0
\$\begingroup\$

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

\$\endgroup\$
  • \$\begingroup\$ How about an analog Mux? MC4053? \$\endgroup\$ – analogsystemsrf Oct 29 '19 at 16:52
  • 1
    \$\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\$ – KingDuken Oct 29 '19 at 16:53
  • 1
    \$\begingroup\$ So how can I do this in the simplest way? \$\endgroup\$ – Goncalo Seruca Oct 29 '19 at 17:01
0
\$\begingroup\$

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

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ Plz see the new image where i put the layout \$\endgroup\$ – Goncalo Seruca Oct 29 '19 at 17:15
  • \$\begingroup\$ Can i level shift the signal with a Voltage divider with 2 resistors? \$\endgroup\$ – Goncalo Seruca Oct 29 '19 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\$ – Spehro Pefhany Oct 29 '19 at 17:18
  • \$\begingroup\$ So how do I do it? \$\endgroup\$ – Goncalo Seruca Oct 29 '19 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\$ – Spehro Pefhany Oct 29 '19 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.