LM317 Current Limited 6v Charger

Discussion in 'General Electronics Chat' started by mbxs3, Apr 1, 2011.

  1. mbxs3

    Thread Starter Active Member

    Oct 14, 2009
    141
    3
    I put together the circuit displayed on page 22 of this datasheet...

    http://www.national.com/ds/LM/LM117.pdf

    All component values match the components in the schematic. I am getting an output voltage of 7vdc. What can I change to make the output voltage become 5vdc? Just to note, my input voltage to the circuit is 19vdc.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Decrease the value of the 1.1k resistor.
    You should use a potentiometer to find the correct value. A 500 Ohm pot in series with a 620 Ohm resistor should work.

    Note that you should test the output voltage with a load resistor on it - say, 1k to 10k Ohms.

    Original schematic:

    [​IMG]
     
  3. mbxs3

    Thread Starter Active Member

    Oct 14, 2009
    141
    3
    Thanks for the reply Sgt. I will set this up later tonight and test it out.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    I hope you have a rather large heat sink on the LM317. If you don't, it will most likely go into shutdown very quickly when you attempt to charge a battery.

    Your input is 19v, the lead-acid battery when charged and "floating" should be about 6.7v to 6.8v at room temperature (yes, that is correct).

    Let's say the battery starts off being 5.5v, and during charging the current is 0.6A, then 0.6a*5.5v= 3.3 Watts of power dissipation in the battery, and (19v-5.5v)*0.6a = 8.1 Watts power dissipation in the regulator. If you don't have a good-sized heat sink on the LM317, it will go into thermal shutdown to keep itself from self-destructing. This is hard on the regulator, and it probably won't last long being so overloaded.

    You could take a good bit of the thermal load off of the LM317 regulator by using a 15 Ohm 10 Watt power resistor between your 19V supply and the input of the LM317.
     
  5. mbxs3

    Thread Starter Active Member

    Oct 14, 2009
    141
    3
    I did as you advised Sgt and put a 1k pot in place of the 1.1k resistor. 710 ohms was what I had to set it at to get my 5vdc output. I was wondering if you would give me a brief run down, if it is possible to be brief, on why that resistor value changes the output value.

    The reason I bumped it down to a 5 vdc output is because I was attempting to charge my iPod with the circuit. At the same time, I replaced my 19vdc input voltage with a 9vdc solar panel.(I took an old iPod usb cord and cut it in half and used the red and black wire to plug into my positive and ground points.) With this configuration my experiment semi worked. The iPod would turn on but after the apple logo screen went away it would cut off. I thought maybe my circuit wasn't providing enough mA so I used the same cord and plugged it into my computer usb with the same result. Then I took my iPod and plugged it into a normal charger and it worked as it should. Anyone know why this is happening? Im wondering if the green and white usb cord wires that I have disconnected are required for the iPod to charge.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Let's start with the resistors, calling the 240 Ohm resistor R1, and the 1.1k or 710 Ohm pot R2.

    The LM317 attempts to keep the voltage between the OUT terminal and the ADJ terminal at a nominal 1.25v. That can vary somewhat, but the target is 1.25v.

    Since R1 is 240 Ohms, it will require I= E/R = 1.25/240 ~= 5.2mA current flow through R1 to achieve 1.25v between the OUT and ADJ terminal.

    Then, the current through R2 determines the output voltage less the 1.25v from OUT to ADJ, plus the current that comes out of the ADJ terminal (around 50 to 80 micro-amperes). You determined 710 Ohms resulted in 5v out.

    So, E=IR, and I = 5.2mA, R=710 Ohms, E=3.692 Volts.
    3.692V + 1.25v= 4.942v. That's 98.9% of 5v. Don't forget, there's also the 50 to 80uA flow from the ADJ terminal. Let's just say 70uA for now.
    E=IR, I=0.07mA, R=710, E=49.7mV. 4.942v+49.7mV = 4.9917 Volts, or 99.8% of the output you received.

    Close enough for you? ;)

    I can't tell you, I've never owned an Ipod. Why are you trying to build a charger? The factory supplied charger is optimal for your Ipod, and it works.

    If you don't charge it properly, you may damage it.
     
    HallMark and mbxs3 like this.
  7. mbxs3

    Thread Starter Active Member

    Oct 14, 2009
    141
    3
    Thanks for the explanation Sgt. I am still trying to grasp a lot of these concepts so your breakdown will definitely help me better understand what is going on.

    As far as the iPod charging goes, I was doing it just to see if I could. It is my old iPod so I won't be heart broken if I melt it. I wanted to see if I could put a circuit together to use my solar panels.
     
  8. HallMark

    Member

    Apr 3, 2011
    89
    5
    Superb Explanation Sir. Salute !
     
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