# How to Solve Decibel dB Volts Formula Calculator

This dB Volts formula calculator can calculate dB (decibels) from input and output voltage and can calculate the voltage from dB. The measurement of dB is always a comparison of the power of two signals. When a dB value is positive it means the output power is larger than the input, and when a dB value is negative it means that the output power is smaller than the input. Although the measurement of dB relates to power, you can still calculate it based on the ratio of voltages. The dB Volts calculator can be used for other signal values too (such as sound volume), as long as you are comparing the signal amplitudes and not the signal power. If you need to calculate dB from the ratio of power, use the dB Watts Formula Calculator instead.

The calculator below finds the decibels (dB) for a given set of output/input voltages.

dB Volts Calculator
 Volts Out MV (megavolts) kV (kilovolts) V (volts) mV (millivolts) uV (microvolts) Volts In MV (megavolts) kV (kilovolts) V (volts) mV (millivolts) uV (microvolts) dB dB

## dB Volts Formula

The formula used to perform this calculation is below. Remember to convert all units to volts when using the dB Volts Formula.

dB Volts Formula:
$\fn_jvn&space;\small&space;dB=20log\left&space;(&space;\frac{V_{out}}{V_{in}}&space;\right&space;)$
where:
• dB = Decibels
• Vout = Output Voltage
• Vin = Input Voltage

## Solution Examples:

We will use the formula above to work out a few real world example problems. You can always check your work with the calculator.

### Example #1:

For this example, we have an amplifier circuit that outputs 2 volts when you input 1 volt. We will be solving for the gain in dB (decibels) of the amplifier. We will enter the values into the formula as shown.

$\fn_jvn&space;\small&space;dB=20log\left&space;(&space;\frac{V_{out}}{V_{in}}&space;\right&space;)=20log\left&space;(&space;\frac{2}{1}&space;\right&space;)\approx&space;20\cdot&space;0.301\approx&space;6.021&space;dB$

The result is that when our amplifier doubles our voltage, it has a gain of 6.021 dB.

### Example #2:

For this example, we have an amplifier with a known gain of 30dB, and you are inputting a voltage of 50mV. What is the expected output voltage?

The first step is to convert everything to the correct units. The input voltage is 50mV (millivolts) so we need to convert it to 0.05 volts. The formula above solves for dB, so we will need to use algebra to rearrange the variables to solve for the output voltage Vout.

$\inline&space;\dpi{150}&space;\fn_jvn&space;\small&space;dB=20log\left(&space;\frac{V_{out}}{V_{in}}&space;\right&space;)&space;\Rightarrow&space;\frac{dB}{20}=log\left(&space;\frac{V_{out}}{V_{in}}&space;\right&space;)\Rightarrow&space;10^{\frac{dB}{20}}=\frac{V_{out}}{V_{in}}\Rightarrow&space;V_{in}\cdot&space;10^{\frac{dB}{20}}=V_{out}$

So now that we have found the formula for for Vout, we can plug in the variables and solve.

$\fn_jvn&space;\small&space;V_{out}=V_{in}\cdot&space;10^{\frac{dB}{20}}=0.05\cdot&space;10^{\frac{30}{20}}\approx&space;0.05\cdot&space;31.62\approx&space;1.58V$

The result is that if you input a 50mV signal into an amplifier with 30dB of gain, you should get an output voltage of 1.58 volts.

### Example #3:

For this example, we have an amplifier that is outputting 2 volts and it has a known gain of -6dB. Note that the gain is negative, so right away you know the output should be smaller than the input. We want to find the expected input voltage to this amplifier.

We need to use algebra to convert the formula for dB Volts to solve for input volts.

$\inline&space;\dpi{200}&space;\fn_jvn&space;\tiny&space;dB=20log\left(&space;\frac{V_{out}}{V_{in}}&space;\right&space;)&space;\Rightarrow&space;\frac{dB}{20}=log\left&space;(&space;\frac{V_{out}}{V_{in}}&space;\right&space;)\Rightarrow&space;10^{\frac{dB}{20}}=\frac{V_{out}}{V_{in}}\Rightarrow&space;\frac{10^{\frac{dB}{20}}}{V_{out}}=\frac{1}{V_{in}}\Rightarrow&space;\frac{V_{out}}{10^{\frac{dB}{20}}}=V_{in}$

So now that we have solved for input voltage, we can plug in the variables and solve for Vin.

$\fn_jvn&space;\small&space;V_{in}=\frac{V_{out}}{10^{\frac{dB}{20}}}=\frac{2}{10^{\frac{-6}{20}}}=\frac{2}{10^{\frac{-6}{20}}}=\frac{2}{10^{-0.3}}\approx&space;\frac{2}{0.5}\approx&space;4V$

This input result is larger than the output, which is what we expected because the dB gain of the amplifier is negative. A rule of thumb to remember is when the output of a system has double the voltage of it’s input, that system has about 6dB of gain, and if the output is half the voltage, the gain is about -6db.

The result is that if you have a 2 volts coming out of a system with -6dB of gain, the input voltage should be about 4 volts.