how this works

Discussion in 'General Electronics Chat' started by Lightfire, Apr 9, 2012.

  1. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
    21
    hiii

    I just want to ask... how this circuit works.. all i know is that diode make the flow in one direction only... [​IMG]

    and the capacitor stores the charge in electric field. [​IMG]

    resistor disipates some power...[​IMG] and LED for indication..

    sorry for my noobish (i know it.)

    so how this really works..

    ok, the capacitor stores the charge. do the cap just store or it also power out... and what about the IN/OUT/ADJ (i forgot the name)............ my plan is to charge my 6 vbattery!!!
    thanks!!!
     
  2. bluebrakes

    Active Member

    Oct 17, 2009
    245
    7
    It's a voltage regulation circuit.

    From my understanding...

    R1/R2 is essentially a voltage divider circuit and this governs the voltage given out by LM317.

    100uf 16V and 10uf 10v are used for voltage stability. So reduce ripples, etc.

    LED1 is simple a power indicator

    D1 is a reverse polarity protection diode.

    D3 seems like its a flyback diode to prevent spikes affecting the LM317

    I could be entirely wrong of course....
     
  3. Markd77

    Senior Member

    Sep 7, 2009
    2,803
    594
    Have a look at the LM317 datasheet:
    http://www.national.com/mpf/LM/LM317.html#Overview

    This isn't a good or safe way to charge batteries, what type of battery is it?
    There's actually a current limited 6V charger at the bottom of the datasheet, but I think you would still have to make sure not to overcharge the battery.
     
  4. Snapester

    New Member

    Apr 9, 2012
    3
    0
    What do the capacitors do in the circuit? Im learning to read schematics currently and i'm trying to see how the current flows through the circuit. :D
     
  5. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    This circuit is a regulated power supply, not a battery charger circuit.
    Look for a battery charger circuit that matches your battery (lead-acid, Ni-Cad or Ni-MH??).
     
  6. BSomer

    Member

    Dec 28, 2011
    433
    106
    The capacitor is used for filtering. It helps prevent oscillation and allows a smooth DC output.
     
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  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    D3 is there to discharge the cap on the ADJ pin, protecting the IC in the case of a short circuit on the output.
     
  8. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
    21
    Ok.


    @ bluebrakes, why voltage divider? does it divide the voltage? :(

    @Audioguru, ok thnx. it's SLA...

    @ BSomer, why DC oscillates or it is just for protection (in any case)?

    will this work, anyway??/
    EDIT: So after the cap 1 charge, it discharges and charge the the LM317 and then discharges again and charge the cap 3???

    thhnx
     
  9. BillB3857

    Senior Member

    Feb 28, 2009
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    Look up the spec sheet on the LM317 and most of your questions will be answered.
     
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  10. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
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    thnx BillB3857!

    I have read it and saw

    Guaranteed 1% output voltage tolerance (LM317A)
    Guaranteed max. 0.01%/V line regulation (LM317A)
    Guaranteed max. 0.3% load regulation (LM117)
    Guaranteed 1.5A output current
    Adjustable output down to 1.2V



    ok it has 1% tolerance but how do i know the total voltage that this lm317a will produce. it says there that it produces voltages ranging from 1.2v to 3.7v (or it is really 37v)


    in ADJ, there were resistor and cap w/ specs of 130 ohms and 10v/10uF respectively.
    would that determine the voltage that will be producd?so how will i calculate



    thanks!
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
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    See attachment. This equation is in all mfr's datasheets.
     
  12. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
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    what is IadjR2?
     
  13. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    That means Iadj multiplied by R2. Iadj is the adjustment pin current. It is a parameter that is specified in the datasheet.
     
  14. Audioguru

    New Member

    Dec 20, 2007
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    The ratio of the two resistors connected to the ADJ terminal (if the input voltage is high enough) determine the output voltage.
    The 130 ohm (or 120 ohm) resistor from the OUT terminal to the ADJ terminal always has about 1.25V across it. (But is from 1.20V to 1.30V in an LM317). Its current is also in the resistor from the ADJ terminal to ground. Then simply use Ohm's Law to calculate the voltage across the resistor to ground then add the two voltages to find the output voltage.
     
  15. BillB3857

    Senior Member

    Feb 28, 2009
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    If you go here....... http://www.whatcircuits.com/lm317-calculator-v2/ you will find a calculator that will show the effects of different resistor sizes. Simply enter the desired voltage and the calculator will show the two different resistor values that will produce that output voltage. You can enter voltage and R1 and the calculator will show the value of R2 needed.
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    BUT the calculator is WRONG!
    It has 240 ohms for the resistor from the OUT terminal to the ADJ terminal that all schematics in the datasheet use on the more expensive LM117.

    The problem is explained in the datasheet. If the LM317 does not have an output current of at least 10mA then some of them will have their output voltage rise. The current in 240 ohms is only 5.2mA so a 120 ohm resistor which will have a current of 10.4mA should be used.
     
  17. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
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    Thanks Audioguru. That's what I get for trusting the internet and Google. This line is from the site I linked. "The example below shows a LM317 used as 11.8 volt regulator. The R1 and R2 use standard ±1% resistor."

    Silly me for believing it!
     
  18. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
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    The purpose of the potential divider between OUT, ADJ, and 0V is to allow the circuit to give more than 1.2V output.

    The potential divider resistors are chosen so that there will be 1.2V between OUT and ADJ when the voltage between OUT and 0V is the value needed at the output.

    Please study the data sheet or application notes if you want to understand more.
     
  19. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
    690
    21
    hi! i calculated...

    for R1 11.999988 @ 0.11166753 amp
    for R2 11.999976 @ 0.09230769 amp
    for the total voltage of 23.999964 volts - 0.7 for diode 23.299964 volts for ampere?? i dont know
    thnx@@@@
     
  20. Audioguru

    New Member

    Dec 20, 2007
    9,411
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    NO.
    The datasheet shows 1.25V across the resistor from the OUT terminal to the ADJ terminal, not 12V. You want its current to be more than 10mA (so the output voltage does not rise without a load) then Ohm's Law sets the resistor to be 1.25V/10mA= 125 ohms. Use 120 ohms. Then its current is 1.25V/120 ohms= 10.42mA.

    If you want an output of 24V then you already have 1.25V in the first resistor so subtract it giving 24V - 1.25V= 22.75V across the resistor from the ADJ terminal to ground. It also has 10.42mA so its value is 22.75V/10.42mA= 2183 ohms. Use 2.2k ohms. The resistor has a voltage of 2.2k x 10.42mA= 22.92V. Its power dissipation is 22.92V x 10.042mA= 239mW so use a 1/2W resistor.
    The 50uA to 100uA current for the ADJ pin reduces the output voltage about 0.15V so the actual output voltage is 24.02V (plus and minus a small tolerance).

    The input diode has nothing to do with the output voltage and the other two diodes are normally reverse-biased and also have nothing to do with the output voltage (the diodes are protection diodes, read the datasheet).
     
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