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1. [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: "Williams, Barry" <Bnwilliams@varco.com>; Date: Wed, 8 Jun 2005 09:15:45 -0500; Hello All, I need to calibrate the watt meters on my Titan, Omni VI+ and 253 Tuner. I have a Drake 2 kw PEP dummy load and a nice 200 mHz storage scope. In order to measure output, should I measure R; /archives//html/TenTec/2005-06/msg00283.html (7,761 bytes)
2. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: "Geoffrey S. Mendelson" <gsm@mendelson.com>; Date: Wed, 8 Jun 2005 17:36:08 +0300; Parts 97 reads: (under definitions) 6) PEP (peak envelope power). The average power supplied to the antenna transmission line by a transmitter during one RF cycle at the crest of the modulation envel; /archives//html/TenTec/2005-06/msg00284.html (8,472 bytes)
3. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: "Martin, AA6E" <martin.ewing@gmail.com>; Date: Wed, 8 Jun 2005 10:53:16 -0400; You may be confusing PEP with RMS and instantaneous voltage. If you have 1500 watts of power at any instant (fraction of a cycle) in a 50 ohm load, you have 274 volts, from P = V**2/R. So far so good; /archives//html/TenTec/2005-06/msg00285.html (9,292 bytes)
4. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: "Martin, AA6E" <martin.ewing@gmail.com>; Date: Wed, 8 Jun 2005 11:12:32 -0400; That was a long-winded way of saying that if you measure 387 volts peak into 50 ohms, you've got 1500 W RMS, assuming you've got a pure sine wave. (If you have visible harmonics, you've got real trou; /archives//html/TenTec/2005-06/msg00286.html (7,591 bytes)
5. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: Robert & Linda McGraw K4TAX <RMcGraw@Blomand.Net>; Date: Wed, 08 Jun 2005 17:53:47 -0500; I measure the peak to peak value using my scope. Then power equals E squared divided by R. This is good enough for "government work". Looking at it another way: 100 watts across 50.0 ohms produces 70; /archives//html/TenTec/2005-06/msg00297.html (9,147 bytes)
6. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: "Martin, AA6E" <martin.ewing@gmail.com>; Date: Wed, 8 Jun 2005 21:04:46 -0400; Bob, If you've got 274 volts sinewave peak-to-peak, that's .707 * (274 / 2) = 96.8 volts RMS. Across 50 ohms, my number is P = 96.8**2/50 = 187 Watts. RMS volts are what you want, and it's the voltag; /archives//html/TenTec/2005-06/msg00299.html (9,924 bytes)
7. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: Duane - N9DG <n9dg@yahoo.com>; Date: Wed, 8 Jun 2005 18:49:06 -0700 (PDT); It's relevant for sizing the RF components of that new direct DSP to RF D to A legal limit Class E amplifier based radio that Ten Tec will be building...:). Duane N9DG ______________________________; /archives//html/TenTec/2005-06/msg00300.html (8,160 bytes)
8. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: Ken Brown <ken.d.brown@verizon.net>; Date: Wed, 08 Jun 2005 18:47:47 -1000; Hi Barry, You want to use the RMS voltage value to calculate the power using Power equals Voltage squared divided by resistance. However measuring the peak to peak AC voltage with an oscilloscope is; /archives//html/TenTec/2005-06/msg00307.html (11,235 bytes)
9. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: Ken Brown <ken.d.brown@verizon.net>; Date: Thu, 09 Jun 2005 14:49:27 -1000; This is wrong! Using peak voltage instead of RMS voltage of the peaks will result in a power miscalculation by a factor of two. Using Peak to Peak voltage instead of RMS voltage of the peaks would r; /archives//html/TenTec/2005-06/msg00310.html (8,827 bytes)
10. Re: [TenTec] Voltages on 50 ohm Dummy Load (score: 1): Author: Robert & Linda McGraw K4TAX <RMcGraw@Blomand.Net>; Date: Thu, 09 Jun 2005 21:55:58 -0500; I'm really surprised at all the folks that missed the entire point. Look at the numbers again that I stated relating to 50 ohms: 70.710 volts x 1.414 amps = ~100 watts and 273.861 volts x 5.477 amps; /archives//html/TenTec/2005-06/msg00313.html (10,284 bytes)

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