0
\$\begingroup\$

As a part of a circuit an optocoupler was driving the gate of a MOSFET as follows:

enter image description here

Now I need to change the above opto with a 4N46 darlington one.

I'm not familiar with these much and having difficulty to replace it. Here is the functional diagram of the opto-darlington I need to use:

enter image description here

How should this be connected as an interface so that it would have the same functionality with previous one?

\$\endgroup\$

2 Answers 2

6
\$\begingroup\$

Connect Vcc to pin 5 and connect pin 4 to the MOSFET gate. Leave pin 6 open.

\$\endgroup\$
14
  • \$\begingroup\$ Thank you, this will not invert the logic of the previous one correct? In this case MOSFET turns ON when opto’s input is HIGH? \$\endgroup\$
    – user1245
    Oct 24, 2018 at 15:46
  • \$\begingroup\$ Yup. Notice that the new truth table is talking about the voltage difference between pins 4 and 5. When the LED is on, the voltage difference, is small. This means that the voltage across the gate resistor MUST be large, and MOSFET will be turned on. The logic would be inverted if pin 4 (labelled GND) were actually connected to ground, but in your circuit that won't happen. \$\endgroup\$ Oct 24, 2018 at 17:36
  • \$\begingroup\$ @WhatRoughBeast I need your help. The reason I want to use 4N46 is because I want Arduinio pin to source less than 10mA with a guarantee turn on. But Im having trouble to size the series resistor for the opto input. Are we sure that the pin 6 will be floating(will it have affect on CTR)? And if so how can I size the resistor? \$\endgroup\$
    – user1245
    Oct 26, 2018 at 23:41
  • \$\begingroup\$ Some say base pin(pin 6) is for setting the gain of transistor. What does that mean? \$\endgroup\$
    – user1245
    Oct 26, 2018 at 23:44
  • 1
    \$\begingroup\$ @atomant - WHY do you think you need to replace the CNY17-1 with a 4N46? What is your Arduino output level (in volts)? "Are we sure that the pin 6 will be floating" - Well, as long as you leave it floating, yes, it will be floating. And what, exactly, is Vcc (in volts)? \$\endgroup\$ Oct 27, 2018 at 4:16
1
\$\begingroup\$

Simply plug the 4N46 in in place of the CNY17. Pin 4 is the emitter of the darlington transistor, and pin 5 is the collector - same pinout as the CNY17.

\$\endgroup\$
3
  • \$\begingroup\$ Will the logic remain the same? \$\endgroup\$
    – user1245
    Oct 24, 2018 at 16:11
  • \$\begingroup\$ Yes - the logic will remain the same (otherwise, I wouldn't have said "just plug it in") \$\endgroup\$ Oct 24, 2018 at 16:14
  • \$\begingroup\$ The reason I want to use 4N46 is because I want Arduinio pin to source less than 10mA with a guarantee turn on. But Im having trouble to size the series resistor for the opto input. Are we sure that the pin 6 will be floating(will it have affect on CTR)? And if so how can I size the resistor? Some say base pin(pin 6) is for setting the gain of transistor. What does that mean? \$\endgroup\$
    – user1245
    Oct 26, 2018 at 23:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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