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1. Re: [Amps] RCA load impedance calculation (score: 1): Author: Xmitters@aol.com; Date: Mon, 10 Apr 2006 21:48:02 EDT; << Gerald, I went ahead and brought the RCA transmitting tube manual TT-5 here to quote what it says. Effective RF load resistance; 2 x (Eb - Ebmin) / Ibo Ibo = Zero signal plate current "Ibo should; /archives//html/Amps/2006-04/msg00159.html (7,309 bytes)
2. Re: [Amps] RCA load impedance calculation (score: 1): Author: "Will Matney" <craxd1@verizon.net>; Date: Mon, 10 Apr 2006 23:56:37 -0400; Jeff, I double checked, and that was what it written, Ibo instead of Ibmax. I couldn't see it either as being what was needed. I guess it was a mis-print. This is at procedure (16) 0n page 52. Mine r; /archives//html/Amps/2006-04/msg00164.html (9,237 bytes)
3. Re: [Amps] RCA load impedance calculation (score: 1): Author: Peter Chadwick <g3rzp@g3rzp.wanadoo.co.uk>; Date: Tue, 11 Apr 2006 09:46:44 +0200 (CEST); Isn't the load impedance Rl = Vs*2/2Po Vs*2 is the peak to peak anode voltage swing squared Po is the max output power Interestingly, if you look at the anode ac resistance curves and figure the effi; /archives//html/Amps/2006-04/msg00166.html (7,079 bytes)
4. Re: [Amps] RCA load impedance calculation (score: 1): Author: Steve Thompson <g8gsq@eltac.co.uk>; Date: Tue, 11 Apr 2006 09:02:09 +0100; I was about to make the same comment - except Vs is the supply voltage, not the peak-peak swing. Where the anode/collector/drain doesn't go down to 0V, use (Vs-Vmin) in place of Vs. It's a first appr; /archives//html/Amps/2006-04/msg00167.html (7,588 bytes)
5. Re: [Amps] RCA load impedance calculation (score: 1): Author: Peter Chadwick <g3rzp@g3rzp.wanadoo.co.uk>; Date: Tue, 11 Apr 2006 10:37:03 +0200 (CEST); Sorry, yes, Vs is half the peak to peak swing. In Class B it's exact, in AB, it's near enough. Few things in radio are that critical - thus my irritation with simulations to 4 and 5 places of decimal; /archives//html/Amps/2006-04/msg00168.html (7,226 bytes)
6. Re: [Amps] RCA load impedance calculation (score: 1): Author: TexasRF@aol.com; Date: Tue, 11 Apr 2006 07:33:19 EDT; Jeff, I misread your example and took your peak current of 1 A as the average current. Your 5400 ohm plate load is exactly correct for the example and I must sincerely apologize for my confusion. I h; /archives//html/Amps/2006-04/msg00169.html (8,927 bytes)

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