Power, voltage and current gain

1. Jan 18, 2014

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!

2. Jan 18, 2014

Staff: Mentor

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

3. Jan 18, 2014

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?

4. Jan 18, 2014

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.

Last edited: Jan 18, 2014
5. Jan 18, 2014

Bromio

I don't understand it.

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

6. Jan 18, 2014

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.