Need BA-330 Schematic for modification

[sma1033]

Hi, I bought a Roland BA-330 last week for my band.

I love the compact design, It takes only 8 AA batteries to make the PA system work and create 40 watt output.

However, There are two points I wanna improve for BA-330.

1. doesn't work with capacitor type MIC. (no phantom power)

2. 30 watt is not enough for street performance.


I am thinking of using 12S8P 18650 Li-ion batteries and a TAS5630 Class-D Amp for power upgrade and adding phantom power to input CH1.

anyone have full schematic circuit file?

really appreciate it.

[J M Fahey]

Now I see this post; so we are talking a compact battery powered Class D amp with serious batteries.

Don't think that you'll be able to fit all this inside the BA330 and even if you did, the speakers wouldn't take it, so I guess the game is still adding external powered cabinets.

Just curious, what power do you plan to achieve?
Driving what speakers?
Using how many batteries?
Supply voltage, series/parallel?
Thanks, looks like an interesting project.

[sma1033]

i tried my high power Class-D amp board TAS5630 on 48V power supply, It consumes 6 amps for satisfying volume to drive four 6 Ohm 60 watt 6.5 inch mid/low range speaker.(loud enough for street usage)

The original Roland BA-330 Speaker is only 7.5 watt/each and unable to handle those plenty power as expect. So I replace it with 60W/each mid/low range speaker

To minimize weight, I plan to use 12S8P Li-ion batteries to power the whole system. I checked empty space in BA-330 cabinet. It seems work fine.

Supply voltage = (3V ~ 4.3V) x 12 = 36V ~ 51.6V
Battery Capacity = 3200 mAH x 8 = 25.6 AH
estimate time usage = ~ 4 Hrs

my main concern is heat dissapation, still considering if i need to make a hole for cooling

=== old roland 7.5W mid/low range speaker ===


=== new 60 watt mid/low range speaker x 4 ===


=== panasonic NCR18650B Li-ion batteries ===


=== TI's TAS5630 Amp Board 600W x 2 ===

[g1]

The batteries will be in series? Each battery is good for continuous 6A ?
Wow!

[sma1033]

I use 12S8P batteries as battery pack.

continuous current for each battery will be 6A / (8 parallel cell) = 0.75A / cell

it should be fine to give 0.75A from one 3200 mAH 18650 Li-ion battery.

[g1]

Sorry but I'm not understanding the battery thing.
If you are putting them in parallel to increase the current capacity, then you get the voltage of one battery (3.6V)
How many batteries are you using and are they in series or parallel?
Or are you using 12 packs of 8 batteries each?

[Jazz P Bass]

Whew.
Guys got over a $100 worth of batteries there.

NCR18650B Li-ion batteries.pdf

[Leo_Gnardo]

I'm a little worried safety wise, on paralleling a number of batteries. Should any start to go flat prematurely, the "hot" ones will try to charge the flat ones instead of delivering power to the amp. With 8 ordinary alkaline AA's not much can go wrong, but with a big stack of Lithium ions... they have a certain reputation. Carry a fire extinguisher & wear a hard hat.

[J M Fahey]

Supply voltage = (3V ~ 4.3V) x 12 = 36V ~ 51.6V
Battery Capacity = 3200 mAH x 8 = 25.6 AH

I use 12S8P batteries as battery pack.

continuous current for each battery will be 6A / (8 parallel cell) = 0.75A / cell

it should be fine to give 0.75A from one 3200 mAH 18650 Li-ion battery.

Sorry but your Math is wrong.

If all batteries are in series, voltage adds up, AH is same as a single cell

If batteries are in parallel, AH adds up, voltage is same as a single cell

Pick one, not both.

EDIT: as of the amplifier, supposing it's bridged , fed 12x3.6V=43V , driven to full power into 8 ohms, it can put out up to :

(41*.707)^2/8=29^2/8=105W RMS ... around 90/95W RMS really, supposing the Class D amplifier is 90% efficient.

Supply load (equivalent) for a bridged amp is (Pi/2)*Rl=1.57*Rl=12.5 ohms.

Supply current will be 43V/12.5 ohms=3.44A , round it up to 3.5A since the amp will also need at least a little idle current for itself.

Those batteries can supply a maximum of 1C current, so 3.2A

Supposing the amp is not fully driven and is happy with 3A (why 3A? ... because there are graphs for that current and it's close enough), say it's used around 60 or 70W average, loud but not harshly clipping, AH value is 3AH so the suggested battery pack should last around 1 hour (3AH/3A=1H) .

Also from same graph (Discharge Characteristics by Temperature) we see that effective available voltage is 3V per cell, so power pack is actually 36V (effective) and for power calculations we must substract 2V for the inevitable drop across semiconductors, math above should be redone with 34V peak available, too sleepy now to redo it, but in any case is close enough for power and life estimates.

If the amp when tested pulled around 6A from the 48V supply, it's because power was somewhat higher (higher voltage available), load was lower , 6 ohms nominal and in fact it behaved as if it were actual 5 ohms.

Original speakers were stamped 8 ohms in the magnets

In a nutshell: the amplifier as described will work, but real power will be somewhat around 60/70W RMS, a huge improvement over Roland original, which I estimate around real 20W RMS , and to boot driving way more efficient speakers, where the real SPL gain is.

But at full volume battery life will be around 1 hour , maybe 2 or 2 1/2 if music is not continuous.

Also from the

[J M Fahey]

I'm a little worried safety wise, on paralleling a number of batteries. Should any start to go flat prematurely, the "hot" ones will try to charge the flat ones instead of delivering power to the amp. With 8 ordinary alkaline AA's not much can go wrong, but with a big stack of Lithium ions... they have a certain reputation. Carry a fire extinguisher & wear a hard hat.

Don't worry , if all batteries are the same chemistry.

You may measure slightly higher or lower voltage **no load** whether battery is fresh out of the charger or almost discharged, but at a chemical reaction level "voltage is always the same" .

One way to see that is to consider that voltage is the same, wht changes is the battery internal resistance, depending on state of charge.

So if you put worn batteries (you measure 3V across terminals) in parallel with fresh ones (you measure 3.6V) they almost instantly even up, at least internally, where it counts, charged ones will feed some current to the discharged ones, because internal voltage instantly rises to 3.6V , and you have the relatively large internal resistance of the weakest ones controlling initial current.

The point is that no dangerous currents flow.

Now, if you put different chemistry battery packs in parallel, suck as 3 NiCd ones (nominally 3.6V) with Li Ion ones (3.6V) , results are unpredictable; even so I suspect no big harm will happen.

[sma1033]

it's heavy duty application, Bro.

it cost me around 300 USD (3 USD / cell x 96 cells ) for just batteries , actually.

Making some loud noise on street is priceless.

[sma1033]

Sorry, but i don't get it.

The power I need to make loud voice is 48V x 6A = 300W (simplified for calculation, let's just assume no power loss on Classs-D amplifier)

each battery contains (240 Watt x Hours / Kg) x 0.05 Kg = 12 Watt x Hours

Now I got 12 x 8 = 96 batteries

total energy contains in battery is (12 Watt x Hours / cell) x 96 cells = 1152 Watt x hours (1152 WH)

Time of usage = 1152 WH / 300 W = 3.84 Hours

Actually, accouding to my calculation, each 12 series(12S) battery cell provide (12 Watt x Hours / cell) x 12 cell = 144 WH Energy, which is, 144 WH / 300W = 0.5 Hour time usage

what do you mean by "Pick one, not both." ??


== NCR18650B official data ==
https://industrial.panasonic.com/cdb...A4000CE417.pdf

[J M Fahey]

You wrote:

i tried my high power Class-D amp board TAS5630 on 48V power supply, It consumes 6 amps for satisfying volume to drive four 6 Ohm 60 watt 6.5 inch mid/low range speaker.(loud enough for street usage)

The original Roland BA-330 Speaker is only 7.5 watt/each and unable to handle those plenty power as expect. So I replace it with 60W/each mid/low range speaker

To minimize weight, I plan to use 12S8P Li-ion batteries to power the whole system. I checked empty space in BA-330 cabinet. It seems work fine.

Supply voltage = (3V ~ 4.3V) x 12 = 36V ~ 51.6V
Battery Capacity = 3200 mAH x 8 = 25.6 AH
estimate time usage = ~ 4 Hrs

WHERE do you mention 96 batteries?

To boot, you show this picture:

I count not even 12 but 11 batteries there, good for 3.6*11=39.6V *no load* or a more realistic 11*3V=33V supplying 3A (forget 6A).
Or 36V (under load) if you actually meant 12 batteries and had one in your hand.

Do you have more batteries?
Cool ... just mention them clearly

First time I (and all others) heard about them is in post #11

and I've been using http://www.batteryspace.com/prod-specs/NCR18650B.pdf .... which is the same suggested by JPBass, just downloaded from another page ... I guess yours should match these?

Power density is an interesting specification, but for accurate dresults let's go to the specific curve , the one I mentioned which shows finer detail:
at 3A current consumption battery has effective capacity of 3000mAH = 3AH and shows 3V voltage, so real world power stored and sent back to load is 3*3=9WH per battery, 12X as much per string (108WH ) and 8X as much if we have 8 strings in parallel: 8*108=864WH

Now I *assume* the 48V supply you mentioned is not a wall powered one but a 96 battery pack, please clear this.

Recalculate available power yourself, discounting 2V for semiconductor losses and (estimated) 2 V loss in batteries respect to no load; unfortunately I only see 65mA and a C (~3200mA) discharge graphs, none for 750mA so I must estimate some intermediate value.

I suggest you feed an audio tone, 400Hz to 1kHz is fine to your power amp, load it with a 8 ohms power resistor, drive it to clipping, and measure:
* real power output
* actual supply voltage under load
* actual supply current under load

nobody argues actual measurements

And please post gut shots, how you fit the amp module and battery pack, etc., this is a very interesting project and we are happy you are sharing it with us .

[sma1033]

Sorry but your Math is wrong.

If all batteries are in series, voltage adds up, AH is same as a single cell

If batteries are in parallel, AH adds up, voltage is same as a single cell

Pick one, not both.

EDIT: as of the amplifier, supposing it's bridged , fed 12x3.6V=43V , driven to full power into 8 ohms, it can put out up to :

(41*.707)^2/8=29^2/8=105W RMS ... around 90/95W RMS really, supposing the Class D amplifier is 90% efficient.

Supply load (equivalent) for a bridged amp is (Pi/2)*Rl=1.57*Rl=12.5 ohms.

Supply current will be 43V/12.5 ohms=3.44A , round it up to 3.5A since the amp will also need at least a little idle current for itself.

Those batteries can supply a maximum of 1C current, so 3.2A

Supposing the amp is not fully driven and is happy with 3A (why 3A? ... because there are graphs for that current and it's close enough), say it's used around 60 or 70W average, loud but not harshly clipping, AH value is 3AH so the suggested battery pack should last around 1 hour (3AH/3A=1H) .

Also from same graph (Discharge Characteristics by Temperature) we see that effective available voltage is 3V per cell, so power pack is actually 36V (effective) and for power calculations we must substract 2V for the inevitable drop across semiconductors, math above should be redone with 34V peak available, too sleepy now to redo it, but in any case is close enough for power and life estimates.

If the amp when tested pulled around 6A from the 48V supply, it's because power was somewhat higher (higher voltage available), load was lower , 6 ohms nominal and in fact it behaved as if it were actual 5 ohms.

Original speakers were stamped 8 ohms in the magnets

In a nutshell: the amplifier as described will work, but real power will be somewhat around 60/70W RMS, a huge improvement over Roland original, which I estimate around real 20W RMS , and to boot driving way more efficient speakers, where the real SPL gain is.

But at full volume battery life will be around 1 hour , maybe 2 or 2 1/2 if music is not continuous.

Also from the

my amplifier output rms viltage is 41v x 0.707 = 29v R.M.S. (TAS5630 under BTL configuration)

however, my load is four 6 ohm speaker all connected in parallel, which means, equavilent 1.5 Ohm resister


I will say the ideal output power will be V(RMS) x V(RMS) / Speaker impedance(Ohm) = 29V x 29V / 1.5 Ohm. = 560 Watt

If we take class-D amplifier efficiency into consideration(Assuming 80% efficiency)

I still can get 560 x 80% = 449 (Watt) output

I've tried those 18650 batteries, most single cell can run under 1 Amp(0.3C) current without significant voltage drop(less than 0.2V)

When taking about 8 cells in parallel configuration, It should give 6 Amp. output current without significant voltage drop.

Therefore, I am confident about giving enough output power(more than 400 Watt capacity)

Comparing to the original Roland 40 Watt(20W+20W) design, 900% power up is insanely loud.

Though the project is still under construction, I guess I will be happy to see the final result.

[sma1033]

Sorry for the misleading batteries picture.



Those batteries are sample batteries just for measuring Panasonic NCR18650B Li-ion cell characteristic.

Since I am really afraid of getting fake batteries from my China website resource, I need to test those cell in advance before I get plenty of cells.

Luckly, those 18650 cells are measured more than 2800 mAH under 0.3C load from my testbench experiment.

Cell load charactics?
Maximum 50 mOhm cell internal resistor (without battery protection circuit) under 1C(3.2A) current condition.

I already ordered another 96 batteries from the same resource.

Can't wait to receive those batteries !!