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Target ID with simple F.E.T analogue multiplier?

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  • Target ID with simple F.E.T analogue multiplier?

    Would this circuit topology work for such a task. In this circuit, Vout is proportional to V1*V2. The F.E.T's conductance are matched to within+\-5%

    Can the circuit can be tweaked for X and R channel inputs to give target information.

    What do you think? This circuit is featured in Wireless world Circard collection 1978. All the relevant math is there.

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  • #2
    I think you need some kind of division function to get a target vector. Multiplying X x R does not let you know which is greater or lesser. One of the early Tesoros used a analog multiplier IC for target ID. If you look at the datasheet these chips also had a division function that was used. Is there any text description with your circuit?.
    Attached Files

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    • #3
      If it helps anyone:

      Page 126 of this pdf shows the circuit ( plus others on previous pages ):
      Circard Collection 1978 | PDF

      There are some nifty things that can be done with analogue multipliers/dividers. I like the one that calculates the vector amplitude A = sqrt ( X*X + Y*Y​ )
      There are some excellent design resources for these IC's out there, mostly older app books from the 70's / 80's, really interesting stuff.
      Update:
      This is one I have in print and pdf:
      Nonlinear Circuits Handbook, 1976 | Analog Devices
      Last edited by Skippy; 05-03-2025, 02:00 PM.

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      • #4

        ATtiny85
        VDI=degrees(atan2(Yy,Xy));
        Magn = sqrt(pow(Xy,2)+pow(Yy,2));
        ​​

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        • #5
          Originally posted by Skippy View Post
          If it helps anyone:
          Page 126 of this pdf shows the circuit ( plus others on previous pages ):
          Circard Collection 1978 | PDF

          https://www.scribd.com/document/6661...ollection-1978
          not the best way to leave links here, leading to a paid site when in 0.2 seconds Google shows that you can download it without any fuss from the magnificent site worldradiohistory.com

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          • #6
            arctan(Y/X) = target ID angle.

            In whites 5900/6000di they do log(A)+log(B) which is the same as log(A*B). I am not sure what is exactly is done in the stage following the summing amp. But after that it goes to the meter via a sample switch.

            In the toltec, RC4200 is used as voltage divider I think, Therefore input pin 8 divided by input pin 5. Pin 4 is the output. pin 1 is also connected via a resistor to VCC. So the math is a little more refined(as per datasheet), but basically that's it. Followed by a gain control stage, then a gated sample and hold before driving the meter.

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            • #7
              Then there was this which I found here on the forum. By Davor.

              Click image for larger version  Name:	US4001602.gif Views:	0 Size:	52.0 KB ID:	437214

              Based on some patent US4001602


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              • #8
                so isn't it just a ratio of 2 vector quantities. The angle subtended being the phase of the target, right. or something like that. Just in two quadrants for the purpose of demonstraing the basis.

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                • #9
                  I am sure Skippy can explain the math. Davor's analog divider sim is interesting. I'll bread board it and see if works the same as the simulation.

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                  • #10
                    here is the sim file

                    The patent explains the basis of this circuit with the 2 diodes. "Advantage is taken of the current-exponential voltage characteristic of a diode".​
                    Attached Files

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                    • #11
                      ... I guess you are gradually describing all the parts of an analogue computer.

                      https://www.analogmuseum.org/english/library.html

                      o
                      r here

                      https://the-analog-thing.org/
                      Last edited by moodz; Today, 05:31 AM.

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                      • #12
                        Altra:
                        If you're breadboarding this diode-divider, it would work best if you used an SMD double-diode, example BAV70. Better opamps would certainly help ... anything is better than 358's. And 1% resistors in most places, maybe even try and match them in pairs for the obvious locations:
                        R5 = R6
                        R8 = R7
                        R1 = R2

                        R10,R11 are intended to be 9:1 ratio to give a non-inverting gain of 10. Better alternative values would be 2k / 18k or 3k / 27k.
                        ------

                        Here's another analogue multiplier application guide, shorter than the book I linked to in my earlier post:
                        Analog Devices : Multiplier Application Guide
                        Last edited by Skippy; Today, 03:28 PM.

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                        • #13
                          Originally posted by Skippy View Post
                          Altra:
                          If you're breadboarding this diode-divider, it would work best if you used an SMD double-diode, example BAV70. Better opamps would certainly help ... anything is better than 358's. And 1% resistors in most places, maybe even try and match them in pairs for the obvious locations:
                          R5 = R6
                          R8 = R7
                          R1 = R2
                          ------

                          Here's another analogue multiplier application guide, shorter than the book I linked to in my earlier post:
                          Analog Devices : Multiplier Application Guide
                          I have BAV99's and BAT54's which should be fine. Thanks

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                          • #14
                            The BAT54 is a Schottky, is it not? And 99's are series-connected diodes, what's needed is common-cathode, though the circuit could be 'flipped' to work with common anode diodes.
                            Dual transistors would seem like a good idea, just use the base-emitter junction as a diode, and connect the unused collector to the base. Trouble is, they are not monolithic, they seem to be two independent transistors, so not obviously well matched.

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                            • #15
                              You are right about the diode configurations, being series connected. I'll just use discrete components to test the theory.

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