dB calculation amplification (gain) and damping (loss) of an amplifier

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Conversion: amplification (gain) and damping (loss) in dB

Enter two values and press the right calculate bar in the line of the missing answer

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In analog audio engineering we deal only with 'voltage' amplification (gain) and damping (loss).
gain and loss - sengpielaudio
V₂ >V₁ = amplification, dB value is positive and V₂< V₁ = damping, dB value is negative.
V₂ / V₁ means the ratio. The amplification or the damping in dB is:
L = 20 × log (voltage ratio V₂ / V₁) V₁ is the reference.

3 dB ≡	1.414 fold the voltage	- 3 dB ≡	damping to the value 0.707
6 dB ≡	2 fold the voltage	- 6 dB ≡	damping to the value 0.5
10 dB ≡	3.162 fold the voltage	-10 dB ≡	damping to the value 0.316
20 dB ≡	10 fold the voltage	-20 dB ≡	damping to the value 0.1

Using voltage we get: Level in dB: L = 20 × log (voltage ratio)

+6 dB = two fold the voltage
+12 dB = four fold the voltage
+20 dB = ten fold the voltage
+40 dB = hundred fold the voltage

Not often one is interested in power, if we consider audio engineering.
Forget to ask what the power amplification is.
Leave it to the telephone company.

Using power we get: Level in dB: L = 10 × log (power ratio)

    +3 dB = two fold the power
    +6 dB = four fold the power
+10 dB = ten fold the power
          +20 dB = hundred fold the power

If you search for the amplification factor, given the dB value,
then go to the program dB calculation

At the cut-off frequency f_c the voltage is always damped to the value of 1/√2
and the voltage level is damped to 20 · log(1/√2) = (−)3,0103 dB.

Enter a value in the left or right box, then press the TAB bar or make
a mouse click at an empty space at the side, to get the solution.
The calculator works in both directions of the ↔ sign.

In audio technique "power or energy amplification " is very unusual:

Voltage/Pressure amplification factor	1	1.414 = √2	2	3.16 = √10	4	10	20	40	100	1000
Increasing of x dB	0	3	6	10	12	20	26	32	40	60

Power/Intensity amplification factor	1	1.414 = √2	2	3.16 = √10	4	10	20	40	100	1000
Increasing of x dB	0	1.5	3	5	6	10	13	16	20	30

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Voltage V: volts	↔	Voltage level L_U: dBV
$V = V_0 \cdot 10^\frac{L_V}{20} \ \mbox{volts}$		$L_V = 20\, \log_{10}\left(\frac{V}{V_0}\right) \ \mbox{dBV}$
Reference voltage V₀ = 1 Volt (0 dBV)

Voltage V (audio): volts	↔	Voltage level L_U: $L_V = 20\, \log_{10}\left(\frac{V}{V_0}\right) \ \mbox{dBu}$ dBu
$V = V_0 \cdot 10^\frac{L_V}{20} \ \mbox{volts}$
Reference voltage V₀ = 0.7746 Volt (0 dBu)

Electric power P: watts	↔	Electric power level L_P: dB

Reference electric power P₀ = 1 W (0 dB)

gain factor v = V₂/V₁: for field quantities, e.g. voltage	↔	amplification level L: voltage level dB

0 dB = voltage gain v = 1

gain factor v = P₂/P₁: for energy quantities, e.g. power	↔	amplification level L: power level dB

0 dB = power gain v = 1

Input: voltage power

input		value 1
output		value 2
level change		dB