Analog multiplier

Discussion in 'General Electronics Chat' started by atferrari, May 23, 2014.

  1. atferrari

    Thread Starter AAC Fanatic!

    Jan 6, 2004
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    I run across the schematic shown in page 13 of the RC4136's datasheet (attached). After 3 days trying to have it "working" in LTSpice (circuit attached) I realized I need help.

    I tend to think I know the basics of log / antilog amps and I understand this circuit should give (hopefully) a Vout = Vy * Vx / Vz. Instead, the prevailing result I get is Vout slightly above Vx, no matter what Vy or Vz are.

    With different op amps, or no matter how many diodes I pile up at the top (to get up to around 3.5V) to polarize the first stages, I always get a result not even close to the expected (and most of the times with >1 MHz noise riding the output).

    I am puzzled by what is wrong: my simulation or the circuit?

    If anyone is tempted to tell about the so many pitfalls of a circuit like that implemented with discretes or how old / lousy the RC4136 is, PLEASE don't. I am asking about a simulation and / or eventual flaw of a design.

    Helps is appreciated.
     
  2. Brownout

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    Jan 10, 2012
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    Sorry, I can see your circuit, as I don't have LTSpice on my laptop. I can only warn that some component models don't work very well, especially diodes. I would maybe try different diodes.

    Sorry for the shot in the dark :)
     
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  3. atferrari

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    Here you have it, Señor Brown!

    Leaving aside the diodes could be not properly modeled, I have no idea what are they for at the output of each amp.

    As long as you do not shoot me...... :D :D :D
     
    Last edited: May 23, 2014
  4. Brownout

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    I'm not exactly sure either, but here's a guess: Each log amp needs to be stabilized for temperature. The diodes might shunt current on a temp dependent bases, which denies current to the log transistors, giving them some temp stability.
     
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  5. t06afre

    AAC Fanatic!

    May 11, 2009
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    Analog multipliers built by discrete components have many issues at least far more than the integrated ones. Analog Devices have a line of analog multipliers still i think. You better check out one of those
     
  6. atferrari

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    Thanks for replying.

    Quoting the OP:

     
  7. crutschow

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    Mar 14, 2008
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    I didn't have much luck simulating the circuit either. I think the design may be questionable.
     
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  8. #12

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    Nov 30, 2010
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    The LT1022 can not work with inputs within 2 or 3 volts of the rails and the outputs miss the rails by 2 to 3 volts. This might be your entire problem. The chip you chose is just not right for the job.
     
  9. Brownout

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    I don't know if this helps but here is a circuit that does simulate. The 1st and 3rd stages are log and anit-log stages, which are the baises of multipler/divider circuits. This circuit would be impractical to build.
     
  10. atferrari

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    Hola Nro. 12,

    Could show one that is right for the job?
     
  11. #12

    Expert

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    Now I see that you gave it enough voltage to work with. Sorry. My bad.
    I thought you were trying to use zero and +5 volts for the chips.
     
  12. t06afre

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    May 11, 2009
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    @Agustín
    It looks like you have forgotten the KISS rule. Just for simulation purpose "a" multiplier or divider may I suggest these simple model files:D
    Code ( (Unknown Language)):
    1. *Multiply Voltages
    2. .SUBCKT MULTV 1 2 3
    3. BX 3 0 V=V(1)*V(2)
    4. .ENDS MULTV
    Code ( (Unknown Language)):
    1. *Divide Voltages
    2. .SUBCKT DIVV 1 2 3
    3. BX 3 0 V=V(1)/V(2)
    4. .ENDS DIVV
     
    Last edited: May 24, 2014
  13. atferrari

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    Thanks to suggestions in another forum, grounding the base of Q2 and Q6, the attached circuit works with positive values at the inputs since the action inside occurs in the negative values realm.

    As you could see, it simulates properly 6*20/10 = 12. Not bad, eh?

    Pending:to understand why the designer polarized those bases with positive values.
     
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  14. Brownout

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    Now you can eliminate R6, R7, R8, R10, D4, D5, D6, D7 and D8
     
  15. absf

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    Dec 29, 2010
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    At least I am learning something new today.;)

    Allen
     
  16. atferrari

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    Yes, time to save on components! :p :p
     
  17. atferrari

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    Same for me, good for both!!

    It took me some time to realize what the op amp does: output the necessary voltage (a Vbe that corresponds to the input current). Matched transistors, a must.

    In this particular circuit the op amps' outputs are all negative.

    What simulator are you using? Free by any chance?
     
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