Convert from
μV with Z = Ω  
 

dBμV

=
\( 20 \cdot log_{10}(\mu V) \)

dBmV

=
\( 20 \cdot log_{10}(\mu V) - 60 \)

dBV

=
\( 20 \cdot log_{10}(\mu V) - 120 \)

dBμA

=
\( 20 \cdot log_{10}(\mu V) - 20 \cdot log_{10}(Z) \)

dBmA

=
\( 20 \cdot log_{10}(\mu V) - 20 \cdot log_{10}(Z) - 60 \)

dBA

=
\( 20 \cdot log_{10}(\mu V) - 20 \cdot log_{10}(Z) - 120 \)

dBpW

=
\( 20 \cdot log_{10}(\mu V) - 10 \cdot log_{10}(Z) \)

dBm

=
\( 20 \cdot log_{10}(\mu V) - 10 \cdot log_{10}(Z) - 90 \)

dBW

=
\( 20 \cdot log_{10}(\mu V) - 10 \cdot log_{10}(Z) - 120 \)
 

mv

=
\( \mu V \cdot 10^{-3} \)

V

=
\( \mu V \cdot 10^{-6} \)

μA

=
\( \frac{ \mu V }{ Z } \)

mA

=
\( \frac{ \mu V }{ Z } \cdot 10^{-3} \)

A

=
\( \frac{ \mu V }{ Z } \cdot 10^{-6} \)

pW

=
\( \frac{ \mu V^2 }{ Z } \)

mW

=
\( \frac{ \mu V^2 }{ Z } \cdot 10^{-9} \)

W

=
\( \frac{ \mu V^2 }{ Z } \cdot 10^{-12}\)
 
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