When designing or working with amplifiers and filter circuits, some of the numbers used in calculations can be very large or very small. For example, if we gradually add the two amplifier stages together with power or voltage Gains of 20 and 36 respectively, the total gain will be 720 (20*36).
Likewise, if we cascade from the first degree to the RC filter circuits with a weakening of 0.7071 each, the total weakening will be 0.5 (0.7071*0.7071). If the output of a circuit is positive, then it produces amplification or gain, and if its output is negative, then it produces weakening or loss.
When analyzing circuits in the frequency field, it is more convenient to compare the amplitude ratio of the output with input values on a logarithmic scale, not on a linear scale. Therefore, if we use the logarithmic ratio of the two quantities, P1 and P2 result in a new quantity or level that can be presented using decibels.
Unlike voltage or current, which is measured in volts and amps respectively, decibels, or simply simple dB, are a ratio of only two values. In fact, it is the ratio of a value against a known or constant value.
Decibels are widely used to indicate the rate (incremental or decreasing) of power change. It is usually defined as a value that is ten times the 10-based logarithm of the two power levels. For example, 1 watt to 10 watts is the same power ratio as 10 watts to 100 watts, that is, 10:1. Therefore, although there is a big difference in the number of watts, compared to 90, the decibel ratio will be exactly the same.
Hopefully then we can see that the decibel (dB) value is a ratio used to compare and calculate the levels of change in power, and it is not the power itself. So if we have two quantities of power, for example: P1 and P2, the ratio of these two values is represented by the equation:
Here, P1 represents input power and P2 represents output power (Pout/Pin).
Since decibels represent base-10 logarithmic change of two power levels, we can further expand this equation by using anthripes to show how much a decibel (1dB) actually changes.
If P2 / P1 = equals 1, that is, P1 = P2:
Thus, a dB change in value equals: 100.1 = 1.259
Obviously, the logarithmic change of the two forces has a ratio of 1,259. This means that a 1db change represents a 25.9% (or 26% rounded) increase (or decrease).
This means that if a circuit or system has a 5 (7dB) gain and increases by 26%, the new power ratio of the circuit will be: 5*1.26 = 6.3, that is, 10log10(6.3) = 8dB. An increase in a gain of +1db proves once again that a change of +1db represents a logarithmic increase of 26%, not a linear change.
An audio amplifier delivers 100 watts to an 8 ohm speaker load when fed with a 100mW input signal. Calculate the amplifier's power gain in decibels.
We can express the power gain of the amplifier in decibel units regardless of input or output values. Because an amplifier that provides 40 watt output for 40mW input will also have a power gain of 30 db.
We can also convert the decibel value of these amplifiers to a linear value by first converting them from Decibels (dB) to Waist, remembering that the decibel is 1/10 of the Waist. Eg:
A 100 watt audio amplifier has a power gain rate of 30db. What will be the maximum input value.
The total gain (+dB) or attenuation (-dB) of a circuit is the sum of individual gains and weight loss for all stages connected between input and output.
For example, if a single-stage amplifier has a power gain of 20db and provides a passive resistant network with a weakening of 2, before the signal is reinforced again using a second amplifier stage with a gain of 200. The total power gain of the circuit between input and output in decibels will be as follows:
For passive circuit, a weakening of 2 is the same as saying that the circuit has a positive gain of 1/2 = 0.5. Therefore, the power gain of the passive section:
dB gain = 10log10[0.5] = – 3dB (note one negative value)
The second-tier amplifier has a gain of 200, so the power gain of this section:
dB gain = 10 log10 = + 23dB
Then the overall gain of the circuit will be as follows:
20 – 3 + 23 = + 40dB
We can double your 40db response by multiplying the individual earnings of each stage normally as follows:
A power gain of 20db in decibels is equal to a gain of 10(20/10) = 100. Like this:
100 x 0.5 x 200 = 10,000 (or 10,000 times larger)
Converting it to a decibel value:
dB gain = 10 log10[10,000] = 40dB
Then we can clearly see that a gain of 10,000 is equal to a power gain rate of +40db, as shown above, and since 40db has a power ratio of 10,000, we can use the decibel value to express large power ratios in much smaller numbers. However, -40dB is a power ratio of 0.0001. That's why using decibels makes math a little easier.
Voltage and Current Decibels
If we know the resistance, any power level can be expressed as voltage or current. According to ohm law, P = V2/R and P = I^2R. Since V and I are related to current and voltage along the same resistance, if (and only) we make R = R = 1, then dB values for voltage ratios (V1 and V2) and current (I1 and I2) will be given as follows:
for this 20log (voltage gain) and current gain:
The ratio of 10 and 20 and dB must be two quantities, both the same units or watts, milli-watts, volts, milli-volts, amps or milli-amps or any other unit for all samples to be correct,
Decibel Gain Statement
From the decibel table above, we can see that the rate gain for power, voltage and current at 0db is equal to "1" (union). This means that the circuit (or system) does not produce gains or losses between input and output signals. That is, zero dB corresponds to a union gain, that is, A = 1, and does not correspond to zero gain.
We can also see that at +3db the output of the circuit (or system) doubles the input value, that is, a positive dB gain (amplification), that is, the > is 1. Similarly – at 3db, the output circuit is half of the input value, that is, a negative dB gain (slimming) that is < 1'dir. This -3dB value is often called the "half power" point and defines the corner frequency in the filter networks.