G
George Herold
Guest
On Friday, April 17, 2020 at 9:41:42 AM UTC-4, amdx wrote:
Yes, 20*log(V1/V2), if you must.
GH
On 4/17/2020 8:15 AM, blocher@columbus.rr.com wrote:
On Friday, April 17, 2020 at 9:10:13 AM UTC-4, amdx wrote:
On 4/17/2020 7:49 AM, Pimpom wrote:
On 4/17/2020 5:32 PM, amdx wrote:
  The fact that I know: a 6db voltage gain is a 3db power gain.
If that is wrong set me straight.
That's where you got it wrong. The same ratio - e.g. 3:1 - expressed in
dB for voltage is twice that for the same power ratio. This convention
arises from the assumption that the two voltages are applied to the same
resistance/impedance, which is not always the case and can create
confusion.
Mathematically, for voltage the dB value is 20log(V1/V2) whereas for
power it's 10log(P1/P2)
Take an amplifier with a gain of 5x and an input voltage of 1V. By
convention, the voltage gain is 20log(5/1) = 14dB.
If the input resistance of the amplifier and the output load resistance
are both 100 ohms, the input power is 1²/100 = 0.01W; output power is
5²/100 = 0.25W. The power ratio is 10log(0.25/0.01) = 14dB.
Same amplifier, same gain.
Voltage gain = 5 or 14dB
Power gain = 25 or 14dB
If the input and output resistances are *not* the same, the gain in dB
will NOT be the same for voltage and power.
I am assuming the same impedance for all calculations.
RE: Voltage gain = 5 or 14dB
Is the voltage gain of 5 expressed as a 14db power gain?
Meaning, when you calculate gain using a voltage ratio, is the answer
in units as power gain?
That would solve my confusion.
Thanks, Mikek
We are trying to teach you to fish and I think you want us to toss you a fish
I do know the 10log and 20log formula, but a simple yes/no to the
20log(V2/V2) Is, the answer expressed in power.
At this point, I think it is, otherwise, why the 20log.
Sorry to be dense, but if I didn't need help, I wouldn't be asking.
Mikek
Yes, 20*log(V1/V2), if you must.
GH