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# Convert 5VDC to 3.7VDC? Answered

Hi everybody !
I'm currently working on a speaker project, and I want it to be powered by 4x 18650 3400mAh 3.7V batteries in parallel, but when they are recharged (with a separated 18650 wall charger), I want to be able to power my speaker from a wall power supply. The idea is to get the 5v power supply voltage down to something between 3.7v and 4.2v, in order to mimic the battery voltage. I already chose this one which is a 10amp-capable power supply in order to be sure to always give the speaker enough current (speaker current consumption is between 0,55 at medium volume and 3 amps a maximum volume). I first thought that i would use this board to convert 5v to 3,7v but it's only able to output 1.25 amps max so I need to figure out a way to convert 5v to 3.7-4.2v without loosing a lot of current. I've heard of low dropout regulator (LDO) but I cannot find one capable of delivering 4amps or more...

Any help is appreciated !
Mahot

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## Comments

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A 3.3V power voltage regulator should work fine, and you can probably get away with powering it with a 5V supply. I would suspect there is enough engineering margin to allow that, I do not know what would burn out or fail from only a 0.8V increase in volts. Although there may be some "smart" circuitry inside it to inform you there is something seriously wrong with the battery, and refuse to operate. Who knows? (In fact, there is likely a 3V LDO regulator internally for digital stuff! so 5V will most likely work.)

If powering it from 5V directly does not work, or you do not want to risk the electronics, You could drop a volt or two with 2 beefy silicon rectifiers in series. Each one will drop approx 0.6V, so 2 of them is series will drop approx. 1.2V, so you will be left with 3.8V, which is right in the range you want to be in for single cell LiPo's and 18650's.

~~~~~Some random power / efficiency / efficacy calculations below:~~~~~~

At 5A, the power loss from it will be the voltage dropped across them, lets assume 1.2V nominal, times the current, 5A. 1.2V*5A = 6W. The power left to power the load is 5V - 1.2V = 3.8V, that times the current, again, 5A, is 19W. So power loss is the total watts that go in, which is 5V*5A or 19W+6W = 25W. Efficiency = output_power / input_power = 19W / 25W = 76%... There is actually a nice shortcut when it comes to calculating efficiency when current that goes in = the current that comes out, and that is to simply divide output volts by output volts, which is 3.8/5 = still 76%, which is not horrible eff. It can be improved with a switching regulator.

Thank you for this answer, I experimented a bit at home yesterday with a 5v XBOX 360 power brick and 2 little salvaged diodes, 1 LED and a resistor and it works, I get 3.8v ! Thank you :) now I just need to find some bigger diodes capable of handling more than 4amps !!
And thank you for the calculations, now I understand better the idea of efficiency :)

boulou, can you give me the the diagram of your circuit and their values, i would really appreciate it if you'd show me, i have a project that I am working with the same problem

It just occurred to me, a gizmo like this bluetooth-speaker might work well with one of these usb external battery packs. I mean when the speaker plugged into a 5 volt usb supply, it will probably run (receive bluetooth and play music) and charge its internal battery at the same time.

That's sort of a guess. I don't know for sure if the speaker is capable of running and charging itself simultaneously, from a 5 volt usb supply, but I know that most cell phones work that way. But if this is true, then the speaker is already capable of being powered from the mains. You know, just plug in its 5 volt usb charger, and it will run all day, or longer, plugged into the wall like this.

Then if that trick worked, a usb external battery, capable of supplying 5 volts, could also be used to power the speaker.

The point is, with just three boxes: the bluetooth-speaker, external usb battery pack, mains powered usb charger, you would achieve all your goals, (ability to run speaker from mains power, longer battery life for speaker, portable power for other usb devices like cell phones) without any soldering.

But with no soldering, maybe that would take all the fun out of it? ;-)

Also, some of these usb external battery packs are capable of other tricks too, like jump starting a car! HTG article on external battery packs:

http://www.howtogeek.com/178374/the-htg-guide-to-e...

As you said, without soldering it would be less fun ! The idea is that I want all of this in a custom cabinet. I want to power it with 4x18650 batteries in parallel. I'll have a DPDT switch alternating between two positions :

- one drawing juice from the batteries and going straight to the pins where the original battery was connected

- another one connected to a 2.1mm DC jack in wich I can connect my power supply and going to the original battery pins too. this position is used when the batteries are out of the cabinet and recharging, in order to power it indefinitely from the wall, or if nothing is connecte to the dc jack, it's considered as the "OFF" mode !

The idea is that I want to find a power supply capable of imitating the batteries voltage (3.3-4.2v) and capable of delivering more than 3 amps of power in this DC jack.

I'vd joined a basic schematic of my project (the speaker represent the whole unit : amplifier, bluetooth module, buttons, etc)

Why would you need to switch it over to a seperate wall power supply, where there is a functional wall adaptor charger keeping some batteries topped off?

It really is not good practice to keep lithium batteries fully charged at there 4.2V-4.3V peek, for storage, they should be at like 3.8V. However, my laptop is plugged in 24/7 and lithium batteries even when fully charged like that just take a while before significant degradion occurs.

It's way easier to remove the batteries when charging them and during this time powering it with a wall supply because if I let them in with a charger etc. it means that I need a protection circuit module and it's a little bit trickier. Moreover, I don't plan to use it a long time without the batteries. I think that I will only use the wall power supply at home !

What I mean is charge the batteries at the same time as powering whatever you want to power with them, it should be fine. Cell phones do that. (they draw power and charge at the same time.)

I am guessing this thing you call "speaker" is something else, perhaps an amplifier and a speaker? Because it does not make sense to me to power a speaker with direct current (DC).

It may be the case that your amplifier (I am guessing that's what it is) will operate comfortably over a range of voltages. Look to the spec sheet to discover what this comfortable range of voltages is.

If that range includes 5.0 volts, then there is no need for a power converter, and the amplifier will be happy supplied with 5.0 V. Or 3.7. Or any value in between.

Hey. Thanks for giving me a clue there. I guess the manufacturer of this gizmo calls it a "speaker", so now everyone else does too.