Gain

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For other uses, see Gain (disambiguation).

In electronics, gain is a measure of the ability of a circuit (often an amplifier) to increase the power or amplitude of a signal. It is usually defined as the mean ratio of the signal output of a system to the signal input of the same system. It may also be defined as the decimal logarithm of the same ratio.

Thus, the term gain on its own is ambiguous. For example, 'a gain of five' may imply that either the voltage, current or the power is increased by a factor of five. Furthermore, the term gain is also applied in systems such as sensors where the input and output have different units; in such cases the gain units must be specified, as in "5 microvolts per photon" for a photosensor.

In laser physics, gain may refer to the increment of power along the beam propagation in a gain medium, and its dimension is m^-1 (inverse meter) or 1/meter.

1 Logarithmic units and decibels
2 See also

[edit] Logarithmic units and decibels

[edit] Power gain

Power gain, in decibels (dB), is defined as follows:

$Gain=10 \log \left( {\frac{P_{out}}{P_{in}}}\right)\ \mathrm{dB}$

where P_in and P_out are the input and output powers respectively.

A similar calculation can be done using a natural logarithm instead of a decimal logarithm. The result is then in nepers instead of decibels.

[edit] Voltage gain

When power gain is calculated using voltage instead of power, making the substitution (P=V ²/R), the formula is:

$Gain=10 \log{\frac{(\frac{{V_{out}}^2}{R_{out}})}{(\frac{{V_{in}}^2}{R_{in}})}}\ \mathrm{dB}$

In many cases, the input and output impedances are equal, so the above equation can be simplified to:

$Gain=10 \log \left( {\frac{V_{out}}{V_{in}}} \right)^2\ \mathrm{dB}$

and then:

$Gain=20 \log \left( {\frac{V_{out}}{V_{in}}} \right)\ \mathrm{dB}$

This simplified formula is used to calculate a voltage gain in decibels, and is equivalent to a power gain only if the impedances at input and output are equal.

[edit] Current gain

In the same way, when power gain is calculated using current instead of power, making the substitution (P=I ²R), the formula is:

$Gain=10 \log { \left( \frac { {I_{out}}^2 R_{out}} { {I_{in}}^2 R_{in} } \right) } \ \mathrm{dB}$

In many cases, the input and output impedances are equal, so the above equation can be simplified to:

$Gain=10 \log \left( {\frac{I_{out}}{I_{in}}} \right)^2\ \mathrm{dB}$

and then:

$Gain=20 \log \left( {\frac{I_{out}}{I_{in}}} \right)\ \mathrm{dB}$

This simplified formula is used to calculate a current gain in decibels, and is equivalent to the power gain only if the impedances at input and output are equal.

[edit] Example

Q. An amplifier has an input impedance of 50 ohms and drives a load of 50 ohms. When its input ( $V i n$ ) is 1 volt, its output ( $V o u t$ ) is 10 volts. What are its voltage gain and power gain?

A. Voltage gain is simply:

$\frac{V_{out}}{V_{in}}=\frac{10}{1}=10\ \mathrm{V/V}.$

The units V/V are optional, but make it clear that this figure is a voltage gain and not a power gain. Using the expression for power, P = V²/R, the power gain is:

$\frac{V_{out}^2/50}{V_{in}^2/50}=\frac{V_{out}^2}{V_{in}^2}=\frac{10^2}{1^2}=100\ \mathrm{W/W}.$