Power, voltage and current gain

  • Thread starter Bromio
  • Start date
  • #1
62
0
Hi.

I know [itex]G_P (dB) = 10log(G_P)[/itex] and [itex]G_V (dB) = 20log(G_V)[/itex] and [itex]G_I (dB) = 20log(G_I)[/itex].

I also know that [itex]G_P (dB) = G_V (dB) + G_I (dB)[/itex].

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

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

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

What's wrong?

Thanks!
 

Answers and Replies

  • #2
berkeman
Mentor
60,005
10,208
Hi.

I know [itex]G_P (dB) = 10log(G_P)[/itex] and [itex]G_V (dB) = 20log(G_V)[/itex] and [itex]G_I (dB) = 20log(G_I)[/itex].

I also know that [itex]G_P (dB) = G_V (dB) + G_I (dB)[/itex].

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

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

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

What's wrong?

Thanks!

Where did your expressions 10^(20/20) come from? That is not part of the definition of dB, AFAIK.
 
  • #3
62
0
Thank you for answering.

They come from [itex]G_V (dB) = 20log(G_V)[/itex]. If [itex]G_V(dB)= 20 dB[/itex], then [itex]G_V = 10^{G_V(dB)/20} = 10^{20/20} = 10[/itex], isn't it?
 
  • #4
1,763
59
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:
  • #5
62
0
I don't understand it.

Isn't the formula [itex]G_P (dB) = G_V (dB) + G_I (dB)[/itex] always valid?
 
  • #6
1,763
59
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.
 

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