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16th April 2004, 12:49 AM  #1 
diyAudio Member
Join Date: Jan 2004
Location: Minnesota

dB into voltage gain????
what is the formula for conversion???
What is 12dB of gain? Also.. what is a good gain to go for in a preamp? Thanks!!! 
16th April 2004, 01:19 AM  #2 
diyAudio Member

A little db tutorial:
Voltage Gain (dB) = 20 log (Volt. Gain) ex: Vin = 0.1V, Vout = 1.0V, Vgain = 1/0.1=10, Vgain (db) = 20 log 10 = 20 dB are logaritmic relational unity. There is a lot of different dB relations, here just a few: dBW=dBWatt, dB difference referenced to 1W, 1W=0dBW, 10W=10dBW, 100W=20dBW equation: dBW = 10 log (x/1W) Used often to express speaker sensitivity. Just a note: In power measurements, 3dB is double. So a speaker of 90dBW is two time more sensitive than one of 87 dBW! In voltage measurements, 6dB is double. dBm = dBmilliwatt, dB difference referenced to 1mW. Also the impedance of the system where the power is measured is often mentionned. For example 0dBm in 600 or 50 ohms. Often used for small signal, example phone line level, receiver maximum input, RF measurements, etc... dBuV = dBmicroVolt, often used to express radio receiver sensitivity. dBA dBB or dBC (also called dB AWeighted, etc...): you see them in acoustic measurements. Refer to Sound Pressure Level measured using different response filters A, B & C. Aweighting takes into account the typical human ear response, reducing the frequency response of the SPL meter to the 500  10,000 Hz range, where our ears are mostly sensitive, while the Bweighting measures the whole range from 32 to 10,000 Hz. The C is like the B but with peak response. 
16th April 2004, 01:30 AM  #3  
diyAudio Member
Join Date: Jul 2003
Location: Wayne, West Virginia

Quote:
Wayne 

16th April 2004, 04:14 AM  #4 
diyAudio Member
Join Date: Dec 2003
Location: new zealand

you may have noticed that the voltage and power db expressions are different for the same ratio, ie. doubling power gives a 3db gain, whereas doubling voltage gives rise to a 6db gain
some find this confusing, allow me to explain. power db ratio is given as 10log(p1/p2) p = V*V/R [this is a clumsy way of saying V squared divided by R] therefore the new ratio = 10log (V1*V1/R)/(V2*V2/R) the R's cancel out giving 10log(V1*V1)/(V2*V2) or put another way 10log(V1/V2)*(V1/V2) or in english ten log (v1 divided by v2) squared using the rules of logarithms, this can be rewritted as 20log(V1/V2) hence the voltage doubling factor being shown as 6db. if you already knew this sorry to teach you to suck eggs. kind regards bill 
17th April 2004, 04:20 AM  #5  
diyAudio Member
Join Date: Apr 2004
Location: Northern California

Quote:
Voltage ratio = 10^(dB/20) Power ratio = 10^(dB/10) Why 10 for power and 20 for Voltage? When you double the Voltage in a circuit you get four times the power. Or put another way to double the power, multiple the Voltage by 1.414. 

17th April 2004, 04:50 AM  #6  
Warp Engineer
On Holiday

Quote:
sqrt(2)*voltage = double the power Double the power vs. Double the voltage. Double the voltage = 4x the power 4x the power = 2x double the power So doubling the voltage is the same as doubling the power twice hence 20 in the formula instead of 10 for power ratio.
__________________
 Dan 

17th April 2004, 01:25 PM  #7 
diyAudio Member
Join Date: Jun 2002
Location: Melbourne, Australia

If a data sheet say a certain opamp has an open loop dc gain of 120dB then that means it has a power gain of 10^12 but a voltage gain of 10^6.

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