The recurring mistake looks like this: you learned that "doubling is 3 dB," so when a voltage goes from 10 V to 20 V you write 10·log10(20/10) = 3 dB and move on. The answer key says 6 dB. It is tempting precisely because "double = 3 dB" feels like a memorized law — and you did apply it faithfully. The catch is that the 3 dB rule is about power, and you fed it a voltage.
Here is the short answer: there is only one decibel formula, and it is always about power. A decibel is 10·log10(P_out / P_in). The "20" you sometimes see is not a second rule — it appears only because power is proportional to voltage squared, and squaring inside a log pulls the exponent out front: 10·log10(V²) = 20·log10(V). So "doubling" is not a fixed number of dB. It depends entirely on what doubled. Double the power, you get about 3 dB. Double the voltage, you get about 6 dB — because doubling the voltage actually quadruples the power.
Why "3 dB = double" feels like the rule
That is the whole trap, and it is worth seeing why so many careful people fall into it.
Every intro lab and audio tutorial drills the same slogan: "3 dB is double the power." It gets repeated until it caches as "double = 3 dB," and somewhere in the repetition the word power quietly drops off the end. What survives in your notes is a bare fact: doubling something is 3 dB.
Then you sit down with an actual meter or an actual circuit problem, and the quantity in front of you is almost never power. It is voltage. Op-amp gains, scope traces, signal levels — all voltage. And the textbook answer for "the voltage doubled" is 6 dB, not 3. So now you have two facts side by side with no rule for choosing between them, and on a timed problem you guess. Pick wrong and you are off by exactly a factor of two.
This exact confusion fills technical forums: why does the dB equation have a 10 in one place and a 20 in another, why did my gain math double up, why did a wireless calculation come out half the answer key? It is one of the most common "I followed the formula and still got it wrong" questions in applied logarithms.
The one definition, and where the 20 comes from
Start from the definition and never leave it:
dB = 10 · log10(P_out / P_in)
Power. Always power. Now bring in the physics that connects voltage to power. Across a resistance R:
P = V² / R
Substitute a voltage ratio into the dB definition (the R cancels if it is the same load):
dB = 10 · log10( (V_out² / R) / (V_in² / R) )
= 10 · log10( (V_out / V_in)² )
= 20 · log10( V_out / V_in )
The 20 is just 10 × 2, and the 2 is the exponent from V². There is no separate "voltage decibel." There is one decibel — a power ratio — and voltage enters it squared.
So the question "how many dB is a doubling?" is incomplete until you say what doubled:
- Power doubles:
10·log10(2) = 3.01 dB - Voltage doubles:
20·log10(2) = 6.02 dB, which is the same thing as the power going up 4×, since10·log10(4) = 6.02 dB
Notice those last two numbers are identical. That is the tell that you have not discovered a new rule — you have applied the one rule to a 4× power change.
Work it through with real numbers
Take a single amplifier and an 8-ohm load, and check every number.
Case 1 — the power doubles. The amp goes from 25 W to 50 W.
10 · log10(50 / 25) = 10 · log10(2) = 10 × 0.3010 = 3.01 dB
Three dB. This is the case the slogan is actually about.
Case 2 — the voltage doubles. Same amp, same load. The voltage across the 8-ohm load goes from 10 V to 20 V. First, what does that do to power?
P_before = V² / R = 10² / 8 = 100 / 8 = 12.5 W
P_after = V² / R = 20² / 8 = 400 / 8 = 50 W
Power went from 12.5 W to 50 W — a factor of 4, not 2. Now the dB, computed honestly from the power ratio:
10 · log10(50 / 12.5) = 10 · log10(4) = 10 × 0.6021 = 6.02 dB
And the 20·log shortcut on the voltage ratio gives the same answer, because it is the same calculation:
20 · log10(20 / 10) = 20 · log10(2) = 20 × 0.3010 = 6.02 dB
Six dB. The voltage doubled, the power quadrupled, and both routes agree.
The trap, made explicit. Suppose you forget the square and feed the voltage ratio straight into the power formula:
10 · log10(20 / 10) = 10 · log10(2) = 3.01 dB ← wrong
That is wrong by exactly half — 3.01 instead of 6.02 — and the missing half is the exponent on V² that you dropped. This is not a small numeric slip. It is the entire 3-vs-6 confusion, sitting in one line of arithmetic.
The rule for picking the rule
When you are unsure, do not try to remember whether this is a "10 problem" or a "20 problem." Convert back to a power ratio and use the one definition:
- Ask what quantity you actually have. Power, or voltage (amplitude, current — anything that gets squared to make power)?
- If you have power, use
10·log10(P_out/P_in)directly. - If you have voltage, either square the ratio first and use
10·log10, or take the shortcut20·log10(V_out/V_in). They are the same arithmetic.
A quick sanity anchor worth memorizing, both pointing back to the same definition:
- +3 dB = ×2 power = ×1.41 voltage (because
√2 ≈ 1.41) - +6 dB = ×4 power = ×2 voltage
These anchors live on a log scale, where ratios — not differences — are what spacing means; the same instinct trips people up when they find the midpoint of a log scale and reach for 50 instead of the geometric mean.
If you ever catch yourself saying "doubling is 3 dB" without finishing the sentence, that is the moment to stop and ask: double what?
Check yourself
A signal's voltage across a fixed load increases from 4 V to 8 V. How many dB is that, and what happened to the power?
A) 3 dB; power doubled. B) 6 dB; power doubled. C) 6 dB; power quadrupled. D) 3 dB; power quadrupled.
Correct answer: C.
The voltage doubled (4 V → 8 V), so 20·log10(8/4) = 20·log10(2) = 6.02 dB. Check it through power on, say, an 8-ohm load: 4²/8 = 2 W and 8²/8 = 8 W — a 4× increase, and 10·log10(4) = 6.02 dB. The voltage doubling and the power quadrupling are the same event described two ways.
A drops the square and gives the wrong dB. B has the right dB but forgets that doubling voltage quadruples power. D mismatches both.
Close the gap
The reason this one sticks is that it was taught as a slogan, not as a definition. Once you hold onto the single fact — dB is 10·log10 of a power ratio, and the 20 is only ever V² peeking through — the 3-vs-6 question stops being a guess and becomes one substitution. Gradual Learning is built to catch exactly this kind of half-remembered rule and rebuild it from the definition up, checking your numbers against your reasoning as you go.