How do you convert dBm to Vrms?

Question

I'm trying to convert dBm to Vrms.

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The formula for dBm is

\$ dBm = 10 * log(P1 / P0) \$ where P0 = 1mW

so
\$ +13dBm = 20mW \$

If I convert this to a voltage using the formula \$ P = Vrms^2/R \$ where R = 50 ohm, I find that
\$ Vrms^2 = 20mW * 50ohm = 1000mV \$
\$ Vrms = 31.62mV \$

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Now if I solve the equation a different way I get a different result.

\$ dBm = 10*log(P1/P0) \$ where \$P0\$ = 1mW
substituting \$P1\$ for \$ Vrms^2/R \$ and \$P0\$ for 1mW we get
\$ dBm = 10*log(Vrms^2/(R * 1mW)) \$

simplifying
\$dBm = 10*log(Vrms^2/SQRT(R * 1mW)^2)\$
\$dBm = 10*log(Vrms/SQRT(R * 1mW))^2\$
\$ dBm = 20 * log (Vrms/SQRT(R*1mW)) \$

plugging in \$ dBm = +13\$ and \$ R = 50 \$ I get
\$ Vrms = 0.999V\$

This is the correct answer according to some online calculators I've used. But where is the inconsistency in the first solution?

Answer 1

Your error stems from a misunderstanding of units. The square root of 1000 mV is not 31 mV. It's either 1 √V or 31 √(mV), which are both strange, non-physical units.

The correct units are actually a bit different: mW * ohms is (mV * A) * (V/A), so you really have a value of 1000 V*mV, or 1 V^2. Taking the square root yields 1 V. 1 V^2 is likewise the same as 1000000 mV*mV, and taking the square root of that yields 1000 mV, which is consistent.

Working in a dimensionally consistent way we can re-do the calculation with embedded dimensions:

\$P = V_\text{RMS}^2/R\ \implies PR = V_\text{RMS}^2\$, hence \$V_\text{RMS} = \sqrt{20\,[\text{mW}] \cdot 50\,[\Omega]} = \sqrt{1\,[\text V^2]} = 1\,[\text{V}]\$.

\$P = V_\text{RMS}^2/R\ \implies PR = V_\text{RMS}^2\$, hence \$V_\text{RMS} = \sqrt{20\,[\text{mW}] \cdot 50\,[\Omega]} = \sqrt{1000000\,[\text{(mV)}^2]} = 1000\,[\text{mV}]\$.

Answer 2

• 13 dBm is 20 mW - correct

$$P = \dfrac{V^2}{50}\text{ hence, } 0.02\times {50} = V^2\text{ hence, } V = 1 \text{ volt RMS} $$

But where is the inconsistency in the first solution?

The square root of (1000 mV)² is not 31.62 mV; it's 1 volt.