I know that my problem with this is some flaw in my logic, I just need someone to point out to me where that flaw is. According to the Wikipedia article, there are different formulas because power gain is proportional to the square of the voltage (amplitude) gain. I can accept that and it makes sense except that the formula for voltage gain is only true when the input impedance is equal to the output impedance. That contradicts what I think I know about gain stages - that is, that the input impedance should be very high and the output impedance should be very low in order to drive the next stage.

So my question is, what's wrong with the way I'm thinking about this? And does the formula for voltage gain in decibels change when input impedance does not equal output impedance?

  • \$\begingroup\$ Like at @olin says - don't get worried about the impedances - it's like mach 1 and mach 2 - only relevant at sea level and nobody really cares about what the actual speed of sound is at 40,000 feet (except those who do of course but most folk don't) \$\endgroup\$ – Andy aka Apr 21 '14 at 20:08
  • \$\begingroup\$ I read that question and all the answers prior to asking my question. I don't think this is a duplicate because nobody mentions the impedance or the disappearance of its terms in the formula. \$\endgroup\$ – nonex Apr 21 '14 at 23:38

Deci-bels always describe a power ratio. Power of a electric signal is proportional to the square of the voltage, and inversely proportional to its impedance, as you say.

However, lots of times the impedance isn't known and we still want the useful properties of expressing the gain between two signals in dB. In that case, we make the simplifying assumption that power is just proportional to the square of the voltage, ignoring differences in impedance.

That could be considered cheating in a strict sense, but there are cases where the impedance isn't that relevant and what you care about is the voltage gain. Or, the impedance is reasonably standardized between the two signal, so that it wouldn't matter even if it was taken into account. In another case, the impedances are all known to be low enough so that under normal use the voltages won't change due to loading.

Yes, these are simplifying assumptions, and you have to look carefully at what is really being said to decide how dB is being used. With some experience, you will get used to which context use just the voltage squared, and which use true power delivery capability.

  • \$\begingroup\$ 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. \$\endgroup\$ – supercat Apr 21 '14 at 20:08
  • \$\begingroup\$ 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. \$\endgroup\$ – nonex Apr 21 '14 at 21:58
  • \$\begingroup\$ @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. \$\endgroup\$ – Olin Lathrop Apr 21 '14 at 22:01

Although the decibels unit apparently first arose to describe a power ratio, and is still used that way, it has since also become well established as a separate unit to describe a voltage ratio, independent of the impedances involved. This is discussed in the Wikipedia article:

The IEC permits the use of the decibel with field quantities as well as power and this recommendation is followed by many national standards bodies, such as NIST, which justifies the use of the decibel for voltage ratios.

For the "field quantity" use of decibel, such as for voltage, the "20" form of the equation is used: 20 log (V1/V2).

If two signals are compared, and the impedance seen by them is equal, then the voltage ratio as reported in dB units will be the same as the power ratio in dB. But it is fine to use dB to report a ratio of voltages where the impedance is different and hence the power ratio in dB would not be the same as the voltage ratio in dB. Obviously it's important to be clear in the surrounding description which ratio is being reported.

(The terminology "field quantity" was found misleading, and was revised to "quantity whose square is, in general, proportional to power".)

To be clear, the answer I'm giving here disagrees with Olin's in the sense that his regards the voltage ratio use of dB as a side effect of the power meaning of dB plus the assumption, even if not true, that the two voltage signals are in the presence of equal impedances. In contrast, my answer regards the power and voltage dBs as two separate and equally legitimate units with their own distinct rules for application.

  • \$\begingroup\$ Thanks for your answer. So is it fair to say that sometimes the two formulas equivalent and sometimes they are used in different scenarios and aren't equivalent? \$\endgroup\$ – nonex Apr 21 '14 at 23:40
  • \$\begingroup\$ The two formulas describe two different units, one reporting a ratio of power, and the other reporting the ratio of voltage. (Or other variables in other areas of engineering, such a sound.) For P and V uses of dB, the numerical result is equal if the impedances of the two compared signals are equal. \$\endgroup\$ – gwideman Apr 21 '14 at 23:52

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