In last week’s post on RF safety, I mentioned a worksheet that the ARRL provides to help you determine your RF exposure. http://www.arrl.org/files/file/Technology/tis/info/pdf/rfex1_2.pdf. I thought it would be useful to show the evaluation of my station on one band.
The worksheet dedicates one page (A) to determining the power at your antenna by looking up the cable you’re using and determining its loss. As I’ve said before, with a 100 W antenna and some unavoidable losses between the radio and the antenna, the 24.9 MHz (or 12 meter) band is the first place many of us will need to determine exposures. My antenna is a Log Periodic Dipole Array (often sarcastically referred to as, “an antenna that operates poorly over a wide range of frequencies”) with minimal – but some – antenna gain on 12m. My cable is RG-213/U, a more recent version of the RG-8/U cables that hams have been using for decades. Here’s what worksheet A looks like for my station:
First, how do I know the loss? Because RG-213/U is a well established standard cable, I find a datasheet from a reputable manufacturer and find the table of losses vs. frequency they provide.
You can see that they list attenuation at 10 MHz and 50 MHz, not at the 25 MHz I need. Time for some linear interpolation.
Given those numbers I’ll round up to 0.6 dB loss at 10 MHz and use 1.3 dB at 50, I’ll solve for the slope assuming it’s a straight line between those points, which is going to be close.
Slope = (1.3 – 0.6) / ( 50-10), which gives and answer of 0.0175 dB/MHz. That means the loss is that much more 15 MHz higher than the 0.6 dB insertion loss. That’s found by multiplying
Loss at 25 MHz = 0.6 + .0175* 15 or 0.863 dB.
I’ll round that to 0.86 dB. Notice that the numbers we used to start (1.3 and 0.6 dB) are per hundred feet. Since my coax is 50 feet, the loss is 0.43 dB.
Now how to I turn that loss into percentage? I’m guessing that’s probably a new thing to people. Remember a dB is a power ratio, where
dB = 10*log (ratio)
is the value in dB and log denotes the base 10 logarithm. We want to find the ratio that gives a loss: loss means negative dB. It’s -0.43 dB not +0.43.
(-.43) = 10 log (ratio), so
(-.43/10) = log (ratio)
10^(-.43/10) = ratio = 0.906 or 90.6% of the power remains.
Out of 100 Watts, 90.6% or 90.6 W makes it to the antenna. Note that in step J, they want to solve for the number of watts lost. That’s (1-.906) or 0.094 times 100 or 9.4 W lost.
Worksheet B takes this process almost to its conclusion. It multiplies the powers found at the end of A and then leaves you to find a few more details.
The Duty Factor is based on recommendations from the ARRL RF Safety page. It considers digital modes as the worst case, and my mental model for this page is operating the FT8 mode, which has taken the ham radio world by storm.
This sheet shows my 100% duty factor and 50% transmit time (built into FT8) means I deliver 45.3 W average to the antenna. By the first table (in the last article) we can see that evaluation is required on 12m if I deliver 75W to the antenna, and I’m delivering 45.3W. The harder question is in the second table in that post, which has required distances for people to be from the antenna, depending on antenna gain.
Antenna gain is a concern of mine because it doesn't mean what people tend to think it means when dealing with powers. I've seen this misconception with professionals, so I'm sensitive to it. Antenna gain doesn't mean that if I put 100 Watts into an antenna with a gain of 6 dB (4 times) that I'm getting 400 watts out. That's impossible. If I put 100 Watts in, I get 100 watts out. What antenna gain means is that one antenna puts more power in some direction than another reference antenna, and it does that by taking power from all the other directions and putting it where the antenna is pointed. That last table looks suspiciously like they're saying there's more power out of the antenna, but there isn't. It's stronger in one direction, compared to another antenna but that's all. I'll cut them some slack because this is work in the extreme near field of the antennas and that's absurdly complex to model.