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Connecting car battery to a portable speaker.

Hi, im about connecting car battery to a portable speaker, as title says.
I'm amateur in electronics so i need help...

I have Kymco GTX4L-BS battery (not sure where it is from)
Specs: 12v
3Ah (10HR)
CCA: 50A
(I have also "VARTA A14" 12v 40Ah 330A(en) wich i want to use for same purpose if is possible)

And i got Step-Down converter from 12v to 4.5v 3A (speaker requires 4.5v)
So i am scared of burning all my electronics beacuse of 50A.
I hope you guys can help me and tell me what CCA is and will it burn everything ???
 
The equipment will only draw what it needs - not what is capable of being delivered.

Just make sure to put a fuse on the 12v feed line. A 5A fuse will be suitable. Also make sure the cable is rated accordingly (not too thin).
 
Yea im 100% sure that red is positive but i want to know where i need to put fuse.. on positive or negative wire ?
And is it better to use Fast Blow fuse or Slow Blow ?
 
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If you want to ensure which is positive and negative, open the casing, then on the PCB near where the supply wires are connected to, look for an electrolytic capacitor, marked (-) on the negative lead. Trace this back via the PCB's copper tracks to the negative supply wire.
The same applies to diodes. The white line on the diode is the cathode (negative or output), so the un-banded side should trace back to your positive. The PCB itself may have screen-printing to denote polarity.
If there is an amplifier IC or transistors, the spec-sheet of the component will tell you which is positive/Vcc and negative/ground.
 
Output power of speaker is 3Watts
Oh yeah?
An amplifier powered from 4.5V might produce an output voltage swing of 3V peak-to-peak, which is 1.06V RMS. Then the power into an 8 ohm speaker is (1.06V squared)/8 ohms= 0.14W.

Maybe the amplifier is bridged using an amplifier for each speaker wire then the voltage swing might be 5.6V peak-to-peak which is 1.98V RMS. Then the power into an 8 ohm speaker is (1.98V squared)/8 ohms= 0.49W.
Maybe the speaker is 4 ohms then the power is almost 1W.
Maybe the amplifier is clipping like crazy producing extremely distorted squarewaves, then the power of the bridged amplifier is almost 2W into a 4 ohm speaker.

How did you wrongly calculate 3W? Is the speaker rated for a maximum input of 3W? Then with an input of only 0.14W its output is very low. You can connect a speaker that is rated for a maximum input of 100W and feed it only 0.14W then its output will also be very low.

To produce 3W into an 8 ohm speaker its voltage swing must be 14V peak-to-peak then its power supply must be about 15.5V or more.
 
Well.... Im not making DIY speaker. The speaker is made by verified manufacturer im just making additional battery because im not happy with 4.5v 2000mAh build in battery(9Wh witch is playing 45 minutes.. With My 12v 3Ah it has 36Wh witch should play for ~3 hours).
 
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If your little amplified speaker has a 4.5V amplifier that produces only 0.14W into 8 ohms then that is the maximum power on the peaks of speech and music. The average continuous power is about 0.03W and its heating is about 0.1W so the battery must provide 0.13W per hour. The battery should last for 9/0.13= 69 hours or 34.5 hours for stereo.

Since your 9Wh battery lasts for only 45 minutes then either it is old and worn out or its load is 12 Watts. The current would be 12W/4.5V= 2.7A which is crazy for a little low power amplifier. Try using a new 4.5V 2000mAh battery that is fully charged.
 
on backside of the speaker is written 3W, battery may be near death but i don't need to know all that, i will just use fuses in case something goes wrong. But first i need to wait for parts to appear from seller.
 

hevans1944

Hop - AC8NS
So i am scared of burning all my electronics beacuse of 50A.
I hope you guys can help me and tell me what CCA is and will it burn everything ???
CCA = Cold Cranking Amperes, maximum amount of current the battery can supply when providing power at some minimal rated temperature. Useful specification for sizing battery to engine starter-motor current requirements. Means absolutely nothing for your application.

As previously stated (@kellys_eye in post #2), the battery will only supply the current demanded by the load, said current always being less than the capability of the battery, so no danger of burning all your electronics unless the battery voltage is too high.

The battery you cite, Kymco GTX4L-BS, is a wimpy little scooter (or motorcycle) sealed lead-acid battery with 3500 mAH (3.5 AH) storage capacity at 12 V. The 3.5 ampere-hour rating for this 12 V battery is only slightly larger than the 2 ampere-hour rating of the built-in speaker battery. So, not accounting for inefficiencies in converting 12 V DC to 4.5 V DC, your proposed battery provides only (3.5 - 2.0) / (2.0) = 0.75 times more ampere-hours energy than the original. But it does this at the wrong voltage (12 V) instead of the correct voltage (4.5 V), so you are unlikely to see any benefit from the substitution.

The replacement 12 V battery with 3.5 AH capacity, has a theoretically equivalent power output of 42 watts for one hour with a current drain of 3.5 A; or 21 watts for two hours with a current drain of 1.75 A; or 3 watts for 14 hours with a current drain of 0.25 A. However, don't expect to reach any such durations, because none of the calculations take into consideration inefficiencies of the 12 V DC-to-4.5 V DC step-down converter, nor the inefficiency of the speaker electronics allegedly providing 3 watts of speaker power from a 4.5 V DC source, nor the inefficiency of the battery in delivering power over such a large range of current.

Perhaps you should consider using a real automobile battery, such as this 35 AH sealed lead-acid battery.
 
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