0
\$\begingroup\$

I'm trying to control a MAX383 analog switch using a Raspberry Pi, through an optocoupler (Sharp PC817), to separate digital logic from the analog section.

On the analog side, I will have a bipolar supply (+6, -6V) from a TLE2426 rail splitter.

I want to connect the positive rail to the logic pin on the MAX383, through the optocoupler which would source current to the pin. If my logic is correct, I will also need a pull-down resistor (I assume somewhere in the 100k range would be ok), from the pin to ground.

Question is: do I need to add a current limiting resistor somewhere in there? From the datasheet I see it needs min 2.4V to register a logic high, but there's not max specified, so I assume 6V from the power rail is OK. However, it does state max current to be 1uA.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

2 Answers 2

2
\$\begingroup\$

If the MAX383 is connected to the positive 6V rail on its Vcc pin then take note of the note underneath the absolute maximum ratings in the data sheet: -

Note 1: Signals on NC, NO, COM, or IN exceeding V+ or V- are clamped by internal diodes. Limit forward diode current to maximum current rating.

This means you are OK because the "IN" input is the digital logic control pin.

You don't need a current limiting resistor but you do need a pull-down as shown on your diagram.

\$\endgroup\$
2
\$\begingroup\$

The logic input pin of MAX383 is an high impedance input port. Even if you don't use a resistor, it will only drain typically a few nano-amperes. Adding a resistor may negatively affect the switching performance on the other hand. So, you shouldn't use a current limiting resistor in your circuit.

enter image description here

\$\endgroup\$

Your Answer

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

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