This is part one ... too big to fit in one post 🙂 ...
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Stage-01 Bulk Charging of a 12V Lead Acid Battery (Constant Current)
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How to Properly Charge a 12V Lead Acid Battery:
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below are some of the recommendations on how to charge a lead acid battery that you may read in the internet:
1. some battery manufacturers recommends that the battery be charge at a voltage which is 17% to 18% higher than the listed voltage of the battery (e.g. 6V, 12V, 24V, etc.).
example: a 12V battery should be charge with a voltage of 14.04V (12V * 1.17 = 14.04V).
2. some battery manufacturers recommends that the charging of the battery should be the same as the recommended discharge use of the battery.
example: a motorcylce battery with a stated capacity of 5AH (ampere hour) for 10 hours (C10) of usage. this means, to use the battery for 10 hours, the current usage should be 0.50A (5A/10 hours =0.50A ) per hour.
3. technical experts recommends that a battery should be charge at 1/10th or 1/20th of its rated AH.
example: if the rated capacity is 5AH, then 1/10th of this is 0.50A (for 1/20th its 0.25A).
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3-Stage Charging Process for Lead Acid Battery:
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good quality chargers in the market for lead acid batteries uses a 3-stage charging process. there are a lot of articles and discussion on forums in the internet about this which you can read. the stages are:
Stage 1 (bulk charging) use of a constant current to charge the battery and bring it up to a certain voltage level.
Stage 2 (absorption charging) use of a constant voltage to charge the battery and gradually bring down and stabilize the current to the level of a fully charge battery.
Stage 3 (float charging) use of a constant voltage to re-charge the battery when needed to keep it at optimum charge level ready to be use at any time. in a normal condition, a lead acid battery (not in use) will self-discharge at a rate of about 4% a week.
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Background on the LM317 Circuit:
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it's basically an adjustable power regulator chip that you can use to control the charging process of a battery. it's maximum current output is about 1.5A and can work in a wide range of voltage from 1.25V to 37V. for it to do its work (control the voltage or current at a certain level), its voltage source (input) should be at least 4.25V (1.25V + 3V headroom) higher than the desired output voltage.
example: if you want to charge a 12V battery, you need the LM317 to output a voltage of around 14V. charging voltage should be higher than the voltage of the battery to be charge so that current can go into the battery and recharge it.
this means, for the LM317 to work properly, its input voltage should be around 18.25V (14V + 4.25V = 18.25V) to charge a 12V battery.
you can learn more about how the LM317 works by downloading its datasheet and reading about it in the internet.
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A Circuit for a Stage-1 Bulk Charging and how the Circuit works:
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Photo: Constant Current 07 - Adjustable Power Regulator LM317.jpg
the circuit on the photo was taken from an LM317 datasheet of TI (Texas Instrument) which basically describe how to charge a 6V battery using an LM317 with 'limited current'.
if your familiar with electronics and understand (e.g. reading the process just like a software program :-)) how the circuit works, you will notice that the 'limited current' here is simply the same 'constant current' describe in a 3-stage charging process, specifically the stage-1 bulk charging.
below is a short description of how the circuit works:
1. output voltage and current of the output pin of LM317 is controlled (regulated) by the voltage that the adjust pin gets (reads) from the voltage divider made up of R1 and R2.
2. the output voltage and current is then fed to the battery being charge (right most with horizontal bars). as such, in the charging process, the voltage (and current) of the battery will increase over time.
3. as the battery's voltage/current level increases, the current above R3 will also increase. why? the negative pole of the battery is connected just above R3. in a hardware circuit, current is the most important thing as it is the one that drives everything to move in a circuit.
4. R3 controls the amount of current that is use by the circuit to charge the battery. why does it have no value? well, it's the one you set for your self based on how much current you want to charge a battery. i'll show later how this is computed.
for now, to complete the explanation of how the circuit works, just assume that you want R3 to allow only 0.5A to pass thru it and goes to VI- to complete the circuit.
5. if the current before R3 becomes > 0.5A as the charging progresses, it cannot pass thru R3, it has to go somewhere. this excess current will then go to the transitor's base pin to start activating the transistor (e.g. the circle with an arrow).
6. once the transistor activates, it will open the circuit connection between the adjust pin of the LM317 and VI-. what happens then? the adjust pin's current will sink into VI- (current flows to where there is less resistance) causing it (adjust pin) to get a lower voltage reading than what it was reading between R1 and R2.
with a lower voltage reading, the adjust pin will also adjust the output voltage to a lower level, thus reducing the voltage and current going into the battery being charge. this will also reduce the current going into R3 until it goes down to 0.5A at which point the transistor will deactivate (no more excess current to turn it on).
7. the whole process repeats its self over and over again while the battery is being charge ... output voltage/current going into the battery going up => when current goes above 0.5A, transistor will activate to force adjust pin to lower the output voltage and current => output voltage/current goes down and transistor will deactivate ... repeat again the whole process.
to sum up whats happening in this simple circuit ... the battery is being charge at a constant current of say, like 0.5A, and making sure that the charging voltage does not exceed a certain level ... its the same as what we expect from a stage 1 bulk charging of a lead acid battery.
this is the last part of the post ...
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Manually Verify if the Circuit 'Really' works:
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this is the most exciting part. if you look at things in a bigger perspective, and manually verify everything thru computation, you will see that this simple circuit is 'really' quite good.
by this ('really'), i mean this simple circuit will charge a lead acid battery in a way that is recommended by manufactures and experts 🙂 he he he ... below are the computation on how to verify it. all that is needed is a simple understanding of basic electronics:
Assumption: the 6V battery is a sealed lead acid (SLA) battery that you normally see being use for an emergency light battery or for whatever purpose.
1. how to get the output voltage of the LM317 that charges the battery?
it's one of the critical info on how the circuit works and its so simple to compute because it's already documented in the datasheet :-). the formula is below:
Vo = Vref * [1 + (R2 / R1)] + [Iadj × R2]
where:
Vo = voltage output
Vref = 1.25V ; it's a constant value that LM317 uses in its normal operation
R2 = 1.1KΩ ; info from the circuit diagram
R1 = 240Ω ; info from the circuit diagram
Iadj = 50 μA (nanoAmpere) = 0.000050A it's also a constant value that LM317 uses
1 nanoAmpere = 1/1000000 of 1A (ampere)
so just plug in all of the values and compute the output voltage of LM317
Vo = Vref * [1 + (R2 / R1)] + [Iadj × R2]
Vo = 1.25V * [1 + (1100Ω / 240Ω)] + [0.000050A × 1100Ω]
Vo = 1.25 * (5.5833) + (0.055)
Vo = 7.0342V
notice the following:
1. for actual charging to take place, Vo (7.0342V) must be higher than 6V.
2. the recommended charging voltage for a battery should be 17% (1.17) to 18% higher than the listed voltage of the battery.
6V * 1.17 = 7.02V Vo is within the 7.02V and 7.08V range 🙂
2. to complete the charging info, we need to determine how much 'constant current' do we need to charge the battery for stage 1 bulk charging. this is R3 in the circuit which has no value. thats because this needs to be computed based on the actual battery that you will be charging.
i went to some shops in marbel to actually look at some 6V SLA batteries being sold. the big ones have a capacity printed on the battery as 5AH at C10 (means 10 hours capacity). it simply means, the battery will deliver a total 5AH if use for 10 hours. this can be express in a simple formula below:
5AH/10 hours = 0.5A
what the battery manufacturer is saying is that they are recommending to you that the proper use of this battery is to use it to power a load that will consume up to 0.5A per hour. if you do this, then you can get 10 hours of usage for this battery before the battery gets fully discharge.
with the above info, you can get the 'ideal' charging current for the battery as:
1. charging current should be = to discharge rate. so from the above, it's obvious that charging current is 0.5A.
2. other experts say to use 1/10th or 1/20th of the rated AH of the battery. so to compute it:
1/10th of 5A is also 0.5A (5A/10=0.50A)
1/20th of 5A is 0.25A (5A/20=0.25A)
so it makes sense to use 0.5A as the 'constant' charging current for stage 1 as it satisfy all the recommendations of the manufacturers and the experts :-).
so using ohms law of V=I*R, we can easily compute for R=V/I:
R = V / I
where:
R = resistor value of R3
V = 7.0342V voltage of the battery being charge
I = 0.50A current of the battery being charge
= 7.0342V / 0.50A
R3 = 14.0684Ω
R3 = 15Ω closes standard resistor value
the closes standard resistor value to 14.0684Ω is 15Ω. so if we use 15Ω instead of 14.0684Ω, the current will be 7.0342/15 = 0.4690Ω. this is already ok as the difference is not that big, about 0.031A or 31mA.
3. last note, the input voltage source to the circuit must be at least 4.25V higher than the output voltage of 7.0342V. this means a minimum input voltage for the circuit of 11.2842V. so any 12V voltage adaptor that can deliver 0.5A will do :-).
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Summary:
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what makes this circuit really great:
1. it's simplicity. if your input voltage is already well regulated (e.g. you have a good 12V or 15V regulated transformer), then you only need this 5 parts to do stage 1 bulk charging. in case the LM317 gets too hot, just add a heat sink to it.
2. to charge a battery of a different voltage (e.g. 12V), all you need to do is to change the values of the 3 resistors and the formula is quite simple to manually compute. if you need a higher output current, you can also change the LM317 to a bigger power regulator.
3. controlling the circuit via an MCU is quite straight forward as well, all you need is a latching relay that receives it's control input from an MCU to switch OFF or ON the charging process.
4. and lastly, indirectly, it also confirmed that what the battery manufacturers and other technical experts familiar with battery charging are saying is correct on how to make a good quality lead acid battery charger.
the tech guy/s who prepared this datasheet works for TI and for them to use this circuit as a sample application in their datasheet means that they also believe that what the other tech guy/s in another industry is saying also makes sense :-).
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cheers
Boo!
Very informative :cool01:
pre, i'm working on a new strategy for creating my circuit board after that bad experience with some of the resistors i bought in marbel. here is how i'm approaching my first circuit board ...
1. use ICs as much as possible instead of passive parts if possible.
2. instead of making a circuit using a prototype board. simply solder the ICs into the prototype board and then separate the area from the rest of the prototype board by breaking the copper wire trace.
3. the copper trace wire should be broken at a point that will leave at least 3-5 holes for each of the pins of the IC or any electronic part for that matter for wire connection.
the idea here is to install the part permanently into a circuit board like a breadboard (e.g. each pin has 5 extra pins for connections).
4. to connect the parts together, use AWG #24 copper solid wire (rated at 577mA). i can use the wires of a CAT5e solid wire cable for this. i found a shop here in marbel that sells the item for P10 per meter, really quite cheap 🙂 which is about 8M since you have 8 wires in a CAT5e cable :-).
the above method hopefully will solve my poor skills in making circuit board 🙂 he he he ... if this is still not easy for my current skills in soldering ... another method i'm looking at is to use a perforated board with no copper wires and then just connect each of the parts using the AWG #24 copper wire.
hopefully, with the 2 methods above, i will be able to make some working circuit boards for my 3-stage automatic charging so that i can start to really test my HAS system in an actual working environment.
i got the idea when i saw a photo in the internet of how they made the first IBM PC prototype in the late 1970's ... they just connected every IC chip using wires - i think they made those wires connections when they found some bugs in the original prototype circuit board they made. it's simply than making a complete circuit board again and going back to square 1 to build another prototype.
any comments on the above will be appreciated 🙂 ... cheers
Boo!
DIYers Innovate ...
go for the first method. mas prone sa error yung 2nd option and mas mahirap mag debug/modify the circuit. for high current traces, fill it with solder. maliliit lang kasi ang mga traces sa gilid ng mga holes sa perfboard.
, thanks for the feedback, i just want to make sure that i'm on the right track. duly noted about the solder tip on the high current ... cheers
Boo!
DIYers Innovate ...
how is your charger now?
pre, i have to stop the things on my charger hardware for now because i'm force to do the wireless functionality of HAS since the start of this year. it was the common feedback i got from the companies that i have approach with my HAS software. it's a bit tough given the limited program space of the MCU ... cheers
Boo!
DIYers Innovate ...
