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Author Topic: Two transmitters, one article.  (Read 2986 times)
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WA4WAX
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« on: July 26, 2016, 09:14:40 AM »

......Take your pick.  It all starts on page 816.

http://www.americanradiohistory.com/Archive-Practical/Wireless/60s/PW-1963-01.pdf
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W3RSW
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Rick & "Roosevelt"


« Reply #1 on: July 26, 2016, 11:13:38 AM »


Perused the circuits, but as usual the rest of the magazine was fun too.
Ads for stuff like gangedd potentiometers, log/log, Log/anti log, log/Lin, etc. for Mullard circuits,  ....just fascinating imagining the audio circuits these fit.
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RICK  *W3RSW*
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« Reply #2 on: July 26, 2016, 01:37:02 PM »

From Page 819:

Quote
The modulating impedance of the P.A. can be found from:
 Z= P.A. H.T. voltage/P.A. current (mA) x 1000.

Assuming that the P.A. draws 40mA at 250V, the
impedance will be approximately 6200.


Actually it's 6250, but where did the x 1000 come from?

Phil - AC0OB
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Charlie Eppes: Dad would be so happy if we married a doctor.
Don Eppes: Yeah, well, Dad would be happy if I married someone with a pulse.NUMB3RS   Smiley
W3GMS
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« Reply #3 on: July 26, 2016, 02:25:45 PM »

From Page 819:

Quote
The modulating impedance of the P.A. can be found from:
 Z= P.A. H.T. voltage/P.A. current (mA) x 1000.

Assuming that the P.A. draws 40mA at 250V, the
impedance will be approximately 6200.


Actually it's 6250, but where did the x 1000 come from?

Phil - AC0OB

Normally you convert the current to ampere's, but if you don't and just use MA then the answer comes to 6.25.  When you multiply that times 1000 your back to the correct answer! 

Joe-GMS
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Simplicity is the Elegance of Design---W3GMS
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« Reply #4 on: July 26, 2016, 02:49:48 PM »

thanks Joe.

Guess I am just to literal. Cheesy

Phil
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« Reply #5 on: July 26, 2016, 11:47:29 PM »

thanks Joe.

Guess I am just to literal. Cheesy

Phil

The 1000 was a giveaway for me.  I though amps vs ma and then 1000 fit into place!   
I understand very well about being literal  Wink.  Been there done that.....

73,
Joe
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Simplicity is the Elegance of Design---W3GMS
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