Confusion in Power gain and voltage gain in decibels

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dexterdev
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Hi all,
I have a very basic doubt. If we have an amplifier with voltage gain A = Vo/Vi , where Vo is output and Vi is input voltages and Ap = Po/Pi which is power gain.

Ap = Po/Pi = Vo^2/Vi^2 = (Vo/Vi)^2

In decibels powergain = 10 log (Po/Pi) = 20 log (Vo/Vi).

In most tech literature 20 log(Vo/Vi) is addressed as voltage gain in decibels and not power gain in decibels. Why is it so?

Actually 20 log(Vo/Vi) is power gain in decibels. Right...

-Devanand T
 
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Actually 20 log(Vo/Vi) is power gain in decibels. Right...

Yes, that's right.

Both forms of this calculation give power gain.

It just that if you have voltages and the impedance is the same, you can work out power just using voltages using the formula Power = Voltage 2 / impedance. Then you take the ratio of the two powers using the 20 log(Vo/Vi) formula.
 
dexterdev said:
Hi all,
I have a very basic doubt. If we have an amplifier with voltage gain A = Vo/Vi , where Vo is output and Vi is input voltages and Ap = Po/Pi which is power gain.

Ap = Po/Pi = Vo^2/Vi^2 = (Vo/Vi)^2

In decibels powergain = 10 log (Po/Pi) = 20 log (Vo/Vi).

In most tech literature 20 log(Vo/Vi) is addressed as voltage gain in decibels and not power gain in decibels. Why is it so?

Actually 20 log(Vo/Vi) is power gain in decibels. Right...

-Devanand T

Power is

[tex]10 \;\log \left(\frac {P_o}{P_i}\right)=10\;\log \left[\left( \frac {V_o}{V_i}\right)^2\right]=20\;\log \frac {V_o}{V_i} \hbox {Only if the resistance R is equal.}[/tex]

Remember [itex]power=\frac{V^2}{R}[/itex]
 
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