zener diode with PNP regulator mod for battery charger

Discussion in 'OnBoard Electronics & Controls' started by sdowney717, Nov 26, 2014.

  1. sdowney717
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    sdowney717 Senior Member

    I have an old working Raritan Charger from 1965 I fixed recently. Had to replace two isolation diodes.
    So when setting the output voltage using the variable 400 ohm center tap sliding 12 watt resistor, it is very difficult to fine tune, a very coarse adjustment. I managed to get it working ok.

    I want to modify this to make it easy to do.
    I bought some minipot 0 - 500 ohms, 1/2 watt rated, part #3296w. They adjust with 25 turns of a screw, which allows for fine control.

    This is the current schematic. The 2N525 transistors are PNP. when the 2N525 attached to zener turns on, it turns off the other 2n525 which turns off the charger. So the zener is reverse biased.
    AFAIK. Somehow it just works, but I dont fully understand how.

    [​IMG]

    This is what might work? pull the adjustment wire off the 400 ohm pot resistor, and run it through the minipot.
    [​IMG]


    this is a close up of the zener regulator part of the schematic.

    [​IMG]

    This is the actual picture of the control board showing that awful 400 ohm pot resistor and the red wire which goes to the zener diode. (gold diode next to resistor in center)
    [​IMG]
     
  2. sdowney717
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    sdowney717 Senior Member

    Thinking more on how this works I found this that says a PNP transistor works as a switch when the base goes to ground, it turns on.
    http://www.rason.org/Projects/transwit/transwit.htm

    All the zener is doing is it's break down voltage exceeds that 6.8v, (grounding the base, there is a current flow from emitter to base charging the zener) and it conducts in reverse bias which turns on Q1 (off transistor), and that turns off Q2 (on transistor) which turns off the charging.
    Seeing there is an oscillating rectified power , DC voltage rises and falls coming from the rectifiers, it just keeps turning on and off and on and off forever.
    So why not just run that zener wire through minipot and maybe extra resistors right to ground? Which is what the original circuit does with those two 300 ohm resistors already in the circuit.

    Positive volts in at top left , negative at bottom

    [​IMG]
     
  3. daiquiri
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    daiquiri Engineering and Design

    I don't think you'll get many technically useful replies to this post on a boat design forum.
    But I might be wrong, who knows. ;)
    Cheers
     
  4. baeckmo
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    baeckmo Hydrodynamics

    Just wait until Cornelis is awake!
     
  5. BertKu
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    BertKu Senior Member

    Lets first go through the working of this charger. I may not have enough time to calculate all the voltage levels. But in large this is what happens.

    1) The 115 Volt is transformed to a lower voltage and with 2 diodes converted to two 50hz halves (or 60 in the States?) i.e. every 8,33 milliSecond a half ( 60 Hz) or every 10 mS (50Hz) . Note: there is no large electrolytic capacitor to smooth the ripple.

    2) Via the Amp meter it goes to the 3 diodes (Shotkey?? diodes) to each of the battery bank.

    3) the minus of the battery bank is connected to thyristor, which act as a switch and regulates the voltage/current

    4) The thermocouple in the collector of the "ON" PNP transistor secures that by overheating of the transformer, the thyristor is switched off by cutting the collector current of the "ON" PNP transistor.

    5) The thyristor is triggered by the 2 PNP transistors which act as a kind of flip flop and controlling the Thyristor. The PNP (ON) get switched, when started by the 400 Ohm and the 100 Ohm resistor which charges the 100 UF capacitor from zero to a higher voltage level. The "OFF" PNP transistor allows to increase the Voltage over the capacitor to such a level that the "ON" PNP transistor switches off. This is achieved by the zener diode of 6,8 Volt and the original potentiometer.

    My guess is that the resistor values due to age have changed value and let you go into circles. Measure first accurately all resistor values. If they are accurate, check the 100 uF capacitor, whether it is still 100 uF or maybe 5 uF due to age. If you are not happy with the regulation, change the old potentiometer to a newer type 12 watt 400 Ohm which regulates better. If not possible, let say they are not available , the way you like to place a 500 Ohm potentiometer with more turns will not help you to regulate the voltage, but you regulate the current flow through the zener diode and the base of the PNP (OFF) transistor.
    As soon the battery is charged, the Voltage level drops at the collector of the "OFF" PNP transistor and then the Electrolytic capacitor drops in Voltage level, this in turn switch via the "ON" PNP transistor the thyristor off.
    Let me sleep a night over and maybe tomorrow morning I will have a solution for you.
    Bert
     
    Last edited: Nov 27, 2014
  6. BertKu
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    BertKu Senior Member

    Well on the way to the bedroom, here is your solution. You place your new 500 Ohm large number of turn potentiometer from the middle point to the point where you have drawn your solution. The middle point from the new 500 Ohm potentiometer goes to the zener diode. The old potentiometer does the course adjustment and the new potentiometer the fine tuning of the voltage level. That will work, as you do regulate the voltage and not the current flow to through the zener diode/basis PNP transistor. While I am sleeping I will calculate whether the low power dissipation of the new potentiometer is enough when your middle point of the old potentiometer is zero (full) and the full voltage is over your new potentiometer.
    P= Voltage x Voltage divided by the resistance. This must be less than the maximum allowed power dissipation of your new potentiometer. Just in case I have a deep sleep and do not worry. Good luck.
    Bert
     
  7. daiquiri
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    daiquiri Engineering and Design

    Well, on my way to the bedroom I'll be swallowing my previous words. :D
     
  8. sdowney717
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    sdowney717 Senior Member

    Thanks Bertku. I need to digest your words some but I followed along pretty well. You got the circuit figured out very well.

    I will draw a mod diagram that shows what I think your suggesting.

    I actually would like to leave the 400 ohm 12 watt resistor just as a resistor and get rid of the center tap adjustable wire hookup.
    Thing is they are hard to find, likely expensive if you can get them, easy to break the little wire windings and offer a very coarse difficult to achieve setting of resistance. I found it was fractions of a millimeter to get a few ohms change which achieved several tenths of a volt change on the output.

    And I looked at the tiny wire windings with a magnifying glass and they are loose on the ceramic core, likely due to 45 years of constant usage, so that resistor is not in the best of shape. The resistors test out fine on resistance, this 400 ohm adjustable resistr is 387 ohms.

    Can you tell me how much current is flowing through it, they list at 12 watts, but cant be that much.
     
  9. sdowney717
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    sdowney717 Senior Member

    Bertku, Is this what you suggested?

    [​IMG]

    What settings to place both adjustable resistors ohmage wise? The current setting of the 400 ohm is 182 ohms between center tap and positive input from rectifier diodes and something like 200 from center tap down.


    The voltage of the charger, to increase that you slide the adjuster center tap clamp up towards the rectified positive input. Lower that resistance, the charger output rises. This due to your increasing the opposite side resistance, means the zener takes a greater voltage rise to overcome the increased resistance set point to fire reverse breakdown and conduct turning on the 2n525 transistor which turns off the other 2n525 transistor which turns off the charger clipping the voltage.

    Why not just sever the center tap red wire from the adjustable 400 ohm resistor and put the zener through the 500 ohm minipot down the point as my original mod drawing? If I need more resistance, I can add it in series with the minipot? I dont know why that wont work? Advantage would be get rid of the trouble prone 400 ohm adjustment that could break the wire old, partially worn out thing.
     
  10. BertKu
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    BertKu Senior Member

    You were right Daiquiri, not many people like to put themselves into the mind of a designer. Whether mechanical, electronics or software. It is always easier to design it from scratch yourself. Bert
     
  11. BertKu
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    BertKu Senior Member

    No, Don't do it that way. You will burn one half of the new pot meter out. The middle point of the OLD potmeter goes to the one end of the NEW pot meter. The middle part of the NEW potmeter goes to the zenerdiode. The other end of the NEW potmeter goes to the other end ol the OLD potmerer i.e where the 100 Ohm resistor is connected to and the PNP transistor.
    Bert
     
  12. BertKu
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    BertKu Senior Member

    Sorry my eye's are not yet woken up. It is 300 Ohm and not 100 Ohm.
    Bert
     
  13. CDK
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    CDK retired engineer

    I am awake, but agree with Daiquiri this is not the place for such questions, so I let this one pass.
     
  14. BertKu
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    BertKu Senior Member

    Hi CDK, why not? He has a problem and we solve it. He must just put the new potentiometer on the correct place. The way he wanted to do it, if he has 10 Volt over the small portion of his new potentiometer. The formula is very clear. 10 x 10 dived by maybe 50 Ohm = 2 watt. His potentiometer will not last. If he does it the way I recommend. i.e. the full resistance is over the middle part/end of the old potentiometer , he will have 10 Volt x 10 Volt divided by 500 Ohm = which is in power dissipation of the new potentiometer, 0,5 watt and he has a fine tuning, that is what he wants.
    Bert
     

  15. BertKu
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    BertKu Senior Member

    Yes, P = E x I = 12 watts E = I x R
    12 = 400 x I x I . The maximum current is thus 173 milli Ampere.
    I have no clue what your Voltages are. Therefore I cannot help you further. You have to do the calculation now yourself.
    Bert
     
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