Relation between voltage gain and dB amplification

[klitgt]

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??!

[peranders]

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/catalo...tName=slod006b

[klitgt]

Originally posted by peranders
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/catalo...tName=slod006b

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?

[peranders]

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.

[klitgt]

Originally posted by peranders
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.

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

[MikeB]

Originally posted by klitgt

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?

Yes, type in 0,45 and press "inv"-log, there is your result !
Same is for reverse, 2.818 -> "log" * 20 -> 9db

Mike

[MikeW]

The formula for voltage gain is 20*Log(Rf/Rg).
The formula for power gain is 10*Log(power out /power in)

If Rf is 10K and Rg is 3.5K you should have a v ltage gain of 9.1 db.

[klitgt]

Thank's to all, I have what I need now!
Regards

[peranders]

MikeW, klitgt said indirectly a non-inverting amp

Non-inverting
A = 1 + Rf/Rg

Inverting
A = -Rf/Rg

[AndrewT]

Hi,
when working in voltage ratios remember to convert from db back to bel by dividing by 20.
But when using power ratios the db is converted to bel by dividing by 10.