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I am a beginner of electronics and mostly fascinated by hobby circuits right now. I do not have any Electronics Degree. Internet and hobby Electronics books are the only places where I learned lill-bit of Electronics. One of the most popular circuits being shared by Experts on internet and in those books is an FM Transmitter.

I've made a so-called 'powerful FM transmitter' and measured the peak current across the Amplifying transistor's collector-emmitter. It reads approx 300mA. The overall circuit consumes approx 400mA. I am operating the circuit at 16V.

Sorry for the 3 questions, all at once. But I guess they are related.

  1. Is this one really a 4.8 Watt FM Transmitter? If not, how do you calculate the Wattage of FM Transmission of a typical FM Transmitter?
  2. How the wattage of the transmitter is related to dB rating of a transmitter?
  3. Is there a formula/rule to calculate how much distance an FM signal can propagate in open and obstructed (populated) areas?
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closed as too broad by Chris Stratton, laptop2d, PeterJ, Dave Tweed Feb 25 '18 at 12:07

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ With less than 25W "you" can go billions of km: science.howstuffworks.com/question431.htm \$\endgroup\$ – Curd Feb 15 '18 at 12:11
  • \$\begingroup\$ Rather obviously, even though performing well below your hopes, the device described is illegal. Were it a legal device, it would have been designed, tested, and certified by someone well knowing the answers to these questions. \$\endgroup\$ – Chris Stratton Feb 18 '18 at 6:57
  • \$\begingroup\$ @ChrisStratton I am taking care of the legal aspect! Thanks for reminding and taking your time to write it down once again in FM Transmitter related thread! I was expecting an "it's not legal" comment on my question, most of the questions related to FM transmitters has a 'non answer' comment like yours!! Lol! :) \$\endgroup\$ – Amanda Miller Feb 23 '18 at 21:54
  • \$\begingroup\$ Rather obviously you are not taking care of the legal aspect, as if you were, this question would have been answered by that process since it is a fundamental part of gaining legal authorization to do something like this. \$\endgroup\$ – Chris Stratton Feb 23 '18 at 22:14
  • \$\begingroup\$ @ChrisStratton Yes, I'm in the process, and see below, some people (part of the process) have actually responded with answers. Thanks again for being the guardian of RF spectrum!! \$\endgroup\$ – Amanda Miller Feb 24 '18 at 18:22
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I've made a 'powerful FM transmitter' and measured the peak current across the Amplifying transistor's collector-emmitter. It reads approx 300mA.

You are measuring current incorrectly. Putting a meter across C and E will just take a large amount of current that is unrelated to the current flowing through C and E in normal operation.

Additionally I suspect that you are measuring DC current and not the carrier wave current - most meters will not happily measure RMS current in the VHF/FM band.

how do you calculate the Wattage of FM Transmission of a typical FM Transmitter?

Measuring the voltage at the antenna is the only direct way. If you antenna is a quarter wave monopole it will have a radiation resistance of about 37 ohms and so if you have 2 V RMS (VHF/FM) then the output power will be 4/37 = 108 mW.

If you are using a badly tuned antenna the power might be just a fraction of the "perfect" scenario. For instance, if your antenna is short of the desired length then its radiation resistance will rapidly fall to a few ohms and more your antenna signal level will fall due to this lower impedance. So if the antenna only presented 4 ohms radiation resistance (because it is too short in length) then it is likely that the 2 V RMS will fall to about one-tenth and then power out is \$0.4^2/4\$ = 40 mW.

How the wattage of the transmitter is related to dB rating of a transmitter?

1 watt is 1000 mW and, in dBm terms this is 10 log (1000) dBm = 30 dBm.

Typically, from your experience please tell me how much distance should I expect this transmitter to cover in a densely populated residential area? Is there a formula/rule to calculate this?

It depends so much on many things and many people have laboured over this for decades. A simple free-space formula is called the Friis path loss model: -

Path loss (dB) = 32.45 + 20\$log_{10}\$(f) + 20\$log_{10}\$(d)

Where f is in MHz and d is in kilometres. This equation tells you how many dB of power loss you can expect at a given distance with a given carrier frequency: -

enter image description here

Here is a model that considers obstacles: -

enter image description here

There is also the Okumura model and the Hata urban propagation model (based on Okumura's model). See my answer here for other details.

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  • \$\begingroup\$ just overwhelmed by this detailed answer. I got most of it. Thanks specially for the Non Line of Sight propagation calculation. \$\endgroup\$ – Amanda Miller Feb 17 '18 at 19:41
  • \$\begingroup\$ Hi @Andy aka do you think this question was too broad? \$\endgroup\$ – Amanda Miller Feb 27 '18 at 20:45
  • \$\begingroup\$ @amanda it’s borderline and i nearly voted to close it myself but decided I could make a decent answer to a clearly written question. It’s also borderline soliciting opinions as you asked for experiences. It’s almost too broad in that your personal details didn’t tell much about your skills hence an answer’s starting point was cloudy. But given I visited Irvine and did some shopping there in 2015 I though.... give the lady a break. Laguna is where I stayed, nice, \$\endgroup\$ – Andy aka Feb 27 '18 at 20:58
  • \$\begingroup\$ Thanks for your answers Sir! I got my doubts clarified and learned from it. My last question was actually related to the formula which you provided, may be by words the first part of the last question sounds like soliciting opinion. I'd love to correct it right away! Laguna is great! By the way, until experts like you give break to newbies like me, it's pretty difficult to learn a topic like Electronics at my age! \$\endgroup\$ – Amanda Miller Feb 27 '18 at 21:11
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  1. No

That is supply current, efficiency will be < 60% and is measured by ratio with Pout.

Try different loads (50~300) such as 50 Ohm load, non-inductive, rated 5W with very short leads and measure Vpp then convert to Pd (rms) by \$\frac{V^2pp}{ 8R_L}\$

  1. 10 log Pd(rms) = [dB] or + 30dB for dBm (0dBm=1mW)

  2. There are many variables such as antenna impedance matching losses , directional gain ( e.g. Yagi) for Tx, Rx and Rx noise level. Look for a Friis Loss calculator and this applies to line of sight.

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  • \$\begingroup\$ Voyager's Radios use extemely low baud rates with high FM deviation ratios in a quiet band at 8GHz so it can communicate with very high gain antenna, billions of miles away with a few minutes of latency. \$\endgroup\$ – Sunnyskyguy EE75 Feb 18 '18 at 1:12
  • \$\begingroup\$ Hi @Tony do you think this question was too broad or unspecific? \$\endgroup\$ – Amanda Miller Feb 27 '18 at 20:46

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