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Lesson 3: Decibels Explained from Scratch

The decibel appears everywhere in electronics. Amplifier gain is measured in decibels. Cable and filter loss is measured in decibels. Speaker sensitivity, microphone output, the noise level of a recording, the dynamic range of an analog-to-digital converter — all expressed in decibels. Understanding what a decibel actually is, and being able to work with dB values in your head, is one of the most practical skills you can develop. This lesson builds that understanding from the ground up.

The decibel (dB) is a logarithmic unit that expresses a ratio — usually a power ratio or a voltage ratio. It compresses enormous ranges onto a manageable scale. +3 dB ≈ double power. −3 dB ≈ half power. +10 dB = ten times power. These three facts cover 90% of practical dB arithmetic.

Why We Need Decibels

A sound system handles signals ranging from the few microvolts of noise at a microphone input to tens of volts at the speaker terminals. That is a voltage ratio of more than a million to one — over 1,000,000:1. The same ratio expressed as a power ratio is over a trillion to one (1012:1). Writing and working with ratios this large on a linear scale is impractical.

The decibel takes the base-10 logarithm of the ratio, compressing enormous ranges into small, easily handled numbers. The ratio 1012:1 becomes 120 dB. The ratio 106:1 becomes 60 dB. The ratio 2:1 becomes approximately 3 dB. And crucially: gains and losses in a signal path can simply be added and subtracted. A 20 dB amplifier followed by a 3 dB attenuator gives 17 dB net gain — no multiplication required.

Linear vs logarithmic scale comparison showing the compression that decibels provide

The decibel scale compresses a trillion-to-one power range into a 120 dB span that is easy to work with.

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The Power Decibel Formula

For a ratio of two power levels P2 and P1:

dB = 10 × log10(P2 / P1)

Where P2 is the output (or measured) power, P1 is the input (or reference) power, and log10 is the base-10 logarithm (the "log" key on a calculator).
Example 1: Amplifier with 10 W output, 1 W input
dB = 10 × log10(10 / 1) = 10 × log10(10) = 10 × 1 = +10 dB

Example 2: A cable passes 25 W, input was 50 W
dB = 10 × log10(25 / 50) = 10 × log10(0.5) = 10 × (−0.301) = −3.01 dB ≈ −3 dB (a loss)

Example 3: Attenuator reduces 1 W to 100 mW (0.1 W)
dB = 10 × log10(0.1 / 1) = 10 × log10(0.1) = 10 × (−1) = −10 dB

A positive dB value means gain (the output is larger than the input). A negative dB value means loss (the output is smaller than the input). Zero dB means unity — the output equals the input.

The Voltage and Current Decibel Formula

For voltage or current ratios, the formula uses 20 instead of 10:

dB = 20 × log10(V2 / V1)
dB = 20 × log10(I2 / I1)

Important: this assumes both voltages (or both currents) are measured across (or through) the same impedance. If impedances differ, use the power formula instead.
Example 1: Amplifier with voltage gain of 10 (Vout/Vin = 10)
dB = 20 × log10(10) = 20 × 1 = +20 dB

Example 2: Voltage divider halves the signal (Vout/Vin = 0.5)
dB = 20 × log10(0.5) = 20 × (−0.301) = −6.02 dB ≈ −6 dB

Why 20 Instead of 10?

Power is proportional to voltage squared divided by resistance: P = V2/R. This means if the voltage doubles, the power quadruples (not doubles). A voltage ratio of 2:1 corresponds to a power ratio of 4:1.

The decibel was originally defined in terms of power. To get the same dB result whether you are measuring voltage or power, the factor of 20 compensates for the squaring relationship:

Voltage doubles (V2/V1 = 2) → power quadruples (P2/P1 = 4)

Using the power formula: dB = 10 × log10(4) = 10 × 0.602 = 6.02 dB
Using the voltage formula: dB = 20 × log10(2) = 20 × 0.301 = 6.02 dB   ← same result

This is why:

  • A 2:1 power ratio = +3 dB
  • A 2:1 voltage ratio = +6 dB

Both represent the same physical increase, just measured differently. The dB system ensures both give the same number when the underlying energy change is the same.

Key Decibel Values to Memorise

The following table shows the most important dB values. The three you must know by heart are highlighted: ±3 dB ≈ ×/÷2 power, ±10 dB = ×/÷10 power, ±20 dB = ×/÷100 power.

dB value Power ratio Voltage ratio Practical meaning
+30 dB 1000× 31.6× Typical small signal amplifier chain gain
+20 dB 100× 10× Typical LNA gain; 10× power increase
+10 dB 10× 3.16× 10× power increase; clearly audible
+6 dB 4× 2× Doubling voltage; quadrupling power
+3 dB ≈ 2× ≈ 1.41× Double power; just noticeable
0 dB 1× 1× No change; unity gain
−3 dB ≈ 0.5× ≈ 0.707× Half power; −3 dB point of filters
−6 dB 0.25× 0.5× Quarter power; half voltage
−10 dB 0.1× 0.316× One tenth of power
−20 dB 0.01× 0.1× One hundredth of power
−30 dB 0.001× 0.0316× One thousandth of power

Adding and Subtracting Decibels

The key practical advantage of decibels: when signals pass through cascaded stages, you add or subtract their dB values instead of multiplying or dividing their power ratios. This makes mental arithmetic in a signal chain much simpler.

Example: Three-stage audio path
A microphone preamplifier with +40 dB gain, followed by a tone-control stage with 6 dB of loss, feeding a power amplifier with 26 dB of gain.

Total net gain = +40 − 6 + 26 = +60 dB

In raw voltage ratios that would be: 100 × 0.5 × 20 = 1,000 — and as a power ratio, 1,000,000. Much harder to compute mentally and easy to get wrong.
Example: Power left after a lossy cable
A 100 W source (50 dBm) drives a load through a long cable with 3 dB of loss.
Power at the load = 50 − 3 = 47 dBm ≈ 50 W
(Every 3 dB halves the power: 100 W → 50 W. Confirmed.)

Decibels in Signal Chains

A typical audio signal chain might look like this, with dB gains and losses tracked at each point. Audio levels are usually given in dBu (decibels relative to 0.775 V), which you will meet in the next lesson:

Worked signal chain example:
  • Microphone produces a −50 dBu signal
  • Long microphone cable: −1 dB loss → signal is now −51 dBu at the mixer
  • Microphone preamplifier: +45 dB gain → signal is now −6 dBu
  • Equaliser: −2 dB insertion loss → signal is now −8 dBu
  • Mixer output sees: −8 dBu
Net gain/loss from microphone to mixer output: −1 + 45 − 2 = +42 dB
Signal at mixer output = −50 + 42 = −8 dBu    ✓
Audio signal chain drawn as blocks with gain or loss in dB: microphone (-50 dBV) into a preamplifier (+40 dB), a volume control (-10 dB), a cable (-0.5 dB) and a power amplifier (+26 dB) driving a speaker, with the running level in dBV under each block

Adding gains and losses in decibels along an audio signal chain: each block simply adds its dB value to the running level.

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Practical Applications

Real dB values you will encounter regularly in electronics:

Application Typical dB value Meaning
Thin coaxial cable loss at 1 GHz ~3 dB per 5 m About half the signal power lost in 5 m of cable
Speaker sensitivity 85–95 dB SPL at 1 W, 1 m A 92 dB speaker is twice as loud per watt (in power terms) as an 89 dB one
16-bit audio dynamic range ~96 dB Ratio between the largest and smallest signals a CD can represent
1 W amplifier output 30 dBm 30 dB above 1 mW reference level
100 W amplifier output 50 dBm 50 dB above 1 mW; 20 dB more than 1 W
400 W amplifier output 56 dBm 6 dB more than 100 W; 4× the power, but only noticeably louder
Typical power amplifier voltage gain ~26 dB Approximately 20× voltage gain (400× power gain)
Good notch filter rejection 40 dB at notch frequency Reduces interference to 1/10,000 of original power

Decibel Calculators

Power Ratio to dB

Enter two power levels (in any consistent unit — both must be in watts, or both in milliwatts, etc.). Calculates the dB difference.

Result will appear here.

dB to Power and Voltage Ratio

Enter a dB value. Calculates the corresponding power ratio (P2/P1) and voltage ratio (V2/V1).

Result will appear here.

Frequently Asked Questions

Why does the formula use ×10 for power but ×20 for voltage?

Because power is proportional to the square of voltage (P = V2/R). If voltage doubles, power quadruples. A 2× voltage ratio gives a 4× power ratio. The decibel was originally defined in terms of power (it is one tenth of a Bel, a unit named for Alexander Graham Bell, used to compare power levels in telephone systems). To get the same dB result from a voltage ratio as you would from the corresponding power ratio, you need to account for the squaring relationship — which means multiplying by 20 instead of 10. Both formulas give the same result for the same physical change: 6 dB is always 6 dB, whether you calculate it from a 4× power ratio (10 × log10(4) = 6.02 dB) or a 2× voltage ratio (20 × log10(2) = 6.02 dB).

Is a 3 dB difference audible or noticeable?

In audio, the human ear can just perceive a 1 dB change in level under careful listening conditions; 3 dB is clearly noticeable. However, 3 dB represents only a doubling of power — suggesting that the ear's perception of loudness scales roughly logarithmically (which is true: the decibel was originally developed partly to match the ear's response). In electronic systems, 3 dB is significant: it doubles or halves the power, so an amplifier must deliver twice the power to sound 3 dB louder, and a measurement instrument with a noise floor 3 dB lower can see signals half as strong. In practice, when comparing two amplifiers, speakers or sensors, a 3 dB difference is a meaningful real-world improvement.

Can I use the dB formula for any ratio, not just electronics?

Yes. The decibel is a dimensionless ratio — it can describe any ratio of the same quantity. dB is used in acoustics (sound pressure level), optics (optical power loss in fiber), seismology (Richter scale, though that uses a slightly different convention), and audio engineering. In every case, the formula is the same: 10 × log10(ratio) for power-like quantities, 20 × log10(ratio) for amplitude-like quantities. The reference level (what you call 0 dB) varies by application. In electronics, when no reference is specified, dB is always a relative quantity — a ratio between two values with no absolute level implied.

Test Your Knowledge

Answer the questions below to check your understanding of this lesson. Every answer can be found in the lesson above.

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