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This is an interface circuit of Quadrature Encoder (A, A', B, B') with MCU. I cannot make sense of the working of OpAmps in this circuit. Is it some standard circuit for doing signal conditioning or is it doing some signal processing on the input signals? And what could be the purpose of OA and OB outputs?

edit: better quality image added

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edit: data for possible encoders added as image.

enter image description here

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    \$\begingroup\$ Those are not op-amps (or at least are not being used as such) those are all comparators. \$\endgroup\$ – Edgar Brown Feb 14 at 22:21
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    \$\begingroup\$ Hard to tell what is the purpose of OA and OB, could be a buffered signal A and B for external use. What you have is a differential encoder signal. Could be A HTL or TTL level. There are better ways to convert them, like RS485 line receivers instead of comparator. \$\endgroup\$ – Marko Buršič Feb 14 at 22:53
  • \$\begingroup\$ Replaced the image with better quality image in the question. \$\endgroup\$ – scico111 Feb 14 at 22:58
  • \$\begingroup\$ Here's some butter - what sort of sandwich did it come from? \$\endgroup\$ – Andy aka Feb 15 at 12:36
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The schematic quality is terrible but we should be able to figure it out.

  • The LM339 is a comparitor (not an op-amp). These are configured to convert the complementary inputs into unbalanced outputs.
  • The inputs are complementary and may not switch fully to V+ or GND (or may be a different voltage than the circuit in question).
  • The output is open-collector which can only pull low so a pull-up resistor is attached to each output. Pull up is to +15.
  • A 2.5 V mid-supply reference is generated by a pair of resistors, R18 and R19 and a stabilizing capacitor, C5. This provides the switching threshold for the inverters.
  • The conditioned outputs are generated by two more comparitors with their associated pull-up resistors and what looks like a current-limiting resistor.

The OA and OB outputs can be fed into some standard logic inputs but CMOS is most likely as it is a 15 V output.

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  • \$\begingroup\$ I wonder if its possible to do this conversion of complementary inputs to unbalanced outputs inside the MCU and get rid of this external circuit? \$\endgroup\$ – scico111 Feb 14 at 22:57
  • \$\begingroup\$ @scico: It might be but you have given us no datasheet link for the encoder. \$\endgroup\$ – Transistor Feb 14 at 23:04
  • \$\begingroup\$ @Transistor I have added an image about possible encoders in the question \$\endgroup\$ – scico111 Feb 14 at 23:27

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