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1) Why does power gain use a coefficient of 10 unlike the other two?

2) Why can't power gain be negative?

  • 1
    \$\begingroup\$ Note that the power gain has the absolute value built in: power is related to the square of voltage, and the square of any real number is positive. \$\endgroup\$
    – Greg d'Eon
    Jan 14 '15 at 20:16
  1. It's just a convention. Decibels always refer to power, and power is proportional to voltage squared and current squared. The math works like this:

$$\begin{align} A_{v,dB} &= 10\cdot\log \frac{V_o^2}{V_i^2} \\ &= 10\cdot\log \left(\frac{V_o}{V_i}\right)^2 \\ &= 10\cdot 2 \cdot\log \frac{V_o}{V_i} \end{align} $$

EDIT: Power is proportional to voltage/current squared for all linear circuits. In AC circuit analysis, voltage, current, and power all become phasors. In nonlinear circuits, power may not be proportional to voltage/current squared, but the convention is still used for decibels. It even holds in an abstract field like signal processing, where the "power" of a signal is defined to be the average of the amplitude squared.

  1. Decibels are used to give the magnitude of the gain. If you want to talk about power loss, you use negative decibels, which represents a fractional gain. For example:

$$A_p = 0.01$$ $$A_{p,dB} = 10\cdot\log 0.01 = -20\:\mathrm{dB}$$

You can also have an actual negative gain, like what you get from an inverting amplifier. The negative there is described as a phase shift. To fully describe an amplifier, you usually need both the magnitude and the phase of the gain. (This might be a bit advanced for you, but you could try looking up Bode Plots to see how this is used in real life.) Anyway, here's how to describe an inverting amplifier with a voltage gain of -2.4:

$$A_v = -2.4$$ $$|A_v| = 20\cdot\log 2.4 \approx 7.6\:\mathrm{dB}$$ $$\angle A_v = 180^\circ$$

  • \$\begingroup\$ Thank you. Power is only proportional to voltage/current squared for resistors. But power can be applied to other devices too can't it? Also, I know you can get negative gain, but can you get negative power gain? You cannot have a phase shift on power can you? Since it's not a phasor? \$\endgroup\$
    – dfg
    Jan 14 '15 at 18:55
  • \$\begingroup\$ Power can indeed be a phasor, although it doesn't work quite the same way. (Look up "reactive power".) I suppose a negative power gain would mean that the input and output are both absorbing power. I can't think of a real-life example of that, though. I'll update my answer to cover P = V^2 a bit more. \$\endgroup\$
    – Adam Haun
    Jan 14 '15 at 19:09
  • \$\begingroup\$ An interesting counter-reference, I was taught in college that voltage and current use 10log(V) or 10log(I), and power used 20log(V). This is consistent with dB's usage in non-power settings, such as dBsm, which is area (in square meters), on a 10log(x) scale. dBsm shows up in a lot of radar calculations. \$\endgroup\$
    – Cort Ammon
    Jan 15 '15 at 7:12
  1. Decibels is always about power. The expressions for voltage and current use a factor of 20 instead of 10 because power ratios are proportional to the square of the voltage or current ratio.

  2. Power gain can be negative, but then it's generally called "loss". The logarithm of any number less than unity is negative.

  • 2
    \$\begingroup\$ @ first bullet: strictly speaking the value of the impedance must not change, but that is not always accounted for. \$\endgroup\$
    – jippie
    Jan 14 '15 at 17:53
  • \$\begingroup\$ But the square law difference is only true for resistors. Power can be applied to other types of devices too can't it? Also, by "power gain", I mean't \$A_p\$, not \$10\log A_p\$. If \$A_p\$ is negative, the log would be imaginary... \$\endgroup\$
    – dfg
    Jan 14 '15 at 18:34
  • 1
    \$\begingroup\$ By definition, "power gain", or \$A_p\$ is \$\frac{|Power_{OUT}|}{|Power_{IN}|}\$. It can't be negative. If the amplifier happens to be an inverting amplifier, it doesn't mean that the power gain is negative. \$\endgroup\$
    – Dave Tweed
    Jan 14 '15 at 18:42
  • \$\begingroup\$ The logarithm of a negative number is undefined. It's the squaring of the voltages (to get power), not the log (to get dB) that makes the ratio of power out to power in positive even if the amplifier is inverting. \$\endgroup\$
    – Phil Frost
    Jan 15 '15 at 11:57

Decibel is a measure of power ratio.

It is 10x (for power ratios) because the unit is bel (deci is one of the 'approved' prefix for 1/10).

It is 20x (ie 2*10) when dealing with voltages and current as the dissipated power is typically proportional to the square of these quantities. Due to log rules a \$10*log(\frac{V_1}{V_2})^2\$ would see the power2 come down as a scalar gain \$2*10*log(\frac{V_1}{V_2})\$

dB is an indication of the ratio of power between two points. If you cannot take credit for a load impedance being held constant then actual power needs to be used

  • \$\begingroup\$ But the square law difference is only true for resistors. Power can be applied to other types of devices too can't it? \$\endgroup\$
    – dfg
    Jan 14 '15 at 18:32
  • \$\begingroup\$ @dfg. Gain=10·log(Powerout/Powerin). Powerout=Vout·Iout and Powerin=Vin·Iin. And you can apply those formulas to any device. \$\endgroup\$
    – Roger C.
    Jan 14 '15 at 19:43
  • \$\begingroup\$ @RogerC. Yes, but neither of those equations have power proportional to the square of voltage. You can get \$P=V^2/R\$ if you substitute Ohm's law into one of your equations, but then Ohm's law doesn't apply to everything. For example, the load could be a diode. \$\endgroup\$
    – Phil Frost
    Jan 15 '15 at 12:00
  • \$\begingroup\$ @PhilFrost, I don't really understand your point. Gain in dB is always about power. In a diode I=f(V). So that power would be V·f(V). There is not a square here. \$\endgroup\$
    – Roger C.
    Jan 15 '15 at 13:09
  • \$\begingroup\$ Yes, exactly. But in your answer you say, unqualified, "It is 20x (for volts and amps) as there is a square law difference between volts, amps and power". You should edit your question to qualify that statement with the conditions where it is true. \$\endgroup\$
    – Phil Frost
    Jan 15 '15 at 14:36

Power decibels are tough because they throw a few concepts at you at once without making clear that's what happening.

Concept 1: Bel is the unit of log ratios (which is really a cancelled out unit of Power/Power).

Concept 2: deci- is a prefix that denotes 1/10th. 1 Bel = 10 * deci-Bel.

Concept 3: logs turn exponents into multiplication, making the math easier to work with and graph. log(x^n) = n*log(x).

Concept 4: logs imply "Power"...even when working with voltage or current. Think Pa/Pb = (Va^2/R) / (Vb^2/R) where the R's cancel out (sort of).

First, break up the word "decibel". "deci-" is the prefix for 1/10th. The "bel" is for the unit "Bel", named after Alexander Graham Bell and is the log ratio of two numbers. Because bel is for the unit "Bel", the the "B" in dB capitalized. Same concept as in kilohertz (kHz = "kilo-" + Hertz).

Assume all logs have base 10 in the examples below.

Example 1 (simple ratio):

value1 = 100,000

value2 = 100

simple_ratio = value1/value2 = 100,000/100 = 1,000

Example 2 (log):

value1 = 100,000

value2 = 100

simple_log = log(value1/value2) = log(100,000/100) = log(1000) = 3 Bel. Same as 10^x = 1,000, x = 3 Bel In English "10 to what power (x) is 1,000? 10^(3) = 10,000. Answer is 3."

Example 3 (decibel):

value1 = 100,000

value2 = 100

simple_log = log(100,000/100) = 3 Bel

To make it a "deci-" Bel... divide by 10: simple_log/10 = simple_deci_log


simple_log = 10*simple_decibel

So.....Why 10x for Power and 20x for Voltage and Current?

The Bel, itself, is not specific to power but, for various reasons, the science and engineering community decided to choose to adhere to it as the convention. Unless otherwise specified dB denotes power (e.g. dBV).


P = I^2 * R

P = V^2 / R

Logs turn exponents into multiplication...which is why we use them. The square relationship of current to power and voltage to power is where the 2 x comes from...so if a Power ratio is expressed in voltage or current it must account for the square relationship.. which means you pull the square from the inside of the log out to the front.

some_log = log(x ^ 2) = 2*log(x) Bel

some_power_ratio_expressed_in_voltage_in_Bel = log(Va^2/Vb^2) = 2*log(Va/Vb) "Bel"

10 * deci-Bel = 1 * Bel

some_power_ratio_expressed_in_voltage_in_deci_Bel = 10 * log(Va^2/Vb^2) = 2 * 10 * log(Va/Vb) = 20 * log(Va/Vb)


Lets take small steps and look at the equations...

Gain (in dB) is given by

gain in dB

Power is given by: enter image description here.

The statement enter image description here holds in linear cicuits.

Thus enter image description here holds linear circuits.

If we substitute power in the first equation we get

enter image description here


enter image description here


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