How to calculate the expected output current being drawn by a speaker?

Question

I'm trying to calculate the current that will be drawn by a 4ohm 3W speaker from a MAX98357a IC amplifier. From my understanding, the amplifier takes in a I2S audio signal (which originates from an ESP32), converts this digital signal to analog then passed through an amplifier with a programmable gain that will be outputted directly to the speaker.

From this I don't fully understand how I can calculate the expected current that will be drawn by the speaker. It is important since this circuit will be powered from a 2500mAh 5V lipo battery and needs to run for a reasonable amount of time.

By searching through the internet, I can see that power output is the important factor in this calculation. Looking at the amplifier IC datasheet, I can see that the output signal level can be calculated by the below equation.

Output signal level (dBV) = input signal level (dBFS) + 2.1dB + selected amplifier gain (dB)

From another stack exchange question, assuming the input signal uses the maximum range, it will result in 0dbFS and choosing a gain of 12dB, the output will result in 14.1dBV. With this number and using the decibel voltage equation 20log10(V/Vo) or 20log10(V) = dbV, the output voltage that will be seen from the speaker is 5.06V.

Using that number and I=V/R I can calculate that the expected current draw will be around 1.26A but that seems to simple and ignores a lot of other things. Overall this all confuses me and don't have any clear ideas on how to tackle this problem.

Any explanation on the principles of how this circuit will play out is best or just any guidance to any other material I can read on.

P.S. This is my very first post on Stack Exchange so sorry for any mistakes in advance.

Answer 1

There is no way to calculate that, because it will depend on volume and what music you play.

There is no audio you can play that would make the speaker consume 1.26A as that would require flat DC, and that is not audio. Generally, expect that when playing the peaks could go up to 1.26A, but you should avoid that as you don't want the audio to clip.

Answer 2

When considering current drawn by a loudspeaker, there are two very different aspects to consider:

1- Maximum current

The speaker is a reactive load with a wiggly impedance curve, especially if it is a multiway speaker with a crossover. Current will not be in phase with voltage, there are impedance dips, and some signals may draw current surprisingly larger than just "max voltage divided by 8 ohms". This is important to make sure the amplifier doesn't run out of available current, or pop an output transistor, or triggers its overcurrent protection which would sound terrible. Also to set your volume and gain to make sure you have enough to reach clipping, or you'll waste headroom.

This is also important for your battery powered use case, because you don't want the BMS to trip on current peaks.

2- Average current and power

Properly recorded, mixed and mastered music has quite high dynamic range: peaks are be 10-20dB above average power.

Maximum amplifier power (also max voltage and current) determine how loud the peaks can be. Crank the volume higher and the peaks will clip, if that doesn't happen too often it will sound okay as long as the speaker doesn't bottom out.

Average power will be a lot lower than this. That makes it very difficult to calculate how much electrical power the amp will use though, because music power fluctuates in such a wide range.

If you listen to music that has been compressed to death as a result of the loudness wars, then it will pretty much draw constant power, but it sounds so bad you'll turn the volume down, which means... it'll use about the same power as normal music.

The only case where a speaker will operate at full power on a near constant basis would be a subwoofer driven hard.

In fact, the main factor is loudspeaker efficiency, ie how loud it goes with one watt. So if you find your battery life to be too short, the solution could be a bigger battery, or it could be a more efficient speaker, or a larger one that produces more bass without needing more watts.

If you want to know how much current it uses on actual music, you have to measure it.

Answer 3

In theory, your battery pack has 12.5 Wh capacity, so factor your average RMS power (W) and divide by efficiency to estimate hours. As a Class D amp, it shall last longer with effic. So estimate 1W average and expect 12h.

Measure DC current with a 50mV shunt on ground side then test and verify performance. Speaker efficiency will affect results, so I might choose a 12” full spectrum speaker. This can be loud with 1W as I have done circa ‘70 with Philips speakers.

Answer 4

You care about battery current and not speaker current, so why do you want to calculate the speaker current? It won't help you all that much. When it comes to audio, it's much simpler to look at power, rather than current. Suppose that the 3W speaker will be operating at constant full power. You then look at the worst-case efficiency of the amplifier: 92% nominal into 8 Ohms, so let's say 85% worst case into 4 Ohms. Thus the amplifier will be consuming 3W/85% = 3.5W.

LiPo batteries don't provide 5V. The nominal output voltage is 3.7V. So, if you have a "pack" that is specified as "2500mAh 5V", then that pack really has a 3.7V cell and a step-up converter with, say, 85% system efficiency at best. It's not clear whether the rating on the pack is 2500mAh at 5V, or is that the rating of the cell inside the pack. Usually the packs are labeled in a stupid way where the mAh is for the raw battery and doesn't include the actual charge you can extract at 5V. It will be necessarily less because the V*charge product must remain constant: voltage goes up, charge goes down - even if everything is 100% efficient - it of course isn't.

So, suppose that the pack is rated in the typical stupid way, so the battery itself holds 2500mAh*3.7V = 33kJ (about 8 kcal, or about 1/30th of a glazed donut's worth of energy). Due to losses at the step-up converter and in the battery's internal resistance, only 33kJ * 85% = 28kJ is available from the output into your circuit.

Assuming that the amplifier consumes most of the power, we can ignore other components. So, a 28kJ energy source can feed a 3.5W load for about 8000s or 130 minutes or 2h15'. That's with the speaker going full blast at its rated 3W power.

That's the estimate you were probably looking for. To get a better estimate, you'd need to measure the average power (in Watts) consumed by subsystems other than the amplifier, and also run some test audio signal at a "reasonable" or "expected" volume and see how much power is really consumed by the amp. That will allow you to refine the estimate.

You'll notice that all that was needed here is elementary math and algebra of units, so you don't even need to know the logarithmic (decibel) scales etc. Easy stuff :)