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Dare4444

Post subject: RF power calculation? Posted: Sep Fri 18, 2020 2:16 am 

Joined: Mar Thu 01, 2018 1:30 am Posts: 398

Attached LtSpice TX out. It's an asymmetrical sine wave with +8V and 10V peaks.
10+8 = 18/2 = 9V peak right?
Then output power is going to be 9*9 /50ohm load *2 = 810mW.
Am I correct?
Attachments: 
IMG_20200918_064210.jpg [ 451.39 KiB  Viewed 744 times ]



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Audioman

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 3:05 am 

Joined: Apr Thu 14, 2016 8:25 pm Posts: 479 Location: pensacola fl

Hi. If you are talking about MW am band part 15 then there are two flavors. The most popular the one where you measure the dc input to the final R.F. amplifier stage (that is the one that drives the load (antenna)). That is dc input voltage times current drawn by that stage. That assures that your actual power will be less than 100mw as the rule goes on to say that the dc input may not exceed 100mw. If you want to read the power into the load then first you must know the impedance of the load then you measure the current into the load and square it then multiply that by the impedance and you have the power. Problem is that unless you are working with a 50 ohm resistance then you may not be able to accurately measure the current. Also if it is reactive and not a resistive load it gets messy with math and measurements. By the way the rest of the rule talks about the antenna + ground lead + transmission line may not exceed 3 meters. So if you are using tubes plate voltage x plate current if you are using solid state collector voltage x collector current or drain voltage x drain current these values may be read at the point where the dc to the stage is applied.


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Mikeinkcmo

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 3:24 am 

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Joined: Oct Sun 15, 2006 12:57 pm Posts: 7292 Location: Liberty, Missouri

If you are presuming a 50 ohm terminating impedance, you are correct.
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https://mikeharrison.smugmug.com/
Mike


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LM386

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 11:24 am 

Joined: Aug Thu 06, 2015 2:20 am Posts: 287

Dare4444 wrote: Attached LtSpice TX out. It's an asymmetrical sine wave with +8V and 10V peaks
Then output power is going to be 9*9 /50ohm load *2 = 810mW. Your answer is correct, but the equation 9*9 /50ohm load *2 does NOT equal 810 mW. The equation should be 9*9 /(50ohm load *2) or 9*9 /50ohm load /2
Last edited by LM386 on Sep Tue 22, 2020 11:21 pm, edited 1 time in total.


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pixellany

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 11:41 am 

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Joined: Jul Mon 26, 2010 8:30 pm Posts: 29054 Location: Annapolis, MD

Power = E^2 / R, where E is the rms voltage The waveform has 2 components: 1 volt DC> equals 1 volt rms 9 volts peak AC> equals 6.36 volts rms total rms voltage: 7.36 volts power: 0.81 watts You can also get the net rms value of the waveform using calculusbut that's another story.....
_________________ Mark "Voltage is fun to watch, but it's the CURRENT that does the work."


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pixellany

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 12:28 pm 

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Joined: Jul Mon 26, 2010 8:30 pm Posts: 29054 Location: Annapolis, MD

pixellany wrote: Power = E^2 / R, where E is the rms voltage
The waveform has 2 components: 1 volt DC> equals 1 volt rms 9 volts peak AC> equals 6.36 volts rms total rms voltage: 7.36 volts power: 0.81 watts
OOPS!! If the rms voltage is 7.36, then the power is 1.08 watts
_________________ Mark "Voltage is fun to watch, but it's the CURRENT that does the work."


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bb.odin

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 2:46 pm 

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Joined: Jun Sun 22, 2008 2:03 am Posts: 943 Location: Burke, VA 22015

The instantaneous power delivered to a resistor R by a timevarying voltage waveform v(t) is: p(t) = v^2(t)/R
By definition, average power, or power for short, is the time average of the instantaneous power over one second. If the waveform is periodic, it is sufficient to compute the time average over one cycle T as P = 1/T integral(0 to T) p(t)dt
Your voltage waveform consists of two components: a sinusoid with amplitude A and angular frequency ω and a DC offset B: v(t) = Acos(ωt) + B
The instantaneous power is: p(t) = 1/R (A^2cos^2(ωt) + 2A Bcos(ωt) + B^2) p(t) = 1/R (A^2/2cos(2ωt) + A^2/2 + 2A Bcos(ωt) + B^2)
Since the cos terms average to 0 in one cycle, the power delivered to R is: P = 1/R (A^2/2 + B^2)
With R = 50Ω, A = 9V and B = 1V, P = 1/50 (9^2/2 + 1) = 830mW
_________________ Binh


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flyboy71

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 5:22 pm 

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Joined: Mar Wed 11, 2015 1:46 am Posts: 1348 Location: Boiling Springs, PA


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bb.odin

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 6:13 pm 

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Joined: Jun Sun 22, 2008 2:03 am Posts: 943 Location: Burke, VA 22015

If your intention is to feed that voltage waveform to a tuned short monopole antenna, the actual resistive load seen by the transmitter is not easy to determine. Unless the antenna is designed for a load resistance, the assumption of 50 Ω load for a short part 15 antenna is incorrect. The load can be roughly lumped into two components: radiation and loss resistances. The radiation resistance accounts for the energy actually transmitted by the antenna. The loss resistance accounts for all the losses in the conductors (mainly in the wire of the tuning inductor) and antennatoground paths. For a 3meter antenna in the MW band, the radiation resistance is about 20 mΩ which is tiny compared to the loss resistance. The ground loss is affected by several factors such as the antenna's height above ground, ground conductivity, surrounding trees and buildings, etc. The antenna is highly capacitive and requires a big inductor to tune it. This equivalent circuit helps to visualize it. Attachment:
Antenna_Equivalent_Circuit.jpg [ 105.84 KiB  Viewed 559 times ]
One easy way to measure the power delivered to the antenna is to insert a 1 Ω resistor in series before the variable inductor for tuning the antenna. The antenna's RMS current can be measured with an AC voltmeter across that resistor. The voltage reading directly corresponds to the RMS current. Adjust the variable inductor for a peak voltage reading which indicates that the antenna is tuned. Since the transmitter feed's RMS voltage is known by design, the power is simply the voltagecurrent product. If it's too high, the feed voltage must be reduced.
_________________ Binh
Last edited by bb.odin on Sep Sun 20, 2020 9:42 am, edited 1 time in total.


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bb.odin

Post subject: Re: RF power calculation? Posted: Sep Fri 18, 2020 8:15 pm 

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Joined: Jun Sun 22, 2008 2:03 am Posts: 943 Location: Burke, VA 22015

pixellany wrote: If the rms voltage is 7.36, then the power is 1.08 watts Sorry, Mark, that is incorrect. RMS voltages do not add up but powers do. If E1 and E2 are two RMS voltages, E^1 and E2^2 are powers delivered to a 1Ω load. The total power is E^2 = E1^2 + E2^2 and the total RMS voltage is E = √(E1^2 + E2^2) = √(6.364^2 + 1^2) = √(40.5 + 1) = 6.442 Vrms
_________________ Binh


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OldWireBender

Post subject: Re: RF power calculation? Posted: Sep Mon 21, 2020 5:24 am 

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Joined: Jan Thu 01, 1970 1:00 am Posts: 4503 Location: Perrysburg, OH, 43551 U.S.A.

If only the AC component of the signal is applied to the load, the DC component has no bearing on the power delivered to the load. This is the case where the antenna is coupled with a transformer or a blocking cap. Then
Vrms=Vpk/sqrt(2) POrms=Vrms^2/RL = Vpk^2/(2*RL) = 9^2/(2*50) = 81/100 = .81W = 810mW
John
_________________ “Never attribute to malice that which can be adequately explained by stupidity.” ― R. A. Heinlein


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ACORNVALVE

Post subject: Re: RF power calculation? Posted: Sep Tue 29, 2020 4:51 am 

Joined: Jan Thu 16, 2020 12:29 am Posts: 816

The DC offset would not normally be coupled into an antenna and a DC blocking cap would be used. Though it might be coupled into a 50R dummy load for testing, but even so its an RF application and the DC should be blocked unless there is some specific application not to.
Since the peak voltage is 9V then the rms voltage is 9 x 0.7071V = 6.364V, and power = V^2/R, so the power delivered to the 50R load is (6.364)^2/50 = 810mW.
If its a MW band antenna application, where the antenna is very small in geometry compared to the wavelength, as pointed out the radiation resistance (and therefore the radiated RF power as EM waves) will be a very low value. One way to improve this situation for a MW band transmitting antenna of small geometry (<3m size) its to use a resonant loop design with 3 to 5 turns, typically with a coupling in at one turn and an impedance matching transformer & capacitance, to lower the input impedance to the region of 50R, all of my AM transmitting loops for "pantry transmitters" are like this and they significantly out perform a linear wire antenna, especially for near field radiation of the magnetic component of the wave which is well received by transistor radios with ferrite rods. The ratio of the electric to magnetic component of the wave doesn't even out to the impedance of free space until you are in the far field a few wavelengths from the transmitting source.


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Dare4444

Post subject: Re: RF power calculation? Posted: Sep Tue 29, 2020 9:28 am 

Joined: Mar Thu 01, 2018 1:30 am Posts: 398

Wow, a wealth of information! Other RF forums pale in comparison when it comes to the sheer knowledge of its members.


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