The decibel (dB) expresses a ratio of two power or amplitude levels on a logarithmic scale. Because our senses and many engineering ranges are logarithmic, decibels turn huge ratios (millions to one) into manageable numbers, and turn multiplication of gains into simple addition. A decibel is always a ratio; when referenced to a fixed level it gains a suffix, such as dBm (relative to 1 mW). This calculator converts ratios to dB and back, and converts between dBm and watts.
Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: Wikipedia: Decibel.
The decibel formulas
The decibel is defined on a power ratio. Take the base-10 logarithm and multiply by ten:
Because power is proportional to the square of voltage or current (P = V²/R), a voltage or current ratio uses a factor of twenty:
To go back from decibels to a ratio, invert the logarithm: power ratio = 10(dB/10), voltage ratio = 10(dB/20).
Why power is 10·log and voltage is 20·log
This is the single most common source of confusion. The decibel is fundamentally about power. Since P = V²/R, the power ratio equals the voltage ratio squared. Taking the log of a square brings the exponent down as a factor of two: 10·log(ratio²) = 20·log(ratio). So +6 dB doubles the voltage but quadruples the power, while +3 dB doubles the power. Mixing up the two factors gives an answer that is out by exactly a factor of two in dB.
dBm — decibels referenced to 1 milliwatt
A plain decibel is only a ratio. To describe an absolute level you reference it to a fixed power; the most common in RF and telecoms is the milliwatt, written dBm:
Handy anchors: 0 dBm = 1 mW, +30 dBm = 1 W, −30 dBm = 1 µW. A typical Wi-Fi transmit level is around +20 dBm (100 mW); a usable received signal might be −70 dBm (0.1 nW).
Worked example 1 — amplifier gain
Scenario: An amplifier delivers 50 W output for 0.5 W input. What is its gain in dB?
A 100× power increase is exactly 20 dB. If the same amplifier doubled the voltage, that would be only +6 dB of voltage gain — a reminder to always state whether a dB figure is power or voltage.
Worked example 2 — an RF link budget
Scenario: A transmitter outputs +20 dBm. The signal passes through 3 dB of cable loss, a +15 dB antenna gain, then 80 dB of free-space path loss. What reaches the receiver?
Because decibels add, a whole signal chain reduces to one sum. −48 dBm is about 16 nW — well above a typical receiver sensitivity of −90 dBm, so the link works. This additive bookkeeping is exactly why RF and audio engineers live in decibels.
Frequently Asked Questions
A decibel is ten times the base-10 log of a power ratio: dB = 10 × log10(P2/P1). For a voltage or current ratio it becomes dB = 20 × log10(V2/V1), because power is proportional to voltage squared.
The decibel is defined on power, and P = V²/R. So log(P2/P1) = log((V2/V1)²) = 2·log(V2/V1); multiplying by 10 gives 20·log for voltage and 10·log for power.
+3 dB is double the power (10^0.3 ≈ 2.0) and −3 dB is half. For voltage, +6 dB doubles and −6 dB halves. The −3 dB (half-power) point defines a filter's cutoff frequency.
dBm is power referenced to 1 mW: 0 dBm = 1 mW. Convert with P(mW) = 10^(dBm/10), so 30 dBm = 1 W and −30 dBm = 1 µW. Reverse: dBm = 10 × log10(P in mW).
Just add them. +20 dB gain, −3 dB cable loss and −1 dB filter loss give a net +16 dB. This additive property is the main reason engineers work in decibels.