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+/-3.5v to 0-100k resistance conversion circuit help please!

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by djpubba, Jan 4, 2006.

  1. Roff

    Roff Well-Known Member

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    What LDR circuit are you testing?
     
  2. djpubba

    djpubba New Member

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    My original:

    [​IMG]
     
  3. djpubba

    djpubba New Member

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  4. dave

    Dave New Member

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  5. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    Looks scarily familiar doesn't it!.
     
  6. djpubba

    djpubba New Member

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    Which op-amp will run on +5? Will the LM324 work?
     
  7. Roff

    Roff Well-Known Member

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    As I said in a previous post, you need a rail-to-rail I/O op amp, which LM324 isn't. Tell me where you are getting your parts and I may be able to suggest one. Can you use SMT parts? A thru-hole rail-to-rail may be hard to find.
    I also have a schematic of the LDR feedback scheme that Len proposed, if you want to look at it. It uses three r-r I/O op amps, so you might want to get a quad if you build it.
    When you have unused op amps, you should keep the unused one(s) active by connecting the output to the inverting input (voltage follower configuration) and connecting the noninverting input to a voltage that is within the common-mode input voltage range. This can be GND in a dual-supply system, or it can usually be the noninverting input node of another, used op amp in the package. You need to do this because some (all?) multiple op amp chips have an internal biasing circuit which is common to all op amps in the package, and floating inputs can upset this circuit.
     
  8. ljcox

    ljcox Well-Known Member

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    MYSTERY SOLVED. I stupidly assumed that the time in the computer is T = k.C.I where I is the output current from Ron's circuit. But, I belatedly did the maths and it is T = k.C/I so it is no wonder the results are non linear.

    So the solution is to use a current amp rather than a trans conductance amp. Then the input current is 5/Rp where Rp = R3 + the pot resistance. R3 is as in Ron's circuit. The current amp could be as drawn or an op amp with current mirror inputs.

    The "-5V" connection can be left open since there will be no current through the R4 resistors.
     

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  9. Roff

    Roff Well-Known Member

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    Len, I need to think about that, but don't you still have a min-to-max ratio problem?
     
  10. ljcox

    ljcox Well-Known Member

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    No, I don't think so. The range is set by Ai and the centre can be set by a DC off set current. However, Ai needs to be <1, ie. about 0.02, so this may be a problem.

    Note that Rp ranges from 2.9k to 12.9k according to my calculkations from Tim's measurenents.

    So if Ai = 0.02, then T = k.C 100 Rp
     
  11. ljcox

    ljcox Well-Known Member

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    No, I don't think so. The range is set by Ai and the centre can be set by a DC off set current. However, Ai needs to be <1, ie. about 0.02, so this may be a problem.

    Note that Rp ranges from 2.9k to 12.9k according to my calculations from Tim's measurenents.

    So if Ai = 0.02, then T = k.C 10 Rp which is roughly equivalent to the 100 k pot.

    So perhaps we need an NPN current mirror in lieu of the amp I drew.
     
  12. djpubba

    djpubba New Member

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    What do you guys think of this?

    http://www.elecdesign.com/Articles/ArticleID/8055/8055.html

    I already have four DS1666-100's, which look very similar to the MAX5160's used there. Would this be simpler? Looks like I'd need a comparator and a oscillator. I'm sure I have oscillators around I can grab, but what's a comparator?
     
  13. Roff

    Roff Well-Known Member

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    Digital pots are quantized, which means they have a finite number of steps. The highest resolution ones I have seen are 10 bits, which means they have 1024 steps. Your apparent pot position would possibly dither randomly by one step, which in the case of a 100k pot would be about 100 ohms. I don't know if this would be acceptable or not. :?:
     
  14. djpubba

    djpubba New Member

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    Hmmm. The DS1666-100s I have are 7 bit... 128 steps. Might be too jumpy. 10 bit might be okay, though.

    The pot position on the Vectrex controller already jumps all over the place at small increments because you're fighting against the centering springs to hold it in one spot. That makes it incredibly far from steady already, Teddy. ;)
     
  15. ljcox

    ljcox Well-Known Member

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    I don't think you need to explore that path. I have to go out for half an hour, when I return, I'll do some calculations on the option I outlined earlier.
     
  16. djpubba

    djpubba New Member

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    I've got Fry's, Vetco and Radio Shack near me. SMT'd be last resort, but my deadline is very near, so I'll take anything that works and is fairly simple.
     
  17. Roff

    Roff Well-Known Member

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    The only one I can find quickly is National's LM6132. It's a dual in an 8 pin DIP. I don't know if Fry's will have it, and I live in the hinterlands. The nearest Fry's is probably 700 miles away. I could look for more, but it would be a crap shoot.
     
  18. djpubba

    djpubba New Member

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    MAX4132? Edit: No, it's not a DIP. Does NTE make one? They're big on NTE parts around here.
     
  19. ljcox

    ljcox Well-Known Member

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    I'm having a problem with the constraints imposed.

    Does the computer have a -5V supply available? That would help me.

    Forget this question, I reviewed your first post and thew answer is no.

    So I'll look into +5 to -5 converters.
     
  20. ljcox

    ljcox Well-Known Member

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    I have tried various options and I think the attached will work.

    It combines my idea of making the result linear and Ron's level shifter.

    The op amp CA3130 will operate from a single 5V supply and the input voltage can be as low as -0.5V.
     

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  21. djpubba

    djpubba New Member

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    CA3130 crosses to NTE930. Hopefully someone will have that in stock.

    Excuse my ignorance, but which of Ron's schematics is the level shifter?
     

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