Convert from dBuA

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Convert from
dBμA with Z = Ω  
 
\(dBμA + 20 \cdot log_{10}(Z) =\)
dBμV
\(dBμA + 20 \cdot log_{10}(Z) - 60 =\)
dBmV
\(dBμA+20 \cdot log_{10}(Z)-120=\)
dBV
\( dBμA-60= \)
dBmA
\( dBμA-120= \)
dBA
\( dBμA+10 \cdot log_{10}(Z)= \)
dBpW
\( dBμA+10 \cdot log_{10}(Z)-90= \)
dBm
\( dBμA+10 \cdot log_{10}(Z)-120= \)
dBW
 
\( 10^{(dBμA+20 \cdot log_{10}(Z))/20}= \)
μV
\( 10^{(dBμA+20 \cdot log_{10}(Z)-60)/20}= \)
mV
\( 10^{(dBμA+20 \cdot log_{10}(Z)-120)/20}= \)
V
\( 10^{dBμA/20}= \)
μA
\( 10^{(dBμA-60)/20}= \)
mA
\( 10^{(dBμA-120)/20}= \)
A
\( 10^{(dBμA+20 \cdot log_{10}(Z))/10}= \)
pW
\( 10^{(dBμA+20 \cdot log_{10}(Z)-90)/10}= \)
mW
\( 10^{(dBμA+20 \cdot log_{10}(Z)-120)/10}= \)
W