We define 0 dBmV as 1 mV across 75 Q impedance. Note that 75 Q is the standard impedance of CATV, coaxial cable, and TV sets. From Appendix A, the electrical power law, we have

where Pw is the power in watts, E the voltage in volts, and R the impedance, 75 Q. Substituting the values from above, we obtain

Pw = (0.001)2/75 0 dBmV = 0.0133 x 10-6 W or 0.0133 |W

By definition, then, 0.0133 W = +60 dBmV

If0 dBmV = 0.0133 x 10-6 Wand0 dBm = 0.001 W,andgainindB = 10 log(P1 /P2), or, in this case, 10 log[0.001/(0.0133 x 10-6)], then 0 dBm = +48.76 dBmV.

Remember that, when working with dB in the voltage domain, we are working with the E2/R relationship, where R = 75 Q. With this in mind the definition of dBmV is voltage in mV

If a signal level is 1 V at a certain point in a circuit, what is the level in dBmV? dBmV = 20 log (1000/1) = +60 dBmV. If we are given a signal level of +6 dBmV, to what voltage level does this correspond? +6 dBmV = 20 log(XmV/1 mV).

Divide through by 20:

These signal voltages are rms (root mean square) volts. For peak voltage, divide by 0.707. If you are given peak signal voltage and wish the rms value, multiply by 0.707.

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