dBf definition?
Robert Igo
I've searched the rec.audio.tech FAQ for this, and I've looked it up in
physics books, and I've asked friends with electronics backgrounds, and
none of these sources seems to have any explanation for dBf and FM
reception.
The only definition of dBf I can find is in the Crutchfield catalog,
where they just say that lower dBf is better for FM reception.
I had several questions:
1) On average, what dBf rating is "good enough", what rating starts to
cause reception problems, and what rating is just plain overkill? I've
seen receivers and radios with dBf ratings from 8 to 17 dBf, and I
really have no concept of what these mean in terms of FM reception.
2) Some companies sell FM signal boosters, and they are rated in terms
of dB gain. How can this dB gain be translated into a change in dBf
rating? I assume that an FM booster lowers the effective dBf of an FM
tuner...
3) What's the technical definition of dBf?
4) Uh, is this a really obscure term? Was I just a moron for not being
able to find it anywhere? I really figured it'd be in the
rec.audio.tech FAQ.
Thanks very much for any clue,
--
Bob Igo, human
John Byrns
In article <34594ED1...@eng.adaptec.com>, Robert De Cruz x1634
<rde...@eng.adaptec.com> wrote:
> dBf is dB femptowatts (10**-15 watts). It is the sensitivity rating
> for a tuner. It has to be referenced to some level of "quieting" (say
> 30 db) to have any relevence.
> In the "old" days, microvolt sensitivities were used, such as 1
> microvolt for 30 db of "quieting". This caused confusion because FM
> tuners usually had both 300 ohm balanced, and 75 ohm unbalanced antenna
> inputs. It was therefore necessary to specify two sensitivities, one
> for 300 ohms, and one for 75 ohms. Mathematically, the microvolt
> sensitivity for a 75 ohm input is one-quarter that for the 300 ohm input
> which in this case would be .25 microvolts.
This is interesting, I always thought the microvolt sensitivity into 75
Ohms was half that into 300 Ohms, not one quarter.
> To eliminate the confusion, (???) the dBf was introduced. It
turns out
> that 1 microvolt across 300 ohms generates the same amount of power as
> .25 microvolt accross 75 ohms, and therefore transfers the same amount
> of energy into the tuners circuits per unit time.
By my math, ".25 microvolt accross 75 ohms", does not generate the same
power as "1 microvolt across 300 ohms", but I could be wrong,
> To convert microvolts sensitivities into dBf sensitivities, do the
> following: 1.) use the formula: power = voltage**2/resistance which at 1
> microvolt and 300 ohms is 10**-6 * 10**-6/300 or 3.33...*10**-15 or 3
> and a third femptowatts. 2). then use the power db equation dBf =
> 10*log(power/1f watt) which is 10*log(3.33...) or 5.2dBf.
> If you do the calculations at .25 microvolt and 75 ohms, the dBf would
> be exactly the same.
I don't think so, could you do the math?
regards,
John Byrns
> The antenna booster increases the dBf of the incomming signal, not
> decreases the dBf of the tuner.
> As to the desirable dBf, it will depend on your location relative to
> the FM stations you want to receive. If you are a long distance from a
> few stations spread around the FM spectrum, the the lower the dBf the
> better. If, however, you live close to many crowded stations,
> selectivity is much more important than sensitivity and a high alternate
> and adjacent selectivity rating is is what to look for.
>
> --
> Robert De Cruz
Robert De Cruz x1634
dBf is dB femptowatts (10**-15 watts). It is the sensitivity rating
for a tuner. It has to be referenced to some level of "quieting" (say
30 db) to have any relevence.
In the "old" days, microvolt sensitivities were used, such as 1
microvolt for 30 db of "quieting". This caused confusion because FM
tuners usually had both 300 ohm balanced, and 75 ohm unbalanced antenna
inputs. It was therefore necessary to specify two sensitivities, one
for 300 ohms, and one for 75 ohms. Mathematically, the microvolt
sensitivity for a 75 ohm input is one-quarter that for the 300 ohm input
which in this case would be .25 microvolts.
To eliminate the confusion, (???) the dBf was introduced. It turns out
that 1 microvolt across 300 ohms generates the same amount of power as
.25 microvolt accross 75 ohms, and therefore transfers the same amount
of energy into the tuners circuits per unit time.
To convert microvolts sensitivities into dBf sensitivities, do the
following: 1.) use the formula: power = voltage**2/resistance which at 1
microvolt and 300 ohms is 10**-6 * 10**-6/300 or 3.33...*10**-15 or 3
and a third femptowatts. 2). then use the power db equation dBf =
10*log(power/1f watt) which is 10*log(3.33...) or 5.2dBf.
If you do the calculations at .25 microvolt and 75 ohms, the dBf would
be exactly the same.
Robert De Cruz x1634
John Byrns wrote:
>
snip
You are correct, of course, John. It's good to see that someone is
reading these posts and doing his math correctly. I think that the FTC
mandated this change because some manufacturers were quoting their 75
ohm sensitivity and unfairly competing against 300 ohm only units. I
lived through this period, and it did seem at the time to be a very
alien way to specify FM sensitivity.
For Robert Igo: I also forgot to mention capture ratio, which is
important if ther are multiple stations transmitting at the same
frequency. Capture ratio is the ratio of signal strengths between two
or among more than two FM stations broadcasting at the same frequency.
The lower the capture ratio, the better the ability to cleanly receieve
the stonger or strongest of the competing stations.
Robert