Convert from mA
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mA with Z = Ω
\( 20 \cdot log_{10}(mA) + 20 \cdot log_{10}(Z) + 60= \)
dBμV
\( 20 \cdot log_{10}(mA) + 20 \cdot log_{10}(Z) = \)
dBmV
\( 20 \cdot log_{10}(mA) + 20 \cdot log_{10}(Z) - 60 = \)
dBV
\( 20 \cdot log_{10}(mA) + 60 = \)
dBμA
\( 20 \cdot log_{10}(mA) = \)
dBmA
\( 20 \cdot log_{10}(mA) - 60 = \)
dBA
\( 20 \cdot log_{10}(mA) + 10 \cdot log_{10}(Z) = \)
dBpW
\( 20 \cdot log_{10}(mA) + 10 \cdot log_{10}(Z) - 30 = \)
dBm
\( 20 \cdot log_{10}(mA) + 10 \cdot log_{10}(Z) - 60 = \)
dBW
\( mA \cdot Z \cdot 10^3 = \)
μV
\( mA \cdot Z = \)
mV
\( mA \cdot Z \cdot 10^{-3} = \)
V
\( mA \cdot 10^3 = \)
μA
\( mA \cdot 10^{-3} = \)
A
\( mA^2 \cdot Z = \)
pW
\( mA^2 \cdot Z \cdot 10^{-3}= \)
mW
\( mA^2 \cdot Z \cdot 10^{-6}= \)
W