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I'm reverse-engineering a CRT anti-pincushion circuit and it uses a complex diode-resistor circuit to approximate the function \$X-k×X×Y^2\$, where \$X\$ and \$Y\$ are the deflection voltages. My question is if there's any way to understand this circuit conceptually. I put it into LTSpice and I get results but is there any way to understand this at a higher level? How did they come up with this circuit in the first place?

I'm familiar with simple resistor-diode function generators that make a piecewise-linear function by treating the diodes as switches that turn on at various points (like this). But this circuit has four resistor-diode "blocks" that are cross-coupled by R14-R17 which mystifies me. Each block has diodes in opposite directions, which sort of makes sense so everything happens in reverse when X and Y are negative.

The blocks on the right (D9-D14 and D15-D20) implement roughly cubic functions, so I tried to understand this as \$(X-Y)^3 - (X+Y)^3 = 6XY^2 +\$ smaller terms, which (scaled) is the desired correction factor, but I couldn't make this explanation work.

Maybe there's a theory of diode function generators that explains this circuit?

Schematic of the circuit

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    \$\begingroup\$ Might start here? Good question though, so +1. \$\endgroup\$
    – jonk
    Commented Aug 10, 2021 at 23:45
  • \$\begingroup\$ @jonk: That patent describes the same problem but a totally different solution, interestingly enough. With a raster-scan CRT, the correction can be a function of time as in the patent. But my circuit is for a vector CRT, where the correction needs to be computed from the X-Y values. \$\endgroup\$
    – Ken Shirriff
    Commented Aug 11, 2021 at 1:31
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    \$\begingroup\$ I was hoping to provide a starting point in time -- a reference point -- at least. It includes a diagram that I also felt might stimulate useful ideas. But I've not thought about the problem, yet, except to note that the needed correction looks hyperbolic in shape. \$\endgroup\$
    – jonk
    Commented Aug 11, 2021 at 1:36
  • \$\begingroup\$ @jonk: thanks for providing it as a starting point. Looking at the patent, maybe you meant parabolic instead of hyperbolic? This would match the Y² contribution I see from the circuit. \$\endgroup\$
    – Ken Shirriff
    Commented Aug 11, 2021 at 1:41
  • \$\begingroup\$ No, I was actually thinking hyperbolic. Parabolic has a unique shape out of an infinity of nearby shapes and is unlikely to actually occur. Hyperbolic is 'everything else' so to speak. And, given some mild experience talking with experts on this decades back, I'm quite sure they used 'hyperbolic' as a general description. I believe, had I bothered to pin them down, that they would have included 'parabolic' in the limit, though. \$\endgroup\$
    – jonk
    Commented Aug 11, 2021 at 1:53

1 Answer 1

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You have come pretty near to answering your own question. In the discussion below, I don't list all of the polarities and parallel paths to keep it brief.

This circuit is a piecewise approximation of the algorithms described. You can see that for small voltages for X and Y, there is no contribution to the Output voltage from the diode arrays because D9, D12, D15 and D18 do not conduct.

The left diode arrays are dominated by Y. As Y increases, the voltage on R13 also increases proportionally to Vy until the voltage across R18 equals two diode drops. D7 and D8 start to conduct, and the slope of the voltage on R13 increases as the resistance of R18 and R4 are in parallel. Since this is happening on R3 as well, the symmetrical nature of the circuit means that the voltage contribution at R9 is zero for X equal to zero.

But as the X voltage changes with respect to the the voltage on R13, D12/D9 will reach one diode drop, and there will be a current path in series with R9, which will affect the Output voltage depending on the voltage on X. This will be affected by the voltage on R13. As the difference between these voltages increase, first R8 will reach one diode drop, and then R5, once again providing two different impedances affecting the output and dependent on the amplitudes of both X and Y.

This is a conceptual description. There are some slight impedance changes, and the left diode arrays are not 100% independent of X. Diodes do not switch instantly. But in general, the circuit performs no multiplication. This is a piecewise-linear approach to your problem, and the resistances must be chosen with knowledge of the X and Y signal levels. It is not a universal circuit and the values shown must be adjusted to match the ranges of X and Y.

Good luck!

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