Convert from
dBpW with Z = Ω  
 

dBμV

=
\( dBpW + 10 \cdot log_{10}(Z) \)

dBmV

=
\( dBpW + 10 \cdot log_{10}(Z) - 60 \)

dBV

=
\( dBpW + 10 \cdot log_{10}(Z) - 120 \)

dBμA

=
\( dBpW - 10 \cdot log_{10}(Z) \)

dBmA

=
\( dBpW - 10 \cdot log_{10}(Z) - 60 \)

dBA

=
\( dBpW - 10 \cdot log_{10}(Z) - 120 \)

dBm

=
\( dBpW - 90 \)

dBW

=
\( dBpW - 120 \)
 

μV

=
\( \sqrt{10^{dBpW/10} \cdot Z \cdot 10^{-9} } \cdot 10^6 \)

mV

=
\( \sqrt{10^{dBpW/10} \cdot Z \cdot 10^{-9} } \cdot 10^3 \)

V

=
\( \sqrt{10^{dBpW/10} \cdot Z \cdot 10^{-12} } \)

μA

=
\( \sqrt{ \frac{10^{dBpW/10}} {Z} \cdot 10^{-9} } \cdot 10^6 \)

mA

=
\( \sqrt{ \frac{10^{dBpW/10}} {Z} \cdot 10^{-9} } \cdot 10^3 \)

A

=
\( \sqrt{ \frac{10^{dBpW/10}} {Z} \cdot 10^{-12} } \)

pW

=
\( 10^{dBpW/10} \)

mW

=
\( 10^{(dBpW-90)/10} \)

W

=
\( 10^{(dBpW-120)/10} \)
 

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