I'm using a DFPlayer Mini to make a little MP3 player, the specs show it can power a speaker up to 3 watts, however, the power supply I'm using (4 AA batteries) could power a speaker with more than 3 watts. How could I still play sound through the speaker from the DFPlayer Mini, but power the speaker with the batteries instead?
The DFPlayer Mini is 5V maximum. Use 3xAA instead, connected to Vcc on the player.
Don't forget to connect all the grounds.
Your speaker only uses as many watts as it needs. This is controlled by the audio content and volume. Lower the volume the longer the batteries will last.
I'm using a resistor, I'm not about to destroy the little thing with 6 volts haha. I only have a 4 AA battery holder which I got off Wish, but it suffices for what I need. I just have to attach a simple 100ohm resistor from the positive of the battery pack into VCC and according to Ohm's law drops it from 6V to a comfy 3.5V (assuming the DFPlayer Mini is using its average current of 35mA) which is nicely within its operating range. However, this doesn't answer my issue about powering the speaker, which this topic is about. Any tips there?
My issue is I'm using a chunky 5W speaker, and if I accidentally raise the volume too high, it'll draw too much and potentially kill the DFP. I want to draw power from the battery pack instead of the DFP (which only has a max of 3W) and still transfer the signal.
That won't work well, but have fun!
Add a resistor approximately the value of the volume pot in series with the CW pin. If it is to loud then increase the resistance.
It's a low current IC, and it's resisted to well below its max operating value just in case of a voltage spike. I don't have a 3AA battery holder just lying around anywhere, and this way I can get more power for my speaker. I'm sure it'll be fine.
You lost me at volume pot and CW pin. Volume is controlled internally by the IC, and I don't see any CW pin on the DFP pinout.
Absolutely fascinating! Where does 3W for the speaker come from?
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Hey friend, sarcasm isn't necessary in this friendly community 
Edit: Also, to answer your question, I don't need 3W for the speaker because, as I said in the original post, I want to get those watts from the batteries and not the IC, that way it doesn't need to draw too much power and fry itself.
Obviously you do not understand the question. With all the information and absent schematic you readily gave us I mistakenly assumed your amp was like mine that has a volume pot. Good Luck!
The simple answer is you can't.
Without batteries, where do you think the IC gets its power from to actually operate, and send the sound output to the battery?
By the way, a resistor lowers the current, not the voltage. You will still fry the IC due to over voltage if you use 4xAA. Use a USB power supply or an old phone charger that produces 5V while you're waiting for your order of a suitable battery holder to arrive.
You will have to add an external battery powered amplifier of 5W capacity in order to drive the speaker at 5W.
The comfy 3.5V is being dropped across the 100 Ohm resistor @ 35 mA leaving a comfy 2.5V to operate your DFP.
Both. 
They are related by Ohm's Law.
As you '"push" power into the speaker it will draw more current and the voltage will drop more. The voltage will probably drop to the point where the amplifier doesn't even work.
You are making a Voltage Divider where the "top" resistor is 100-Ohms and the "bottom" resistor is the amplifier, including the speaker.
With a 100 Ohm resistor and an 8-Ohm speaker the maximum voltage you can get into the speaker is 8/108 x 5V, which is 0.37V peak = 0.26V RMS.
And, that's 0.26V squared / 8 Ohms = 0.085 Watts = 85 mW.
...The DFPLayer has bridge amplifier which can put-out a theoretical maximum of 10V peak-to-peak with a 5V supply. A "regular" singled-ended amplifier (with a ground connection to the speaker) can only put-out 5V peak-to-peak for 1/4 of the wattage.
Okay, so clearly I was wrong with using a resistor, I'll take full fault for that. But, I still have two questions that remain-
- How could I safely lower the 6V battery pack to use for the DFP if not with a resistor, and,
- How can I then safely drive my 5W speaker?
- Some battery packs have the series connections of the batteries exposed and if yours is like that you could tap it at the 4.5 volt point.
- The 5W speaker is designed to be driven at 5W safely. It doesn't "draw" 5W from the amplifier if the amplifier is not capable of delivering it. The DFP has a drive capability of a few milliwatts and can't be coaxed into supplying any more no matter what 8 Ohm speaker you hook up to it. If you have the module with the DAC outputs, you can feed those to an external power amplifier that you can run off a higher voltage and that can drive a 5W speaker.
