Timeline for I'm confused by the formulas for power gain decibels and voltage gain decibels
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 21, 2014 at 22:01 | comment | added | Olin Lathrop | @nonex: It depends on what matters to you or the context. Yes, the output impedance of a opamp amplifier circuit is usually lower than its input, and in a strict sense you have to take that into account when computing dB gain. However, there are many cases where the impedance is low enough so that it doesn't matter, and all you really care about is voltage gain. In that case, you'd simply use 10x the log of the square of the voltage gain (or 20x the log of the voltage gain, same thing). It all depends on context. | |
| Apr 21, 2014 at 21:58 | comment | added | nonex | Thanks @olin, this helps explain it. I can take it on faith for now but still have a related question. If I take for example a non-inverting op-amp circuit and set the voltage gain using the feedback voltage divider, do the impedances still not make a difference in the decibel calculation? Because if I understand correctly, the input impedance would be much greater than the output impedance. | |
| Apr 21, 2014 at 21:47 | vote | accept | nonex | ||
| Apr 21, 2014 at 20:08 | comment | added | supercat | If two signals are driving the same resistive impedance, then the voltage ratio will be the same as the current ratio, and the power ratio will be proportional to the square of the voltage ratios. In many cases, what is being compared are signals that fed to a non-changing load at different times, so impedance is guaranteed to be equal. Showing that impedances are equal will make their exact values irrelevant, but that's not the same thing as "ignoring" them, since one must still demonstrate that the impedances are, in fact, equal. | |
| Apr 21, 2014 at 19:25 | history | answered | Olin Lathrop | CC BY-SA 3.0 |