Welcome to the FlexRadio Community! Please review the new Community Rules and other important new Community information on the Message Board.
The latest SmartSDR Software:
SmartSDR v4.1.5 | SmartSDR v4.1.5 Release Notes
SmartSDR v3.10.15 | SmartSDR v3.10.15 Release Notes

The latest 4O3A Genius Product Software:
The latest 4O3A Genius Product Software and Firmware
How to Receive Technical Support::
If you are needing assistance with FlexRadio products, please refer to the product documentation or check the Help Center for known solutions. Need technical support from FlexRadio? It's as simple as creating a HelpDesk ticket.

Automated Antenna Connection

Bob G W1GLV
Bob G W1GLV Member ✭✭
edited January 2017 in New Ideas
I need to automate my antenna connections. I have a network centric relay board. I need to know the amperage of 500 - 600 watt RF signal. Does anyone know how to figure that out?
0
Up Down
0 votes

In Review · Last Updated

Comments

  • Jim Gilliam
    Jim Gilliam Member ✭✭
    edited November 2016
    The current is a function of the impedance. One needs to know that before Ohm's law can be applied.
  • Walt - KZ1F
    Walt - KZ1F Member ✭✭
    edited November 2016
    Jim, wouldn't the impedence be the standardly 50 ohms?, isn't the hv drop across the antenna?
  • Jim Gilliam
    Jim Gilliam Member ✭✭
    edited September 2015
    Well if the impedance is 50+j0 and the AVERAGE power is 500 Watts, the RMS current would be  3.162 Amperes.
  • Burt Fisher
    Burt Fisher Member ✭✭
    edited August 2016
    If the impedance was always 50 ohms we wouldn't need SWR meters.
  • Barry-W4TGA
    Barry-W4TGA Member ✭✭
    edited September 2015
    Assuming the impedance is 50 ohms, the Ohms Law formula says at 600 watts, the current will be 3.46 amps with a voltage of 173.21 volts.  Let's assume you don't switch the antenna with the transmitter keyed, and then a relay with 5 amp or better contact rating should get the job done.

    Good luck,
    Barry (W4TGA)
  • Bill Turner
    Bill Turner Member ✭✭
    edited January 2017

    Divide the power by the impedance and take the square root of the answer.

    73, Bill W6WRT


  • Jim Gilliam
    Jim Gilliam Member ✭✭
    edited September 2015
    Youi answer is correct only if the impedance is real. If it is complex, you will get both average real and reactive power.
  • KY6LA_Howard
    KY6LA_Howard Member ✭✭✭
    edited January 2017
    Jim is correct. The RMS current of 3.162 A is only valid if your SWR is 1:1. So you really need to design your system for a worst case SWR. When you have a SWR greater than 1:1 you have both a resistive current and a reactive current So please give us your worst case SWR and whether or not it's capacitive or inductive so we can calculate your worst case current THEN follow the usual Engineering safety factor so that your relay is able to handle at LEAST TWICE your worst case current just to be safe because RMS currents are Only 70.7% Peak currents.
  • Jim Gilliam
    Jim Gilliam Member ✭✭
    edited September 2015
    Power seems to be a misunderstood subject. When I hear someone use the term RMS power I want to go somewhere and puke. Power is a squared function and why in the world would one want to square a squared function, takes its average over a period and, then, take the square root. Oh well life goes on.

Leave a Comment

Rich Text Editor. To edit a paragraph's style, hit tab to get to the paragraph menu. From there you will be able to pick one style. Nothing defaults to paragraph. An inline formatting menu will show up when you select text. Hit tab to get into that menu. Some elements, such as rich link embeds, images, loading indicators, and error messages may get inserted into the editor. You may navigate to these using the arrow keys inside of the editor and delete them with the delete or backspace key.

Comment As ...

Categories