>> Neil,
>> I think something like this makes it easier to estimate what
>>types of power one will expect to get in terms of db to
>>watts.
>> www.minicircuits.com/dg03-110.pdf
>> Radiopilot
Neil,
Nonsense, you are correct! Anyway if you look at the table in the minicircuits site, you'll see that 47uV equates to -74dbm approx. and the power output is somewhere around .05nw, does this sound familiar in my arguments in the previous posts about how much power the PCFM transmitter outputs! Not very much power but even with that amount of power on a well installed antenna will give you 200-300 feet of coverage, maybe more but for a high school radio broadcast system it's not too bad, as I said and it should comform to the FCC as far as compliance, I just don't understand sometimes why people make harsh issues of things when all they have to do is sit back and have a cup of Jo...
Radiopilot
_____________
Radiopilot,
You found a chart that says 45 µV is the voltage across 50 ohms at a power level of -74 dBm. The chart is right, but that chart is not directly useful to you to make your point. Here's why.
The output level of the PCFM transmitter is given in terms of dBmV, but the units in the column you used in the chart are in µV. In order for the chart to apply directly, the units in it would have to be shown in dBmV, not µV.
Here's how to use the chart to find the approximate output power of your PCFM tx. First, calculate the output voltage itself. I'll use +35 dBmV as its output level, as I think this is the spec shown for it on the Pico MACOM website.
Millivolts = 10^(35/20) = 56.23, or ~0.056 volts
Now look up the closest voltage value to this in the voltage column of the chart. It appears as the voltage across 50 ohms at a power level of -12 dBm.
To convert dBm to milliwatts...
mW = 10^(-12/10) = 0.063 mW, which is over 5,000 times more power than it takes to generate the maximum Part 15 FM field using a simple 1/2-wave dipole. So the output power of your PCFM tx is not ".05nw" as you concluded in your clip above.
Of course, this chart shows values for 50 ohm systems, and the PCFM output level spec is referred to 75 ohm systems (a cable industry standard). So using this chart doesn't strictly apply in this situation, but at least it will give you a rough idea.
And... I'm not wishing to be "harsh," just wanting people to have valid information. Part 15 issues are hard enough, even then.
//
>> I think something like this makes it easier to estimate what
>>types of power one will expect to get in terms of db to
>>watts.
>> www.minicircuits.com/dg03-110.pdf
>> Radiopilot
Neil,
Nonsense, you are correct! Anyway if you look at the table in the minicircuits site, you'll see that 47uV equates to -74dbm approx. and the power output is somewhere around .05nw, does this sound familiar in my arguments in the previous posts about how much power the PCFM transmitter outputs! Not very much power but even with that amount of power on a well installed antenna will give you 200-300 feet of coverage, maybe more but for a high school radio broadcast system it's not too bad, as I said and it should comform to the FCC as far as compliance, I just don't understand sometimes why people make harsh issues of things when all they have to do is sit back and have a cup of Jo...
Radiopilot
_____________
Radiopilot,
You found a chart that says 45 µV is the voltage across 50 ohms at a power level of -74 dBm. The chart is right, but that chart is not directly useful to you to make your point. Here's why.
The output level of the PCFM transmitter is given in terms of dBmV, but the units in the column you used in the chart are in µV. In order for the chart to apply directly, the units in it would have to be shown in dBmV, not µV.
Here's how to use the chart to find the approximate output power of your PCFM tx. First, calculate the output voltage itself. I'll use +35 dBmV as its output level, as I think this is the spec shown for it on the Pico MACOM website.
Millivolts = 10^(35/20) = 56.23, or ~0.056 volts
Now look up the closest voltage value to this in the voltage column of the chart. It appears as the voltage across 50 ohms at a power level of -12 dBm.
To convert dBm to milliwatts...
mW = 10^(-12/10) = 0.063 mW, which is over 5,000 times more power than it takes to generate the maximum Part 15 FM field using a simple 1/2-wave dipole. So the output power of your PCFM tx is not ".05nw" as you concluded in your clip above.
Of course, this chart shows values for 50 ohm systems, and the PCFM output level spec is referred to 75 ohm systems (a cable industry standard). So using this chart doesn't strictly apply in this situation, but at least it will give you a rough idea.
And... I'm not wishing to be "harsh," just wanting people to have valid information. Part 15 issues are hard enough, even then.
//