# Best method to power my home-built sound bar.

#### GeorgeBC

Joined Mar 22, 2019
13
Background: I am building my own wireless bluetooth connected sound bar using some really excellent stereo speakers I had laying around. With that, I bought a 20Watt stereo amp that runs off of an external 12Volt AC adapter. I also bought a Bluetooth Receiver to feed the audio from my Iphone into the amplifier. It runs off of 5Volts. As I want to use this outdoors, it will be battery powered. I will be using 4 18650 Lithium batteries.

Info: Measurements indicate the amplifier draws about 200ma on average @ 12V. The Bluetooth receiver draws about 35ma @ 5V. The 18650 batteries at full charge put out 4.2Volts and are empty at about 2.75Volts. They are 3500maHour capacity.

Question: What would be the most efficient means to generate the required voltages? My first thought (using 4 batteries at about 17Volts full charge) is to use a linear regulator to generate the 12Volts (thinking of a 7812 regulator IC). And then down regulate the 12Volts to 5Volts using another regulator IC. Seems like I might be wasting a lot of power using this method. Would it be significantly better to just use 3 batteries (12.6 Volts at full charge, but less then 9 volts at discharge) to generate the 12 Volts? Seems like crossing over from reducing the initial output voltage at full charge to increasing it as the batteries deplete might be tricky. Other means might be to just use all 4 batteries in parallel and just use an up converter for the required voltages.

Thanks everyone for your suggestions. I have thick skin so feel free to bash away if this is a terrible idea.

#### wayneh

Joined Sep 9, 2010
17,152
What would be the most efficient means to generate the required voltages?
My choice would be a buck converter to get your 12V. This would be far more efficient than the 7812. It's a tossup on the 5V. Another buck converter would be more efficient than a 7805 but for 35mA I don't think it matters that much. You could use the 7805 after the 12V to lessen heat and power loss in the 7805. A buck converter to output 5V is pretty cheap, though, so that would be tempting.

Take a look at DC-DC converters on eBay. You'll find tons of them.

#### crutschow

Joined Mar 14, 2008
27,177
I'm basically with Wayneh.
Use the 4 batteries in series with a buck regulator to get 12V.
Then a 7805 to get the 5V from the 12V since it's only 35mA.
A switching regulator for the Bluetooth receiver could introduce noise from its switching action.

#### GeorgeBC

Joined Mar 22, 2019
13
Thanks for your input guys. I decided to order a buck/boost dc-dc converter from Amazon for like \$5.99 to generate the 12volts. Way cheaper then I could build it for myself. I'm going to use two series connected batteries (input range would be 5v - 8.4v). And use the other two as backup power when the others are depleted. I'm not sure how long the charge will last yet. Sort of depends on volume of the amplifier (140ma at low volume and 300ma at really high volume).

https://www.amazon.com/gp/product/B07NTXSJHB/ref=ppx_yo_dt_b_asin_title_o00_s00?ie=UTF8&psc=1

#### crutschow

Joined Mar 14, 2008
27,177
I'm not sure how long the charge will last yet.
The battery draw from the converter will typically be 1.5 to 2 times the amp current draw, depending upon the battery voltage..

#### wayneh

Joined Sep 9, 2010
17,152
FWIW, you’ll get longer total service by using your batteries together in one larger pack than splitting them in two. Size and weight are good reasons to go smaller but service time isn’t.

#### GeorgeBC

Joined Mar 22, 2019
13
The battery draw from the converter will typically be 1.5 to 2 times the amp current draw, depending upon the battery voltage..
I'm hoping that is not the case. The spec sheet of the IC on the buck/boost converter promises 80 to 90% efficiency. That's a big hmmmm! I'll post my results once I get all the parts in and can prototype it.

#### wayneh

Joined Sep 9, 2010
17,152
I'm hoping that is not the case. The spec sheet of the IC on the buck/boost converter promises 80 to 90% efficiency. That's a big hmmmm! I'll post my results once I get all the parts in and can prototype it.
You can’t create power from nothing. The higher current is required to make up for the lower voltage.

#### GeorgeBC

Joined Mar 22, 2019
13
You can’t create power from nothing. The higher current is required to make up for the lower voltage.
Oh. I see what you are saying. Can my system be analyzed from an energy point of view? Ultimately I'll be burning 2.4 watts (12V x 200ma) at nominal volume levels. If the dc-dc converter is only 80% efficient then the batteries on the front end will need to supply 3 watts. I'm primarily interested in the run time. Average voltage of the battery from full charge to depleted is 3.5 volts. And the batteries are rated at 3500maH. Can I then say that any particular battery would supply 12.25 Watt hours. If I am consuming 3 watts then the run time for one battery is about 4 hours and for 4 batteries would be 16 hours independent of the battery configuration that I use.

#### wayneh

Joined Sep 9, 2010
17,152
Oh. I see what you are saying. Can my system be analyzed from an energy point of view? Ultimately I'll be burning 2.4 watts (12V x 200ma) at nominal volume levels. If the dc-dc converter is only 80% efficient then the batteries on the front end will need to supply 3 watts. I'm primarily interested in the run time. Average voltage of the battery from full charge to depleted is 3.5 volts. And the batteries are rated at 3500maH. Can I then say that any particular battery would supply 12.25 Watt hours. If I am consuming 3 watts then the run time for one battery is about 4 hours and for 4 batteries would be 16 hours independent of the battery configuration that I use.

#### PhilTilson

Joined Nov 29, 2009
96
I would echo your proposal to use all four cells in parallel and use the up-converter. I have been carrying out some tests recently on a similar thing and these very small - and very cheap! - up-converters seem to be extremely effective. A big advantage is that they will continue to put out a constant 12 Volts even as the LiPo cells discharge. Just a couple of thoughts: I would avoid letting the cell voltage drop below 3 Volts - you'll get longer battery life that way and the extra time to get from there to 2.75V is very short; running the cells in parallel makes charging them a lot simpler (no need for balancing) but just be sure that they all have a similar level of charge when you start, or you can get some pretty hefty currents flowing from one to another; and I support the idea of using a simple voltage regulator to get your 5V 35mA for the Bluetooth. Little to be lost there.

#### BobTPH

Joined Jun 5, 2013
3,306
Another possibility is to use a power bank. These are cheap for what they do, and have all the charging and protection built in. Plus, you can easily keep two of them if you need extra capacity.

They would provide the 5V directly and the 12V through a boost converter. I believe I have seen some that actually have both outputs.

Bob

#### GeorgeBC

Joined Mar 22, 2019
13
The buck/boost dc-dc converter I bought requires a minimum input voltage of 5V. So that is why I was doing a "2x2" battery arrangement. Had I known of the power bank product that would of the way to go as I am not building a charging system into the soundbar. My plan is to remove the batteries and charge them with an external charger. All great ideas. Thanks!

#### bassbindevil

Joined Jan 23, 2014
158
I would look at what voltage range the amp can actually tolerate, and build a battery pack that can power it directly. The TPA3110 amp that's currently in my portable speaker thing claims to be OK from 8V to 26V, but, I went with a 3S4P pack in case I need to use a TA2024 amp which isn't rated to handle 16.8 V. Also, these little class-D amps may not be rated to drive 4 ohm speakers at the higher end of the voltage range (refer to the data sheet for the amp chip, because vendor provided specs can be misleading or wrong).

#### GeorgeBC

Joined Mar 22, 2019
13
The Class D amplifier I am using (Lepai LP-2020TI) uses a Texas Instruments TPA3118 amp which has an operational range of 4.5 - 26 volts. Supposed to be 20 watts RMS per channel however that is only obtained with 4 ohm speakers and the higher supply voltage. I think my general operational range is closer to 5 watts total. Plenty loud at that point. Question here. If I use the battery power directly without an intervening regulator, won't I always be adjusting the volume as the batteries discharge? Also I need a means to keep track of the battery voltage so that the batteries don't discharge below 3V as that can be destructive.

As a side note, these Class D amplifiers are very impressive with their efficiency, frequency response and very low distortion. They were developed by a company by the name of Tripath Semiconductor.

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#### GeorgeBC

Joined Mar 22, 2019
13
Finished my Soundbar. The most fun was actually the woodworking part. I have a slight problem with regulator switching noise getting into the audio. At least I think that is the source. It shows up a 5-10KHz noise (random, not consistent, whistling is the way I would describe it). Not sure how to filter this out. Much worse with RCA input source vs Bluetooth input. Oh .... by the way, it weighs about 30 pounds!!!

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#### Dodgydave

Joined Jun 22, 2012
9,951
The whistling is from the buck regulator..

#### bassbindevil

Joined Jan 23, 2014
158
If you wired the 4 cells in series, a 4S battery pack should have a "BMS" or protection board that protects it from over-charge, over-current, over-discharge, and keeps the cells balanced. Search ebay for 4S (pcm,bms,protection). If it's a 4P pack, there are protection boards for 1S packs that do all of the above except balancing.
If the loudness of a class-D amp decreases as the batteries discharge, I really doubt that will be enough to be noticeable over a period of hours.