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How can I build a CMOS current generator ?

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by malc9141, Sep 28, 2011.

  1. malc9141

    malc9141 Member

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    if you know the DC resistance of the injector that number would help

    This value has been lost over time - it is very low. 4 ohm comes to mind. I'll see if I can check it with a simple multi meter.
     
  2. alec_t

    alec_t Well-Known Member Most Helpful Member

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    That's one way. Apply a known voltage to the injector and monitor the rate of rise of the current. Or use an inductance meter.
    Is that 18V a brief pulse (1mS) or is it the steady-state voltage needed to fire it (i.e. with the current limited by the coil resistance)?
    Was the 15A current drawn when 18V was applied (which would suggest a resistance of ~ 1 Ohm)?
    By 'in place' do you mean actually connected to your circuit, or in the engine?
     
  3. malc9141

    malc9141 Member

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    When I take a run-of-the-mill multimeter, set it to Resistance, join its terminals, it reproducibly reads 4 ohm.
    When I apply terminals to Injector terminals, It is exactly the same (tested on two similar injectors).

    Is that 18V a brief pulse (1mS) or is it the steady-state voltage needed to fire it (i.e. with the current limited by the coil resistance)?
    Was the 15A current drawn when 18V was applied (which would suggest a resistance of ~ 1 Ohm)?
    By 'in place' do you mean actually connected to your circuit, or in the engine?


    I would say the voltage is brief (we use 24 V and it briefly drops to 16 V with the igen) and I believe that's when giving the ~15A. I think this is with the injector and in the electronics workshop. Not in the engine.
     
  4. dave

    Dave New Member

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  5. ronv

    ronv Well-Known Member Most Helpful Member

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    Injector

    Can't get there with those numbers. ;) Current = voltage / resistance, so with 4 ohm the most you could hope for is 6 amps. The fact that it is 4 ohms would also suggest a lot of wire and pretty high inductance. Any part number on the injector?
     
  6. alec_t

    alec_t Well-Known Member Most Helpful Member

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    4 ohm is quarter the resistance quoted by TI. If the same wire gauge is used in your injector and in the TI ones (a big IF !) then the number of coil turns is also a quarter so the inductance will be 1/16 of the TI injector inductance, i.e. about 8.5/16 = 0.5mH.
    It would be interesting to know if that calculated value agrees with the actual value.

    I agree with Ronv. No way will you get 15A through that 4 ohm injector from 24V. Apparently, though, 6A is enough to fire the injector.
    It's beginning to look as though the current-limiting part of your igen circuit will be more for protecting the switching power transistor than for controlling injector current.
     
  7. malc9141

    malc9141 Member

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    Even with the 'ifs' these are amazing insights. I am out of my depth but I get the drift and others can fill in the details.
    Is this too much? : I could mail an injector to you and you could check it. But you must act quickly because I'm going overseas for months tomorrow a.m.
    SMS me on 07887946529 with an address if you agree.
     
    Last edited: Oct 12, 2011
  8. malc9141

    malc9141 Member

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    Delphi EJBR 02101Z (0 = zero)
     
  9. malc9141

    malc9141 Member

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    Of course, I don't know if 4 ohms is correct. If the Multimeter said 0 for zero, not 4 ohms, I'd be more certain. The meter says 4 and so does the coil - the coil might be close to zero ohm.
     
  10. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Sorry, malc; don't have the test facilities. I've Googled around but the web seems strangely silent as to details of that particular injector.
     
  11. malc9141

    malc9141 Member

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    These things really are industrial secrets. I've found that.
    I'll see if I can get my hands on a meter. My pal could do it but it's an odd friendship and he would only do it as part of some personal gain. He was interested in the beefier item (circuit + physical layout) and might step up. But if he felt undermined, he wouldn't. Bit sad. Yet he was the only guy who could help initially after masses of false promises, so he remains my hero ΩΨΞφ :eek: He's too poor to have email!!!

    When I'm more settled, I'll try to master all of this correspondence and might have ago at building it.
     
    Last edited: Oct 12, 2011
  12. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Ah, so that doesn't get us much further forward. Can you zero your meter ohms range (there should be an adjuster somewhere)? Or connect the coil in series with a known resistor (say 100 Ohm, 2W) across a 12V battery and measure the voltage drop across the coil, then calculate its resistance from the measured volts.
     
  13. malc9141

    malc9141 Member

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    Using two 1.25 ohm resistors and (resisting the chance of destroying it with a big 12 v battery) a small solid 12v battery (checked), I get 0.055 ohm for the coil.
     
  14. alec_t

    alec_t Well-Known Member Most Helpful Member

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    That seems remarkably low. A limited amount of Googling indicates that injectors come in two flavours: 'low' impedance (1-5 Ohm) and 'high' impedance (10-20 Ohm). So yours is way below the normal 'low' range. Is it a custom-built one (let's hope it's not got a shorted turn) ?

    Edit: I'm a bit concerned about that measurement. If indeed the coil resistance is only .055 then the current through it in your test would have been 12/(2*1.25+.055) = ~ 5A. But from what I read the low impedance injectors would get overheated at 5A continuous, which is why the peak-and-hold method of firing them is normally used (~5A for ~ <1ms followed by ~1A for ~1ms). I hope your ~5A was for a sufficiently brief period to prevent coil damage.
     
    Last edited: Oct 13, 2011
  15. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Have you considered using an LM1949 chip to control your injector peak/hold firing? That needs only half a dozen external components and so would be 'neater'. The datasheet for that chip is quite informative.
     
  16. malc9141

    malc9141 Member

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    The litle djry battery could,t produce a high current. But the FIRE to HOLD. Is to drop the current, I agree. As for the transistor, All this will have to go before my colleague! 8 I don,t think I,m going to be able to put all this together as I,d intended. but it IS all useful.
     
  17. malc9141

    malc9141 Member

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    As I said, the injector was from a Renault Clio, year 2002, 1.5 turbodiesel.
    I believe in action, it takes avery high charge to get it to open ultra fast - they go even faster now but I think that's with a ceramic pile.
    I wouldn't have been running much current through it because I used a tiny 12v battery. It held 12 v but the division with the 2.5 ohm left a calculated 0.055 ohm in the coil.

    Would it be reasonable to draw for me your modified circuit incl the LM1945 chip? It's got quite modified from the start. Actually, I haven't caught on to what V-ign is.

    Cheers
    Malc
     
  18. alec_t

    alec_t Well-Known Member Most Helpful Member

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  19. malc9141

    malc9141 Member

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    Hello:

    May I ask: concerning the solenoid in a fuel injection unit, where the solenoid lifts a ferrite piston, allowing the ultra-high pressure fuel to inject, what happens when the current is cut?

    In my circuit, about 16 amps at 20 v DC briefly (about 2 ms) activate the coil. The current comes from a Current Generator.
    On switch-off, (I presume) there should be an induced spike pushing "current" from the "bottom" of the coil in the same direction to earth. We have a big diode permitting this current to return to the top of the coil.

    I'm wondering how this might delay the valve closing. Given that the coil still goes to earth, maybe it
     
  20. moffy

    moffy Member

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    Yes, while current flows through the solenoid and is above the hold open threshold, the injector will remain open. The inductive energy is dissipated by the resistance of the coil and the voltage drop across the diode.
     
  21. malc9141

    malc9141 Member

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    the diode return

    Hi - thanx. This was posted accidentally before I'd quite finished.

    The +ve is at the "top" of the coil and the "bottom" goes to earth. The diode allows current to flow from the bottom, back to top. I'm wondering what purpose this serves.

    Might it slow the rate of current collapse in the coil? I want max rate of collapse of the field when the switch-off occurs.

    Thanks

    Malc
     

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