Basics
A decibel is a ratio on a log scale
The decibel expresses the ratio of two powers as ten times its base-10 logarithm. Doubling power is +3 dB, ten times is +10 dB, a hundred times is +20 dB, and half is −3 dB. Because logs turn multiplication into addition, a signal chain with a +20 dB amplifier, −3 dB of cable and a −1.5 dB filter simply sums. For voltage or current ratios you use 20·log, since power goes with the square of voltage (P = V²/R): twice the voltage is +6 dB, ten times is +20 dB. That 20·log rule only equals the power ratio in dB when both points see the same impedance.
Add a reference after 'dB' and you get an absolute unit: dBm is referenced to 1 mW (0 dBm = 1 mW, 30 dBm = 1 W), dBW to 1 W, dBµV to 1 µV and dBmV to 1 mV. Moving between power and voltage units needs an impedance. RF and test gear standardise on 50 Ω, video and cable TV on 75 Ω, so 0 dBm is 223.6 mV rms (107 dBµV) at 50 Ω and 273.9 mV rms (108.75 dBµV) at 75 Ω.
Power ratio: dB = 10 × log10(P2 / P1) Voltage ratio: dB = 20 × log10(V2 / V1) dBm = 10 × log10(P / 1 mW) P[W] = 10^((dBm − 30) / 10) V_rms = √(P × R) dBµV = 20 × log10(V_rms / 1 µV) At 50 Ω: dBµV = dBm + 107.0, Vpp = 2√2 × V_rms (sine wave) Reflection: Γ = (Z_L − Z₀) / (Z_L + Z₀) VSWR = (1 + |Γ|) / (1 − |Γ|) Return loss: RL = −20 × log10|Γ| Mismatch loss = −10 × log10(1 − |Γ|²)
dBm values you'll meet every day
| dBm | Power | Voltage at 50 Ω (rms) | Where you see it |
|---|---|---|---|
| +30 | 1 W | 7.07 V | FCC Part 15 conducted limit for many 2.4 GHz systems |
| +20 | 100 mW | 2.24 V | Typical Wi-Fi router transmit power |
| +10 | 10 mW | 707 mV | Upper end of BLE and many LoRa modules |
| 0 | 1 mW | 224 mV | Signal generator default, reference level |
| −30 | 1 µW | 7.07 mV | Strong signal at short range |
| −70 | 100 pW | 70.7 µV | 'Good' Wi-Fi RSSI |
| −90 | 1 pW | 7.07 µV | Around Wi-Fi/BLE receiver sensitivity |
| −174 dBm/Hz | 4×10⁻²¹ W/Hz | — | Thermal noise floor kT at 290 K |
Mismatch
VSWR, return loss and mismatch loss are one number in four outfits
When the load impedance at the end of a line differs from the line's characteristic impedance, part of the power reflects. Know the magnitude of the reflection coefficient |Γ| and you know everything else: VSWR, return loss, the percentage of power reflected and the mismatch loss. An antenna spec of 'VSWR ≤ 2:1' means 9.54 dB return loss, 11.1 % reflected power and only 0.51 dB less power delivered. Contrary to intuition, 2:1 is not a performance disaster — it's commonly used simply to define an antenna's usable bandwidth.
| VSWR | |Γ| | Return loss | Reflected power | Mismatch loss |
|---|---|---|---|---|
| 1.1 : 1 | 0.048 | 26.4 dB | 0.23 % | 0.010 dB |
| 1.2 : 1 | 0.091 | 20.8 dB | 0.83 % | 0.036 dB |
| 1.5 : 1 | 0.200 | 14.0 dB | 4.0 % | 0.177 dB |
| 2.0 : 1 | 0.333 | 9.54 dB | 11.1 % | 0.512 dB |
| 3.0 : 1 | 0.500 | 6.02 dB | 25.0 % | 1.25 dB |
Worked examples
Link budgets and converting measurements
- Transmitter +20 dBm, 3 dB of cable and connector loss, +2 dBi antenna: EIRP = 20 − 3 + 2 = 19 dBm ≈ 79 mW. Addition replaces multiplication.
- A 0.5 Vpp sine on a 50 Ω-terminated scope: V_rms = 0.5 ÷ (2√2) = 0.177 V, P = 0.177² ÷ 50 = 0.625 mW = −2.04 dBm.
- Two uncorrelated noise sources at −90 dBm each sum to −87 dBm (+3 dB) — not −180 dBm.
- A 75 Ω load on a 50 Ω line: |Γ| = 25/125 = 0.2, so VSWR 1.5:1, 14 dB return loss and 0.18 dB mismatch loss.
Pitfalls
Common mistakes
- Using 10·log for voltage: voltage and current ratios take 20·log. Mixing them up doubles or halves your answer.
- Taking dB of a voltage ratio across different impedances: 20·log(V2/V1) equals the power gain only when impedances match. A 50 Ω input and 1 kΩ output give a voltage gain, not a power gain.
- Adding dBm arithmetically: 0 dBm + 0 dBm = 3.01 dBm. Convert absolute levels to watts, add, then convert back (section ④ does this).
- Confusing Vpp and Vrms: for a sine, Vpp = 2√2 × Vrms ≈ 2.83 × Vrms. Scopes default to peak-to-peak; power math uses rms.
- Mixing dBi and dBd: dBi is relative to an isotropic radiator, dBd to a half-wave dipole; dBi = dBd + 2.15.
- Reading a 50 Ω circuit with a 1 MΩ scope input: the termination changes the voltage you measure. Confirm 50 Ω termination before converting to dBm.
In practice
Tips from KEMI's bench
None of this dB math means much unless the line on your board actually is 50 Ω. Antenna feeds and RF module traces must be sized to the stack-up (dielectric thickness and Dk), and the fab order needs controlled impedance specified, or your VSWR won't land anywhere near the design value. Chip and PCB antennas detune easily with nearby ground, the enclosure and the battery, so the industry habit is to leave a pi matching network (three 0402 footprints) on the layout and pick values from VNA measurements on the real assembly. Size the trace with our PCB impedance calculator and see the docs guide on ordering controlled-impedance boards.