# How to Solve Decibel dB Watts Power Formula Calculator

This dB Watts Power formula calculator can calculate dB (decibels) from input and output power and can calculate the watts power 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. The dB Watts calculator can be used for any system where the power is amplified or attenuated, as long as you are comparing the signal power and not just the signal amplitudes. If you are looking for the power measurement of dBm (decibel-milliwatts) then you should use the dBm Watts Power Formula Calculator. If you need to calculate dB from the ratio of volts or amplitudes, use the dB Volts Formula Calculator instead.

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

dB Watts Calculator
 Watts Out MW (megawatts) kW (kilowatts) W (watts) mW (milliwatts) uW (microwatts) Watts In MW (megawatts) kW (kilowatts) W (watts) mW (milliwatts) uW (microwatts) dB dB

## dB Watts Formula

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

dB Watts Formula:
$\fn_jvn&space;\small&space;dB=10log\left&space;(&space;\frac{P_{out}}{P_{in}}&space;\right&space;)$
where:
• dB = Decibels
• Pout = Output Power (watts)
• Pin = Input Power (watts)

## 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 watts when you input 1 watt. We will be solving for the gain in dB (decibels) of the amplifier. We will enter the values into the formula as shown.

$\fn_jvn&space;\small&space;dB=10log\left&space;(&space;\frac{P_{out}}{P_{in}}&space;\right&space;)=10log\left&space;(&space;\frac{2}{1}&space;\right&space;)\approx&space;10\cdot&space;0.301\approx&space;3.01&space;dB$

The result is that when our amplifier doubles our wattage, it has a gain of 3.01 dB.

### Example #2:

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

The first step is to convert everything to the correct units. The input voltage is 50mW (milliwatts) so we need to convert it to 0.05 watts. The formula above solves for dB, so we will need to use algebra to rearrange the variables to solve for the output wattage Pout.

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

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

$\fn_jvn&space;\small&space;P_{out}=P_{in}\cdot&space;10^{\frac{dB}{10}}=0.05\cdot&space;10^{\frac{30}{10}}\approx&space;0.05\cdot&space;1000\approx&space;50W$

The result is that if you input a 50mW signal into an amplifier with 30dB of gain, you should get an output power of 50 watts.

### Example #3:

For this example, we have an amplifier that is outputting 2 watts 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 power to this amplifier.

We need to use algebra to convert the formula for dB Watts to solve for input watts.

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

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

$\fn_jvn&space;\small&space;P_{in}=\frac{P_{out}}{10^{\frac{dB}{10}}}=\frac{2}{10^{\frac{-6}{10}}}=\frac{2}{10^{-0.6}}\approx&space;\frac{2}{0.251}\approx&space;7.96V$

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 power of its input, that system has about 3dB of gain, and if the output is half the power, the gain is about -3db.

The result is that if you have a 2 watts coming out of a system with -6dB of gain, the input voltage should be about 8 volts.