I'm looking at a datasheet for a bit of hardware I am adding to a design. However, one of the pins on the chip is listed as:

ON/OFF - 1.8V must be driven to open drain/collector

It does not specify if it is an input or output.

I don't fully understand what open drain/collector means in this situation, however what I understand from my research is that it typically refers to an internal connection to a transistor. They usually seem to be used as outputs (as far as I understood).

open collector output

However, what is confusing me is the use of "driven" in the pin description. This appears to imply that I should be powering the pin with an open drain/collector. Is this what I am supposed to be doing? How would I achieve this?

  • 1
    \$\begingroup\$ What hardware is it? Make/model, link to datasheet? What does the pin do? \$\endgroup\$
    – Justme
    Jun 17, 2020 at 10:34
  • \$\begingroup\$ yep, linking to that datasheet seems ... more than sensible. \$\endgroup\$ Jun 17, 2020 at 10:38
  • \$\begingroup\$ Regard it like a little switch that pulls the line to ground hence, if you had a pull-up resistor to 12 volts, you'd get an output that could be 0 volts or 12 volts (data sheet permitting of course). \$\endgroup\$
    – Andy aka
    Jun 17, 2020 at 10:40
  • \$\begingroup\$ @Justme This is the datasheet \$\endgroup\$
    – JolonB
    Jun 17, 2020 at 11:30

1 Answer 1


The transistors going to an output pin are referred to as the pin driver stage. So the term driven here means that the driver must do the thing described, in this case go high-impedance ('tri-state' for a logic gate output driver).

Your datasheet text states that this pin must be driven by an open-drain or open-collector output. These refer to the use of a FET output transistor or a BJT respectively in the driver. You have correctly deduced that this is a single output transistor that can drive a (logic) low connection to ground or a connection to 'nothing at all'. The latter is the high-impedance connection where the driver transistor is off and presents only a very high impedance to ground as the transistor will still let through a tiny 'leakage' current.

The '1.8 V' part means that the driver must be capable of withstanding 1.8 V across its transistor's drain or collector when the driver is high-impedance.

(Please expand your question to explain your situation, and what you have, for an answer on 'how can I achieve this'.)

  • \$\begingroup\$ Here a link to the datasheet. Unfortunately, I don't know what else I can add as that's pretty much all I understand about it. I've just spotted an image on pg 35 of the datasheet for the modem on the board. Should I be creating my own open collector circuit and feed that into the ON/OFF input? \$\endgroup\$
    – JolonB
    Jun 17, 2020 at 11:40
  • \$\begingroup\$ @JolonB, it's not just info on the chip - it's: what you're designing, what it's for, how many you're making and why, what else is in the circuit around it, what sort of cost limits you have. It's what you're doing and why. Paragraphs rather than one-liners. At the moment, your question just says cryptically 'a bit of hardware I am adding to a design'. The better the quality of your question, the better the quality of the answers you will attract. Thanks. \$\endgroup\$
    – TonyM
    Jun 17, 2020 at 11:45

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.