What is the power gain in dB for a voltage amplifier with a ratio of 200 between output voltage and input voltage?

[MissP.25_5]

An amplitude of output voltage Vo of a voltage amplifier was compared to the amplitude of input voltage Vi. The ratio Vo/Vi was 200.

Find the power gain in dB.

Can someone explain to me how to find this? The solution is given as below, but I don't understand how can Po/Pi = 200^2

 

[rude man]

Power goes as the square of voltage, assuming equal resistances: P = V^2/R.
The solution is correct.

 

[MissP.25_5]

Power goes as the square of voltage, assuming equal resistances: P = V^2/R.
The solution is correct.

Ah, I got it. There's another subquestion that I really don't understand. It says to find the output voltage when Vi is -6 dBV. I don't understand the solution. If Vo is 200 times larger than Vi, then how come Vo = 40 dB?

 

[MissP.25_5]

Disregard my question. I have understood it :)
The solution seems to be written in a descending order.

 

[rezeew]

To find the power gain in dB, we first need to understand the relationship between power and voltage. Power is proportional to the square of voltage, meaning that if we double the voltage, the power will increase by a factor of four. This is expressed in the equation P = V^2/R, where P is power, V is voltage, and R is resistance.

In this case, the ratio of output voltage to input voltage is 200, or Vo/Vi = 200. Using the power equation, we can express the ratio of output power to input power as (Vo^2)/(Vi^2) = (200^2). This is because we are squaring both the output and input voltage values.

Next, we can take the logarithm of both sides of the equation to convert the ratio to a logarithmic scale. This gives us log[(Vo^2)/(Vi^2)] = log[(200^2)]. Using the logarithm property log(a^b) = b*log(a), we can simplify the equation to 2*log(Vo/Vi) = 2*log(200).

Finally, we can use the definition of decibels (dB) as 10*log(Po/Pi) to express the power gain in dB. Substituting in the values from our equation, we get 10*log[(Vo^2)/(Vi^2)] = 10*log[(200^2)]. Simplifying further, we get 20*log(Vo/Vi) = 20*log(200). And since the power gain is expressed in dB, we can drop the factor of 20, leaving us with the final equation for power gain in dB: 20*log(Vo/Vi) = 20*log(200).

Therefore, the power gain in dB for a voltage amplifier with a ratio of 200 between output voltage and input voltage would be 20*log(200) = 46.02 dB. This means that the output power is 46.02 dB higher than the input power, or about 200 times greater.