How much power can I receive from a FM station

Usually FM receivers are designed with excess gain, since there is no need for amplitude linearity. You didn't say what sort of FM station you want to receive. FM signals vary greatly in the bandwidth they occupy: communications channels (police, fire, pagers, etc) are typically only a few kHz wide, and FM broadcast is typically about

150kHz wide. That makes a difference: since you'll pick up about the same amount of noise per unit bandwidth (atmospheric noise; galactic noise; ...) for signals at approximately the same carrier frequency, the total noise in the channel varies with bandwidth. So for wider bandwidth, it takes more signal to rise above the noise and be useable. A reasonable goal for FM broadcast at 100MHz would be to be able to receive signals greater than -165dBm/Hz bandwidth, or about

-114dBm. That's about half a microvolt at 75 ohms, I believe.

You may wish to think about a bunch of other things before going very far with your project: how will you avoid distortion from big signals when you are trying to listen to small ones? How will you filter out frequencies (channels) other than the one you want to listen to? How will you maintain phase linearity so you achieve low distortion?

If you don't get discouraged along the way, and you keep your eyes and mind open, you'll discover that there's a lot more to designing a good receiver than initially meets the eye. And I want to make it clear that I do NOT want to discourage you, because it's by doing things like that that you will learn, possibly learn a whole lot.

Cheers, Tom

Define "acceptable range". Define how much power you want. I know this Q regards an FM station, but in the SF bay area, near the Dumbarton bridge, an AM station (KGO) broadcasts (i think) 100KW, a "clear channel" station. In the Los Altos hills, one can string up a 100 foot horizontal antenna and connect it to a (partly tuned circuit may not be needed) flashlight lamp and watch it flicker according to the amplitude (but always lit). Why i said "partly tuned circuit may not be needed", is that i do not know exactly how that lamp was connected to the antenna.

Well, you don't care as much about power received as voltage applied to the receiver's antenna terminals. Another poster asked some other relevant questions, so I'll wait to hear those answers before proceeding.

-mpm

Antennas gather microvolts. The resulting power, if one wants to call it that would be in femtowatts.

ha, haven't seen that spelt out in years! use to seeing as "f" reference :)

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Yeah, I was trying to save everyone the whole uV/m discussion again. Some people think they can charge 9-volt batteries using FM Radio Waves a half mile from the tower.

But in the OP's case, unless one knows the antenna efficiency, and the field (power) into which it will be placed, it's really not possible to accurately answer his question directly. I do know the FCC Blanketing rules for FM extend to about 115 dBu, but that's not likely to help him very much.

Also, depending the bandwidth, Q might be more important than voltage anyway. Depends on the app.

Allen hath wroth:

Easy enough to calculate. The maximum EIRP that FCC will tolerate for FM broadcasters is 100,000 watts EIRP or +80dBm. That's your starting point.

As you move way from the transmitting antenna, the field strength goes down by the square of the distance. Assuming your out of the near field, the free space path loss is:

FSPL = 20 log(distance) + 20 log(freq) + 32.44 where: free space path loss = dB distance = km frequency = MHz

So, at 100MHz this reduces to: FSPL = 20 log(distance) + 72.44

Since you didn't bother supplying any numbers with your question, I'll assume that you're about 1km away from the transmitter. That conveniently works out to: FSPL = 72.44 dB

Therefore, the signal received by a handy dandy isotropic antenna will be: +80dBm - 72.44 dB = +7.6dBm

Unfortunately, my stock of isotropic antennas are somewhat depleted. However, I do have a 1/2 wave dipole with a gain of 2.15dBi. That will yield a signal level at the antenna terminals of: +80dBm - 72.44 + 2.15 = +9.7dBm

Converting to milliwatts across 50 ohms, 9.7dBm = 9.3 milliwatts.

There are some additional losses involved. Most big FM stations broadcast about 75% of their power horizontally polarized, and 25% of their power vertically polarized. You'll see anywhere between -1.25 and -6.0 additional cross polarization loss. There will also be some minor transmission line losses between your dipole and whatever you're trying to drive.

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Jeff Liebermann hath wroth:

(...)

Oh crap. I forgot to check if this is someones homework assignment. Yep. Boston University. RDNS = edns03.bu.edu. [192.12.188.130]

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Jeff Liebermann     jeffl@cruzio.com
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Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann     AE6KS    831-336-2558

That does not mean that it is someone's homework assignment, presumptuous dipshit.

I once had a singleboard square inch FM transmitter with say 10mW on the output and it reached 2 miles easily. That makes everything minus 70dB

Rene

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It takes a hell of a lot of femtowatts to illuminate a flashlight bulb...

It would likely take a good deal of them to even get an LED to give off photons.

It's posted via Google. you will never hear from "allen" again!

I wish they'd teach you kids to disguise your homework questions better.

Good Luck! Rich

Be Nice Rich,

we all know the answer is 42.

donald