Toning down a transmitter

Have you looked at the CCrane FM transmitter?

I'm gonna get some flak on this but...The FCC has bigger fish to fry. 10 Mw on a channel that causes no interference to licensed services, nobody complaining that your signal is blocking something they are try to hear, etc...etc.. You get the picture.

WDCX AM1610 Part 15
John
Owner-Operator-Chief Engineer-Program Manager

A commonly used method to cut signal power is called a "pi attenuator". It is basically three resistors chosen to give the correct input and output impedance and the desired attenuation. A web search will turn up a lot of information on this.

As a rough start, we can calculate the attenuation needed to drop 10 mW. down to 11 nW. The 11 nW. number gives about the legal field strength using a 1/4 wave gnd. plane antenna.

Atten = 10 log (10 mW./ 11 nW.) = 59 dB.

You will probably not be happy with the range which results from this much attenuation, nor from any system operating within the rules since you are used to a block range. Try it to determine if it works for you. Use your judgement on how to proceed.

Neil

I'd agree in principle. But if we're going with Rich's estimate for the 11 nw, I *think* that was into a dipole, which has slightly less gain than a groundplane. Also there was no discussion in that estimate of losses due to coax, possible swr mismatch, less than optimal antenna construction practices and so on. Also one has to consider that proximity of metal objects within the first few wavelengths of the antenna may act as antenna elements, and larger metal objects further away can also reflect, all of which can add up to some areas that might have considerable cancellation or reinforcement of the field strength at a given spot.

A dipole or groundplane may be an efficient antenna, but that doesn't mean automatically that a given installation when viewing the antenna, coax, and antenna as a system will be. If it was all professionally designed, built and installed? Yeah, it could be pretty low loss, but I'm assuming most hobby stations are not.

In other words, even though one can calculate the transmitter's output to the coax/antenna and use math to figure out precisely how much attenuation to use to theoretically bring it down to the 11.4 nw into a dipole level, I don't personally feel it would be a method 100% sure of yielding the allowed 250 uV/m @ 3 M distance. It'd really be nice if it could all be done with calculator, but I just don't think it will hold up in practice. Even just between different brands or lengths of coax, the difference in losses can be considerable and I don't think I've ever seen an swr meter that can work with a signal as small as 11 nw. In the other direction, if some RF was escaping the transmitter via the audio or power supply wires, there could be more radiation occurring there than from the presumed antenna.

I'd agree that attenuation is going to be necessary in any case, however. My thought on the matter as stated in other threads on this forum, is that the more practical approach would be to make a variable attenuator and adjust it down until range/levels that are not exceeding what would be expected with the ranges which the 250 uV specification for part15 FM can provide.

The FCC's estimate of that range is about 200 ft. One can consider that there are about 12 average city blocks (lnear, not area) to a mile. That works out to 440 ft. So if your station was in the center of the block on a street, and you could pick it up at both ends of the block, that'd be about what they're talking about. 200 ft out would be close to a block of coverage in terms of actual area. Car radios are sometimes quite sensitive, so yeah, maybe a block in all directions to a good one, depending on where the transmitter is and the amount of noise and objects like buildings in the way.

But if you were being received clear as a bell on something like a small cheap radio in a house that was a whole block away? Almost certainly putting out more than the allowed amount. Quite a bit more. It'd have to be exceeding the allowed amount of field strength to do that, so far as I understand the matter.

Daniel

Daniel has posted some excellent information on this subject. Just a couple of points to add:

The gain of a 1/4 wave dipole is essentially the same as a 1/2 wave dipole since electrically the radiator is mirrored by the ground plane so it appears as two 1/4 wave elements center fed. Practically there may be a slight difference but it is not worth worrying about.

The city blocks I am used to are about 1600 feet long so my comment about block range stands corrected.

The loss for genuine RG58/U coax is .05 dB per foot at FM frequencies. Here's how to use this info. If as in my previous post you need to attenuate your signal 59 dB and you are using 50 feet of this coax, the coax will add 50 x .05 = 2.5 dB of loss. Your attenuator then can be 59 dB - 2.5 dB = 56.5 dB. I assume that if you are not familiar with decibels, this example may help.

You will not be able to find the exact resistances given in the pi design table. Just get the closest standard value. For example, one chart I saw had two 49 ohm and one 7120 ohm resistor. Two 47ohm and one 6.8 kohm are close enough for what you are doing.

Also beware of "RG58 type" cable, especially the stuff sold for CB work. A garden hose will as well as this at 100 MHz.

Losses due to SWR will not amount to much, even if using a 72 ohm cable with a 50 ohm antenna so you might do OK with video cable which is easier and cheaper to get. After all, you can compensate for cable and SWR losses by changing your attenuator since you are starting with a lot of power.

I hope this deluge of information doesn't discourage you from some experiments. As Daniel suggested, put your station together and go for about 200 foot range (maybe a bit more on a car radio). I don't think you will have any problems with this "one block" range. The trouble comes with mile range.

Once you go on the air in stereo with a frequency stabile transmitter, you will never go back.

Neil

10 mw will cover a long ways on a high ground plane on FM!!!
How about 5 to 8 miles!!! Do you feel lucky?
I guarantee an FCC "visit" real soon!
Best to put the described attenuator inline and be compliant.
Range with proper attenuator,1 to 2 blocks. BT,DT!

Regards,Lee
http://www.freewebs.com/wilcomlabs/index.htm

What a lot of information! Thanks, that's far more than I expected. I guess I'll start at the top.

Mojoe, I looked at the CCrane transmitter, nice price, but it doesn't look like it will accept an external antenna. I don't really want the antenna in the house as I want even coverage. Though it sure would be a lot safer, lightning wise.

John, I really don't want to take a chance. Maybe if I lived in a rural environment, I'd go for the full 10 mw, but as I said, I live 10 minutes out of Boston. There really is no such thing as a vacant frequency here. Even on my 89.3, when I drive around town listening outside the range of my station, I can hear weak signals from other stations popping in and out.

Neil, thanks for the tip on the pi attenuator. I looked on Google, and the first result I found was a calculator with instructions. This is exactly what I need. And your formula helps a lot as well.

Daniel, I think a variable attenuator would be the right approach. I'll have to see if I can modify the pi attenuator with variable resistors.

And with the talk about the size of city blocks, I went to Google maps, and measured the size of mine. As it turns out, my signal is more even than I thought it was. I live on a corner lot, and one street is longer than the other. The direction I was getting a full block is about 400 feet, and the direction I was getting a half a block is about 600. And yes, that is with the car radio. When I use my cheap walkman, I get about 200 feet, tops.

As for the cable, I was thinking of ordering about 20 feet of RG8 from a ham shop. Though at these power levels and frequencies, that would probably be overkill, and the cheap RG58 from radio shack would work as well. That's what I'm using now. And don't worry Neil, I'm a ham, experimenting is part of the fun.

Thanks again for the great information!

Brian

No worries Lee, I have no plans of using the full 10 mw. Though I may connect it for a minute, just to see how far it goes. 5 to 8 miles, yeah, I'd believe that. I'm always amazed at what I can do with my VX-2 1.5 watt handheld ham radio connected to a roof mounted groundplane antenna I built.

By the way Lee, I went to your site, nice ham shack you have. My shack is my VX-2 right now, but I'll be getting more.

Thanks,
Brian

Brian,

I suspected you might be a ham but it was also possible that you just mistyped "wolves"....I wasn't sure.

As you ponder making the pi net variable consider that you can get quite a difference in attenuation and still have a decent match by keeping the shunt Rs near 47 ohms and making the series R variable. As the attenuation gets smaller, this approximation fails, but it can possibly save you a lot of work over trying to fool with three pots.

On a related theme, I had a Q5 QSO using a 2 m HT with a rubber duck and 250 milliwatts over a distance of 100 miles. My ham buddy was a ham and a commercial pilot running time critical freight in a single engine plane. He was airborn at 10,000 feet so coditions were ideal. A little power goes a long way with the right circumstances.

Neil

Neil,

Yeah, I understand about the call sign. Actually, I've gotten some interesting reactions to it on the air.

Thanks for the idea, I'll use the 47 ohm resistors for the shunts, and I'll see what I can get for a good variable resistor. I assume it would be a good idea to make sure the attenuator is in a metal enclosure of some sort. That's a better way to go. At least then I won't have to sit there with an ohm meter to see if the shunts match.

Wow, 100 miles. I guess when you have a clear line of sight, anything is possible. That's still pretty amazing though.

Brian

For those who want to attenuate without the math, one of our advertisers offers a wide selection of attenuators from 1 dB to 10 dB, priced at $27.05 (around $6,972.95 less than an NAL 🙂

www.pcs-electronics.com/mounted-attenuators

Experimental broadcasting for a better tomorrow!

Fish of all sizes are welcome to the FCC's FM fish fry. (Actually, they will release those that do not exceed the Part 15 field strength limit of 250 uV/m at 3 meters.) I did some calculations on the last ten FM NOUOs on the FCC Enforcement Bureau web page to find out what radiated power from a half wave dipole corresponds to the field strength measurements reported in the NOUOs. Here are the results:

172 mW, 0.402 mW, 8.44 mW, 0.277 mW, 4.05 microwatts!, 38 mW, 1.98 watts, 256 mW, 436 mW, and 479 watts!

4.05 microwatts seems small, but it exceeds the FCC limit by 25.5 dB. The half kilowatt transmitter came as a surprise to me. I read about pirates using large commercial FM transmitters a few years ago, but I thought they had been shut down by now. It looks like they still pop up every now and then. These stations were operating like any other commercial station--charging for advertising, and so forth.

From this sample, we see that 10 mW is not too small to attract the FCC's attention.

Hi All,

Ermi has presented data which should be food for thought. I posted a similar article about a year ago and mention it because it confirms the ranges of powers as reported by Ermi. He and I draw the same conclusion.

www.part15.us/node/716#comment-1578

I presume some may not be familiar with micros and nanos so I did a little work on this. The 4.05 microwatt datum seems that it is insignificantly small, yet it is not. The often cited "Rich's constant" predicts a legal field using 11.8 nanowatts. 4.05 microwatts is 343 times this power. (4.05 uW. / 11.8 nW. = 343). I ignored antenna assumptions here because I only wanted to point out the difference between micro and nano.

Neil

Neil wrote: The often cited "Rich's constant" predicts a legal field using 11.8 nanowatts.

Thanks, Neil, but that ~ 11 nanowatts doesn't belong to me, really. I am one person putting a spotlight on it, but this value may be calculated using standard equations found in antenna engineering texts, and from this simple expression shown in FCC OET Bulletin 63 about unlicensed transmitters:

P = 0.3*E^2

where P is the radiated power (EIRP) in watts, and E is the field strength at 3 meters expressed in volts/meter.

The FCC equation gives an answer of about 18.75 nanowatts to generate the maximum Part 15 FM field, but that applies to an isotropic radiator -- which doesn't exist in the real world.

A 1/2-wave dipole has a peak gain that is 2.15 dB above, or 1.64 times that of an isotropic radiator.

So the power applied to the feedpoint of a 1/2-wave dipole that is needed to generate the legal, peak Part 15 FM field in free space is 18.75/1.64 = 11.43... nanowatts.

BUT, reflections of the transmitted signal can become significant when measuring VHF field strength some dozens, hundreds or thousands of meters from the transmit antenna. Those reflections can either increase or decrease those measured fields compared to a path in free space.

The theoretical range of the direct+reflected signal(s) might vary from ~ twice the free-space value to ~ zero the free-space value.

Probably the FCC chose the 3-meter distance for Part 15 FM field strength limits because the affect of reflections on the free-space value of the radiated field is relatively small at that range.

Maybe the FCC has a policy that recognizes this reality, where they tend to accept a measured field quite distant from the transmit antenna that is not more than 6 dB above its calculated and legal free-space value? This is only speculation, so please process this statement accordingly.
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