[TenTec] 5 mV = -67 dBm

[George, W5YR]

I am late getting in on this one, but if the question is how to relate 5 mV
of voltage to a power level of -67 dBm, one answer is 125.3 kilohms.  <:}

I have no idea what the original question was but I am just going by the
subject line.

Try this:

First, find the absolute power level represented by -67 dBm.

Use -67 dB = 10 log (P/10^-3) to find the power level P which is 67 dB lower
than one milliwatt.

Do the math and you get 1.9952 x 10^-10 watts unless I fumbled-fingered
again.

Now, what value of resistance would show 5 mV across it if it were
dissipating -67 dBm or 1.9952x 10^-10 watts?

>From  P = E^2/R, solve for R = E^2/P

R = [(5x10^-3)^2]/1.9952 x 10^-10

I get 125.3 x 10^3 ohms or 125.3 Kilohms.

As a check on the arithmetic, plug the numbers back in   and work the
problem backwards.

What power is represented by 5 mV across 125.3 Kohms?

P = E^2/R  or P = [(5x10^-3)^2]/125.3 x10^3

P = 1.9952 x 10^-10 watts

Now, what is that power level expressed in dBm?

dB = 10 log(P/10^-3) = 10 log (1.9952x10^-10)/10^-3

      = -67 dBm

since the power was taken referred to 10^-3 watts or one milliwatt.

I think that all this is correct but someone else please get a similar
answer!   <:} I have a drawer filled with misplaced decimal points!

73/72, George
Amateur Radio W5YR -  the Yellow Rose of Texas
Fairview, TX 30 mi NE of Dallas in Collin county EM13QE
"In the 57th year and it just keeps getting better!"
<mailto:w5yr at att.net>