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simple precharging circuit will this work?

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terramir

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Okies I have a bunch (2000+) of never used but dropped down to 0V LiFePO4's now I dun expect all of them to revive, matter of fact I already got my money's worth with the ones that were at a higher than 0V and most of those seem to be holding a charge fine. Anyways I attached a picture of my circuit. this is what I'm thinking the Q1 MPSA06 is happily letting through 107mA which the 47 ohm resistor will let through. when the battery being charged rises in voltage to 3.31V the input into the base of Q2 rises to .7V at that time the base of Q1 get's starved until the voltage in the li-FePO4 cell drops below 3.3V then q2 turns back off and the charging will continue again untill the cell rises past 3.31V again so the LED will flash when the cell is precharged to 3.3V and the LED will keep flashing and the cell won't be able to charge past this point, which means it's protected as well.
The when I see them flashing I can put them in my real charging circuit CC/CV to 3.65V it's just the cells being stored that long shouldn't be exposed to .5A of current till there in the normal voltage range for the cell. .1C is enough for the precharging.
Will this circuit work am I missing something did I get something wrong?
BTW I'm kinda married to the NPN's cause they were 4.5c each and these circuits (I'm gonna make 20 of em, they are only gonna be used for a few times till I have charged all the cells once then shelved, so cheap and dirty is the goal here.
terramir
 

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MikeMl

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I had to futz with the resistor values to make it work. Battery is modeled as a 5 FARAD capacitor, with an initial voltage of 2V. Final voltage is dependent on the forward drop across the LED, which for the one I picked is about 1.8V. Constant current behavior prior to battery reaching 3.3V isn't too good.
 

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Roff

Well-Known Member
Q2 won't turn on until the LED is on, so you have to add the LED fwd voltage to Vbe in your calculations. If you want the charging to stop at 3.3V, then the LED current will be about 4.5mA as your circuit now stands. You need to measure the LED forward voltage at this current, and add that to the Q2 Vbe in order to calculate your R3, R4 voltage divider.
As the cell approaches your target voltage, the LED will slowly transition from off to on. It won't flash. If the LED isn't bright enough, reduce the value of R2. This will require an adjustment of your voltage divider, because the LED fwd voltage will change.
The best design approach is to first decide the LED current you want, depending on the desired brightness. Measure or look up the LED fwd voltage at that current.
R2=(Vcc-Vbe1-Vtarget)/Iled.
For example, if Vcc=5, Vtarget=3.3,Vbe1=0.7, and Iled=10mA, then
R2=(5-0.7-3.3)/.01.
R2=100Ω.
If Vled=2V at 10mA, then Vb2=2+Vbe2≈2.7V
R4/(R3+R4)=2.7/3.3
Let R4=2.7k
R3=600Ω.

Since Vled will vary slightly from one unit to the next, you should incorporate a pot into the voltage divider if you want to set the target voltage precisely. If it can safely vary by a few tenths of a volt, then you probably don't need a pot.
 

Roff

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Mike modeled the battery the same way I did, and got similar results.

Mike, I'm a little miffed that you never commented on the solar cell solution I offered.:eek:
 
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dougy83

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Okies I have a bunch (2000+) of never used but dropped down to 0V LiFePO4's
I'm jealous.
when the battery being charged rises in voltage to 3.31V the input into the base of Q2 rises to .7V
You've forgotten to take the forward voltage drop of the LED into account. The base voltage will have to be more like 2.7V.
charged rises in voltage to 3.31V the input into the base of Q2 rises to .7V at that time the base of Q1 get's starved until the voltage in the li-FePO4 cell drops below 3.3V then q2 turns back off and the charging will continue again untill the cell rises past 3.31V again so the LED will flash when the cell is precharged to 3.3V and the LED will keep flashing and the cell won't be able to charge past this point, which means it's protected as well.
Even with the correct resistor values, I can't imagine the LED flashing, it would just come on and stay on; there is no positive feedback. The current to the cell would be reduced though when the set point was reached.
Will this circuit work am I missing something did I get something wrong?
Not as is.
so cheap and dirty is the goal here.
Why not just use a 3.3V power supply and a single resistor to each LiPO cell? It's definitely cheaper, and won't charge your cell past 3.3V.
 

MikeMl

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...Mike, I'm a little miffed that you never commented on the solar cell solution I offered.:eek:
I'm building it. I'll report on how it works.
 

terramir

New Member
okies I'm a little confused :s but then again I'm new at this, Ive dabbled a little but I'm not really good at this yet :p. I thought I was using the transistors as switches, I saw his graph and it seems like the current stops rising but still continues flowing once the led goes on :( this is not what I wanted to accomplish :( I wanted to short the base to ground through the led so Q1 would effectively turn off.
I figured once the current to the battery stops the voltage of the battery would drop (which is natural behavior for liFePO4 cells). Then the q2 would turn off and q1 would resume to conduct current and this would cycle until the voltage drop in the cell was so minute as not to immediatly turn off the led. So somewhere I went wrong here anyone got a hint or two? Cause this is just a precharger once it's charged I can use the attached charger to fully charge the cell (see attached file) if I add a resistor to the base of q1 on my original schematic would that allow for starving q1 and effectively turning the current to the battery off? Cause my goal is to turn the current off and have a light go on so I can see when it's done so I can put it in the other charger.
Thanks for all your comments so far.
terramir
 

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terramir

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I'm jealous.
You've forgotten to take the forward voltage drop of the LED into account. The base voltage will have to be more like 2.7V.
Even with the correct resistor values, I can't imagine the LED flashing, it would just come on and stay on; there is no positive feedback. The current to the cell would be reduced though when the set point was reached.
Not as is.
Why not just use a 3.3V power supply and a single resistor to each LiPO cell? It's definitely cheaper, and won't charge your cell past 3.3V.
Okies 3.3V power supplies are easy enough to come by, but there is a caveat in the li-ion family chemistry that would seriously screw me up all lithium based rechargable chemistries can be charged to a lower than max voltage as the top voltage. So this is what would happen the cell would charge to about 40 % of capacity (liFePO4 max is 3.6V to 3.65V when it reaches this voltage it's 60% charged) so at 3.31 V it would be charged to 40% and then eventually the current would drop to .05 of C and it would be "full" at 3.31V albeit it would only have a capacity of about 60% of C this would waste a bunch of time and not really achive my goal to normalize these over stored cells. While the Li-FePO4 cells are much more hardy than traditional li-ion and li-polymer chemistries I need to "train the internal chemistry back to normal" so to speak.
 

MikeMl

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... I thought I was using the transistors as switches, I saw his graph and it seems like the current stops rising but still continues flowing once the led goes on :( this is not what I wanted to accomplish :( I wanted to short the base to ground through the led so Q1 would effectively turn off.
I figured once the current to the battery stops the voltage of the battery would drop (which is natural behavior for liFePO4 cells). Then the q2 would turn off and q1 would resume to conduct current and this would cycle until the voltage drop in the cell was so minute as not to immediatly turn off the led. So somewhere I went wrong here anyone got a hint or two? ...
The GRAY trace I(C1) in the upper plot pane in my sim is the current into the battery. Notice that when the PINK trace V(bat) is at 1V, the charging current is 130mA. As the battery charges toward 3.3V, the current decreases linearly to 45mA. This is exactly the same behavior as you would get if you simply connected the battery to a 5V supply through R4 (33Ω). Upon reaching 3.3V, the battery voltage is held constant, and the current rapidly decreases to zero.

The LED current I(D1), BLACK trace, is in the bottom pane. Note that the LED turns ON just as the charging current begins to taper to zero.

The circuit does what you want it to do: namely charge rapidly at first, slowly towards the end, and then it stops charging when the battery voltage reaches 3.3V. The only thing that would make it better is if it maintained the ≈100mA charging current right up to the point when the battery reaches 3.3V

There is no "foldback" or LED blinking behavior!Ω
 
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terramir

New Member
Ahhh yes Nvm I understand why the current decreases, it's the potential difference that count's and not the actually 5V voltage that's where I went wrong. 1.7V/33 ohms=.051A got a question could you plug this back through that program you have with r4=47 ohm because 135 mA when the cell is at 0V would be shaking things up a bit too much in my opinion. Anyways I'm starting to understand the mistake I made with the voltage divider.
Thanks
terramir
BTW: In order to do constant current how would I go about it? short of using a variable power supply and increase the voltage linearly with the battery voltage or somehow decrease the resistance as the battery voltage rises I dun know how I would do that.
 
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MikeMl

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...Is there a problem with saturation of q1 that is causing the current to decrease? I thought that when transistor is on there is virtually no voltage drop. Maybe I'm still thinking to close to a relay model and maybe another factor is the voltage drop over the resistor ? So what is causing the current to drop ? ...
The current into the battery is equal to the current through the 33Ω resistor. By Ohms Law, the current in the resistor is (5-Vbat)/33. As Vbat increases, the current decreases...
 

Roff

Well-Known Member
I thought a transistor only has two states on or off?
You apparently have never looked at the schematic of a linear circuit such as an amplifier.:D
Or if you have, did you think all those transistors were switching on and off?:confused:

It's not clear to me what you actually want the circuit to do. Try to explain, and we can probably design something that will work.
 

terramir

New Member
You apparently have never looked at the schematic of a linear circuit such as an amplifier.:D
Or if you have, did you think all those transistors were switching on and off?:confused:

It's not clear to me what you actually want the circuit to do. Try to explain, and we can probably design something that will work.
Actually it does work, but it doesn't have constant current on the simple. just want it to put through 100 mA to the cell untill 3.3V is reached and then shut the current to the battery dead off until such time when the cell goes below 3.3V, but actually this circuit will do. The constant current it not a requirement it would have just been nice to have (cause it would have been faster).
These chargers will cost me a total of about 25 cents each, plus the battery holders, but those I can reuse so unless I can find a solution that will cost me not much more these will do, I bought 100 mpsa06's for 4.50 yesterday and I just bread-boarded (mike's revised one) the circuit it is behaving as expected three changes: I made i'm using a 47 ohm r4 and a 150R1 and well I'm using the mpsa06 transistors instead of the ones in his model)I expect it to turn off a bit earlier like around 3.2V but that is ok. so this will trickle charge the cells until they reach around 3.2V and then the l;ight will go on then I know it's time to put them into the other charger to fill them (that charger could deliver like an amp of current to the cell if it was at 0V). So this will do unless someone knows how I could add a transistor or so more to make it better this will do.
Thanks for all your help
I will keep the jelous one updated on how many cells in the end hold a charge :D
Ty
terramir
 
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terramir

New Member
oops got a problem the thing seems to terminate at 3.05V and I actually went out and bought 180Ohm resistors @ r1 being 150 ohms it was like 2.9 something could it be that this transistor MPSA 06 has a lower trigger voltage than 0.7V so the calculations are off?
I tried to read the data sheet but I can't make heads or tails of the thing (help)
terramir
 

MikeMl

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Most Helpful Member
The cutoff voltage is effected by four things: the forward drop across the LED, the Vbe of the lower transistor, and the voltage divider.
 

Roff

Well-Known Member
oops got a problem the thing seems to terminate at 3.05V and I actually went out and bought 180Ohm resistors @ r1 being 150 ohms it was like 2.9 something could it be that this transistor MPSA 06 has a lower trigger voltage than 0.7V so the calculations are off?
I tried to read the data sheet but I can't make heads or tails of the thing (help)
terramir
I hate to say I told you so, but you should use a pot.:rolleyes:
 

terramir

New Member
The cutoff voltage is effected by four things: the forward drop across the LED, the Vbe of the lower transistor, and the voltage divider.
Yeah well I saw the transistor spec vbe is like 1.2 but it seems like I figured it out a 100 and 470 seem to work like a charm even though mathamatically that would be 3.517V it ends up being like 3.35V with a 1.7V drop from a diode it's strange but I think I'm gonna have to play around with every circuit I make since the diodes are all different. but there all red so they should be between 1.7 and 2V good thing I decided to breadboard all these circuits because it's less wasteful in the end.
I remembered I have this END (electronic network designer) can't get boards for it anymore the company is defunct but it's got a good breadboard under it.
one of these day's I'll have some manufaturer make me some boards for it :D but for now it makes a real good bread board.

I hate to say I told you so, but you should use a pot.:rolleyes:
yeah the only pots I got laying around are 1k trimmers one turn so they would be a pain to adjust, and I need them for my chargers, someone is etching me a board to make 6 (on one 6 1/4 x 4") and then I'm gonna get 3 (so 18 more) more later in the month.
terramir
 

terramir

New Member
Okies I got 10 circuits running but it's not as clean as I would like it to be. the voltage cut-off point is to fluid for my taste. Someone on another forum suggested using the lm339 which is a voltage comparitor. This would probably be optimal if I know how to work it. But honestly I dun have a clue.
Any ideas where the circuit would do the same thing but with a very distinct cutoff point set by let's say one lm317 for all 20 charging slots, let's say 5 of those lm339's for the 20 battery slots and well I can reuse those mpsa06 for the switching operations.
terramir
 

Roff

Well-Known Member
Okies I got 10 circuits running but it's not as clean as I would like it to be. the voltage cut-off point is to fluid for my taste. Someone on another forum suggested using the lm339 which is a voltage comparitor. This would probably be optimal if I know how to work it. But honestly I dun have a clue.
Any ideas where the circuit would do the same thing but with a very distinct cutoff point set by let's say one lm317 for all 20 charging slots, let's say 5 of those lm339's for the 20 battery slots and well I can reuse those mpsa06 for the switching operations.
terramir
Do you want a true constant current "precharger", with an adjustable cutoff voltage and a little hysteresis? If so, tell us the charging current, the cutoff voltage adjustment range, and the hysteresis.
 

terramir

New Member
Do you want a true constant current "precharger", with an adjustable cutoff voltage and a little hysteresis? If so, tell us the charging current, the cutoff voltage adjustment range, and the hysteresis.
What's hysteresis? cutoff voltage range is 3.2V - 3.4V constant current is 100mA but a selectable range of 50mA to 200mA would be great for experimenting, on what does the 0V cells better in terms of revival rates.
terramir
 
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