Convert from pW

Enter a value for the units below and press calculate.

Convert from
pW with Z = Ω  
 
\( 10 \cdot log_{10}(pW) + 10 \cdot log_{10}(Z) = \)
dBμV
\( 10 \cdot log_{10}(pW) + 10 \cdot log_{10}(Z) - 60 = \)
dBmV
\( 10 \cdot log_{10}(pW) + 10 \cdot log_{10}(Z) - 120 = \)
dBV
\( 10 \cdot log_{10}(pW) - 10 \cdot log_{10}(Z) = \)
dBμA
\( 10 \cdot log_{10}(pW) - 10 \cdot log_{10}(Z) - 60 = \)
dBmA
\( 10 \cdot log_{10}(pW) - 10 \cdot log_{10}(Z) - 120 = \)
dBA
\( 10 \cdot log_{10}(pW) = \)
dBpW
\( 10 \cdot log_{10}(pW) - 60 = \)
dBm
\( 10 \cdot log_{10}(pW) - 90 = \)
dBW
 
\( \sqrt{pW \cdot Z \cdot 10^{-12} } \cdot 10^6 = \)
μV
\( \sqrt{pW \cdot Z \cdot 10^{-12} } \cdot 10^3 = \)
mV
\( \sqrt{pW \cdot Z \cdot 10^{-12} } = \)
V
\( \sqrt{ \frac{ pW }{ Z } \cdot 10^{-12} } \cdot 10^6 = \)
μA
\( \sqrt{ \frac{ pW }{ Z } \cdot 10^{-12} } \cdot 10^3 = \)
mA
\( \sqrt{ \frac{ pW }{ Z } \cdot 10^{-12} } = \)
A
\( pW \cdot 10^{-9} = \)
mW
\( pW \cdot 10^{-12} = \)
W