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Help me find the right amplifier

audioguru

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If the larger speaker has the same impedance as the small speaker then at the same output voltage level the battery consumption will be the same.
But the larger speaker will be able to be much louder (10dB is produced by 10 times the power) then people might turn down its loudness which will make the battery charge last longer.

50W into 4 ohms from an amplifier needs a sinewave of 14.2V RMS which is 40.2V p-p minus a loss of 2V but the amplifier is bridged so the total p-p output and battery voltage must be 19.1VDC.
6 Lithium cells is 25.2V fully charged or 22.2V nominal.
5 Lithium cells is 21V fully charged or 18.5V nominal.
Using 5 cells at 18.5V then the output power into 4 ohms will be 47.5W which sounds the same as 50W.

The peak current for 50W RMS into 4 ohms is 4.2A then the battery voltage might drop a little.
 
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Thank you ! I feel better now, because the guy will be really happy about this speakers performance and i would hate for him to play for an hour and the speaker stops. Awesome.

Ok i have another question, actualy 2 questions:

1) I am making a speaker that must be able to handle the rain. Thats no problem BUT, just to be sure, just to sleep more calm, i would like to glue some fabric to the inside of the speaker case. Material from an old t-shirt or something so just incase there is something strange, like condens, water drop or divine intervention ... the fabric will soak it up (again, the enclosure will be made airtight, this is just for my calm sleep). So .. if i do that, take very thin t-shirt and glue it to inside of enclosure, what effect will that have on the sound ? Will it be huge, will i lose a lot of dB or not ?

2) It is fairly clear that typical song consumes less power than sinewave. Why ? Well ... lets say it consumed 2A (25V x 2A = 50W, lets say woofer takes 42W, tweeter takes 8W). Well since my batteries are around 1.6Ah at 2A current ... that means the speaker will only work like 1.6 hours at max volume if using sinewave, while in reality using popular music (usualy some techno) at max volume, it plays many many many many many many hours. Ok, so ... without fancy equipment, all i have is multimeter and watt meter ... how the hell could i measure the average power consumption of my speaker at top volume ? I know i wont be able to get the absolute exact number but, at least an estimate, like, is it 0.5A or 1A ? Is it 0.3A or 0.7A ... Because people ask me how long it plays at max volume but i cant test it ... neighbours willcall police if i leave it at max volume playing all day .. not to mention i wont be able to do any work
 

audioguru

Well-Known Member
Most Helpful Member
1) I am making a speaker that must be able to handle the rain. Thats no problem BUT, just to be sure, just to sleep more calm, i would like to glue some fabric to the inside of the speaker case. Material from an old t-shirt or something so just incase there is something strange, like condens, water drop or divine intervention ... the fabric will soak it up (again, the enclosure will be made airtight, this is just for my calm sleep). So .. if i do that, take very thin t-shirt and glue it to inside of enclosure, what effect will that have on the sound ? Will it be huge, will i lose a lot of dB or not ?
A piece of cloth onside the enclosure will not affect the sound.
Most speaker enclosures are filled or lined with sound-absorbing material to absorb reflections of sounds inside. The sound-absorbent material also converts the sound waves inside the enclosure into heat waves that makes the enclosure seem a little larger deepening the bass response a little.

2) It is fairly clear that typical song consumes less power than sinewave. Why ? Well ... lets say it consumed 2A (25V x 2A = 50W, lets say woofer takes 42W, tweeter takes 8W). Well since my batteries are around 1.6Ah at 2A current ... that means the speaker will only work like 1.6 hours at max volume if using sinewave, while in reality using popular music (usualy some techno) at max volume, it plays many many many many many many hours. Ok, so ... without fancy equipment, all i have is multimeter and watt meter ... how the hell could i measure the average power consumption of my speaker at top volume ? I know i wont be able to get the absolute exact number but, at least an estimate, like, is it 0.5A or 1A ? Is it 0.3A or 0.7A ... Because people ask me how long it plays at max volume but i cant test it ... neighbours willcall police if i leave it at max volume playing all day .. not to mention i wont be able to do any work
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Your battery can probably produce 10A. The cells are rated at 2Ah which is energy vs time, not max current. Then it could produce 0.2A for 10 hours or 2A for maybe 3/4 of an hour.
Music has a crest factor because only the drum beat and a few other sounds are at max power unless it is severely clipping all the time.
If the peaks are at full power but without clipping distortion then the average power is 1/10th.
Additional power from the battery is wasted by making heat.
 
I have some more questions. For now ... i am looking at my BMS, as i said, its 6S and i really like it. But one thing i dont understand is ... it is passive balancing board, meaning when cell is over 4.2V, a resistor will activate to reduce the amount of current that flows into that cell. But what i dont understand is, why is this resistor such, that 45mA. I can verify that it is working and turns on at 4.2V, i can see the little bugger heat up. But what i dont understand is, why such weak current/resistance ? My charger charges at 0.5A constant current so you can see my problem ... this 0.045mA is not enough to do anything serious ... I mean why not add a freakin 0.5A resistor to activate at 4.20V ? The cell is full, we want to completely remove current flowing into it, not a measly 45mA.


Optional question:what happens if i replace this resistor with some other resistor ?
 

audioguru

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Most Helpful Member
Read about Lithium batteries at www.batteryuniversity.com and see that when the voltage of as cell reaches 4.2V then it is far from fully charged. A cell is fully charged when its voltage is 4.2V AND its charging current has dropped to a low amount.
The charger limits the charging voltage to 4.2V per cell to avoid an explosion and/or fire.

The 45mA balancing circuit you have is too simple.
 

rjenkinsgb

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Most Helpful Member
But what i dont understand is, why is this resistor such, that 45mA. I can verify that it is working and turns on at 4.2V, i can see the little bugger heat up. But what i dont understand is, why such weak current/resistance ?
As the name describes, it is to ensure the cells are balanced when the whole pack is fully charged.

The battery protection circuits in the BMS should cut charge completely when any one cell reaches its maximum voltage.
As each cell gets near that threshold voltage, the balance section will apply a load to it - slightly reducing that cell voltage and allowing overall charging to commence again.

That will happen until all cells have the same voltage and all are being bypassed a 45mA (or whatever the balance current is).

If the whole battery were perfectly balanced and stayed that way, all the cells would reach full charge at the same time and no balance current would be needed.

The balance circuit is to make up for real-world imperfections and not to control the bulk charge.
As long as all the cells are good, it should hardly do anything, just maintain an equal full charge voltage.
 
The problem is .. few days ago i put together 6 batteries of similar mAh capacity (real tests show a max of 100mAh difference between them). I usualy assemble empty cells, just to limit the damage if i connect something together. But as it turns out, i accidently put together 5 empty cells and 1 of them was full. So i connected the charger and sadly, the full cell hit 4.25V at which BMS cut off the charging, leaving other cells at 3.45V. My point being ... that 40mA balance load is not nearly enough to stop the full cell from going over 4.20V, since the charging current is constant 0.5A. So i do not understand why that balance load is only 40mA instead of 400mA or 1000mA. If it was, it would work perfectly in my case and i could load even completely unbalanced cells and it would balance ok. But since its only 40mA ... i need to limit the charging current to 40mA if i want the pack to balance itself. I dont understand why
 

rjenkinsgb

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Most Helpful Member
i need to limit the charging current to 40mA if i want the pack to balance itself.
No; with the correct voltage charger connected, it will eventually sort itself out - just taking a lot longer than if you had equally charged the cells first. That "full" cell will be bypassed at 65mA and that current will gradually charge the other cells.

It will work perfectly when used as it is intended to be.

The battery pack will take the balance current forever once all the cells reached equal voltage. It needs to be low enough to be a practical indication that charging is complete, and not waste energy.

Just charge your cells equally before assembling a battery and stop worrying about it; the balance system is designed as it is for very good reasons.
 
I will do that dont worry. But i am interested in how it works. You are talking about active balancing, my board doesn't have balancing. The resistor will simply discharge the cell from 4.25V to 4.20V ... but that is not enough for the board to allow charging again, it will only happen when cell will go down to 4.15V or even lower, which means .. well, i told you the results of yesterdays charging .. it should be 25.2V but in reality, stopped charging at 21.5V. Btw this is the board i have:

 

audioguru

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Most Helpful Member
You have a simple "protection board" with a charging current limit of 25A!!! If your battery cells are powerful enough then the explosion will deafen you.
It is for making a pack with a protection circuit, it is not a charger circuit. Many batteries have a protection circuit built into them then the entire battery is charged from a proper balanced charger circuit. The online store in India does not say or does not know who makes it and say it is 22.2V in one spot and says 25.2V in another spot. Maybe the 25A is a misprint and is actually 2.5A? Maybe the 45mA is actually 450mA? Happy Diwali to them.
 
Actualy max amperage allowed is about 8A. My batteries are 10A so yes, it would be a mighty explosion.

As for the board, i did some testing on it. For instance, if the battery pack is 24.8V and i connect it directly to 25.2V DC power, it will not charge, the BMS simply blocks it completely. I use a 25.2V liion cheap charger, which limits the current to 0.5A and with it, it works. So it does detect when there is to high voltage or to high current trying to charge it, thats good. As for 45mA ... sadly it is 45mA, otherwise my unbalanced pack would be charged but it isnt.

Btw i am testing something now. So i connected the charger with 15 ohm resistor in series. Now the charger light is green, as opposed to red when its charging. So green light, but, multimeter shows 0.1A current. I will go and check now to see if it actualy charged it. Because that would be 1 solution to the problem. 0.5A charging is a lot, considering voltage is like 25V, that would mean over 10W charging. There is no need for that. The device i make is meant to play for a long long time and only be charged over night in which case 10W or 5W wont matter since it will charge fully anyway over night. But if charging current was 0.25A instead of 0.5A ... well then suddenly the 45mA balance is more important. That means over many cycles, the batteries would be more and more balanced
 
I have a question i need solved fast, i hope you two will see this message.


Ok, so i have 2 identical speakers in series on a single mono channel. Speakers are 8 ohm so since i am using two, thats 16 ohm combined. Now i want to add a capacitor to make it only play frequencies i want it to. I plug 16 ohms into calculator and get the right value. Now how do i actualy connect it to speakers, same as if i only used 1 16 ohm speaker, in series with the two speakers ? Or is there some other more advanced way i have to do it ?
 
You have a simple "protection board" with a charging current limit of 25A!!! If your battery cells are powerful enough then the explosion will deafen you.
It is for making a pack with a protection circuit, it is not a charger circuit. Many batteries have a protection circuit built into them then the entire battery is charged from a proper balanced charger circuit. The online store in India does not say or does not know who makes it and say it is 22.2V in one spot and says 25.2V in another spot. Maybe the 25A is a misprint and is actually 2.5A? Maybe the 45mA is actually 450mA? Happy Diwali to them.
Just hoping u see this if i quote u.
 

audioguru

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Most Helpful Member
A speaker resonates. It should be connected directly (or through a coupling capacitor) to the output of a modern amplifier since the extremely low output impedance (0.04 ohms or less) "shorts" away the resonance. Resistance or another speaker in series also spoils the frequency response.
Speakers in series violates the damping provided by the amplifier then you have speakers behaving like resonating bongo drums. Read about Amplifier Damping Factor in Google. One article misprints Ohms and shows W wrongly.

I know that Acid Rock guitar players have speakers in series that are driven from an old vacuum tubes amplifier that has a fairly high output impedance but they like the Baddd bonging and spoiled frequency response effects.

EDIT: I forgot to say that using a series capacitor to cut low frequencies should be at the input of an amplifier, not at the output where it affects damping. But a speaker crossover network is different
 

throbscottle

Well-Known Member
Going back to your original project, instead of removing the pot and replacing it with resistors, why not just remove the knob and hacksaw off most of the spindle? With a little bit more effort you could even give it a screwdriver slot to make it pre-settable.
 
Going back to your original project, instead of removing the pot and replacing it with resistors, why not just remove the knob and hacksaw off most of the spindle? With a little bit more effort you could even give it a screwdriver slot to make it pre-settable.
Not sure what u mean ?

A speaker resonates. It should be connected directly (or through a coupling capacitor) to the output of a modern amplifier since the extremely low output impedance (0.04 ohms or less) "shorts" away the resonance. Resistance or another speaker in series also spoils the frequency response.
Speakers in series violates the damping provided by the amplifier then you have speakers behaving like resonating bongo drums. Read about Amplifier Damping Factor in Google. One article misprints Ohms and shows W wrongly.

I know that Acid Rock guitar players have speakers in series that are driven from an old vacuum tubes amplifier that has a fairly high output impedance but they like the Baddd bonging and spoiled frequency response effects.

EDIT: I forgot to say that using a series capacitor to cut low frequencies should be at the input of an amplifier, not at the output where it affects damping. But a speaker crossover network is different
Well to be honest, i just want that speaker to leave my house cause its driving me nuts. I had to put both full ranges in series cause putting them in parallel made them way to loud and i didnt wanna use resistors to lower their power, i hate wasting battery live. Anyway what i did is connect both speakers in series and put 20uF in series on the + side that goes into amplifier.Since both speakers are 8 ohms, thats 16 ohms together so 20uF is supposed to give me -6dB at 500Hz. We will see if it works or not, i dont even care much, all i care is minimise the distortion the speaker produces under 100Hz when driven to high. Like i said, i just want this speaker to leave the house cause its made my life a living hell. The next one i make, i will make it how i want it to be, not how my friend wants it to be. Its impossible to make a good speaker when a friend says how he wants it to be build, even if it goes against physics ...
 

audioguru

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20uF into 16 ohms is 500Hz where the level is reduced -3dB, not -6dB due to the phase shift. That is much higher than the resonant frequency so they will be fine.
Reducing the battery voltage also reduces loudness and battery power.
 
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throbscottle

Well-Known Member
Only skimmed the thread since the first couple of pages really. So I looked at the pic where you'd un-soldered the pot and soldered in some resistors. So I am suggesting that instead of removing the pot and fitting resistors, you could cut off its shaft with a saw, leaving a tiny (like, 5mm) stump you can still adjust up to max or whatever with your fingertips, or as a refinement, cut a little groove into the end of it to take a screwdriver. And still bridge the wiper pins as per rjenkinsgb. I just thought it would be a simpler way to remove the external control.
 

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