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Convert decibels (dB) to voltage ratio and vice versa. Essential for audio amplifier gain, signal processing, and electronics design.
Voltage Ratio = 10^(dB/20)dB = 20 × log₁₀(Voltage Ratio)Note: Voltage uses factor of 20 (not 10) because power is proportional to voltage squared (P ∝ V²). To maintain consistency with power dB, voltage dB uses 20×log₁₀.
| Decibels (dB) | Voltage Ratio | Description |
|---|---|---|
| -40 dB | 0.01 (1/100) | 1% of original voltage |
| -20 dB | 0.1 (1/10) | 10% of original voltage |
| -6 dB | 0.501 (~1/2) | Half voltage (half power = -3 dB) |
| -3 dB | 0.708 (1/√2) | Half power point |
| 0 dB | 1 | Unity gain (no change) |
| 3 dB | 1.413 (√2) | Double power point |
| 6 dB | 2 | Double voltage (4× power) |
| 12 dB | 4 | Quadruple voltage |
| 20 dB | 10 | 10× voltage (100× power) |
| 40 dB | 100 | 100× voltage (10,000× power) |
| 60 dB | 1,000 | 1,000× voltage (1,000,000× power) |
| 80 dB | 10,000 | Typical preamp + power amp gain |
Voltage ratio is the ratio of output voltage to input voltage (V_out/V_in) in an electronic circuit or system. It represents the voltage gain or attenuation through a component, amplifier, or signal path. A voltage ratio of 1 means unity gain (no change), ratios greater than 1 indicate amplification, and ratios less than 1 indicate attenuation. Voltage ratios are fundamental in electronics because they directly describe how signals are scaled through various stages. Unlike power ratios, voltage ratios require the factor of 20 (not 10) when converting to/from decibels due to the quadratic relationship between voltage and power. Understanding voltage ratios is essential for analyzing amplifier circuits, filter responses, signal processing chains, and impedance matching networks in audio and RF applications.
When decibels are used to express voltage ratios, they provide a logarithmic measure of voltage gain or loss. The key difference from power dB is the use of factor 20 instead of 10 in the formula: dB = 20×log₁₀(V_ratio). This factor of 20 exists because power is proportional to voltage squared (P = V²/R in a resistive load). To maintain consistency between voltage and power dB measurements, voltage uses 20: since P ∝ V², we have 10×log₁₀(P_ratio) = 10×log₁₀(V²_ratio) = 20×log₁₀(V_ratio). This means that a 6 dB voltage gain represents a doubling of voltage (but quadrupling of power), while a 20 dB voltage gain represents a tenfold increase in voltage. Voltage dB is commonly used in audio, RF, and telecommunications to specify amplifier gain, attenuator loss, and signal level changes throughout a system.
Determine the decibel value you want to convert. For example, 20 dB voltage gain.
Divide your decibel value by 20 (not 10 - this is for voltage, not power). For 20 dB: 20/20 = 1.
Raise 10 to the power of the result from step 2. For our example: 10^1 = 10.
The result is your voltage ratio. For 20 dB, the voltage ratio is 10, meaning the output voltage is 10 times the input voltage.
Check: every 6 dB should double voltage, every 20 dB should multiply voltage by 10. For 20 dB: 10× voltage ✓
The factor of 20 in voltage dB calculations (instead of 10 used for power) stems from the fundamental relationship between voltage and power. In a resistive circuit, power is proportional to voltage squared: P = V²/R. This quadratic relationship is the key to understanding why voltage uses factor 20.
Consider an amplifier that doubles the voltage (voltage ratio = 2). The power ratio is 2² = 4 (quadrupling the power). When we express power gain in dB, we get: Power_dB = 10×log₁₀(4) = 10×log₁₀(2²) = 10×2×log₁₀(2) = 20×log₁₀(2) ≈ 6 dB. Notice that 10×log₁₀(V²) = 20×log₁₀(V). This is why voltage gain must use the factor 20 to give the same dB value as the corresponding power gain.
This consistency is crucial in practice. If an amplifier has 6 dB gain, you can interpret this as either "4× power gain" or "2× voltage gain" - both are correct and both give 6 dB. The formulas are: Power: 10×log₁₀(4) = 6 dB, and Voltage: 20×log₁₀(2) = 6 dB. The factor of 20 ensures that voltage dB and power dB remain consistent when describing the same physical phenomenon.
It's important to remember that this only applies when comparing quantities of the same type. You can add/subtract dB values for gains in cascade (6 dB + 6 dB = 12 dB), but you cannot directly convert between voltage dB and power dB without understanding the circuit impedance and power relationships. The factor of 20 for voltage (and for pressure, SPL, and other amplitude quantities) and factor of 10 for power (and intensity) are fundamental to maintaining mathematical consistency in the decibel system.
In audio engineering, voltage gain expressed in dB is fundamental to understanding signal flow and system design. Microphone preamps typically provide 40-60 dB of voltage gain (100× to 1,000× voltage amplification) to boost the tiny millivolt-level signals from microphones up to the line level of 1-2 volts. This massive voltage gain is necessary because microphones produce such weak signals, and the gain must be achieved with minimal noise addition.
Mixing consoles and audio interfaces specify their input and output levels in dBV or dBu (decibels relative to 1V or 0.775V respectively). Understanding voltage ratios helps engineers properly set gain staging throughout the signal chain. For example, if a mixer's input expects +4 dBu (1.228V) but receives a signal at -10 dBV (0.316V), the gain difference is about 12 dB, requiring the input gain to be increased by that amount to achieve proper operating level.
Dynamic range processors like compressors and limiters work based on voltage ratio thresholds and ratio controls expressed in dB. A compressor with a 4:1 ratio means that for every 4 dB of input level increase above the threshold, the output increases by only 1 dB. This is a voltage ratio operation - when the input voltage doubles (+6 dB), the output increases by only 1.5 dB (voltage ratio of 1.189×) instead of the full 6 dB doubling.
Equalizers and filters specify their boost/cut ranges in dB, which represents voltage gain or attenuation. A +3 dB shelf filter at 10 kHz increases the voltage of those frequencies by a factor of √2 (1.413×), which corresponds to doubling the power at those frequencies. Understanding the voltage ratio helps when designing cascaded EQ stages or when compensating for frequency response issues in speakers and rooms.
In electronics and RF (radio frequency) design, voltage gain calculations in dB are essential for analyzing amplifier stages, attenuators, and transmission line losses. Op-amp circuits are commonly analyzed using voltage gain in dB. For example, a non-inverting op-amp with a gain of 101 (set by resistor ratios) has a voltage gain of 20×log₁₀(101) = 40.09 dB. This dB representation makes it easier to cascade multiple stages and predict overall system gain.
RF attenuators are specified by their attenuation in dB, which represents voltage reduction. A 6 dB attenuator reduces voltage by half (ratio = 0.5) and power by one-quarter (ratio = 0.25). In RF systems, matching networks and filters are characterized by their insertion loss in dB, which describes how much voltage (and power) is lost when the component is inserted into the signal path. A filter with 0.5 dB insertion loss at the passband frequency has a voltage transmission ratio of 10^(-0.5/20) = 0.944, or about 94.4% voltage transmission.
Transmission line calculations heavily use voltage ratios and dB. A coaxial cable with 3 dB loss per 100 feet at a given frequency has a voltage transmission ratio of 10^(-3/20) = 0.708 per 100 feet. For a 200-foot run, the loss is 6 dB (voltage ratio = 0.5), meaning only half the voltage reaches the end. This is why amplifiers and repeaters are needed in long cable runs - to compensate for the cumulative voltage loss represented by increasing negative dB values.
Signal-to-noise ratio (SNR) in communications systems is expressed in dB and represents the voltage (or power) ratio between the desired signal and the background noise. An SNR of 60 dB means the signal voltage is 1,000 times larger than the noise voltage (voltage ratio = 10^(60/20) = 1,000). This high SNR is necessary for high-fidelity audio and reliable data transmission. Understanding voltage ratios helps engineers design systems that maintain adequate SNR through all processing stages, from input to output.
6 dB represents a voltage ratio of 2 (exactly 1.995, but typically rounded to 2). This means the output voltage is double the input voltage. You can verify: 20×log₁₀(2) = 20×0.301 = 6.02 dB ≈ 6 dB. This is an important reference point: every 6 dB increase doubles the voltage, and every 6 dB decrease halves the voltage. Note that doubling voltage quadruples power, which is why 6 dB voltage gain equals approximately 6 dB power gain (since power is proportional to voltage squared).
3 dB represents half power because 10×log₁₀(0.5) = -3.01 dB. However, half power corresponds to voltage divided by √2 (not half voltage). The voltage ratio at -3 dB is 1/√2 = 0.707, which you can verify: 20×log₁₀(0.707) = -3.01 dB. This is why filter cutoff frequencies are defined at the -3 dB point - it's where power drops to half, but voltage only drops to 70.7% of maximum. If voltage actually halved (-6 dB), power would drop to one-quarter (-6 dB power).
No, you must use factor 10 for power ratios and factor 20 for voltage (or current, pressure) ratios. Using the wrong factor gives incorrect results. Power dB: 10×log₁₀(P₂/P₁). Voltage dB: 20×log₁₀(V₂/V₁). The factor of 20 for voltage exists specifically because P ∝ V². If you try to use 20 for power or 10 for voltage, your dB values will be off by a factor of 2. Always identify whether you're working with a power-like quantity (use 10) or an amplitude-like quantity (use 20).
20 dB gain represents a voltage ratio of 10 (tenfold voltage increase). Calculate: 10^(20/20) = 10^1 = 10. This is a fundamental reference: every 20 dB is a decade (factor of 10) in voltage. So 40 dB = 100× voltage, 60 dB = 1,000× voltage, etc. Similarly, -20 dB = 0.1× voltage (one-tenth). The 20 dB/decade relationship makes it easy to estimate voltage ratios: just count how many 20 dB increments you have and that's how many zeros to add (or remove for negative dB).
dBV and dBu use different reference voltages: dBV uses 1V reference, dBu uses 0.775V reference. The conversion is: dBu = dBV + 2.21. For example, 0 dBV = +2.21 dBu (both equal 1V). The 0.775V reference in dBu comes from the historical telephone standard (600 ohm load, 1 mW power gives 0.775V). In professional audio, +4 dBu (1.228V) is the standard operating level, which equals +1.78 dBV. Consumer audio typically uses -10 dBV (0.316V) as the operating level, which equals -7.78 dBu.
Voltage ratios as a numeric value are always positive (you can't have negative ratio of magnitudes). However, the dB representation can be negative, indicating attenuation rather than gain. Negative dB means the voltage ratio is less than 1 (output smaller than input). For example, -20 dB means voltage ratio = 10^(-20/20) = 0.1 (one-tenth of input voltage). In circuits with inverting amplifiers, the voltage polarity may flip (180° phase shift), but the voltage gain magnitude is still expressed as a positive ratio or positive/negative dB.
When stages are cascaded, voltage ratios multiply, but dB values add. If stage 1 has 20 dB gain (10× voltage) and stage 2 has 26 dB gain (20× voltage), the total gain is 20+26 = 46 dB, which equals 10×20 = 200× voltage ratio. This additive property of dB makes cascade calculations much simpler than multiplying ratios. For example, a microphone preamp (40 dB) → mixer channel (0 dB) → mixer master (6 dB) → power amp (30 dB) gives total gain of 40+0+6+30 = 76 dB, or about 6,310× voltage gain.