Decibel dB calculation amplification gain and damping loss ratio - sengpielaudio
 
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Calculation: amplification (gain) and damping (loss)
as
level in decibels (dB)
 
Enter two values and press the right calculate bar in the line of the missing answer

 
Input: voltage power        
           
input   value 1    
output   value 2    
level change   dB        
 

In analog audio engineering we deal only with 'voltage' amplification (gain) and damping (loss).
gain and loss - sengpielaudio
V2 > V1 means amplification. The dB value is positive.
V2 < V1 means damping. The 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 times the voltage    (−3) dB ≡  damping to the value 0.707
 6 dB ≡          2 times the voltage    (−6) dB ≡  damping to the value 0.5
10 dB ≡  3.162 times the voltage    (−10) dB ≡  damping to the value 0.316
20 dB ≡       10 times 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 times the voltage
+12 dB = four rimes the voltage
+20 dB = ten times the voltage
+40 dB = hundred times the voltage

Not often one is interested in power, if we consider audio engineering.
Do not ask what the power amplification is.
Leave that to the telephone company.
Is there really a power amplification?
Read the text at the bottom.

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

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

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

Aha!
 
At the cut-off frequency fc the dropped voltage is always fallen to the value of
1/√2 and the voltage level L is damped to 20 × log (1/√2) = (−)3,0103 dB.
 

Enter a value in the left or right box.
The calculator works in both directions of the sign.

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

The voltage is always given as RMS value - but not the electric power.

Electric Power P
watts
 ↔  Electric Power level LP
dB
Formula1   Formula2
Reference electric power P0 = 1 W ≡ 0 dB

There is also the reference power P0 = 1 milliwatt or 0,001 watt ≡ 0 dBm

Electric power (telephone) P:
watts
 ↔  Electric Power level (phone) LP
dBm
Start   Start
Reference power P0 = 1 milliwatt (mW) = 0.001 W ≡ 0 dBm

Amplification and Damping

gain ratio v = V2/V1:
for field quantities, e.g. voltage

 ↔  amplification level L:
voltage level
dB
Formula Factor   Formula Amplification
Voltage gain v = 1 ≡ 0 dB

In audio technique the following "power or energy amplification " is rather unusual.

gain factor v = P2/P1:
for energy quantities, e.g. power

 ↔  amplification level  L:
power level
dB
Formula1   Formula2
Power gain v = 1 ≡ 0 dB
  Voltage/Pressure 
  amplification ratio  
  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 ratio  
  1  
 
 1.414 
= √2
  2  
 
 3.16
= √10
  4  
 
 10 
 
 20 
 
 40 
 
 100 
 
 1000 
 
  Increasing of y dB  0   1.5   3   5   6   10   13   16   20   30 

Level in psycho acoustics as subjectivly perceived loudness (volume)
 
Amplifier conversion - Convert decibels to voltage gain / loss

Aha!
 
The word "power amplifier" is a misnomer. Power is not really something
that can be "amplified". Voltage and current can be amplified. The term
"power amplifier" although technically incorrect has become understood
to mean an amplifier that is intended to drive a load such as a loudspeaker.

We call the product of current and voltage gain just "power amplification".
The total energy within a closed system due to the conservation of energy
is neither increased nor decreased.

 
 
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