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Lm317 battery charger

Discussion in 'General Electronics Chat' started by baxterdmutt, Dec 21, 2016.

  1. baxterdmutt

    baxterdmutt Member

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    So the one in this pic says1.4Ah but I have another identical one that says 1.2Ah.
     

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  2. spec

    spec Well-Known Member Most Helpful Member

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    Ah I see, the batteries are three-cell lead acid with a relatively high terminal voltage.

    1.4 Ah and 1.2Ah lead acid batteries are for all intents and purposes the same. The Ah rating is not precise and will vary from one battery, of the same type, to another battery.

    spec
     
  3. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    OK I see my error. I was trying to keep 0.6 across 4R7.
    When actually the current will go up some until the emitter of the transistor get negative enough to pull the IC down.
     
  4. dave

    Dave New Member

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  5. spec

    spec Well-Known Member Most Helpful Member

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    I do not see any error- your reasoning was correct.

    By the way, as a result of this power supply discussion, I have done a completely new design but it has turned into a precision voltage and current bench supply and I am not sure what to do with the design- maybe material for a new post or even an article. It does not use an LM317.

    spec
     
  6. baxterdmutt

    baxterdmutt Member

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    In the schematic of message #76, am I correct that R7 regulates the current?

    If you don't mind doing a bit of education, I'm having difficulty understanding how current gets regulated in the charger. I understand how voltage is regulated but don't grasp the amps. I thought that the higher the current regulators resistance setting (in the previous drawing) would be the higher the alps would be but you said ...adjusted to minimum it would give about 49ma. So I'm clearly understanding this backwards!??? Thanks
     
  7. spec

    spec Well-Known Member Most Helpful Member

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    Yes, that is exactly right. The formula for the maximum current is 0.6V/R7 which in the circuit of post #76 works out to be 0.6V/4.7R = 0.128 Amps
    No problem BM.

    From the formula, Imax =0.6V/R7 you can see that the higher resistance R7 is the less current will flow into the battery.

    How does this work:
    (1) With no current flowing into and out of the battery on charge there is no voltage drop across R7, so Q2 is turned off (Q2 will only turn on when its base/emitter voltage reaches 0.6V).
    (2) As a result Q2 has no affect on the LM317 regulator output voltage. Thus the LM317 acts like a normal voltage regulator.
    (3) But if you connect a discharged battery across the charger output terminals, the battery will try to take a large current.
    (4) But as soon as the current flowing through R7 reaches 0.6V, Q2 turns on and the current from Q2 collector drags the LM317 ADJ terminal down with the result that the charger voltage will drop and thus reduce the charging current.
    (5) After a while of being charged with this constant current, the battery voltage will start to rise until it reaches the charger output voltage that you have set.
    (6) The battery will continue to charge, but the current will gradually taper off as the battery reaches full charge.

    I hope that helps to explain how the circuit operates.:)

    spec
     
  8. baxterdmutt

    baxterdmutt Member

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    That's perfect. Thanks.
     
  9. baxterdmutt

    baxterdmutt Member

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    So I've put the designs from post 76,78 & 79 together on a breadboard. None of them work properly. Plan #76 was my final attempt (I went up the list backwards). On that one I replaced R7 with a 100R pot because I could not get sufficient current. I was unable to get more than 20-30 ma on a battery with just 3.5 volts left. (Am I wrong, shouldn't I need more than that on a battery that is drained that far?) So I thought I would adjust the current to see what happened. It doesn't adjust like the 12v charger Spec designed early on during this project. On this one adjusting R7, the voltage changes but the current does not. I noticed that the schematic still shows that the output is 14 volts and the input is 20. The plan was to just use a small 1 amp 12v wall charger as the mains source. During testing I'm using an adjustable bench top power supply, so I did try increasing the input to 20VDC. That didn't change much. The good news is that when there is no current draw the red LED goes out! I'm starting to feel that my desire for an LED to indicate when a charge is complete is just a big pain in the butt. I was able to get an adjustable charger made on my own without the LED and I should maybe have done that instead of bothering you good people just because I wanted an indicator.
     
  10. chemelec

    chemelec Well-Known Member

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    Even a Charge Indicator Won't really tell you When you have a FULL CHARGE.
    That LED Might Indicate up to 95%.

    The Final Charge rate is VERY LOW Current. (Just a Few Milliamps)
    Assuming the Battery is Really Good, Typically it can take 24 hours to Fully charge a Lead Acid Battery.
     
  11. baxterdmutt

    baxterdmutt Member

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    Hi,
    Thanks, I did know that already but I appreciate your help. It shouldn't be that low on a 6v battery that only has 3.5v though. That's an effectively discharged battery at that voltage, correct?
     
  12. chemelec

    chemelec Well-Known Member

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    That sounds like a BADLY SULFATED BATTERY!
     
  13. spec

    spec Well-Known Member Most Helpful Member

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    There are quite a few regimes for charging lead acid batteries where you trade battery life for capacity- like for most battery chemistries. I tend to favor 14.1V terminal voltage as a good compromise. I would be happy with 95% charge.:)

    Hmm, that was my impression.:eek:

    I have done quite a bit with lead acid batteries over the years, in the 7AH to 20AH range, for portable applications and have found lead acid batteries generally to be troublesome things, especially the lower capacity types.

    spec
     
  14. spec

    spec Well-Known Member Most Helpful Member

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    Hi BM,

    Sorry to hear about your problems.

    As Chemelec says, there may be a problem with the batteries.

    If you have a bench power supply with a current control can you set the current control to around 1A and the voltage of the power supply to maximum.
    Then connect the bench power supply to the battery and note the current for both of your batteries.

    If you have not got a bench power supply use a power supply with as higher voltage as available, but no more than 50V and connect a high power resistor of around 50 Ohms 50W and and an ammeter in series. Then note the current.

    If the current is around 0.75A, or higher, the battery may be OK. But if the battery only takes a few miliamps, the battery is badly sulphated.

    spec
     
  15. chemelec

    chemelec Well-Known Member

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    Hi Spec, Even 13.6 is acceptable.

    I usually charge my batteries at 14.6 Volts from my supply, but when the current gets down below 15 mA, I take them off charge.
    And I set my Maximum Current Limit to 1/10 Capacity of the Battery.
    So they are really not charged up to that voltage.

    There are Good and Poor Quality SLA Batteries.
    Most of the ones I use are the 1.2 and 2.4 A/H.
    I have some over 10 years old, Still in great shape.
    Others that died after 3 years.

    My only point was: Is that Charged up LED is Not really Accurate.
     
  16. spec

    spec Well-Known Member Most Helpful Member

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    Hi Chemelec,

    You sound like someone who has a way with LA batteries. Where I worked the other lads used them a lot without too much bother.

    There are many types of LA batteries from the high current automobile batteries to the huge static battery banks used in back-up installations, and it very much depends on balancing the cost/weight/size/life.

    In automobiles the alternator just slaps a voltage across the battery (with a current limit of around 20A to 100A upward) and that is that.

    Yes, that is the dilemma with a single LED indicator- when is the battery charged? I would much prefer a current indicator, either digital or, better still, a good old fashioned analog meter- the bigger the better.:D

    spec
     
  17. baxterdmutt

    baxterdmutt Member

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    You do know it's just a 6v 1.4ah battery? The come up to 7v just fine if I do it manually. I didn't think I was suppose to put that much amperage through a battery that small. I know the charge led isn't accurate but it gives me a reminder that they are close to finished.
     
    Last edited: Jan 8, 2017
  18. baxterdmutt

    baxterdmutt Member

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    The batteries will take the whole 1Amp. If I set my power supply to 14.6v like Chemlec says he charges his at then if I don't control the current, they will draw 1.72 amps.
     
  19. spec

    spec Well-Known Member Most Helpful Member

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    14.6V is hellish high for a 6V battery. Do you mean 14.6V?

    spec
     
  20. baxterdmutt

    baxterdmutt Member

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    Exactly! That's what I thought but Chemlec said that's where he charged his! I guess he didn't realize these are 6v 1.4ah batteries. I thought it was high but wanted to answer the suggestion that my batteries are not sulphated. I only allowed them to take that current for 2-3 seconds. I would never charge one of these at that rate.
     
  21. baxterdmutt

    baxterdmutt Member

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    Spec,
    Just to be clear, your previous design with the low amperage could be adjusted to work well at 120ma but the led wouldn't work. On your new design, the led works. But I can't get any current out of it. If I set the voltage to 7.2v with a 1k resistor (I assumed that's how I was to set it on this design also) the current is 2ma. If I then adjust R7 for more current I can get a bit more but that's because the voltage increases. Your previous design did not do that. So since basic concept is the same, something is not working correctly now and I thought you'd like to know. Does that make sense?
     

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