I need to build a controllable output shutdown for the OPA 564 so I can force the amp to shut down. The solution in the datasheet uses an optocoupler but I rather use it with some pull-down resistor. My build uses dual-supply with +/- 6,2 V. In this link you have the datasheet of OP 564: http://www.ti.com/lit/ds/symlink/opa564.pdf

enter image description here Could this be a solution? enter image description here

  • \$\begingroup\$ How about an analog Mux? MC4053? \$\endgroup\$ Oct 29, 2019 at 16:52
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    \$\begingroup\$ The datasheet says above that picture, ... to shut down the OPA564 the voltage level of the logic signal must be level-shifted by some means. Do you believe a pull-down resistor will accomplish this? \$\endgroup\$
    – user103380
    Oct 29, 2019 at 16:53
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    \$\begingroup\$ So how can I do this in the simplest way? \$\endgroup\$ Oct 29, 2019 at 17:01

1 Answer 1


Something like this might work (one zener and one resistor):


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Plz see the new image where i put the layout \$\endgroup\$ Oct 29, 2019 at 17:15
  • \$\begingroup\$ Can i level shift the signal with a Voltage divider with 2 resistors? \$\endgroup\$ Oct 29, 2019 at 17:18
  • \$\begingroup\$ No that won't work, there is already a pullup resistor inside the chip (the nominal 100K I show above). The absolute simplest way is to use a CD4053 as @analogsystemsrf suggests in a comment, only a single component (because it has built-in level shifting). \$\endgroup\$ Oct 29, 2019 at 17:18
  • \$\begingroup\$ So how do I do it? \$\endgroup\$ Oct 29, 2019 at 17:20
  • \$\begingroup\$ 4053? Power it from +/-6.2, make sure the logic '1' input is more than 0.7 * 6.2 = 4.3V and connect a switch of the 4503 (your choice of N.O. or N.C.) between the E/bS input to -6.2V. \$\endgroup\$ Oct 29, 2019 at 17:25

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