DC to AC rail splitter

ericgibbs

Joined Jan 29, 2010
13,650
hi mv,
Depending upon the OPA circuit application, the Passive splitter is very limited..
The Active OPA based splitter is a better choice, but again it has some limitations.

What are the required 3, voltages and current that you Amp requires.

E
 

ericgibbs

Joined Jan 29, 2010
13,650
hi,
Looking over that PCB image, I would say that the 'black' rectifier and the two large capacitors, suggests a dual power supply of approx +/-18V
I would say that a Active splitter is not going to be a solution.
Divide your 36V battery pack into two 19V supplies, one as +V and one as -V. do you know how to do that.?
E
 

Thread Starter

missile villegas

Joined Jan 7, 2016
10
hi,
Looking over that PCB image, I would say that the 'black' rectifier and the two large capacitors, suggests a dual power supply of approx +/-18V
I would say that a Active splitter is not going to be a solution.
Divide your 36V battery pack into two 19V supplies, one as +V and one as -V. do you know how to do that.?
E
Oh ok
Thank you very much
One last question
Would it work just plugging it in straight like that???
Being that the battery is DC and the board requires ac
 

DickCappels

Joined Aug 21, 2008
7,736
It is ok, but you will loose 1 to 2 volts in the rectifiers, or if you are real careful to always connect the batteries in the correct polarity you can remove the diode bridge and connect the batteries directly to the bridge's outputs marked + and - .

Keep in mind that as you use the batteries the voltage will drop and the specification in the listing says " - Working voltage: double AC18-24V " so maybe you want to start at a higher voltage so you can get more running time out of your batteries.

To get much volume (the amplifier is rated at hundreds of watts!) you will probably want to use a much larger battery than the 18650 unless you plan to run at very low volume.
 

Thread Starter

missile villegas

Joined Jan 7, 2016
10
It is ok, but you will loose 1 to 2 volts in the rectifiers, or if you are real careful to always connect the batteries in the correct polarity you can remove the diode bridge and connect the batteries directly to the bridge's outputs marked + and - .

Keep in mind that as you use the batteries the voltage will drop and the specification in the listing says " - Working voltage: double AC18-24V " so maybe you want to start at a higher voltage so you can get more running time out of your batteries.

To get much volume (the amplifier is rated at hundreds of watts!) you will probably want to use a much larger battery than the 18650 unless you plan to run at very low volume.
I see
Screenshot_20200324-135859.png
Initially this is what I was planning on doing since it seemed like the easiest thing to do
Being that I don't really know much about electronics
I just don't know what size capacitors and resistors I would use
Thanks for your help I really appreciate it
 

Audioguru again

Joined Oct 21, 2019
3,542
Why would anybody want to buy that cheeep, no-name-brand audio amplifier that has no detailed audio spec's??
An 18650 battery pack is rated for about 3A max then the 38V x 3A= 114W where half (57W) can be used for the 3 audio amplifiers and the other 57W is heating.

The Chinglish spec's say 80W, 100W and 120W. Whats? If the total supply voltage is 38V - 2V for the rectifiers then the total is 36V and the stereo amps will probably produce 33V maximum peak output which is 23.3V RMS. Then the stereo output max power per channel is 23.3V squared/8 ohms= 68W and the subwoofer output is probably bridged for 108W output.
Then the total maximum output power is 68W + 68W + 108W= 244W and the heating is another 244W so the battery must produce 488W. Don't even think about 4 ohm speakers.
488W from a 38V battery is an average maximum current of 488/38= 12.8A and the peaks will be 18A. The little 18650 3A battery will explode.
 

DickCappels

Joined Aug 21, 2008
7,736
Your resistor divider will not work for this kind of circuit. That is because the current needed to flow from the positive power supply to "ground" will not be the same amount that is needed from the negative supply. The currents are constantly changing when the speakers are being driven and much higher than what will be drawn by the 100 ohm resistors.

Eric gave a very good solution

(Some text removed for clarity)
I would say that a Active splitter is not going to be a solution.
Divide your 36V battery pack into two 19V supplies, one as +V and one as -V. do you know how to do that.?
E
And by that I am pretty sure he meant to make two battery packs, one for the positive supply and one for the negative supply.

Going back and looking at the specification, it says " - Working voltage: double AC18-24V ". When an AC voltage is referred to without otherwise specifying the type of measurement, it is taken to mean RMS (root mean squared), which is 0.707 less than the peak voltage. Your DC (battery) power supply should therefore be ±18V/0.707 to ±24V/0.707, or 25 to 34 volts DC, so the battery pack supplies DC that is the same voltage as the peak. Is that right @Audioguru again ?
 

Thread Starter

missile villegas

Joined Jan 7, 2016
10
Your resistor divider will not work for this kind of circuit. That is because the current needed to flow from the positive power supply to "ground" will not be the same amount that is needed from the negative supply. The currents are constantly changing when the speakers are being driven and much higher than what will be drawn by the 100 ohm resistors.

Eric gave a very good solution



And by that I am pretty sure he meant to make two battery packs, one for the positive supply and one for the negative supply.

Going back and looking at the specification, it says " - Working voltage: double AC18-24V ". When an AC voltage is referred to without otherwise specifying the type of measurement, it is taken to mean RMS (root mean squared), which is 0.707 less than the peak voltage. Your DC (battery) power supply should therefore be ±18V/0.707 to ±24V/0.707, or 25 to 34 volts DC, so the battery pack supplies DC that is the same voltage as the peak. Is that right @Audioguru again ?
Ok Thanks again for your help
 

Audioguru again

Joined Oct 21, 2019
3,542
I agree that resistor rail splitter will not work and if the amplifier was made properly then a DC supply of +34V/-34V can be used but probably then it might explode with its very cheap price and tiny heatsink. The fan will blow the smoke around. Its main filter capacitors are shown as 40V ones so they might be OK.
 
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