dBmV?
Dave Rosenbloom
If an RF device has a maximum output of +44dBmV, how would you convert
that to p-p or RMS volts so that one could calculate the wattage into a
given load?
Thanks
--
Dave Rosenbloom
http://www.dancetronics.com
"Enhance Your Environment With Electronic Entertainment"
ANTI-SPAM measures in effect!! To reply, remove capital letters
Robert R. Koblish
Dave Rosenbloom wrote:
>
> If an RF device has a maximum output of +44dBmV, how would you convert
> that to p-p or RMS volts so that one could calculate the wattage into a
> given load?
If the impedance is unchanged from the reference level,
dB = 20 * log10 (V2 / V1).
Rearranging this equation, V2 / V1 = 10^(dB / 20).
Plugging in your numbers, the output is at 158 millivolts.
As a check, and assuming that we're dealing with the 75-ohm cable
commonly used in the CATV bidness in which the dBmV reference is used,
1 dBmV @ 75 ohms is 13.3e-9 watts (from P = V^2 / Z) using 1mV and 75
ohms.
Amplifying that 44 dB means multiplying it by 10^(4.4) to give a power
level of 335 microwatts.
Converting back to volts at 75 ohms by plugging into P = V^2 / Z and
solving for V, gives 158 millivolts.
It's Electronics 101.
--
regards
-rrk
To reply, remove the leading X in my return address.
Clifton T. Sharp, Jr.
Dave Rosenbloom wrote:
> If an RF device has a maximum output of +44dBmV, how would you convert
> that to p-p or RMS volts so that one could calculate the wattage into a
> given load?
I get just over 25 volts. (?) 44 = 10 * log (unknown voltage / .001V);
4.4 = the log; 25118 = 10^4.4 (the antilog); 25118 * 0.001 = 25.118V.
--
+---------------------------------------------------------------------------+
| Cliff Sharp | If tin whistles are made of tin, what do they make |
| WA9PDM | foghorns out of? --Lonnie Donegan |
+---------------------------------------------------------------------------+
Robert R. Koblish
Clifton T. Sharp, Jr. wrote:
>
> Dave Rosenbloom wrote:
> > If an RF device has a maximum output of +44dBmV, how would you
> > convert that to p-p or RMS volts so that one could calculate the
> > wattage into a given load?
>
> I get just over 25 volts. (?) 44 = 10 * log (unknown voltage / .001V);
> 4.4 = the log; 25118 = 10^4.4 (the antilog); 25118 * 0.001 = 25.118V.
Sorry, Cliff.
dB is a POWER ratio; the reference level was given in dBmV.
Since P = V^2 / R, the formula for power in dB, given the voltage and
given that the impedance is constant, is dB = 20 * log10 (V2 / V1).
Therefore,
2.2 = the log; 158 = 10^2.2 (the antilog); the voltage is 158 mV; the
power is 335 microwatts at 75 ohms.
Dave Rosenbloom
Well, the fact is that this device is outputting about 10-12 watts or so
into 50 ohms with its max rating of +44dBmV.
This thread should get really interesting!
Clifton T. Sharp, Jr.
Robert R. Koblish wrote:
> Clifton T. Sharp, Jr. wrote:
> > Dave Rosenbloom wrote:
> > > If an RF device has a maximum output of +44dBmV, how would you
> > > convert that to p-p or RMS volts so that one could calculate the
> > > wattage into a given load?
> >
> > I get just over 25 volts. (?) 44 = 10 * log (unknown voltage / .001V);
> > 4.4 = the log; 25118 = 10^4.4 (the antilog); 25118 * 0.001 = 25.118V.
>
> Sorry, Cliff.
> dB is a POWER ratio; the reference level was given in dBmV.
> Since P = V^2 / R, the formula for power in dB, given the voltage and
> given that the impedance is constant, is dB = 20 * log10 (V2 / V1).
>
> Therefore,
> 2.2 = the log; 158 = 10^2.2 (the antilog); the voltage is 158 mV; the
> power is 335 microwatts at 75 ohms.
I stand corrected. I have had more trouble remembering who gets a 10 and
who gets a 20 over the years than I can possibly express.
Nice address
On Tue, 24 Jun 1997 12:14:01 -0400,
Robert R. Koblish <Xn3...@erols.com> wrote:
>Dave Rosenbloom wrote:
>> If an RF device has a maximum output of +44dBmV, how would you convert
>> that to p-p or RMS volts so that one could calculate the wattage into a
>> given load?
>Plugging in your numbers, the output is at 158 millivolts.
>As a check, and assuming that we're dealing with the 75-ohm cable
>commonly used in the CATV bidness in which the dBmV reference is used,
>1 dBmV @ 75 ohms is 13.3e-9 watts (from P = V^2 / Z) using 1mV and 75
>ohms.
If we assume that we are dealing with 75 ohm cable and that
0 db is equal to 1000 micro volts, then 60 db is equal to 1 volt.
20 db is exactly a ratio of 10 to 1 when dealing with voltage.
6 db is about a ratio of 2 to 1.
So 44 db would be almost exactly 1/2 volt.
j...@blkbox.com
P O Box 5865
Kingwood, Texas 77325
Robert R. Koblish
Nice address wrote:
> If we assume that we are dealing with 75 ohm cable and that
Irrelevant, as long as the impedance doesn't change.
> 0 db is equal to 1000 micro volts, then 60 db is equal to 1 volt.
Agreed.
> 20 db is exactly a ratio of 10 to 1 when dealing with voltage.
> 6 dB is about a ratio of 2 to 1.
So far, so good.
And 10 db is a voltage ratio of sqrt(10), or 3.16 if I read my slide
rule correctly.
> So 44 db would be almost exactly 1/2 volt.
Only if 44 dB is almost exactly equal to 46 dB.
It helps to add and subtract your dB accurately.
44 dBmV = (20 + 20 + 10 - 6) dB, so the voltage is
1e-3 volts [reference level] * 10 * 10 * 3.16 / 2,
or 1.58e-1 volts, or 158 millivolts.
You want to back it down from a volt? OK.
44 dBmV = 60 dBmV - 10 - 6.
1 volt / 3.16 / 2 = 0.158 volts.
I hope that this helps.
--
regards
-rrk
To reply, remove the leading X in my return address, if there is one.
Nice address
OK, so I am wrong. That was not so hard to say.
It helps when the the db is added correctly.
54 db is about 1/2 volt, not 44 db.
On Fri, 27 Jun 97 01:42:11 CST,
<j...@blkbox.com> wrote:
>On Tue, 24 Jun 1997 12:14:01 -0400,
>Robert R. Koblish <Xn3...@erols.com> wrote:
>>Dave Rosenbloom wrote:
>>> If an RF device has a maximum output of +44dBmV, how would you convert
>>> that to p-p or RMS volts so that one could calculate the wattage into a
>>> given load?
>>If the impedance is unchanged from the reference level,
>>dB = 20 * log10 (V2 / V1).
>>Rearranging this equation, V2 / V1 = 10^(dB / 20).
>>Plugging in your numbers, the output is at 158 millivolts.
>>As a check, and assuming that we're dealing with the 75-ohm cable
>>commonly used in the CATV bidness in which the dBmV reference is used,
>>1 dBmV @ 75 ohms is 13.3e-9 watts (from P = V^2 / Z) using 1mV and 75
>>ohms.
>
> If we assume that we are dealing with 75 ohm cable and that
> 0 db is equal to 1000 micro volts, then 60 db is equal to 1 volt.
>
> 20 db is exactly a ratio of 10 to 1 when dealing with voltage.
> 6 db is about a ratio of 2 to 1.
> So 44 db would be almost exactly 1/2 volt.
>
Clifton T. Sharp, Jr.
Nice address wrote:
> Robert R. Koblish <Xn3...@erols.com> wrote:
> >Dave Rosenbloom wrote:
> >> If an RF device has a maximum output of +44dBmV, how would you convert
> >> that to p-p or RMS volts so that one could calculate the wattage into a
> >> given load?
> >If the impedance is unchanged from the reference level,
> >dB = 20 * log10 (V2 / V1).
> >Rearranging this equation, V2 / V1 = 10^(dB / 20).
> >Plugging in your numbers, the output is at 158 millivolts.
> >As a check, and assuming that we're dealing with the 75-ohm cable
> >commonly used in the CATV bidness in which the dBmV reference is used,
> >1 dBmV @ 75 ohms is 13.3e-9 watts (from P = V^2 / Z) using 1mV and 75
> >ohms.
>
> If we assume that we are dealing with 75 ohm cable and that
> 0 db is equal to 1000 micro volts, then 60 db is equal to 1 volt.
>
> 20 db is exactly a ratio of 10 to 1 when dealing with voltage.
> 6 db is about a ratio of 2 to 1.
> So 44 db would be almost exactly 1/2 volt.
How do you figure? We started with 1 mV; +20 dB is 10 mV; +20 dB more is
100 mV; +6 dB more would be 200 mV; by your rules of thumb, it's going
to
be less than 200 mV.
Dave
Clifton T. Sharp, Jr. wrote:
>
> > Robert R. Koblish <Xn3...@erols.com> wrote:
> > >Dave Rosenbloom wrote:
> > >> If an RF device has a maximum output of +44dBmV, how would you convert
> > >> that to p-p or RMS volts so that one could calculate the wattage into a
> > >> given load?
> > >If the impedance is unchanged from the reference level,
> > >dB = 20 * log10 (V2 / V1).
> > >Rearranging this equation, V2 / V1 = 10^(dB / 20).
> > >Plugging in your numbers, the output is at 158 millivolts.
> > >As a check, and assuming that we're dealing with the 75-ohm cable
> > >commonly used in the CATV bidness in which the dBmV reference is used,
> > >1 dBmV @ 75 ohms is 13.3e-9 watts (from P = V^2 / Z) using 1mV and 75
> > >ohms.
> >
> > If we assume that we are dealing with 75 ohm cable and that
> > 0 db is equal to 1000 micro volts, then 60 db is equal to 1 volt.
> >
> > 20 db is exactly a ratio of 10 to 1 when dealing with voltage.
> > 6 db is about a ratio of 2 to 1.
> > So 44 db would be almost exactly 1/2 volt.
>
> How do you figure? We started with 1 mV; +20 dB is 10 mV; +20 dB more is
> 100 mV; +6 dB more would be 200 mV; by your rules of thumb, it's going
> to
> be less than 200 mV.
The ARRL handbook says that at 50 ohms 40 dBmV=100mV RMS and 46.02
dBmV=200mV RMS therefore +44dBmV falls somewhere inbetween 100 and 200
mV RMS.
Ed Price
Dave wrote:
>
> Clifton T. Sharp, Jr. wrote:
> >
> > Nice address wrote:
> > > Robert R. Koblish <Xn3...@erols.com> wrote:
> > > >Dave Rosenbloom wrote:
> > > >> If an RF device has a maximum output of +44dBmV, how would you convert
> > > >> that to p-p or RMS volts
> > > >Plugging in your numbers, the output is at 158 millivolts.
Dave R is right!*********************************************** :)
> > > >As a check, and assuming that we're dealing with the 75-ohm cable
> > > >commonly used in the CATV bidness in which the dBmV reference is used,
> > > >1 dBmV @ 75 ohms is 13.3e-9 watts (from P = V^2 / Z) using 1mV and 75
> > > >ohms.
> > > If we assume that we are dealing with 75 ohm cable and that
> > > 0 db is equal to 1000 micro volts, then 60 db is equal to 1 volt.
> > > 20 db is exactly a ratio of 10 to 1 when dealing with voltage.
> > > 6 db is about a ratio of 2 to 1.
Robert R has logical failure at this point.*************** :(
> > > So 44 db would be almost exactly 1/2 volt.
I think he meant to say "about 1/2 way between 100 & 200 mV".
> > How do you figure?
>
> The ARRL handbook says that at 50 ohms 40 dBmV=100mV RMS and 46.02
> dBmV=200mV RMS therefore +44dBmV falls somewhere inbetween 100 and 200
> mV RMS.
44dBmV=158.5mV RMS and in a 75 ohm system, P=335 microwatts
44dBmV=158.5mV RMS and in a 50 ohm system, P=502 microwatts
Let's move on.
Robert R. Koblish
Ed Price quoted a bunch of multiply-quoted messages
and got the attributions wrong, to wit:
>
> Dave wrote:
> >
> > Clifton T. Sharp, Jr. wrote:
> > >
> > > Nice address wrote:
> > > > Robert R. Koblish <Xn3...@erols.com> wrote:
> > > > >Dave Rosenbloom wrote:
> > > > >> If an RF device has a maximum output of +44dBmV, how would you convert
> > > > >> that to p-p or RMS volts
>
> > > > >Plugging in your numbers, the output is at 158 millivolts.
>
> Dave R is right!*********************************************** :)
Dave R asked the original question, and also posted the answer
to which Ed was replying. Count your >'s and spaces!
-rrk ***
>
> > > > >As a check, and assuming that we're dealing with the 75-ohm cable
> > > > >commonly used in the CATV bidness in which the dBmV reference is used,
> > > > >1 dBmV @ 75 ohms is 13.3e-9 watts (from P = V^2 / Z) using 1mV and 75
> > > > >ohms.
> > > > If we assume that we are dealing with 75 ohm cable and that
> > > > 0 db is equal to 1000 micro volts, then 60 db is equal to 1 volt.
> > > > 20 db is exactly a ratio of 10 to 1 when dealing with voltage.
> > > > 6 db is about a ratio of 2 to 1.
>
> Robert R has logical failure at this point.*************** :(
I didn't say it, I quoted it. It's correct so far. -rrk ***
>
*** begin words of "Nice address" <j...@blkbox.com> -rrk ***
> > > > So 44 db would be almost exactly 1/2 volt.
>
> I think he meant to say "about 1/2 way between 100 & 200 mV".
*** That's not the way I read the words.
Anyway, he made a simple math error in saying that 44 = 60 - 6.
He was off by 10 dB, as I pointed out in another post.
And he already has acknowleged that. -rrk ***
>
> > > How do you figure?
*** Begin words of Dave <da...@HAVEYOUKILLEDASPAMMERTODAYdancetronics.com> -rrk ***
> >
> > The ARRL handbook says that at 50 ohms 40 dBmV=100mV RMS and 46.02
> > dBmV=200mV RMS therefore +44dBmV falls somewhere inbetween 100 and 200
> > mV RMS.
*** end words of Dave <da...@HAVEYOUKILLEDASPAMMERTODAYdancetronics.com> -rrk ***
>
> 44dBmV=158.5mV RMS and in a 75 ohm system, P=335 microwatts
> 44dBmV=158.5mV RMS and in a 50 ohm system, P=502 microwatts
> Let's move on.
*** I agree. It's been beaten to death.
But I'd appreciate not being misquoted.
Check http://www.dejanews.com if you doubt me.
-rrk ***
Dave
> *** I agree. It's been beaten to death.
> But I'd appreciate not being misquoted.
> Check http://www.dejanews.com if you doubt me.
> -rrk ***
>
> --
> regards
> -rrk
>
> To reply, remove the leading X in my return address, if there is one.
***********************************************************************************
Thank you to everyone who helped me to arrive at an answer to my
question.
Someone suggested to me that I refer to the ARRL handbook, I did that
and found the answer to my own question. Hopefully we will put this
thread to rest.
--