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Using a Current Transformer to charge a NiCD

Discussion in 'Electronic Design' started by [email protected], Sep 17, 2003.

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  1. Guest

    I'm trying to design a circuit that utilises a Current Transformer to
    create a low current charging circuit on the secondary, for a 2.4V
    NiCD with the addition of a few components .

    primary 4 - 6A load
    1 : 100 = Isec approx 40 to 60mA into a short circuit

    Isec ________
    -->----O | O--->--| |-----||+
    ILoad O | O |Low I | ||NiCD
    O | O |Charger |-----||-
    --<-----O | O------|__cct___|

    Any advice appreciated!



    Jim
     
  2. happyhobit

    happyhobit Guest

  3. mike

    mike Guest

    Use a diode and a light bulb to limit the current.
    But there are issues. Do you really need the current
    transformer thing?
    mike

    --
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  4. Guest

    Jay

    Thanks for this.

    I've tried to simulate a similar circuit with little success. I think
    it is due to rectifier diodes' forward voltage combined with CT
    (remember only one turn on the primary) saturation?
     
  5. Guest

    Thanks mike

    Yes I do need to use a CT since all I have is the live feed to the
    load on the primary side. Interesting idea, but since this has only
    one diode will the secondary effectively be open cct every half cycle
    with potentially disastrous results? What are the issues you have
    thought about?
     
  6. Fred Bloggs

    Fred Bloggs Guest

    If by "live feed" you mean the AC line voltage then a CT is the last
    thing to use. Did you ask yourself about the 4A in and 0.04A out? This
    means something like 4x 120VAC=480 Watts in to charge 0.04A x 2.4=0.096
    Watts out. Does this strike you as ridiculous?
     
  7. happyhobit

    happyhobit Guest

    Hi Jim,

    Have you measured the output of the CT into a load (100-ohms maybe?).

    Jay
     
  8. mike

    mike Guest

    Yep, thems is the issues.
    Use a center tap and two diodes, or a voltage doubler with 2 diodes.
    You're gonna need about a hundred to 1 turns ratio. Put back to back
    zeners across the secondary. If the secondary load
    opens, you've got a big pile of smoking electronics.

    The whole idea seems generally....ill conceived.
    mike



    --
    Bunch of stuff For Sale and Wanted at the link below.
    laptops and parts Test Equipment
    4in/400Wout ham linear amp.
    Honda CB-125S
    400cc Dirt Bike 2003 miles $550
    Police Scanner, Color LCD overhead projector
    Tek 2465 $800, ham radio, 30pS pulser
    Tektronix Concept Books, spot welding head...
    http://www.geocities.com/SiliconValley/Monitor/4710/
     
  9. Chris1

    Chris1 Guest

    I assumed that 4A would go on to do some work in some other load. His
    charger would only drop the voltage going to the other load by some tiny
    amount. I sure don't know why he needs to do it this way, though.

    Chris
     
  10. John Fields

    John Fields Guest

    ---
    Actually, I think it's pretty clever. He's got an ostensibly constant 4
    to 6 amp load on the mains, (I don't think it's a 500 watt resistor he's
    hooked up just to be able to charge his NiCd cell!^) and by using the
    CT he can get make an isolated 40 to 60mA supply to charge his NiCd
    cell. As long as he full-wave rectifies the output of the CT and puts a
    Zener across the DC output of the bridge to limit the voltage rise when
    he disconnects the battery, what's wrong with it?
     
  11. Fred Bloggs

    Fred Bloggs Guest

    Sounds like a PIC project shoved into a switch box.
     
  12. It does sound like saturation.

    You don't say what core you are using, what type of material and
    the dimensions. It is possible for a 1/100 turn CT to work, but
    only when Vdc is designed to be a minimum, with a magnetic
    material that has a high saturation flux density, and a certain
    minimum (magnetic) cross-sectional area (csa).

    In this app, the sec is connected to a bridge rectifier and then
    clamped or smoothed to produce a Vdc. Therefore the sec waveform
    is going to look like a heavily truncated sine, for which the
    nearest approximation is a square wave. The transformer sum to use
    here then is the "volts-seconds sustaining" version.

    -8
    V.t = 2.N.A.Bmax.10

    V is the peak secondary volts, (Vdc + bridge drop).
    t is the half-cycle time of the mains, = 1/2F.
    N is the number of turns.
    A is the csa of the core, in square-cm.
    Bmax is the peak flux density, in gauss.

    Let's see what A we need for a fixed Nsec= 100 turns.

    Suppose you are going to go straight into the 2.4v battery,
    unregulated. Secondary peak volts, V= (2.4 + 1.6) = 4v.
    To get the high Bsat it has to be silicon steel material,
    and assume we are going to work it up to 10000 gauss.
    I don't know which country you are in, but assume 50Hz.

    8
    A = 10 *4/(2*50*2*100*12000) = 2 sq-cms.

    That is a *big* core for this job. It strongly suggests that
    instead of 1/100 turns it might be better to go for something
    like 5/500 (or even 10/1000) turns on a much smaller core.
     
  13. Guest

    John

    Spot on

    cheers
     
  14. Guest

    Thank you to all for your replies

    Jim
     
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