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What did I do wrong?

Discussion in 'General Electronics Discussion' started by 1-3-2-4, Feb 24, 2013.

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  1. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    I built this here using this person's design..

    http://shdesigns.org/gcellchg.html

    I get 13.5V on the output, but tiny loads like say a white LED has the voltage going from 13.5 to 2.8V and as soon as that load is released it goes back up to 13.5V again.


    I don't have a 5k pot yet so I have a 100k in it's place now but still I have it set for a 13.5 V output and somehow my voltage output gets cut in half on the smallest possible loads.. my power supply is 17 V
     
  2. davenn

    davenn Moderator

    13,710
    1,911
    Sep 5, 2009
    what is your input voltage ?

    you shouldnt have anything like a 100k instead of a 5 k it wont give you the same adjustment capabilities have you got a10k or a 47k pot ?


    ohhhh and lets see a couple of sharp well lit pics of your construction to see if there are any wiring errors :)


    Dave
     
    Last edited: Feb 24, 2013
  3. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    My input is 17 v and all I have are 100k here, I'm about to order some 5k potentiometers but that can't be the cause of the issue could it?
     
  4. Electrobrains

    Electrobrains

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    5
    Jan 2, 2012
    Just a question: Have you added any series resistor to that white LED?
     
  5. john monks

    john monks

    693
    1
    Mar 9, 2012
    Two items:
    1. With the power off can you check the value of R1 with an ohmmeter and report back?
    2. With the power off can you check the output resistance with an ohmmeter and report back?
     
  6. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    No I connected it direct to the output with no resistor.

    1.) 1.7 Ohm
    2.) 93.1k
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,270
    Nov 28, 2011
    That circuit should have a resistor in series with the base of Q1. Otherwise Q1 can easily be damaged if the load current exceeds the limit even briefly. I suggest 1~10k.

    As others have said, you need a 5k trimpot for the voltage adjustment.

    You should not connect this circuit directly across an LED unless the LED is designed to operate at 650 mA continuously. This circuit is a battery charger, not an LED driver. If you connect an LED to it, it will go into current limit, and provide its maximum output current (650 mA) to the LED; the output voltage will drop to the LED's forward voltage at that current, which could be 2.8V. If the LED is not designed to operate at 650 mA, it will probably lose its smoke, and go either short or open circuit.

    I'm not sure what "issue" you are having problems with. The behaviour you describe, of the output voltage dropping while an LED is connected across it, is the behaviour you will normally see if you connect an LED across a current-limited battery charger. But why would you do that? It's a battery charger, not an LED driver. What were you expecting it to do? Charge up the LED? ;-)
     
  8. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010

    So I noticed one of the leads for R4 had a bad connection so I fixed it so I can now adjust from 0-15 V it's somehow still dropping too far.. since I have over 50 of these led's I don't mind getting rid of some. So what I expect is the led to burn out because of the current at least, plus the led never lights for the max charge current.
     
  9. john monks

    john monks

    693
    1
    Mar 9, 2012
    Your output should be a little less than 7.67kohms.
    If you don't have a 5k pot try substituting The pot with a 2.7kohm resistor and see what happens. The pot is hooked up as a restate that adjusts between 0 and 5k.
     
  10. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    This is going to sound like a dumb question but switching to a 2.7k resistor (best I have is 2.2k) it would prevent it from limiting the current right? I thought a trim pot while both ends are connected the wiper is what determines what the resistance valve will be?

    I mean it must be a reason why most sheets I see with a Lm317 datasheet show a 5k trimpot but what I'm trying to say is adjusting is one leg of the trimpot still going to be way higher then 5K?

    I'm not too used to using trimpots yet.

    From my understanding just the wiper and one end is just a variable resistor but all three is a voltage divider.
     
    Last edited: Feb 24, 2013
  11. john monks

    john monks

    693
    1
    Mar 9, 2012
    No. A 5k trimpot can be used with only two legs. In that case it is a 0 to 5kohm variable resistor. What I was suggesting is that you ruffly split the difference and replace the pot with about 2.7kOhm resistor. 2.2k should work fine. One lead should go to R2 and the other lead should go to pin 1 of the LM317 and R3. Then we can see what happens.
    What I am perplexed about is that you are getting 93.1k on the output, way to high.
     
    Last edited: Feb 25, 2013
  12. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    I think I'm going to strip the whole board.. still getting some issues, I hate to do this but..

    I do have some questions before I start.. D1 and R4 the position of the wires wont matter as long as they are on pin 1 of the LM317?
     
  13. john monks

    john monks

    693
    1
    Mar 9, 2012
    I can't imagine the position of the wires making any difference.
     
  14. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    hmm.. ok this is how I have it connected the resistors.. I have the two 2.2k in series from pin 1 of the LM317 which goes to ground, Pin 2 I have the 470 ohm connected to the output.

    when I test the ohm for the output leads I get some stupid high reading like 26 Mohms and plugging it in I get nothing on the output leads.
     
  15. john monks

    john monks

    693
    1
    Mar 9, 2012
    You should have R2, a 2,2k resistor in series with another 2,2k resistor. And the other end of that resistor should be connected to pin 1 of U1 and R3, a 470ohm resistor. The other end of that resistor should go to U1 pin 2 and the output.
    Now just looking at the output you should be going into R2 and then into another 2,2k resistor and then into another resistor, 470ohms, and to the output. The total resistance should be 2200 + 2200 + 470 = 4870 ohms.
    I suspected that something is not hooked up.
    Look and try again.
     
  16. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    What's confusing me to what you are saying is, on my board from Pin 1 of U1 goes into the two 2.2K resistors, Pin 2 of U1 is going into R3 in front of the two 2.2k resistors, then to the output.

    My output resistance is 4.8Kohm one thing funny is the Led is not lighting up.. I understand it should when at max current but testing it on some old SLA batteries as in like 7 years old they show a fake charge level of like 11.8V but it has hardly any current.. So I don't know if that's the reason why the LED wont light or not.. Whenever I connect it to the charge my output voltage of 15.3V cuts out to the battery voltage.

    Not sure what would happen with the good battery connected, I assume with the right trim pot I wont be outputting that high of a voltage.

    *edit I was wondering when checking a 12V LED the voltage of the output drops to around 6V from the 15.. why does that keep happening?
     
    Last edited: Feb 26, 2013
  17. john monks

    john monks

    693
    1
    Mar 9, 2012
    R3 goes between pins 1 & 2 Of U1. So there should be three resistors in a string across the output. And the resulting output should be about 4870 ohms as measured with an ohmmeter with the power sully shut off. Next you want to turn on the supply and see what the voltage is. Just try that and let's see what happens.
     
  18. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    ah jesus finally got it! I checked the output leads and sure enough I got 4.8k, I tried a 10 LED high powered module and it lit brightly and the led came on, it seems like the first way I had it before I stripped half the board down was right but the R3 was not between pins 1 & 2 of U1

    My 5k trim pot should be here sometime this week, thanks for taking the time to help me out on this one I really appreciate it!

    One other question so I'm sitting here trying it out with a simulated load.. a 12V hi power LED with a 10 ohm resistor the temp of the heatsink after about 5 min on is 90C is that a little too hot? should I think about adding a fan for ventilation?

    Also another thing at what voltage is a float charge and which would be at max current?
     
  19. john monks

    john monks

    693
    1
    Mar 9, 2012
    If your output is 12 volts and the LED draws 2 volts then the 10 ohm resistor has 10 volts across it. So the current through the 10 ohm resistor is 1 amp and the current through U1 is 1 amp. And if your input voltage is 18 volts the the power dissipation of U1 is (18-12)*1 watts or 6 watts. You might confided a better heatsink or a fan.
    For the trickle current you need to look up the maximum charge voltage of your battery and adjust your output to that voltage and the trickle current should take care of itself.
    If you battery is a lead acid battery the maximum voltage is probably around 13.5 volts and that is what you should set the voltage to.
     
    Last edited: Feb 27, 2013
  20. 1-3-2-4

    1-3-2-4

    53
    0
    May 7, 2010
    well that's the extreme end of things, that's what I'm trying to find out.. with the battery with a load on it I connected it direct to the LED for a few min ontop of a 67 mm tall heatsink it ended up still being over saturated with heat.. but anyways I wanted to say was say the battery was at 50% or better yet 10% charge level how long would the charger stay at the maximum rate before slowly going down to float?

    As for a fan I was thinking of building a temperature controlled fan, but at the float charge rate the heatsink temp is only around 32 C which is not much of anything.
     
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