Relation between voltage gain and dB amplification

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klitgt

klitgt

Member
Joined 2004
Opamps have voltage gain = 1 (0 dB) if the output is fed back to the inverting input without a resistor.
If the feedback loop is made with a resistor (Rf) e.g. 10K and a resistor (Rg) from inverting input to ground of the same value, the gain is 2
(Gain = 1+(Rf/Rg).

I need a gain of 9dB. Which value for voltage gain corresponds to 9dB amplification?
There must be a calculation for this, I just do not find anything that I understand when searching Google??!:confused:
 
klitgt

klitgt

Member
Joined 2004
peranders said:
A = 10^(dB/20)

10^ means that you use 10 power to the thing in the paranthesis

A = 10EXP(9/20) = 2.818 => Rf = 18 k, Rg = 10 k

A good source of knowledge is here:
http://focus.ti.com/docs/apps/catalog/resources/appnoteabstract.jhtml?abstractName=slod006b
Click to expand...


Thank's peranders. I understand that if I need 9 dB amplification I need a voltage gain of 2.818 which can be obtained by resistors Rf=18K and Rg=10K.

I just do not understand the "10 power" issue. How do you come from 10EXP(9/20) = 10EXP(0,45) to the value 2.818?
Can the Windows calculator in "scientific" mode be used?
 
klitgt

klitgt

Member
Joined 2004
peranders said:
10 powered to X, I'll guess you have such button on your calculator.

Take 0.45 and push the 10x button => 2.818

Normal functions are:

e powered to X (e = 2.718)
logaritm of e usually "ln" at the button.

10 powered to X
logaritm of 10 usually "log" at the button.
Click to expand...


Thank's again. I found this site in the meantime. Select dB and enter 9 in the input field and hit calculate. The result is 2.8183829312644537

http://www.sengpielaudio.com/calculator-db.htm
 
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