## Average vs peak power

Sep 21st 2011, 19:18 | |

## K9BFJoined: Apr 4th 1998, 00:00Total Topics: 0Total Posts: 0 |
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 |

Sep 21st 2011, 22:45 | |

## W1VT
Joined: Apr 4th 1998, 00:00Total Topics: 0Total Posts: 0 |
It may help to do some exercises with DC voltages. Q1 What is the average voltage of a sine wave over the first 180 degrees or PI radians? Answer 2/PI, which is approximately 0.637I. If you take calculus, you will learn how to compute the area under the hump of a sine wave. Q2 What is the power dissipated by a 50 ohm resistor connected to a 10 volt battery Answer 2 watts=10V*10V/50 ohms Q3 What is the power dissipated by a 50 ohm resistor connected to a 5 volt battery Answer 1/2 watt=5V*5V/50ohms Q4 Suppose we only connect the 50 ohm load to a 10 volt battery 50% of the time, what is the average amount of power dissipated by the resistor? Answer 1 watt=2 watts *((0.5) Q5 If both the 5V battery and the 10V battery connected half the time have the same average voltage across the load, why aren't the powers equal? If you have gotten this far, a useful but tedious exercise is to approximate a sine wave with a series of stepped DC voltages, so you can compute the power across the resistor for a sine wave. As you add more and more steps, you will a more accurate value for a sine wave. Similarly, you can get am approximation for the area under the hump of a sine wave by drawing a sine wave on a piece of cardboard, cutting it out, and measuring it on a sensitive scale. Zack Lau W1VT ARRL Senior Lab Engineer |

Sep 22nd 2011, 13:30 | |

## K9BFJoined: Apr 4th 1998, 00:00Total Topics: 0Total Posts: 0 |
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 |

Sep 22nd 2011, 14:14 | |

## W1VT
Joined: Apr 4th 1998, 00:00Total Topics: 0Total Posts: 0 |
Hi Ben--glad you got an explanation that helps--that is the purpose of these forums! The 43 Bird wattmeter does not read average power! It has a simple diode detector and a meter that is heavily damped. If you hold down the key long enough it will measure CW or continuous wave. If you try to measure SSB it will significantly underestimate the average power. Similarly, cheap AC voltmeters will just measure the peak voltage, and scale the reading as if it were a perfect sine wave. What should the scaling be? Well, we have at least two choices. One, is that we can scale for the average value of the sine wave. The other is that we can scale it for what we call the RMS value, or the effective power or heating value. Contrary to “common” sense, the two scaling constants aren’t the same—they differ by the 1.11 factor you mention. What does your 100 watt CW transmitter read when you measure it on the Bird 43? If it measures 100 watts on the Bird 43 and 200 watts on the 4391A in PEP, this suggests you have an ALC glitch that momentarily allows transmission of a 200 watt signal that isn’t picked up by the 43. We also have a Bird 4391--it measures CW or PEP, but has no circuitry for measuring the actual average power of your voice or modulated CW signal. Zack Lau W1VT Senior Lab Engineer |

Sep 22nd 2011, 19:33 | |

## K9BFJoined: Apr 4th 1998, 00:00Total Topics: 0Total Posts: 0 |
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 |

Sep 22nd 2011, 22:30 | |

## W1VT
Joined: Apr 4th 1998, 00:00Total Topics: 0Total Posts: 0 |
In practice, neither formula is actually used. What you do is you generate a high power signal of a known power, and adjust the potentiometer inside the Bird slug to read the proper value of RF. In our case, we would use a kilobuck Agilent/HP Thermistor based wattmeter and a series of attenuators to accurately measure or set up the standard high power signal. In theory, if we had directional couplers with a precise coupling factor, we would measure the peak voltage. Then, knowing we have a clean sine wave signal, we would use the 0.707 factor to convert the peak value into an rms voltage suitable for calculating the average power of a sine wave. If we try to calculate the average power of a sine wave using the average voltage (Vav^2/R), we will be off by over 20% (1.11 squared). This technique is more useable with couplers using transformers, such as the Tandem Match by John Grebenkemper or the QRP Wattmeter by Roy LeWallen. But, these designs use couplers with a direct connection to the transmission line--something quite different from the Bird wattmeter/removable slug concept. Vrms is chosen to allow accurate power calculations without having to throw in a correction factor. If you know the waveform of the signal and the peak voltage, you could calculate a correction factor to get accurate power calculations. Zack Lau W1VT ARRL Senior Lab Engineer |

Sep 22nd 2011, 22:45 | |

## K9BFJoined: Apr 4th 1998, 00:00Total Topics: 0Total Posts: 0 |
Thanks Zack for all your help. I think my curiosity is satified. 73 and take care Ben K9BF |