For the longest time, decibels felt like a strange kind of magic to me. I’d see specs like “-3dB bandwidth” or “gain of 20dB” and just accept them without really understanding what they meant in terms of actual voltage or power. It wasn’t until I sat down and worked through the math myself — and made a few embarrassing mistakes along the way — that decibels finally clicked. In this article, I want to walk through exactly how voltage and decibels relate, the formulas involved, worked examples, and the common pitfalls people run into when converting between the two.
Why Decibels Exist in the First Place
Before diving into formulas, it helps to understand why engineers use decibels at all instead of just working directly with volts or watts. Electronic and communication systems often deal with an enormous range of signal levels — from picowatts of received optical power in a fiber link to kilowatts of transmitted RF power. Working with raw linear numbers across such a huge range is unwieldy; you end up writing values like 0.000000001 or 50,000 in the same document.
Decibels compress this huge dynamic range into much more manageable numbers using a logarithmic scale. A 100x change in power becomes a change of just 20dB. A million-to-one power ratio becomes 60dB. This logarithmic compression also happens to align well with how humans perceive changes in loudness and brightness, which is part of why decibels became standard in audio engineering originally, before spreading to RF, telecom, and general electronics.
Another huge practical benefit: when you cascade multiple stages in a system (amplifiers, attenuators, filters), their individual gains or losses multiply in linear terms, but in decibels, you simply add or subtract. This makes system-level budget calculations — like a link budget for a fiber optic or RF communication system — dramatically simpler.
The Fundamental Decibel Formula
Decibels are always a ratio between two values, not an absolute unit on their own (except when referenced to a specific standard value, which I’ll cover later). The general formula for power ratio in decibels is:
$$dB = 10 \times \log_{10}\left(\frac{P_2}{P_1}\right)$$
Where $P_2$ and $P_1$ are two power values being compared.
Why Voltage Uses a Different Multiplier (20, Not 10)
This is the part that trips up almost everyone at first, including me. Power is proportional to the square of voltage (assuming a constant impedance):
$$P = \frac{V^2}{R}$$
So if I want to express a voltage ratio in decibels, I need to account for that squaring relationship. Starting from the power formula:
$$dB = 10 \times \log_{10}\left(\frac{P_2}{P_1}\right) = 10 \times \log_{10}\left(\frac{V_2^2/R}{V_1^2/R}\right) = 10 \times \log_{10}\left(\frac{V_2^2}{V_1^2}\right) = 10 \times \log_{10}\left(\left(\frac{V_2}{V_1}\right)^2\right)$$
Using the logarithm power rule ($\log(x^n) = n \times \log(x)$):
$$dB = 20 \times \log_{10}\left(\frac{V_2}{V_1}\right)$$
This is the key formula for converting a voltage ratio to decibels. Note that this assumes the impedance $R$ is the same for both $V_1$ and $V_2$ — this assumption is critical and is one of the most common sources of error, which I’ll come back to.
Converting From Decibels Back to Voltage Ratio
To reverse the process — going from a dB value back to a linear voltage ratio — you invert the formula:
$$\frac{V_2}{V_1} = 10^{(dB / 20)}$$
For power:
$$\frac{P_2}{P_1} = 10^{(dB / 10)}$$
These two formulas (and their inverses) are really the entire toolkit you need for converting between decibels and linear voltage or power values.
Worked Examples: Voltage to dB
Example 1: Suppose an amplifier’s input voltage is 0.1V and its output voltage is 1V. What’s the gain in dB?
$$dB = 20 \times \log_{10}(1 / 0.1) = 20 \times \log_{10}(10) = 20 \times 1 = 20dB$$
So a voltage gain of 10x corresponds to exactly 20dB — a number worth memorizing because it comes up constantly.
Example 2: An attenuator reduces a 2V signal down to 0.2V. What’s the loss in dB?
$$dB = 20 \times \log_{10}(0.2 / 2) = 20 \times \log_{10}(0.1) = 20 \times (-1) = -20dB$$
The negative sign indicates attenuation (loss) rather than gain — this is an important convention: positive dB means gain, negative dB means loss.
Example 3: A signal doubles in voltage. What’s the dB change?
$$dB = 20 \times \log_{10}(2) = 20 \times 0.301 \approx 6.02dB$$
This is another number worth memorizing: doubling voltage is approximately +6dB, and doubling power is approximately +3dB.
Worked Examples: dB to Voltage
Example 4: A datasheet specifies a gain of 26dB. If the input voltage is 5mV, what’s the output voltage?
$$\frac{V_2}{V_1} = 10^{(26/20)} = 10^{1.3} \approx 19.95$$
$$V_2 = 19.95 \times 5mV \approx 99.76mV$$
Example 5: A cable introduces -3dB of loss. If the signal enters at 1V, what’s the voltage at the output?
$$\frac{V_2}{V_1} = 10^{(-3/20)} = 10^{(-0.15)} \approx 0.708$$
$$V_2 = 0.708 \times 1V \approx 0.708V$$
This -3dB point is extremely important in electronics — it represents the “half-power point,” where output power drops to 50% of its original value, even though voltage drops to about 70.7% (since power scales with voltage squared). This is exactly why the term “-3dB bandwidth” is used to define the frequency range of a filter or amplifier: it marks where the output power has dropped to half its maximum value.
Absolute Decibel References: dBm, dBV, dBu
So far I’ve talked about decibels purely as a ratio between two values. But you’ll often see decibel values used as if they were absolute measurements — like “the signal is at -20dBm” — and this works because they’re referenced to a standardized value.
- dBm: Power relative to 1 milliwatt. $dBm = 10 \times \log_{10}(P / 1mW)$. Extremely common in RF and telecom for specifying absolute power levels.
- dBV: Voltage relative to 1 volt RMS. $dBV = 20 \times \log_{10}(V / 1V)$.
- dBu (or dBv, lowercase): Voltage relative to 0.775V RMS, a reference historically tied to the voltage that produces 1mW across a 600-ohm load (a standard impedance in old telephone and audio systems). $dBu = 20 \times \log_{10}(V / 0.775V)$.
Example 6: What is 5V expressed in dBV?
$$dBV = 20 \times \log_{10}(5 / 1) = 20 \times \log_{10}(5) \approx 20 \times 0.699 \approx 13.98dBV$$
Example 7: What power in mW does 0dBm represent? By definition, 0dBm = 1mW exactly, since $\log_{10}(1) = 0$.
The Impedance Trap: A Critical Warning
Here’s a mistake I made early on that I want to flag clearly: the $20 \times \log_{10}(V_2/V_1)$ formula for voltage-to-dB conversion is only valid when the two voltages are measured across the same impedance. If the input and output impedances of a system differ (which is common in RF systems using impedance-matching networks), then a straightforward voltage ratio in dB does NOT directly correspond to a power ratio in dB.
In such cases, if you actually want the true power gain in dB, you have to explicitly calculate power at each point using $P = V^2/R$ with the correct respective impedances, and then use the standard power dB formula. Voltage dB and power dB are only interchangeable representations of the same thing when impedance is constant across the comparison — this single point causes more confusion in RF and audio engineering than almost anything else related to decibels.
Quick Reference Table
I keep a mental table like this handy for quick estimates:
- +3dB ≈ voltage × 1.41 (√2), power × 2
- +6dB ≈ voltage × 2, power × 4
- +10dB = voltage × 3.162, power × 10
- +20dB = voltage × 10, power × 100
- +40dB = voltage × 100, power × 10,000
- -3dB ≈ voltage × 0.708, power × 0.5
- -6dB ≈ voltage × 0.5, power × 0.25
- -20dB = voltage × 0.1, power × 0.01
Real-World Communication Applications
- Link budgets: Fiber optic and RF link budgets are calculated almost entirely in dB, adding transmitter power (dBm), subtracting cable/connector losses (dB), adding amplifier gains (dB), and comparing the result against receiver sensitivity (dBm) to determine if a link will work reliably.
- Filter and amplifier specifications: Bandwidth is defined by the -3dB points; stopband attenuation is specified in dB.
- Signal-to-noise ratio (SNR): Almost universally expressed in dB, since it’s a ratio of signal power to noise power.
- Audio mixing consoles: Faders and meters are calibrated in dB (often dBu or dBFS for digital systems) rather than raw voltage.
- Cable and connector loss specs: RF coaxial cable datasheets specify attenuation in dB per 100 feet or dB per meter at various frequencies.
Common Mistakes
- Using the voltage formula (20×log) when power was actually intended, or vice versa, without checking whether the underlying quantity is power or voltage.
- Ignoring impedance mismatches when converting voltage ratios to dB, as discussed above.
- Forgetting the sign convention — treating a loss as if it were a positive dB gain, which flips the entire calculation.
- Adding absolute dB references incorrectly — for example, you cannot simply “add” two dBm values together the way you can add dB gain/loss values; dBm values represent absolute power levels, and combining two power sources requires converting back to linear watts, adding, and converting back to dBm.
- Rounding log values too early in multi-step calculations, which compounds error in cascaded system budgets.
Troubleshooting Tips
- When a calculated dB value looks unreasonable, double check whether you used log₁₀ (correct for dB) versus natural log (ln), which is a very common calculator or spreadsheet mistake.
- If your voltage-based dB calculation doesn’t match a power-based measurement, check for impedance differences between measurement points.
- When summing gains and losses through a system, always work entirely in dB (adding/subtracting) rather than mixing dB and linear values mid-calculation.
- Use the memorized reference points (+3dB, +6dB, +20dB, etc.) as sanity checks against your calculator output.
Key Takeaways
The relationship between voltage and decibels comes down to one core formula: $dB = 20 \times \log_{10}(V_2/V_1)$, which exists because power is proportional to voltage squared, and the base power formula uses a factor of 10. Reversing the formula gives you $V_2/V_1 = 10^{(dB/20)}$. The most important caveat is that this voltage formula is only equivalent to a true power-in-dB calculation when impedance is held constant — otherwise you need to calculate power directly. Once these formulas are second nature, reading datasheets, calculating link budgets, and understanding gain/loss specifications throughout electronics and communication systems becomes dramatically easier.
