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Sony SQD-2020 wavematching logic,

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Joined
Oct 12, 2018
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Location
Brazil
Can someone please help me understand this logic circuit from the Sony SQD-2020, service manual attached. I marked point a1 and point a2 as I'm considering them to be different nodal voltages, although the manual states they are the same, circled as |Lf - Rf|. I did conclude a1 = |c| - |f| and a2 = |f| - |c| as c and f are the output signals coming straight from the Modified SQ Decoder into the full wave rectfiers. Then, a1 and a2 go to a negative rectifier which I assume will be the sum of the negative semicycles of a1 and a2, resulting in what I called at, whice has nothing above 0. Am I missing something that would make a1 and a2 the same thus |Lf - Rf|?
sony wavematching.jpg
 

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par4ken

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When Left and Right (front) are equal the voltages match, negative from left and positive from right, and vice versa. The output is zero. If one channel is more dominant than the other then the summed voltages will be more than zero (positive or negative), the diodes pass the negative voltage to the differential amp. The same is done with the rear channels. The output of the differential amp is a control voltage to vary the gain of the front and rear output amplifiers.
 

par4ken

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At point a just before the half wave negative rectifier? Equal in amplitude?
As shown in the block diagram each output is rectified, the rectifier puts out a positive and a negative voltage of the same value. The positive of each channel and the negative of the opposite channel are added together, as such the output will depend on the difference in level between the two. If the two channels are equal in level the voltages cancel each other out, giving no output. The output before the half wave rectifiers could be positive or negative, they are using only the negative portion, the positive being blocked by the diodes.
 

Sonik Wiz

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As shown in the block diagram each output is rectified, the rectifier puts out a positive and a negative voltage of the same value. The positive of each channel and the negative of the opposite channel are added together, as such the output will depend on the difference in level between the two. If the two channels are equal in level the voltages cancel each other out, giving no output. The output before the half wave rectifiers could be positive or negative, they are using only the negative portion, the positive being blocked by the diodes.
Yup.
I can't really add anything.
 
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