Power, voltage and current gain

[Bromio]

Hi.

I know $G_P (dB) = 10log(G_P)$ and $G_V (dB) = 20log(G_V)$ and $G_I (dB) = 20log(G_I)$.

I also know that $G_P (dB) = G_V (dB) + G_I (dB)$.

So, if I have a common base amplifier whose current gain is $G_I (dB) = 0$, then $G_P (dB) = G_V(dB)$.

Suppose
$G_V (dB) = 20 dB$. So $G_V = 10^{20/20} = 10$
$G_I (dB) = 0 dB$. So $G_I = 10^{0/20} = 1$
As I've written above, then $G_P (dB)= G_V (dB) = 20 dB$. So $G_P = 10^{20/10} = 100$.

But, in linear scale (not in dB), $G_P = G_V\cdot G_I$, so $G_P = 10\cdot 1 = 10$, which is not $100$.

What's wrong?

Thanks!

 

[berkeman]

Hi.

I know $G_P (dB) = 10log(G_P)$ and $G_V (dB) = 20log(G_V)$ and $G_I (dB) = 20log(G_I)$.

I also know that $G_P (dB) = G_V (dB) + G_I (dB)$.

So, if I have a common base amplifier whose current gain is $G_I (dB) = 0$, then $G_P (dB) = G_V(dB)$.

Suppose
$G_V (dB) = 20 dB$. So $G_V = 10^{20/20} = 10$
$G_I (dB) = 0 dB$. So $G_I = 10^{0/20} = 1$
As I've written above, then $G_P (dB)= G_V (dB) = 20 dB$. So $G_P = 10^{20/10} = 100$.

But, in linear scale (not in dB), $G_P = G_V\cdot G_I$, so $G_P = 10\cdot 1 = 10$, which is not $100$.

What's wrong?

Thanks!

Where did your expressions 10^(20/20) come from? That is not part of the definition of dB, AFAIK.

 

[Bromio]

Thank you for answering.

They come from $G_V (dB) = 20log(G_V)$. If $G_V(dB)= 20 dB$, then $G_V = 10^{G_V(dB)/20} = 10^{20/20} = 10$, isn't it?

 

[skeptic2]

The voltage gain formula GV(dB)=20log(GV) assumes the input and output impedances are the same. If the impedances are the same, then a 10 times increase in voltage will result in a 10 times increase in current and 10 x 10 = 100. In your example you specify a current gain of 1 or 0dB. The requires that the output impedance be 10 times that of the input. That increase in impedance reduces the power gain to 10.

 

[Bromio]

I don't understand it.

Isn't the formula $G_P (dB) = G_V (dB) + G_I (dB)$ always valid?

 

[skeptic2]

No, it's only valid if the input and output impedances are the same. Your example proves it's not valid if the impedances are different.