I have a currently dead Bluetooth speaker that had a Micro-USB port that fell off. The battery can no longer be charged and I'm not skilled enough with a soldering iron to re-solder the Micro-USB back on to the board.

However, I am capable of re-soldering the large battery leads with that of the positive and negative leads of a stripped power brick. Since the speaker doesn't move all that often in the first place, I figure this is an adequate solution.

I'm curious what kind of pitfalls I should avoid when doing this process. I have basic knowledge of electronic components, so I'm not so worried about that. I just want to clarify my assumptions before I go and set the board on fire.

Currently, the board is powered by a (dead) 3.7 V 300 mAh battery, as read off of the back of the battery. I want to replace this with a 3.7 V power supply.

However, I cannot seem to find one that matches these specs exactly. Since the power output is so small, and since it was a battery powered device, my assumption is the specs don't need to be exact. The board should have been designed with some redundancy in mind for voltage drops and such. Thus I was curious if a 3 V 1 A power brick would be sufficient to power this thing and not completely fry it.

Is my understanding of electronics right here or will I run the risk of breaking everything going this route?

  • 2
    \$\begingroup\$ 3 V is probably too low if there is any hard undervoltage lockout other than low battery indication, but if you have it available, it's highly unlikley you will damage anything from trying. 3.3 V might be easier to come by. I would not hestiate to connect 4 V either. \$\endgroup\$
    – winny
    Apr 6, 2018 at 7:49
  • \$\begingroup\$ How do you know your power brick is 3V output? Seems unlikely to me as it would be useless for charging li-ion cells \$\endgroup\$
    – Pop24
    Apr 6, 2018 at 7:53
  • \$\begingroup\$ @Pop24 I just did a quick Google Search for 3.7v power brick and upon finding nothing the next closest thing that I could find was a 3v power brick. \$\endgroup\$
    – duper51
    Apr 6, 2018 at 7:56
  • 1
    \$\begingroup\$ @winny could I potentially test this off the 3.3v jack on a Raspi, my understanding is pulling from the power pins (not the GPIO) should supply a reliable 500mA current at 3.3, if not more. \$\endgroup\$
    – duper51
    Apr 6, 2018 at 8:00
  • \$\begingroup\$ No idea about Raspi, but assuming some normal LDO or SMPS with 3.3 V and 500 mA, I see no issues unless you play dubstep at close to clipping volume. \$\endgroup\$
    – winny
    Apr 6, 2018 at 8:07

1 Answer 1


A fully charged Lithium is 4.2V.

Just power it from 5V USB, with a standard 1n4001 diode in series to drop a volt.

You might need an electrolytic capacitor across the battery place e.g. 1000uF to reduce noise and supply peaks. (batteries have a low impedance, and it has only 3" of wire)

  • \$\begingroup\$ Brilliant! Why didn't I think about this? ;-) I'm sure I've suggested something very similar elsewhere. Or a 1.2 V zener and drop down to 3.8 V. \$\endgroup\$
    – winny
    Apr 6, 2018 at 9:12
  • \$\begingroup\$ Can you explain why I'd need the electrolytic capacitor? Also, when powering from the 5V USB, will I need to use a specific USB cable/power brick? From recollection, the USB spec dictates a negotiation of amperage output. \$\endgroup\$
    – duper51
    Apr 6, 2018 at 21:14
  • 1
    \$\begingroup\$ If you device enters a very deep sleep, then the voltage drop across the diode might be as low as 0.5V. I can't really imagine this being a problem, but you could always put an led+r power light at the battery point as well to ensure the drop is 0.7V min... \$\endgroup\$
    – Henry Crun
    Apr 6, 2018 at 21:32

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