Convert from uV
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μV with Z = Ω
\( 20 \cdot log_{10}(\mu V) =\)
dBμV
\( 20 \cdot log_{10}(\mu V) - 60 =\)
dBmV
\( 20 \cdot log_{10}(\mu V) - 120 = \)
dBV
\( 20 \cdot log_{10}(\mu V) - 20 \cdot log_{10}(Z) = \)
dBμA
\( 20 \cdot log_{10}(\mu V) - 20 \cdot log_{10}(Z) - 60 = \)
dBmA
\( 20 \cdot log_{10}(\mu V) - 20 \cdot log_{10}(Z) - 120 = \)
dBA
\( 20 \cdot log_{10}(\mu V) - 10 \cdot log_{10}(Z) = \)
dBpW
\( 20 \cdot log_{10}(\mu V) - 10 \cdot log_{10}(Z) - 90 = \)
dBm
\( 20 \cdot log_{10}(\mu V) - 10 \cdot log_{10}(Z) - 120 = \)
dBW
\( \mu V \cdot 10^{-3} = \)
mV
\( \mu V \cdot 10^{-6} = \)
V
\( \frac{ \mu V }{ Z } = \)
μA
\( \frac{ \mu V }{ Z } \cdot 10^{-3}= \)
mA
\( \frac{ \mu V }{ Z } \cdot 10^{-6}= \)
A
\( \frac{ \mu V^2 }{ Z } = \)
pW
\( \frac{ \mu V^2 }{ Z } \cdot 10^{-9}= \)
mW
\( \frac{ \mu V^2 }{ Z } \cdot 10^{-12}= \)
W