## K9BF

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### Latest Topics

Topic | Created | Posts | Views | Last Activity |
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Average vs peak power | Sep 21st 2011, 19:18 | 7 | 2,651 | on 22/9/11 |

### Latest Posts

Topic | Author | Posted On |
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Average vs peak power |
K9BF | on 22/9/11 |

Thanks Zack for all your help. I think my curiosity is satified. 73 and take care Ben K9BF |
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Average vs peak power |
K9BF | on 22/9/11 |

Hi Zack Boy this posting method sure limits the ability for one to fully express themselves. hihi. So is the scaling on the Bird meter average power using the .63 formula or rms power using the .707 formula. This is what I want to know. Thanks Ben K9BF |
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Average vs peak power |
K9BF | on 22/9/11 |

Here is the answer I was looking for: Hi Ben, The answer is that Pave is not (v)ave times (i)ave. Average v is zero for a full cycle of a sine wave for example and so is the average i. Note that the voltage (or current) is positive half the time and therefore the average is zero. The thing is that we want to calculate the power that does work. The average of v or i is zero over a whole cycle and we want the power that does work in a whole cycle (note that the Vave of a half cycle of a sine wave is .6366 times peak...but that is not relevant to what we want to calculate here). We know the power in a full cycle does work and is not zero. So we want to find a value of voltage or current by considering the instantaneous power at many points along the waveform cycle. Instantaneous power is i squared times R (or v squared divided by R). Average power then is average of (i) squared times R or [average of the i's squared] times R. Average power is also the average of (v) squared divided by R or [average of v's squared] divided by R. Kind of working backwards, but you can see that a current or voltage value needed to calculate power are the square roots of these means of squared i's and v's. So now...the Irms is the square root of [average of i's squared] or the square root of the mean of all the squared i's. And the Vrms is the square root of [average of v's squared] or the mean of all the squared v's. So therefor Pave = (Irms) squared x R and Pave = (Vrms) squared / R And you can show that...Average power equals Irms x Vrms. Also if the waveform is a sinewave and we take instaneous values of v and i and square them and take the mean value of them and take the square root of that...we get about .707 times the peak value.. It's unfortunate that ARRL discusses Vave and Iave of a sine wave right before they jump into PEP for a sinewave since it has nothing to do with calculating the average power of a full RF cycle. Chuck, KE9UW |
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Average vs peak power |
K9BF | on 21/9/11 |

A bright EE friend of mine keeps reminding me that the devil is in the details. It is my understanding that my Bird 43 watt meter is reading average power. In chapter 4 of the 2009 handbook this is discussed. What is confusing me is that .63 X peak is used to find average voltage and current levels. Only makes sense that average power is then average voltage times average current. And it makes sense that PEP is the average power of the highest peak. So PEP is really peak average power. In CW, PEP would be the same as the average power. In the book they use the RMS value when calculating PEP. What they are saying at one point is that RMS is used to compute average power and yet they refer to using .63 to find average. The factor between RMS power and Average power is 1.11. Not much I know but is my Bird meter reading average average power or RMS average power???? And when a transmitter is rated at 100 watts CW and 200 watts PEP it seems they are only taking into consideration the duty cycle difference between the two modes. If I key down my transmitter I get 200 watts out and if I use my Bird 4391A peak reading meter, I also get 200 watts PEP in sideband. Ben |