Convert from mW

Enter a value for the units below and press calculate.

Convert from
mW with Z = Ω  
 
\( 10 \cdot log_{10}(mW) + 10 \cdot log_{10}(Z) + 90= \)
dBμV
\( 10 \cdot log_{10}(mW) + 10 \cdot log_{10}(Z) + 30 = \)
dBmV
\( 10 \cdot log_{10}(mW) + 10 \cdot log_{10}(Z) - 30 = \)
dBV
\( 10 \cdot log_{10}(mW) - 10 \cdot log_{10}(Z) + 90 = \)
dBμA
\( 10 \cdot log_{10}(mW) - 10 \cdot log_{10}(Z) + 30 = \)
dBmA
\( 10 \cdot log_{10}(mW) - 10 \cdot log_{10}(Z) - 30 = \)
dBA
\( 10 \cdot log_{10}(mW) + 90 = \)
dBpW
\( 10 \cdot log_{10}(mW) = \)
dBm
\( 10 \cdot log_{10}(mW) - 30 = \)
dBW
 
\( \sqrt{mW \cdot Z \cdot 10^{-3} } \cdot 10^6 = \)
μV
\( \sqrt{mW \cdot Z \cdot 10^{-3} } \cdot 10^3 = \)
mV
\( \sqrt{mW \cdot Z \cdot 10^{-3} } = \)
V
\( \sqrt{ \frac{ mW }{ Z } \cdot 10^{-3} } \cdot 10^6 = \)
μA
\( \sqrt{ \frac{ mW }{ Z } \cdot 10^{-3} } \cdot 10^3 = \)
mA
\( \sqrt{ \frac{ mW }{ Z } \cdot 10^{-3} } = \)
A
\( mW \cdot 10^9 = \)
pW
\( mW \cdot 10^{-3} = \)
W