# Why is power gain a meaningful quantity at all?

I'm quite confused at a very basic concept: power gain.

An amplifier can be said to have a power gain of x decibels. The gain of an amp in decibels is never voltage or current, it is always power. But let's say I was given some black box amplifier with some power gain. Since I'm only given the power gain, I don't really know the output voltage of the amplifier. I can feed the amp with some known input voltage, and given the input impedance of the amp I can calculate the input power and therefore the output power. But aren't we usually more interested in how much output voltage we are getting from the amp?

For example, if I wanted to use the amp to amplify some weak audio signal to feed it to some speakers. The amplitude of the speaker vibration is not really a function of how much power is dissipated in the speaker (let's assume a simple dynamic coil speaker), but rather how much current is flowing through it (right?).

I find it quite difficult to express my confusion as a question but I hope somebody understands.

• amplitude reflect power more than current because different speakers have different resistances, so 80V can be 10W or 20W. – dandavis Oct 5 '20 at 18:44
• "An amplifier can be said to have a power gain of x decibels. The gain of an amp in decibels is never voltage or current, it is always power." No it's not. Audio amplifier always specify voltage gain and for most amplifer it's adjustable anyway. They specifiy MAX power but not power gain. Where did you read that ? – Hilmar Oct 5 '20 at 19:19
• In audio, voltage gain is often more relevant since the input impedance is high and output impedance is low and impedance matching is not done. In RF it is more convenient to work with power gain because it simplifies link budget calculations, etc. It is understood most of the time in RF that everything is 50 Ohms, so power levels (dBm and such) can be easily converted to voltage if need be. – mkeith Oct 6 '20 at 1:27

The gain of an amp in decibels is never voltage or current, it is always power.

Nope. A 30 dB power gain is a 30 dB amplitude gain. See: how decibels are defined for amplitude and power quantities (namely: differently, so that this works).

But aren't we usually more interested in how much output voltage we are getting from the amp?

That fully depends on what you're doing, so I'll go with a "no, not usually".

The amplitude of the speaker vibration is not really a function of how much power is dissipated in the speaker (let's assume a simple dynamic coil speaker),

of course it is!

but rather how much current is flowing through it (right?).

which is the same thing, but square-rooted (and proportional).

Also, for you as human listener, the amplitude of the membrane movement isn't more important than the power; both are linked inherently, and with human perception being logarithmic to intensity, whether or not we square the current or not makes only a difference of a constant factor. So, you're wrong to put emphasis on that!

• This is only true for input Z = load Z. A unity gain buffer can have power gain. – user_1818839 Oct 5 '20 at 18:35
• @BrianDrummond you're right, I brain-slipped there. Changed to "amplitude gain", because that captures it, no matter whether we're looking at voltage through the same Z or current through a lower Z. – Marcus Müller Oct 5 '20 at 18:36
• 30 dB power gain is 30 dB amplitude gain. But if you start specifying amplitude gains in dB when the input and output don't have the same impedance you're going to confuse people, and the 20*log10(Vout/Vin) formula won't be correct any more. I'd much rather learners were learning "a gain in dB specifies a power ratio" than "dB are dB, for both power and amplitude". – The Photon Oct 5 '20 at 20:16

The amplitude of the speaker vibration is not really a function of how much power is dissipated in the speaker (let's assume a simple dynamic coil speaker), but rather how much current is flowing through it (right?).

Nope. For a given speaker impedance this is true, yes. But a $$\4\Omega\$$ speaker has half the power output for a given current (and twice the power output for a given voltage) than an $$\8\Omega\$$ speaker. This is why car stereo speakers are often $$\4\Omega\$$, and sometimes $$\2\Omega\$$ (or why you'll see two $$\4\Omega\$$ speakers in parallel). It's so you can get the power you want within the capabilities of the car's 12V rails.

But let's say I was given some black box amplifier with some power gain. Since I'm only given the power gain, I don't really know the output voltage of the amplifier.

This is true -- and if you're given an amplifier that only specifies the power gain, then either it's underspecified, or it's being sold into a market where the input and output impedances are understood.

But aren't we usually more interested in how much output voltage we are getting from the amp?

Not entirely. If you go shopping for speakers -- or for motors -- you'll find that for roughly the same size and expense of the actuator, the power handling capability is the same. However, within a given basic electromagnetic actuator design, there's usually some (sometimes considerable) freedom to choose the operating voltage.

Also - roughly - a power amplifier that's a given amount of difficult to design and build (and expensive to buy) can have a wide range of operating voltages that it can be designed for.

If you're building something to shove the physical world around, whether it is by emitting light or sound or turning motors, you care about power first -- so you care about power gain. Only after you have that part satisfied do you get down to finer detail like supply voltages &c.

• a very fine answer hightlighting the acoustic/physical aspects of this. I prefer this over my own answer! – Marcus Müller Oct 5 '20 at 19:06

Whether power amplification, voltage amplification or current amplification matters is a question of the amplifier's application.

Different applications require different ways of specifying the amp. It matters whether you are driving a motor, an antenna, a pair of audio speakers, LED lighting, piezo transducers, etc...

I'll attempt your question from the perspective of an audio amplifier, since that seems to be your primary concern:

An amplifier can be said to have a power gain of x decibels. The gain of an amp in decibels is never voltage or current, it is always power.

An audio amplifier is sometimes called a "power amplifier" but that is a misnomer. Usually the amount of power entering the amplifier is no factor in determining the output power or voltage.

What matter with audio "end stage" amplifiers is how much power is delivered into a specified load of, say, 8 Ohm or 4 Ohm.

The input voltage level that produces that maximum output power is often also specified, so that the pre-amp or the source output signal levels can be set to deliver the desired output power.

Since I'm only given the power gain, I don't really know the output voltage of the amplifier. I can feed the amp with some known input voltage, and given the input impedance of the amp I can calculate the input power and therefore the output power.

The input impedance matters only in order to determine the input voltage once it is connected to a source or a pre-amp output. It does not matter what input power that input voltage corresponds to.

With an input impedance of, say, 10 kOhm, any pre-amp with an output of 100 Ohm to 1K Ohm will be fine as the loading by the input impedance will have little effect on the voltage level.

But aren't we usually more interested in how much output voltage we are getting from the amp?

Yes and no. We are actually interested in how much power an amp can deliver into a target load. Once we know the target load, we can determine the voltage required over that load. Once we know that required output voltage we can calculate the voltage gain based on the input voltage level that would produce that required output voltage.

I chose the word "level" to once more emphasize that with audio amplifiers the amount of power delivered at the input does not matter. What matter is the signal level, expressed in V.

Take the following example from an amplifier spec sheet:

and

One of the quoted models, model "800" produces 200W for an 8 Ohm speaker load.

The voltage over that load, at 200W, will be 40V.

Also according to the specs below, the input required to produce a 200W sinus is 1.4 V.

The voltage gain is 40V / 1.4V = 28.6 = 29dB (coincidentally, the factor and the dB number are close, so don't be confused)

That matches the spec above, quoting it at "29dB" voltage amplification. with audio amplifiers, that number is tightly connected to the specified load, 8 Ohm in this calculation.

Change the load, and the power will change, because the amplifier will deliver a different amount of power into a different load. Current and voltage will both change. This is the topic of load matching, a.k.a impedance matching.