centre tap transformer output

Discussion in 'The Projects Forum' started by jody, Jan 6, 2015.

  1. jody

    Thread Starter Member

    Nov 12, 2012
    39
    0
    hi
    im repairing my simple nicad charger that melted its current output resistor due to a bad battery pack. the transformer i believe is a 12-0-12 center tap and is full wave rectified using 2 diodes. the battery pack is 9.6v 1.3ah. the sticker on the case says output 11.6v. how do i calculate the resistor. is it done to get 11.6v@1.3a
    or is it done at the batteries voltage of 9.6v @'1.3a. what im unsure about is whether the battery is charged at its rated voltage or does it have to be 2 volts above at 11.6 v.
     
  2. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,548
    2,373
    A 12-0-12 CT Tfmr with full wave (2 rect) would only be around 8v output, or <6v without capacitor?
    Max.
     
  3. MrChips

    Moderator

    Oct 2, 2009
    12,442
    3,361
    An accepted practice is to charge nicads at C/10. Hence if the capacity is 1.3Ah, C/10 is 130mA for 10-12 hours.

    So if the transformer outputs 12VAC, the rectified voltage is 16VDC. You can check this with a DVM.
    If you assume a 4V drop across the resistor, then the resistance required is 4V/0.13A = 33Ω @0.5W
    Use a 33Ω 1W resistor.
     
  4. jody

    Thread Starter Member

    Nov 12, 2012
    39
    0
    i have had the multimeter on the outer two leads of the transformer with no load and it read 24.2vac so i assume it is a 12-0-12 or have i got that bit wrong.
     
  5. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,548
    2,373
    Yes correct I made a simple blunder, it has happened before!o_O
    Max.
     
  6. jody

    Thread Starter Member

    Nov 12, 2012
    39
    0
    im good at blunders,clever people call it science.
    i have a question about the transformer output.when i measure the voltage after the diodes so its the dc bit i get 11.6 volt. i have a 1.4ohm resistor and there is no capacitor. now when on charge the battery gets up to nearly 13v. do i assume from this the transformer max output is higher than the no load reading?
     
  7. #12

    Expert

    Nov 30, 2010
    16,298
    6,808
    Without a capacitor you're measuring an AC component that is only active in the positive half cycles. This confuses some meters and most humans. Add a capacitor, any capacitor, and remove the load to read the peak voltage on a DC meter setting. That peak voltage will be inflicted on your battery, but it's not the average voltage. The average voltage is peak minus battery voltage times something I don't remember. Do what MrChips says. He's usually top dog in this department.
     
  8. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,548
    2,373
    Your 11.6 would be about right, the battery is also going to experience the peak which is 16v, a fully charged automotive battery is around 12.6 volts.
    An automotive alternator usually charges at a mean 14.5v.
    Max.
     
  9. jody

    Thread Starter Member

    Nov 12, 2012
    39
    0
    yes im on that track just wondering how he came up with a 12vac trans outputs 16vdc.is that with a capacitor
     
  10. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,548
    2,373
    The peak will be 16v regardless, without a capacitor you will measure the RMS, not the peak, IOW 11.6v.
    Connected to the battery it will vary depending on the inherent charge.
    Max.
     
  11. jody

    Thread Starter Member

    Nov 12, 2012
    39
    0
    perhaps i should add that the charger has a fast charge circuit and then a trickle charge circuit.its crude in that a relay cuts the fast charge and becomes the trickle charge. at the moment i have the trickle charge at the right current but a volt too high. the feed for the trickle comes from the main feed so im guessing i havnt quite got it set to 11.6 volts max.
     
  12. #12

    Expert

    Nov 30, 2010
    16,298
    6,808
    Volts peak = 1.414 times volts RMS. RMS is what transformers are rated for. Then subtract the voltage loss in the diode or diodes.

    12 x 1.414 = 16.968 volts. If one diode per leg, V out is about 16.3 volts peak...assuming your power line voltage is the same as the transformer label. 120 VAC?

    To arrive at 9.6 volts and .13 amps,
    16.3 - 9.6 = 6.7V
    But 6..7 is the peak voltage.
    You're operating at about half that on the average.
    3.35V/.13A = 25.7 ohms, and we're right about where MrChips was.
     
  13. jody

    Thread Starter Member

    Nov 12, 2012
    39
    0
    right im getting it. yeh line voltage is same as transformer,240v here. i shall try it and see what happens.
     
  14. #12

    Expert

    Nov 30, 2010
    16,298
    6,808
    Just had to mention that because the voltage at my house is 250 VAC +/- 1%
     
  15. jody

    Thread Starter Member

    Nov 12, 2012
    39
    0
    one more question. the fast part of this charger is crude basically a resistor after the rectifier and then a relay to cut the fast charging at a set voltage. should the maximum current be set for a flat batteries voltage ie the discharged pack will be 8 volts so as i read it this is when it will draw most current to charge which tappers of as the battery charge level increases. i am assuming that as a nicd cell charges up it will draw less current like a lead acid cell??
     
  16. jody

    Thread Starter Member

    Nov 12, 2012
    39
    0
    ah just done some sums and seen that at 11.6v if the battery pack wanted it it could draw 2.9amp @11.6 volt. i guess the limit is set for the higher voltage in case of peak demand if the voltage is high.
     
  17. #12

    Expert

    Nov 30, 2010
    16,298
    6,808
    Batteries don't discharge to zero volts and survive.
     
Loading...