PAM8610 amplifier - speakers and power, a bothering question ...

Hello, I'm going to build a portable speaker consisting of an amplifier on the PAM8610 chip and two 15W 8? speakers. In the amp's note it says that the maximum output power is 2x15W with 15W 4? speakers, and 10W 8?, it means that if I connect my two 15W 8? speakers, I will get 10W and 8? amplifier power, yes? Forgive me for such a banal question, but as they say, "No stupid questions .." Thank you in advance for each answer.

From this chip with a supply voltage of 14V, you can get 11W / 8ohm at THD = 1%, less than 15W at THD10% and you have to take into account that with this output power, the loss power is about 5-6W, so you need to equip the circuit with a small heat sink. Max supply voltage 15V during operation, 16V during rest. From what I see at the beginning of the note, 13V and 10W / 8ohm at THD = 10% are given, although the characteristics show 12W, it is probably caused by a voltage drop.
And the answer to yours above is 2x 10W / 8ohm with 1 speaker 8 ohms per channel. it should be understood that it is a two-channel amplifier.

A two-channel amplifier, if I connect 2 speakers, I will have 10W on each, with a power supply of about 11.1V, right? (cells from a laptop battery connected in series 3x assuming a voltage of 3.7V)

At 11.1V, the power will drop to 6.5W 1% and 8W 10% THD, but with a fully charged 8W-11W package with a discharged 5W wheel. However, it is also worth using step-up converters to increase the voltage to a constant value and to keep the power at a constant level while using the entire package capacity. Hope you are aware of the S3 serial packet loading and related problems.

Could you speak a little easier? That's at 11.1V and with 2 speakers, 6.5 W on each and a loss of 40dB or 6.5 on one and 8 on the other, I really don't understand: /

THD - Total Harmonic Distortion, i.e. harmonic distortion power with 1% harmonic distortion and 10% harmonic distortion of the output signal. You can google these things! And the voltage on the packet will vary between 8V discharged and 12.6V fully charged, so it is worth using a lifting converter, keeping the voltage at one level, e.g. 13V. I didn't say anything about the loss in dB, so I don't know what you mean, although it will appear very audible while listening and gradually discharging the package without the converter. To charge the S3 packet, you need balancers for each link / parallel packet in series.

I see no problem with charging the cells, I take each one out, connect it to the charger and that's it. In my opinion, it will be fine without the converter, only when discharged the speaker will sound quieter. This is how, I can go ahead and connect 2 15W 8? speakers to this amplifier, and I will get each speaker at 14V, as you wrote, will be 11W / 8?. Unless you've read about a different amplifier and not the one I mean.

maximum output power 2 x 15W (stereo)
2 x 10W into 8? load
2 x 15W into a 4? load

Oh god, I wrote it all down. You will receive

udon_bukakke wrote:

At 11.1V, the power will drop to 6.5W 1% and 8W 10% THD, but with a fully charged 8W-11W package with a discharged 5W wheel

8-11W with a charged, similar to the 6.5 and 8W and 11.1V power supply, I did not write lower THD 1 and 10% because it is a difference of 0.5W. The dip you will notice is the 3dB circle between charged and discharged.
In brief:
Charged 12.6V package, the boost will give 8W of power into 8 ohms impedance with 1% distortion
At a nominal voltage of 11.1V, it will give 6.5W into 8 ohms impedance with a distortion of 1%
On a discharged 8V, it will give 5W at 8 ohms impedance with a distortion of 1%
The drop in volume from a charged packet to a discharged packet is around 3dB

However, I decided on the converter, now I assume that I will make a parallel connection of 2 cells, the converter has a range of 2-24V, if I set 15V (the maximum voltage for the amplifier), I should get about 10W at 8?. I think I got it right.

Even over 10W. Better serial, unless you have additional cells that you connect in parallel to the serial ones, it would be better, because you will unload the cells, which will extend their life. Why private? Smaller input voltage difference to the output converter, and this results in higher efficiency of the converter itself, there is no difference in the amount of energy in series and there is no parallel, the difference is in voltage and capacity, or you have low voltage and high capacity, or higher voltage and lower capacity, accidentally comes out the same.

I will have a maximum of 4 cells at my disposal, I can connect 2 in series and I can put another 2 in parallel, and I think I will. Then I will get a voltage of about 8V and a capacity of 2 cells. If I am saying something wrong, correct me. Thanks for the help.

You speak well. It does not change the fact that the peak current for the cell at a voltage close to discharge will be 5A, a lot for lithium-ion. Nevertheless, it is the most optimal version that you propose, having only 4 cells.

This is where the problem arises, the maximum current that I can treat the converter is 2 amps. The inverter is MT3608. You need to give a resistor for the inverter input only if you choose it right. I am not a professional electronics / electrician and if something breaks in the design, it will break the converter According to my calculations, I should give a 4? resistor. I counted from the formula I = U / R where the initial resistance is 2 ?. Then it comes out clean 4.


I did these calculations at 3:30, so I do not know what I wrote here, with discharged cells and about 5A on the output, the resistance is 1.6 ?, I do not know where to get the resistance of the cell, because unfortunately I do not have a meter, so it will be hard to calculate how many amps this one has. But since these are cells from a laptop battery, it's probably about 1-2 A. So by connecting 2 cells in series and 2 more in parallel, I will get about 7.4 V and 1.5 A. So I don't need a resistor.

Sorry, I was wrong, the current will be 2.5A per cell, maybe 3A per cell, it will not be bad, knowing the large capacity of laptop cells and their high efficiency. Since this, a converter, it will slightly limit the current that can be obtained, but if you use a slightly larger output capacity, then in the case of a greater demand for electricity, it will provide the necessary energy for a short period of time, while the capacitor will simply recharge. You do not need to use any resistor at the inverter input, it will take as much as it can. The converter system has an internal resistor limiting the maximum peak current and it works with the measuring system, but you can give a 4A fuse at the input? It would be worthwhile to measure the current during operation.

Well, and this inverter will be treated with 4A, it will not burn? Where should the remaining 2A of current go? It won't evaporate like this. Put this fuse on the plus wire in front of the inverter, right?

The peak current, it will be up to 4A, it is internally limited, it will simply not take the current anymore, the stabilizing system will turn on and probably turn off the converters for a while or lower the filling, when the current drops, it will restart and it will happen many times in a row seconds .. What's the matter with the remaining 2A of current ?? Yes, the fuse on the positive wire in front of the inverter.

OK I understand. If something happened, I will write, but for now I'm waiting for parts.

Hello, maybe someone else will come back to this topic, I built this loudspeaker on the PAM8610 amplifier, but there is a problem, at the maximum volume and stronger "bass" the loudspeaker cuts off the sound and turns on again, the voltage is ~ 14.80 (I did not want to give equal 15V), when I turn the volume down by 1 degree on the phone it works fine. What could be causing this?

Too weak power supply, the amplifier lacks power to power itself, so it does not work sometimes