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I'm currently working on a project and it consists of amplifying a 5V signal from an Arduino to 24V. I made my circuit and hooked it up to a oscilloscope and it shows that the output of the optocoupler is active low. I just wanted to know if there's a way I can make it active high.

Thanks in advance.

schematic

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Yes. Swap the positions of the 1.6 kΩ resistor and the opto-transistor. That will default to pull-down and the opto-isolator will pull up.

It's not clear why you're using the opto-isolator other than as an inverter (which you seem to not want) since you have CH connected to the micro-controller's D7 so you have lost all isolation.

Your CH1, if it's a 5 V input, should be fed directly from D7, not from the top of the infrared LED where it will be clamped by the 1.4 V forward voltage, Vf of the infrared LED.

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    \$\begingroup\$ I assumed the box on the right was an oscilloscope just used for looking at the signals during prototyping, and not some external circuit. If that's the case, then the output of the opto could still be isolated (assuming it uses a different ground). \$\endgroup\$ – newothegreat Aug 1 at 15:30
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    \$\begingroup\$ It can't be isolated or there's no return path for the CH1 current. OP has used the same earth symbol for both sides. \$\endgroup\$ – Transistor Aug 1 at 15:34
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    \$\begingroup\$ Like then answer shows, in this case, swapping the output transistor side is possible. There is another option too which is used with logic output optocouplers where the output side cannot be swapped. In this case the input LED side is changed to be driven active low. Of course it depends on what is the best idle state for power consumption. \$\endgroup\$ – Justme Aug 1 at 17:25

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