# How do you convert dBm to Vrms?

I'm trying to convert dBm to Vrms.

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 \$$

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?

• Because the SQRT(1000 mV) is NOT 31.62 mV. Square 31.62 mV (0.03162) and see what you get. Dec 8 '21 at 16:37
• "some squared volatage = some number of volts" is obviously incorrect. Dec 9 '21 at 3:31

## 2 Answers

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}]\$$.

• 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.

• oh... duh. 1000 mW = 1W. SQRT(1) = 1 Dec 8 '21 at 16:37
• "The square root of 1000 mV is not 31.62 mV; it's 1 volt." – The square root of $(1000 \textrm{ mV})^2$ is $1000 \textrm{ mV}$, which is 1 volt. Technically, the square root of $1000 \textrm{ mV}$ is approximately $31.62 \ \sqrt{\textrm{mV}}$, but $\sqrt{\textrm{mV}}$ isn't a unit you see particularly often. Dec 8 '21 at 16:39
• A nit - "The square root of 1000 mV is not 31.62 mV; it's 1 volt." - it's neither. Dimensionally speaking it should be either 1 √V or 31.62 √mV, which are non-physical (but can be used for calculation similarly to how V/√Hz is sensibly used to describe power spectral densities) Dec 8 '21 at 16:40
• Thanks guys for the obvious typo warning. Dec 8 '21 at 16:56