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I'm using two Visaton FRS 5X-8 Speakers in a project. The amplifier being used is a Dayton KAB-215. Powering the whole thing from my power supply at 12V, I see a current draw of ~50 mA even at full volume. I'm a bit confused by this result.

My math suggests that for the two 5 W RMS (8 W Max) speakers I should see around 10 W of power draw or ~0.8 A @ 12 V. While I realize they may not operate continuously at this power during music, there is a big spread between 600 mW and 10 W.

It seems like both the power supply (150 W) and the amplifier board (2× 15 W @ 8 Ω) have tons of headroom here, so I don't really understand the low power draw. The Bluetooth board doesn't have a volume knob on it currently, which the datasheet indicates should result in "full volume" on the amplifier side. My phone volume is maxed (when testing), and yet the power draw is super low. The speakers are fairly loud but not insanely so. I do have SPL measuring equipment that I can use to try to validate if the output is close to what it should be, but I'm still a bit perplexed.

I look forward to your thoughts on what is limiting the current draw, and whether this is a surprising result or not.

Update 1

I tested this with a couple other devices, and agree that it my phone Bluetooth output is quite low.

Source:
Galaxy S10, Via Bluetooth, "Audio Tools" App 100 Hz Sine Wave

Drivers:
2 Visaton Drivers, one in a 3D printed enclosure, one with no enclosure

Power Supply: In constant voltage mode at 12V, supplying power to both speakers
Voltage: 12 V
Current: 0.313 A

Single driver measurements + output:
Voltage (measured with true RMS meter in AC mode): 4.1 V
Driver impedance from datasheet: ~22 Ω
Volume: 81 dB @ (roughly) 1 m, measured via the same phone app

I'm not quite sure how to translate the AC voltage at the speaker into wattage, given that the impedance is likely complex (right?). I read the EE post Max Voltage for Speaker of Given Power Rating, but didn't gain a lot of insight from it.

The more I work on this the more I realize I don't understand, but I suppose that's part of the process. I also have a better measurement mic/setup, but I feel like I'm chasing bigger issues rather than small optimizations. I'm going to trace the graph, put it into WinISD and see what it expects from an SPL standpoint. I think that probably I just don't have that much input voltage from the Bluetooth chip, and that everything else is behaving as expected, but I also don't understand the system well enough to conclude that.

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    \$\begingroup\$ Get a tone generator app for your phone, inject a low frequency that your meter will be capable of measuring - 100 or 200 Hz, say - and measure the current into and voltage across the loudspeaker. Update your post with the results. \$\endgroup\$
    – Transistor
    Commented Jan 18, 2020 at 21:56
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    \$\begingroup\$ There are a lot of unknown variables here. You cannot assume that using 8 Ohm speakers, 12V power supply, and your phone on full volume, playing who knows what, will drive 5 watts into each speaker. You don't know what gain the D-class amp chip is set to, and you don't know the total gain from bluetooth receiver to D-class amp input. The board can use a wide range of supply voltages (12-24), and have some headroom for people who put 4 ohm speakers there (the amp chip can do it, but not necessarily this board). Chip is also not capable of more than 10W per channel at 13V supply. \$\endgroup\$
    – Justme
    Commented Jan 18, 2020 at 22:20
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    \$\begingroup\$ "My math suggests that for the two 5W RMS (8W Max) speakers I should see around 10W of power draw" - was it loud enough to make your ears ring? If not then it wasn't at full power (10W into these speakers would be 96dBSPL at 1 meter). sengpielaudio.com/PermissibleExposureTime.htm \$\endgroup\$ Commented Jan 19, 2020 at 1:27
  • \$\begingroup\$ @Transistor, Just did, thanks. I've tried to control more variables this time and added the measurements. \$\endgroup\$ Commented Mar 7, 2020 at 18:52

3 Answers 3

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My phone volume is maxed (when testing), and yet the power draw is super low.

If you're using headphone out, that is usually less than line level (and in the EU a lot less than line level), so unless you have some make up gain somewhere in that system you won't get anywhere near the max output.

Test again with the line out from a CD player or similar device.

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Design Flaw

The Class D amp works best with a small battery or massive capacitor > 1 Farad with low ESR. A small Li-Ion 18650 cell can have 10 kF then reduces with cells in series. Any time you put a boost DC-DC regulator on a Buck regulator source, you can expect problems unless there is a good large C low ESR buffer in between. A class D amp is a variable DC-AC inverter and has similar demands on a low source impedance.

Alternatively, a supply that can drive 10× the load you expect so you don't need to rely on large capacitors or a small battery. Try it on an HDD port from a PC with 12 V.

There are many links to advance your understanding of concepts and test methods in my profile.

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  • \$\begingroup\$ The graph for the tiny speaker shows 10 ohms at 100Hz, not 22 ohms. The datasheet for the amplifier with a 12V supply shows full output is 8.1V RMS because the output is bridged, not 4.1V. Then the power in one speaker is (8.1V squared)/10 ohms= 6.6W and the amplifier current will be a little more than 6.6V/12V= 0.55A. The power supply current will probably be 0.55A x 1.11 (because the amplifier class-D efficiency is 90%)= 0.61A. You will not hear much because the tiny speaker can barely produce 100Hz. \$\endgroup\$
    – Audioguru
    Commented Dec 19, 2021 at 3:07
  • \$\begingroup\$ The tiny speaker is a shrieker that produces no bass. \$\endgroup\$
    – Audioguru
    Commented Aug 24, 2022 at 3:37
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There are too many unknowns here, you need to control the variables.

1. Load.
Replace the speaker with a resitive load, ideally 8Ω. Difficult to get a 10W 8Ω resistor, so try many parallel resistors, eg: 15 x 120Ω 1W resistors will give you a robust 8Ω 15W load, and 59 x 470Ω 1W resistors will be a good 50W load. I've seen these arrangements mounted in a coffee jar filled with vegetable oil (low conductivity) to help remove the heat from the resistors. This will also save your ears, and not annoy your roommates and neighbours.

2. Input signal.
Use a signal generator. Many phone apps for this. Use a standard signal, eg: sine-wave, 1kHz, and increase amplitude while watching the output voltage (across the load resistor) until either the target power level is reached, or the onset of clipping occurs.

3. Measurement of output power.
Infer output power from the measurement of output voltage. Use a scope if available. Be aware that the output negative may not always be connected to earth, so use a probe on both terminals of the output port and subtract them (scope math function useful for this). Otherwise, use a circuit to convert the AC voltage across the load to a form suitable for measuring with a simple volt-meter, eg: an RMS converter. The link below will give you some good circuits for this purpose:
https://sound-au.com/appnotes/an012.htm

Convert output voltage to output power by the simple formula: P=Vrms^2 / R.

4. Measurement of input power.
Place a very stable voltage source, such as a battery, or a large capacitor bank, between the DC power supply and the amplifiers input power port. Measure voltage and current simultaneously.

Input voltage: Place the probes for input voltage very close to the input port of the amplifier. Observe this on a scope to confirm voltage is stable during the test. If stable, then this can be measured with simple DC voltmeter.

Input current: Measure input current using a "burden resistor" placed in the negative line at the output of the power supply, located between the DC power supply the battery (or cap bank), to get as much filtering of the signal as possible. The voltage drop across the current measurement device must be low, less than, say 2% of the supply voltage.

Even with the extra energy storage at the power amp input, the current can be highly dynamic with the input signal, so the signal from the current burden resistor should also go via a precision averaging/RMS circuit, similar to what is used to measure RMS voltage at the output.

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