mA
> (unknown)
>
> The unknown choke is 1/3 larger than the 16Hy @ 450mA
choke and the
> unknown choke is the largest in the chain,
physically.  I'm going to
> 'assume' it's a minimum of 10Hy.  Eventually, I will
take it out of the
> lineup, feed about 10vAC through it, and a 1k
potentiometer, and adjust
> the pot until the voltage across it, matches the
voltage dropped across
> the choke.    Thanks to John/WA5BXO for help in that
area:
>
> __________________
>
> by equalizing the voltage across the resistor and
choke results in the creation of a resistance that is
equal to XL of the choke at 60 CPS.
>
> Measure the resistance of the variable resistor after
equalizing the voltage drops and you have the XL of the
choke at 60 cps.
>
> L = XL / 2 * pi * F
>
> L = XL / 6.28 * 60
>
> L = XL / 376.8
>
> Divide the resistance by 376.8 and you have the
inductance.
> ___________________
>
>
> I will do that, soon.
>
> Still using 9uF of coupling capacitance @ 4kVDC
(because those two caps
> were still paralelled together and it was easier) and
this morning, was
> able to work (among several other stations)
John/W5GI, on Lake Travis
> near Austin, TX.
>
> John is working with the guys who are continuing
development on the
> 'Flex Radio' software-based, computer-controlled
radio transciever system.
> (http://www.flex-radio.com/)
>
> He sent me a screen-shot of how my audio currently
appears on the air.
> A Lot of information is on this screen.
>
http://w5omr.shacknet.nu:81/~w5omr/pictures/w5omr_frm_w5gi.jpg
>
> Would appreciate your comments.
>
> ---
> 73 = Best Regards,
> -Geoff/W5OMR
>
>
>
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