We read that the formula for expressing microvolts in decibels is different from that for milliwatts by a factor of 2:

\$U = 20\log_{10}{\frac{u}{u_0}}\$

\$P = 10\log_{10}{\frac{p}{p_0}}\$

(Note the leading factor of 20 versus 10.)

Why is this the case? Why do we not use the factor 10 everywhere?

  • 1
    \$\begingroup\$ We use "20" because P = (V^2)/R for a constant R, hence log(a^2) = 2log(a) \$\endgroup\$
    – G36
    Jun 6, 2017 at 17:52
  • \$\begingroup\$ \$P = VI = \frac{V^2}{Z}\$ where \$I = \frac{V}{R}\$ \$\endgroup\$
    – Shamtam
    Jun 6, 2017 at 17:52
  • \$\begingroup\$ $$P = 10\log_{10}{\frac{p}{p_0}}= 10\log_{10}{\frac{\frac{V^2}{R}}{\frac{V_0^2}{R}}}=10\log_{10}{(\frac{V}{V_0})^2} = 20\log_{10}{\frac{V}{V_0}}$$ \$\endgroup\$
    – G36
    Jun 6, 2017 at 18:02

1 Answer 1


When talking about power quantities, we use the following formula:

\$ 10\log_{10}(P/P_0)dB \$

Which defines the decibel ratio in terms of power units.

However, when we deal with voltages, we actually compare the ratio of the squares of the voltages (because power is proportional to voltage squared, and it is useful for both decibel formulations to give the same results) which gives us:

\$10\log_{10}(V^{2}/V_0^{2})dB = 10\log_{10}[(V/V_0)^{2}] \$

According to the list of logarithmic identities, \$\log(a^{x}) = x\log(a) \$, so we get \$2*10\log_{10}(V/V_0) = 20\log_{10}(V/V_0) \$

  • \$\begingroup\$ Thank you! I had not seen before anyone explicitly call out that it is the ratio of the squares being compared. \$\endgroup\$ Jun 6, 2017 at 19:47
  • \$\begingroup\$ The algebra seen in G36's comment more explicitly explains why we use the squares of voltages and the normal power values. Doing it that way ensures the decibel value of a system's output remains the same, regardless of whether you measure voltage or power. \$\endgroup\$
    – Chris M.
    Jun 6, 2017 at 19:55

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