When designing or working with amplifier and filter circuits, some of the numbers used in the calculations can be very large or very small. For example, if we cascade two amplifier stages together with power or voltage gains of say 20 and 36, respectively, then the total gain would be 720 (20*36).

Likewise if we cascaded together to first-order RC filter circuits with attenuations of 0.7071 each, the the total attenuation would be 0.5 (0.7071*0.7071). Remembering of course that if a circuits output is positive, then it produces amplification or gain, and if its output is negative, then it produces attenuation or loss.

When analysing circuits in the *frequency domain*, it is more convenient to compare the amplitude ratio of the output to input values on a logarithmic scale rather than on a linear scale. So if we use the logarithmic ratio of two quantities, P_{1} and P_{2} we end up with a new quantity or level which can be presented using *Decibels*.

Unlike voltage or current which is measured in volts and amperes respectively, the **decibel**, or simple **dB** for short, is just a ratio of two values, well actually the ratio of one value against another known or fixed value, so therefore the decible is a dimensionless quantity, but does have the “Bel” as its units after the telephone inventor, Alexander Graham Bell.

The ratio of any two values, where one is fixed or known and of the same qunatity or units, whether power, voltage or current, can be represented using decibels (dB) where “deci” means one tenth (1/10th) of a Bel. Clearly then there are 10 decibels (10dB) per Bel or 1 Bel = 10 decibels.

The decibel is commonly used to show the ratio of power change (increasing or decreasing) and is commonly defined as the value which is ten times the Base-10 logarithm of two power levels. So for example, 1 watt to 10 watts is the same power ratio as 10 watts to 100 watts, that is 10:1, so while there is a large difference in the number of watts, 9 compared to 90, the decibel ratio would be exactly the same.

Hopefully then we can see that the decibel (dB) value is a ratio used for comparing and calculating levels of change in power and is not the power itself. So if we have two quantities of power, for example: P_{1} and P_{2}, the ratio of these two values is represented by the equation:

_{10}[P

_{2}/P

_{1}]

Where, P_{1} represents the input power and P_{2} represents the output power, (P_{OUT}/P_{IN}).

As the decibel represents the Base-10 logarithmic change of two power levels, we can expand this equation further by using antilogarithms to show by how much change one decibel (1dB) really is.

dB = 10log_{10}[P_{2}/P_{1}]

If P_{2}/P_{1} is equal to 1, that is P_{1} = P_{2} then:

dB = 10log_{10}[1] = log_{10}[1/10] = log_{10}[0.1] = antilog[0.1]

Thus a dB change in value equals: 10^{0.1} = 1.259

Clearly then the logarithmic change of two powers has a ratio of 1.259, meaning that a 1dB change represents an increase (or decrease) in power of 25.9% (or 26% rounded-off).

So if a circuit or system has a gain of say 5 (7dB), and it is increased by 26%, then the new power ratio of the circuit will be: 5*1.26 = 6.3, so 10log_{10}(6.3) = 8dB. An increase in gain of +1dB, proving again that a +1dB change represents a logarithmic increase in power of 26% and not a linear change.

## Decibel Example No1

An audio amplifier delivers 100 watts into an 8 ohm speaker load when fed by a 100mW input signal. Calculate the power gain of the amplifier in decibels.

We can express the power gain of the amplifier in units of decibels regardless of its input or output values, as an amplifier delivering 40 watts output for 40mW input will also have a power gain of 30 dB, and so on.

We could also, if we so wished, convert this amplifiers decibels value back into a linear value by first converting from decibels (dB) to a Bel remembering that a decibel is 1/10th of a Bel. For example:

A 100 watt audio amplifier has a power gain ratio of 30dB. What will be its maximum input value.

So the result is 100mW as declared in example No1.

One of the advantages of using the base 10 logarithm ratio of two powers is that when dealing with multiple amplifier, filter or attenuator stages cascaded together, we can simply add or subtract their decibel values instead of multiplying or dividing their linear values. In other words, a circuits overall gain (+dB), or attenuation (-dB) is the sum of the individual gains and attenuations for all stages connected between the input and output.

For example, if a single stage amplfier has a power gain of 20dB and it supplies a passive resistive network that has an attenuation of 2, before the signal is amplified again using a second amplifier stage with a gain of 200. Then the total power gain of the circuit between the input and output in decibels would be:

For the passive circuit, an attenuation of 2 is the same as saying the circuit has a positive gain of 1/2 = 0.5, thus the power gain of the passive section is:

dB Gain = 10log_{10}[0.5] = -3dB (note a negative value)

The second stage amplifier has a gain of 200, thus the power gain of this section is:

dB Gain = 10log_{10}[200] = +23dB

Then the overall gain of the circuit will be:

20 – 3 + 23 = +40dB

We can double check our answer of 40dB by multiplying the individual gains of each stage in the usual way as follows:

A power gain of 20dB in decibels is equal to a gain of 100, as 10^{(20/10)} = 100. So:

100 x 0.5 x 200 = 10,000 (or 10,000 times greater)

Converting this back to a decibel value gives:

dB Gain = 10log_{10}[10,000] = 40dB

Then clearly we can see that a gain of 10,000 is equal to a power gain ratio of +40dB as shown above and that we can use the decibel value to express large ratios of powers with much smaller numbers as 40dB is a power ratio of 10,000, whereas -40dB is a power ratio of 0.0001. So using decibels makes the maths a little easier.

## Decibels of Voltage and Current

Any power level can be expressed as a voltage or current if we know the resistance. According to Ohms Law, P = V^{2}/R and P = I^{2}R. As V and I relate to the current through and the voltage across the same resistance, if (and only if) we make R = R = 1, then the dB values for the ratios of voltage (V_{1}, and V_{2}) as well as for current (I_{1}, and I_{2}) will be given as:

that is 20log(voltage gain), and for the current gain would be:

Thus the only difference between defining the power, voltage, and current decibel (dB) calculations is the constant of 10 and 20, and that for the dB ratio to be correct in all instancies the two quantities must both have the same units, either watts, milli-watts, volts, milli-volts, amperes or milli-amperes, or any other unit.

## Decibel Example No2

A passive resistive network is used to provide an attenuation (loss) of 10dB, with an input voltage is 12V. What will be the networks output voltage value.

As *decibels* represents a logarithmic change in terms of power, voltage or current, we can construct a table to show the specific gains and their equivalent decibel values below.

### Decibel Table of Gains

dB Value | Power Ratio 10log(A) |
Voltage/Current Ratio 20log(A) |

-20dB | 0.01 | 0.1 |

-10dB | 0.1 | 0.3162 |

-6dB | 0.25 | 1/2 = 0.5 |

-3dB | 1/2 = 0.5 | 1/√2 = 0.707 |

-1dB | 0.79 | 0.89 |

0dB | 1 | 1 |

1dB | 1.26 | 1.1 |

3dB | 2 | √2 = 1.414 |

6dB | 4 | 2 |

10dB | 10 | √10 = 3.162 |

20dB | 100 | 10 |

30dB | 1000 | 31.62 |

We can see from the above decibel table that at 0dB the ratio gain for power, voltage and current is equal to “1” (unity). This means that the circuit (or system) produces no gain or loss between the input and output signals. So zero dB corresponds to a unity gain i.e. A = 1 and not zero gain.

We can also see that at +3dB the output of the circuit (or system) has doubled its input value, meaning a positive dB gain (amplification) so A > 1. Likewise, at -3dB the output the circuit is at half its input value, meaning a negative dB gain (attenuation) so A < 1. This -3dB value is commonly called the “half-power” point and defines the corner frequency in filter networks.

It is all well and good tabulating the power gains against decibels in a reference table, but when dealing with amplifier and filters, Electrical Engineers prefer to use Bode Plots, charts or graphs as a visual display of the circuits (or systems) frequency response characteristics. Then using the data values in the table above we can create the following “decibel” Bode plot showing the various positions of the power points.

### Decibel Power Bode Plot

Then we can clearly see that the power curve is not linear but follows the logarithmic ratio of 1.259.

## Decibel Tutorial Summary

We have seen in this tutorial about the **Decibel** (dB) that it is a Base-10 logarithmic unit of power change and that the decibel unit is a 1/10th dimensionless value of a Bel (1 Bel = 10 decibels or 1dB = 0.1B). The decibel allows us to present large ratios of powers using small numbers and we have seen above that 30dB is equivalent to a power ratio of 1000 with the most commonly used decibel values being: 3dB, 6dB, 10dB and 20dB (and their negative equivalents). However, 20dB is not double the power of 10dB.

The decibel also shows us that any change in power by the same ratio will have the same decibel ratio. For example, doubling the power from 1 watt to 2 watts is the same ratio as 10 watts to 20 watts, that is a +3dB change, while a -3dB change means that the power ratio will be halved.

If the dB ratio is positive in value, then it means amplification or gain is present as the output power is greater than the input power (P_{OUT} > P_{IN}). If however the dB power ratio is of a negative value, then this means an attenuation or loss is affecting the circuit as the output power will be less than the circuits input power (P_{OUT} < P_{IN}). Clearly then 0dB means the power ratio is one with no reduction or gain of the signal.