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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.

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  • \$\begingroup\$ You care about battery current and not speaker current, so why do you want to calculate the speaker current? ;) \$\endgroup\$ – Kuba hasn't forgotten Monica Jan 16 at 0:45
  • \$\begingroup\$ With a 5V supply and a 4 ohm speaker, the datasheet for the amplifier shows maximum undistorted RMS power of 2.5W. If you play a continuous tone at 2.5W then your hearing and your battery will not last long. Speech or music is played with an average power of 1/10th to 1/20th the max so the battery will provide 0.125W to 0.25W plus the amplifier's efficiency. But nobody makes a 5V Li-Po, it is an average of 3.7V or 7.4V. \$\endgroup\$ – Audioguru Jan 16 at 1:18
  • \$\begingroup\$ @Audioguru yeah you're right I meant to put 3.7V for the battery. I have a booster though that is used to power an ESP32 and also provide the voltage supply for the amplifier at 5V. So from what I understand if outputting speech the worst case of 0.25W RMS and an amplifier efficiency of 85%, the power that the speaker will be drawing will be 0.2125W? \$\endgroup\$ – Ricky Hernandez Jan 16 at 20:47
  • \$\begingroup\$ Your amplifier with an average output of 0.25W has heating of (1 / 0.85) x 0.25= 0.044W so the total average 5V current will be 58.8mA. Boosting the voltage results in boosting the battery current, plus a voltage booster uses some current all the time. \$\endgroup\$ – Audioguru Jan 18 at 1:25
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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.

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  • \$\begingroup\$ Yeah I realize the audio signal will determine the amount of power being outputted and as long as I keep the volume relatively low I can get a reasonable amount of current being drawn. I just got to make sure the audio signal peaks don't exceed the amplifiers maximum ratings I assume. \$\endgroup\$ – Ricky Hernandez Jan 15 at 22:38
  • \$\begingroup\$ Like many amplifiers, this one has two power ratings: with horrible-sounding 10% of clipping distortion so that the power number is high or a power where the distortion is less at 1% but is still heard. \$\endgroup\$ – Audioguru Jan 19 at 20:00
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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.

enter image description here

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.

enter image description here

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.

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  • \$\begingroup\$ Thank you @bobflux I think I got a grasp on what you're saying. So it really comes down to the audio signal and determining if the max power of the amplifier can handle the peaks. In theory if I wanted to get an estimate of the current drawn, I'd have to know the average audio range and the resulting average gain to get a rough estimate of the current? I'm aware a speaker is a reactive component so its not a straight I=V/R but in general that could work right? \$\endgroup\$ – Ricky Hernandez Jan 15 at 22:35
  • \$\begingroup\$ Yes you got it right. Also since our ears have logarithmic sensitivity, subjectively "twice as loud" requires a lot more actual watts than 2x, so the volume setting will matter a lot. \$\endgroup\$ – bobflux Jan 15 at 23:09
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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.

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    \$\begingroup\$ OK thank you that makes sense! \$\endgroup\$ – Ricky Hernandez Jan 15 at 22:41
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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 :)

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  • \$\begingroup\$ Yes you're correct that I have a step up converter. I forgot about that and stupidly said it was the battery that is rated 5V but the converter does power an ESP32 and the speaker. The worst case of a 3.5W load is important for my calculations but your second option of measuring the power dissipation using an expected audio volume would probably be the best. Thank you for the advice! \$\endgroup\$ – Ricky Hernandez Jan 16 at 20:54
  • \$\begingroup\$ Nobody except acid rock players use an amplifier that is at full blast and is clipping like crazy all the time. \$\endgroup\$ – Audioguru Jan 18 at 1:32
  • \$\begingroup\$ @Audioguru Who said the amplifier would be clipping? P = U^2/R, so U = sqrt(P/R). For a 4 Ohm speaker, the input into the speaker needs to be about 0.9Vrms, and the class D amplifier used by the OP, running from 5V, will have no trouble with that. As for how said speaker would behave being driven at 3.5W - it's hard to say. Good speakers would be rated such that they provide about the same distortion at 1/10th rated power as they do at the rated power. A cheap little speaker of course won't perform anywhere near so well, so it'll sound even crappier than it normally does - I agree :) \$\endgroup\$ – Kuba hasn't forgotten Monica Jan 18 at 18:11

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