WEBVTT NOTE Power, RMS, and Decibels at HF 1 00:00:00.000 --> 00:00:22.053 The three power forms. P equals E times I P equals I to the power of 2 R P equals (E to the power of 2 ) divided by (R) The exam gives you two of {voltage, current, resistance} and wants power. 2 00:00:22.053 --> 00:00:47.190 Pick the form that uses what you were given: 400 VDC across 800 ohms. Voltage and resistance, so P = E to the power of 2/R = 160000/800 = 200 watts. 12 VDC, 0.2 amperes. Voltage and current, so P = E times I = 2.4 watts. 3 00:00:47.190 --> 00:01:11.985 7.0 mA through 1250 ohms. Current and resistance, so P = I to the power of 2R = (0.007) to the power of 2 times 1250 = 0.061 W = 61 milliwatts. Convert to base units first, amps, volts, ohms, and back only at the end. 4 00:01:11.985 --> 00:01:30.724 The milliampere in that last one is where the mistakes happen. RMS, again, and why it is the one that matters. There is a diagram here. A sine wave over two cycles. 5 00:01:30.724 --> 00:01:46.263 Vertical markers show peak amplitude from the zero line to the crest, peak-to-peak from trough to crest, and RMS at about 0.707 of peak. 6 00:01:46.263 --> 00:01:56.090 A horizontal marker shows one period between successive positive-going zero crossings. 7 00:01:56.090 --> 00:02:06.970 Question. What value of an AC signal produces the same power dissipation in a resistor as a DC voltage of the same value? 8 00:02:07.870 --> 00:02:11.080 Answer. The RMS value. 9 00:02:11.680 --> 00:02:22.639 That is the definition of RMS, and it is why every power calculation uses it. 10 00:02:22.639 --> 00:02:48.543 For a sine wave: V RMS equals (V peak ) divided by (the square root of (2)) equals 0.707 V peak V pp equals 2 V peak Two conversions the pool asks directly: RMS 120 V → peak-to-peak? 11 00:02:48.543 --> 00:03:05.480 peak = 120 times 1.414 = 169.7; peak-to-peak = 2 times 169.7 = 339.4 volts. Peak 17 V → RMS? 17 times 0.707 = 12 volts. 12 00:03:05.480 --> 00:03:30.835 PEP, in the five shapes the exam uses. Peak envelope power is the average power during one RF cycle at the crest of the modulation envelope. Every PEP question is one of these. Peak-to-peak voltage into a known load. 13 00:03:30.835 --> 00:03:49.733 P equals ((V pp divided by 2the square root of (2)) to the power of 2 ) divided by (R) Two steps: peak-to-peak → RMS (divide by 2, then by √2), then RMS → power. 14 00:03:49.733 --> 00:04:13.562 200 V peak-to-peak into 50 ohms: RMS = 200 / 2 / 1.414 = 70.7 V; P = 70.7 to the power of 2 / 50 = 100 watts. 500 V peak-to-peak into 50 ohms: RMS = 176.8 V; P = 176.8 to the power of 2 / 50 = 625 watts. 15 00:04:13.562 --> 00:04:35.160 Power into a known load, backwards to voltage. V RMS equals the square root of (P R) 1200 W into a 50-ohm dummy load: √(1200 times 50) = √60000 = 245 volts RMS. An unmodulated carrier. 16 00:04:35.160 --> 00:04:43.096 Question. What is the ratio of PEP to average power for an unmodulated carrier? 17 00:04:43.996 --> 00:04:47.772 Answer. 1.00. 18 00:04:48.372 --> 00:04:58.692 Question. What is the output PEP of an unmodulated carrier if the average power is 1060 watts? 19 00:04:59.592 --> 00:05:03.392 Answer. 1060 watts. 20 00:05:03.992 --> 00:05:14.366 A steady carrier has no envelope to peak, its amplitude never changes, so peak and average are the same number. 21 00:05:14.366 --> 00:05:33.152 This is the reason a 100 W carrier and a 100 W PEP SSB signal are very different things in practice: the SSB signal's average power is a fraction of its PEP, while the carrier's is all of it. Decibels. 22 00:05:33.152 --> 00:05:41.321 Question. What dB change represents a factor of two increase or decrease in power? 23 00:05:42.221 --> 00:05:45.857 Answer. Approximately 3 dB. 24 00:05:46.457 --> 00:05:54.156 The anchor from the foundations track, unchanged. And one General-level refinement: 25 00:05:54.156 --> 00:06:00.904 Question. What percentage of power loss is equivalent to a loss of 1 dB? 26 00:06:01.804 --> 00:06:05.386 Answer. 20.6 percent. 27 00:06:05.986 --> 00:06:31.370 One decibel is a factor of 10^(−0.1) = 0.794, so 79.4% gets through and 20.6% is lost. Worth knowing because feed-line loss is quoted in decibels per hundred feet, and "only 1 dB" turns out to be a fifth of your power. 28 00:06:31.370 --> 00:06:33.466 Parallel currents. 29 00:06:33.466 --> 00:06:41.826 Question. How does the total current relate to the individual currents in a circuit of parallel resistors? 30 00:06:42.726 --> 00:06:48.246 Answer. It equals the sum of the currents through each branch. 31 00:06:48.846 --> 00:07:02.003 Same voltage across each branch, currents add. The complement of the series rule, where the current is common and the voltages add. Check yourself. 32 00:07:02.003 --> 00:07:21.516 What is the PEP of a signal showing 100 V peak-to-peak on a scope across a 50-ohm load? Your feed line is rated at 2 dB loss. What fraction of your power reaches the antenna? An unmodulated carrier reads 500 W average. 33 00:07:21.516 --> 00:07:22.948 What is its PEP? 34 00:07:28.948 --> 00:08:04.658 RMS = 100 / 2 / 1.414 = 35.4 V; P = 35.4 to the power of 2 / 50 = 25 watts. 2 dB is two 1 dB losses: 0.794 times 0.794 = 0.63, so about 63% arrives, a third of your power gone. (Equivalently, 10^(−0.2).). 500 W. 35 00:08:04.658 --> 00:08:13.628 For an unmodulated carrier, PEP equals average power.