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Batter Charger AGM 12V 4 Amp/hr

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voodoo5293

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I want to build this battery charger:



To charge this battery:

Model: YTX5L-BS


I want to limit the current to about .75A. So I guess that means I need to use a .8 ohm resistor for the resistor that it doesn't have a value for, unless I am mistaken. I also want to add an LED that will light up to indicate the battery is fully charged. I have no idea how to go about doing that. Also, what voltage should I set it to charge at? 13.8? One other thing, what does the transistor do in this circuit? I have seen many similar circuits using the same LM317, and none of them used any transistors. Thanks in advance for any help.
 
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ericgibbs

Well-Known Member
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I want to build this battery charger:
I want to limit the current to about .75A. So I guess that means I need to use a .8 ohm resistor for the resistor that it doesn't have a value for, unless I am mistaken. I also want to add an LED that will light up to indicate the battery is fully charged. I have no idea how to go about doing that. Also, what voltage should I set it to charge at? 13.8? One other thing, what does the transistor do in this circuit? I have seen many similar circuits using the same LM317, and none of them used any transistors. Thanks in advance for any help.
hi,
The voltage drop across that resistor has to be about 0.65V at the 0.75amp , so thats 0.86R.

Just adding an LED to indicate full charge will not give a good result. It really requires a voltage sensing circuit.

The transistor, the base is connected to the high side of the 0.86R, so when the current thru the resistor causes a voltage drop across the resistor, the transistor will be turned on at 0.65V. As the collector of the transistor is connected to the top of the SET pot, it will pull down the set voltage, so the current into the battery will be limited to 0.75A.

OK.?:)

EDIT:
if the battery is going to be left connected to the charger for long periods of time, set the Vchg at 13.8V

If you want a faster charge then 14.0V, BUT dont leave the battery connected after its fully charged.
 
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MikeMl

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Eric did the heavy lifting.

Most 12V AGMs can be floated at 13.5 to 13.8V, and provided that the current is limited like your circuit does, can be charged to 14.5 to 14.7V. During charge at the higher voltage, watch the charging current; when it naturally tapers to less than 0.25A, switch your charger from 14.7V to less than 13.7V. You can do this with a switch and second 5K trimpot in your circuit. If you want to get fancy, you can add a controller which will switch from "charge" to "float" automatically. Do not leave it in charge mode after the charging current tapers.
 

voodoo5293

New Member
Eric- Thank you very much. I am still a novice in electronics, so I don't quite understand how you got some of your numbers. Like how did you get the .65V voltage drop across the resistor? I also don't get how transistors work in general, so your explanation, while I am sure is correct, makes no sense to me. I have read, and read, and read, and transistors are still mysterious to me. I need to do some more reading about them I guess. Any advice on where to go to read up on them or any books to buy?

Also, you mentioned adding a voltage sensing circuit, for the LED. What would that consist of? Would a transistor and resistor(or potentiometer) suffice for voltage sensing to turn on the LED when the battery reaches float voltage? I don't know, I am kind of shooting in the dark, because like I said before I don't fully understand transistors, but I think I have seen them used in that type of application before.

And yes, I just want this to be a float charger, so 13.8Vsounds good. I was just worried, as I have read a bunch of different suggestions as to what voltage to use for float charging an AGM battery.

Mike- I don't want to get to complicated, so I will just leave it floating.
 

ericgibbs

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Most Helpful Member
Eric- Thank you very much. I am still a novice in electronics, so I don't quite understand how you got some of your numbers. Like how did you get the .65V voltage drop across the resistor?
The base to emitter voltage to turn ON a transistor is about 0.65V
'turn ON', means that current starts to flow from the collector to the emitter.
I'll keep it that simple at this stage.


I also don't get how transistors work in general, so your explanation, while I am sure is correct, makes no sense to me. I have read, and read, and read, and transistors are still mysterious to me. I need to do some more reading about them I guess. Any advice on where to go to read up on them or any books to buy?
There are loads of good tutorial ebooks on the web

Also, you mentioned adding a voltage sensing circuit, for the LED. What would that consist of? Would a transistor and resistor(or potentiometer) suffice for voltage sensing to turn on the LED when the battery reaches float voltage? I don't know,
A transistor could be configured to give a rough indication of the 13.8V terminal battery voltage.

I am kind of shooting in the dark, because like I said before I don't fully understand transistors, but I think I have seen them used in that type of application before.

And yes, I just want this to be a float charger, so 13.8Vsounds good.
Hi,
I'll look thru my ebook links.


EDIT:
http://www.electronicsteacher.com/tutorial/

http://www.allaboutcircuits.com/
 
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MikeMl

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Here is an "almost charged" indicator you can add across the battery terminals. The LED lights up as the voltage increases. The trimpot adjusts the turn-on voltage. In the simulation, the X axis is the increasing battery voltage. The vertical axis shows the current through the LED at different values of the trimpot...

The reason I call it an "almost charged" indicator is because it will turn on before the battery is fully charged. The only way to determine that the battery is "fully charged" is to monitor the current into it, and then light the LED when the current drops below a preset low value. That requires a more complex circuit.

btw- you could put an LED in series with the collector of the BC547 (LED cathode connected to BC547 collector). The LED would start out lit, and when the battery is charged enough to where the current drops below the current-limit threshold, it will go out. Again, this will happen when the battery is only ≈75% charged. Detecting that the battery has accumulated the last 25% is harder...
 

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voodoo5293

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Thank you so much. I have a question. In your first post you said the high side of the .86R is connected to the base of the transistor. Isn't the base of the transistor connected to the 100R? I am just trying to figure out how everything is working together.

I believe this is how most of the stuff is working. The 220R and 5K pot. seem to be making a voltage divider which the collector of the transistor is connected to.Why it is done that way, I have no idea. Correct me if I am wrong.

The 100R and .86R I am not to sure about. I am assuming the 100R is to bias the base of the transistor to get current to flow. Or something like that. I don't get how the .86R fits into this circuit at all.

Sorry for asking so many questions. I just really want to understand this so I can design this kind of stuff on my own.


Also, does anyone have any ideas on how to implement the LED? Thnks again guys. I really appreciate all the help.
 

MikeMl

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Thank you so much. I have a question. In your first post you said the high side of the .86R is connected to the base of the transistor. Isn't the base of the transistor connected to the 100R? I am just trying to figure out how everything is working together.

I believe this is how most of the stuff is working. The 220R and 5K pot. seem to be making a voltage divider which the collector of the transistor is connected to.Why it is done that way, I have no idea. Correct me if I am wrong.

The 100R and .86R I am not to sure about. I am assuming the 100R is to bias the base of the transistor to get current to flow. Or something like that. I don't get how the .86R fits into this circuit at all.

...
I entered your circuit into LTSpice and ran a simulation. I had to substitute some of the parts because I didn't want to take time to find the ones specified.

In my version, Q1, R2 and R5 create the current limiter which I will discuss later. Without the current limiter in the circuit, you would have just the LM317 whose output is adjusted to 13.7V (by voltage divider R1 & R6). 13.7V is the recommended float charging voltage of the battery. However, if the battery is initially discharged, its terminal voltage is much lower than 13.7V so that excessive current would flow through the LM317, overheating it, and the maximum recommended charging rate for the battery would be exceeded.

The way that the current limiter works is by draining current (to ground) via Q1. This lowers the voltage at node A (V(a) in the plot) such that the current that flows into the battery (and through R5, I(R5) in the plot) is reduced to a safe limit. As the battery voltage approaches the LM317 set voltage, Q1 turns off so that it no longer steals any current from node A, allowing the LM317's output voltage to be held at 13.7V.

Stare at the simulation. Notice how the battery voltage V(plus) climbs and then levels out. Note how I(R5), which is the charging current, starts out constant, and then goes to almost zero. In the lower plot panel, note the collector current of Q1 I(Q1) which reduces the voltage at node A V(a) in the upper panel. The voltage developed across R5 turns on Q1 when it exceeds about 600mV. R2 is there only to limit Q1's base current to a safe value.
 

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voodoo5293

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I was getting ready to order parts for this, and I wasn't sure about the transformer. What should the VA ratting of it be?
 

MikeMl

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Depends on what you set the current limit to. Suppose you choose 1A. Power into the battery would be 14*1=14W. To derate the transformer for the low-duty current pulses from the secondary winding into the rectifier, de-rate it by a factor of two, so 28VA.
 
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