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Setup: A digital output from a device is connected to an output buffer to drive an optocoupler. If device power is lost the output will go in high-impedance mode.

Problem: I want to make sure that the buffer output is low when device power is lost. How do I dimension a pulldown resistor for the input of the 74HC244N buffer?

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2 Answers

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10 kΩ is a typical value for a pullup/pulldown, and the 500 µA loss at 5 V is usually not a problem; the optocoupler's LED will usually need a multiple of that. For low power applications you can increase the value, and the upper limit is determined by the 74HC244's input leakage current. The datasheet says that is maximum 1 µA, then a 1 MΩ resistor may cause a 1 V drop across it. That's a rather large value, still OK at 5 V supply voltage, but I would choose a lower value.

A 100 kΩ resistor will cause a 50 µA leakage from the device's output, and a maximum input voltage of 100 mV when floating. This looks like a good solution.

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The higher the resistance, the more sensitive it will become against EMC though. It is not unlikely that the purpose of the optocoupler is to shut out a noisy environment. To make a detailed suggestion of a value, we'd need to know the nature of the application. – Lundin Oct 23 '12 at 13:51
up vote 5 down vote

Typically, the pull down resistors for CMOS logic are between 10kΩ and 100kΩ, up to 1MΩ for battery powered devices with low power consumption.

At the same time, the LED in the optocoupler is a current-driven device, as opposed to a voltage-driven device. If an LED is connected to high impedance, it will not produce light. Effectively, an LED works as a pull-down by itself.

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Thanks for your answer. You say typically between 10kΩ and 100kΩ - this is a security feature so I have to be absolutely sure that this is working at all time. Do you know a way of determining the optimal value, rather than relying on what usually works? The optocoupler input is never high impedance since it is driven by the output buffer. – mola Oct 23 '12 at 8:00

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