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Not being able to understand simple LM317 lead acid battery charger

Discussion in 'General Electronics Chat' started by Willen, Nov 25, 2015.

  1. Willen

    Willen Well-Known Member

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    Here's a simple 12V battery charger. I simulated first to understand it but became MORE confuse than before! :)

    I used to think that when battery gets full power then the transistor starts conduct and will short the ADJ pin of the Reg IC to the Gnd then the charging will stop. But how transistor conducts when bettery gets around 13.8V. Because when bettery starts to get 13.8V, the 1 ohm current limiting resistor has just few mV (voltage across it) or, upper side of the resistor is just few mV. It means the base of transistor gets just few mV too. Which means the voltage difference between emitter and base of the transistor is just few mV. The main question is how the transistor is conducting and stopping the IC, with just few mV base voltage? (base needs at least 600mV to conduct)

    Another again, when bettery has around 12V (or less), voltage across 1 ohms resistor is around 650mV. It means 650mV is in the base of the transistor. Now why transistor is not conducting when it is getting 650mV base voltage? Which thing making me silly?
     

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    Last edited: Nov 25, 2015
  2. JimB

    JimB Super Moderator Most Helpful Member

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    ??

    Please attach the schematic!

    JimB
     
  3. Willen

    Willen Well-Known Member

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    Ops sorry, I was just about sleep so. Look at once, please!
     
  4. dave

    Dave New Member

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  5. JimB

    JimB Super Moderator Most Helpful Member

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    Where did you find that circuit?

    The datasheet which I found for the LT1083 describes it as a fixed voltage regulator.
    However it does show an application where the output voltage is adjustable, but does not give detail to allow me to make calculations.
    http://www.farnell.com/datasheets/1941079.pdf

    In your circuit:
    The output voltage of the regulator is the regulator voltage plus the ADJ terminal voltage.
    If U2 is a 12v regulator and the voltage across R2 is 1.5v, the voltage at the OUT terminal will be 13.5v.

    R4 is a current sense resistor, when the battery current is 700mA, there will be 700mV across R4. This will turn on Q1.
    As Q1 turns on the current through R2 and hence the voltage across it reduces, so reducing the output voltage of the regulator.

    JimB
     
  6. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    More than you wanted to know about this circuit...

    I have redrawn the circuit you posted, and modified the simulation a bit. I changed where the ground symbol is placed to make it easier to understand what is going on. I used a huge capacitor (200 Farads) to model the battery under charge. I use the .IC statement to precharge the "battery" to 11V. I reduced the current sensing resistor from 1Ω to 0.66Ω to make the initial charging current ~1A.

    Now you can see that the circuit is basically a voltage regulator (output voltage set by R4 and R1) coupled with a current limiter (current limits when the drop across R2 turns on Q1, so it can shunt R1).

    I simulate for 1000 sec to show how initially the circuit acts a ~1A current source [see I(C1)] as the "battery" charges from 11V to near 13.2V, at which the voltage regulator takes over, and limits the final "battery" voltage to ~13.7V [see V(pos)].

    307.gif

    I am also including the .asc file so you can try it.
     

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    Last edited: Nov 25, 2015
  7. Willen

    Willen Well-Known Member

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    Lets say I am using LM317, Actually my main question is conduction through the transistor. When it conducts?

    I used think when bettery is in heavy charging state (low voltage across battery), the trasistor does not conduct. When battery gets full charging voltage, the transistor conducts and shorts the ADJ then the charging stops through LM317. Overall circuit operation making me confuse.

    But simulation shows the transistor conducts when bettery is in heavy charging state. Probably I made wrong concept.
     
    Last edited: Nov 26, 2015
  8. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    You made the wrong conclusion about how the circuit works. It is just the opposite.

    Refer to my schematic (not yours). The transistor Q1 conducts (lowering the output of the LM317/LT1083) only when the current through R2 (same as the battery current) is >=1A, which causes Q1's Vbe to exceed ~0.66V, turning on Q1. This happens early in the charging cycle, only when the low battery voltage would cause a very high current flow if it were not for the the current limiter circuit. This is a negative feedback circuit, where if the output current exceeds a set limit, the current-limiter circuit reduces the output voltage of the regulator by shunting current away from R1, thereby lowering the regulator's output voltage.

    The final voltage on the battery many hours into the charging cycle is set only by the voltage divider consisting of R4 and R1. By then, the current in R2 is so low that Vbe is low enough so that Q1 is totally off, meaning that the collector current of Q1 is not affecting the voltage divider.
     
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  9. Mosaic

    Mosaic Well-Known Member

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    With a 17V supply and a possible 6V drop across the regulator at 1A load we have 6W of regulator heat to be handled. You're going to need a heatsink for the LM317. A To220 pkg has a 19°C/W spec. Thus you can hit almost 120°C over ambient (30°C?) without a heat sink. A max 150°C Tj is the rating..so you'll probably go into thermal shutdown w/o a heatsink.

    BTW, u can add a pass transistor to share the load and avoid a heatsink.
    http://www.ee.teihal.gr/labs/electronics/web/downloads/LM317T_Variable_Voltage_Regulator.pdf

    Edit:
    I note@MikeMl has 'disliked' my post. It seems he believes his posts/circuits cannot be improved upon.
     
    Last edited: Nov 27, 2015
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  10. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    Irrelevant to Willen's question. The 17V was just used because it is well above the DropOut Voltage for the LM317 or whatever regulator...
     
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  11. Mosaic

    Mosaic Well-Known Member

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    If a member asks for advice and is shown a circuit that has thermal failure staring it in the face with no mention of such issues, further advice is required to prevent poor results if such a circuit is attempted.
    Such is the benefit of a forum. Now the OP has more than one POV on that sample circuit and that is to his benefit.
     
  12. Willen

    Willen Well-Known Member

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    Which thing sets the voltage across R2 (current limiter) around 0.6V? And how 0.6V maximum voltage drop fixed across it?
     
  13. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    The regulator, the battery and R2 are in series. The battery current flows through R2. By Ohms law, the voltage across R2 is proportional to the battery current.

    If the voltage across R2 exceeds ~0.6V (the turn-on voltage of of Q1's Base to Emitter junction), then Q1 begins sinking current from its collector to its emitter.

    Look at this simplified sim of just the current limiter part of the circuit. Notice that as the current through R2 increases from Zero to 0.9A, nothing much is happening. As it increases from 0.9A to 1.3A, things get interesting.

    At the point where V(b) exceeds 0.6V (the turn on threshold of Q1), Ic(Q1) begins sinking current, causing the voltage at V(c) to decrease. There is a rather sharp knee at I1=~1A, past which V(c) decreases rapidly, as Ic(Q1) increases rapidly.

    That is the point at which current-limiting occurs in the battery charger circuit of post #5, because Ic(Q1) reduces the voltage at the Ref pin of the regulator, causing a reduction of its natural 13.65V (set by the voltage divider R4/R1) output voltage, but only while the current through R2 exceeds 0.9A (Vbe of Q1 > 0.6V)!

    307c.gif
     

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  14. chemelec

    chemelec Well-Known Member

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    Here is a Schematic for a charger. It also includes a Diode to Prevent Reverse Battery Discharge.
    And Formulas to calculate the Output Current and Voltage.

    Charger.png
     
  15. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    To help understand LM317 , I made a simplified equivalent circuit. that you can change at will , (enable) JAVA in browser.
    upload_2015-11-28_11-20-49.png

    I did not add bypass current booster but that can easily be done using PNP or Pch FET so that current sense biases the bypass switch.

    The 16V input is an ideal source with a switch add +/-2V of ripple or sag. I chose 0.1Hz but it could be 100Hz just to show effect of dropout when Vin-out<2V or so.

    I chose fixed values of hFE, so they are shown as ideal transistors not real.

    RED parts dissipate heat.

    Any questions for the forum?

    Tony
     
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  16. Willen

    Willen Well-Known Member

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    Hi all,

    Posts by Mike, Chelemec, Mosaic...helping me to visualize the circuit in nice way. I have poor cellphone (NOKIA 201) as internet source so content from Tony (maybe animation in link?) is not being displayed. Just I am seeing a more complicated PNG image. :)

    Circuit in internet making me silly again. Look at this image: http://www.electronicshub.org/wp-co...cuit-Diagram-of-Automatic-Battery-Charger.jpg

    I recently taught that the transistor conducts when battery is in under heavy charging, but not around full charging. Transistor of the circuit of the link above just conducts when battery gets full charging around 13.8V (11V ZD+Vbe drop+1.8v LED drop). What about it?
     
    Last edited: Nov 28, 2015
  17. chemelec

    chemelec Well-Known Member

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    The Transistor is Comparing the Voltage Difference between the voltage drop of an 11 volt Zener at the battery voltage and the voltage at the Adjustment terminal.
    This than Turns on the Transistor.
     
  18. Mosaic

    Mosaic Well-Known Member

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    Transistors are interesting devices that can behave in different ways and deliver different results when connected differently.
    If you can see you tube, this might provide a bit of fundamentals that can help.

     
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  19. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    Last edited: Nov 28, 2015
  20. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    I showed exactly how LM317 works with external current limit so it does not shutdown from overtemp.
    If you want ot understand how to boost LM317 with bypass, ask!
    If you have a specific question how to perform any task, ask with details!
    Vague questions get vague eanswers.
     
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  21. Willen

    Willen Well-Known Member

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    Yes, actually I am planning to ask about boosting current through LM317 and planning to make a simple adjustable bench power supply. From some circuit I made the diagram (attached). What about it?

    And I don't know how it's generating ripple supply at output! I have 6A PNP transistor.
     

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