I know that open collector outputs are effectively either 0 V or high-Z.

I also know that TTL chips tend to read a disconnected output as high, but that really you should have something like a 1 kΩ pull-up resistor for safety.

But does the warning about a pull-up resistor still apply if you're linking an open-collector output to an input?

As a concrete example, I want to implement OR logic with an open-collector NOR chip (74LS33) by connecting pins 12, 11, and 10: pins 8 and 9 are the inputs and pin 13 should be the (open collector) OR of 8 and 9.

74LS33 data sheet segment

Do I need to have a pull up resistor attached to the pin 12-11-10 net? Or will the open collector output for pin 10 still reliably drive the logic value for 11 and 12 high?

  • \$\begingroup\$ You need a pull-up or the signal won't go to a logic 1. It can function with current sinking logic such as TTL (but not CMOS) but will be unreliable, slow and with no noise immunity. \$\endgroup\$ Commented Jan 5, 2022 at 2:15

3 Answers 3


The open-collector output cannot drive the inputs high because the output does not produce a drive current.

So you need a pull-up resistor. The value depends on your switching frequency but 10K should be suitable.

Typically, logic gate inputs will float high due to input leakage current flowing out of the inputs. With an LS chip, this is a strong current because of the LS input circuit design.

However, it is not wise to rely on this by trying to go without the pull-up resistor. It will be more susceptible to noise, have a slower rise time and not rise right up to the supply rail. The low cost resistor removes these weaknesses.


It'll work without a pull-up, but it will be slow(er) than normal logic for the 0..1 transition, and it will be more sensitive to noise pickup -- it might have spurious transitions if there are nearby noisy signal lines.

  • 1
    \$\begingroup\$ That doesn't sound like working to me. \$\endgroup\$
    – AnalogKid
    Commented Jan 5, 2022 at 3:06
  • 2
    \$\begingroup\$ The point is that it will actually function and provide the logic behaviour expected; not that it is necessarily a good idea. \$\endgroup\$
    – jp314
    Commented Jan 5, 2022 at 4:10
  • \$\begingroup\$ I lived through the 60's and 70's. That whole "TTL inputs float high" thing never was as consistent, let alone reliable, as people say today. Connecting an unused pin to the one next door takes 1/1052th of a square inch of copper. What is the advantage (or allure) of not going the extra 1.66 millionth of a mile? \$\endgroup\$
    – AnalogKid
    Commented Jan 5, 2022 at 13:26

The 74LS33 datasheet specifications do not include a minimum value for the input pull-up current, and the maximum for the output high-Z leakage is quite high:

Motorola 74LS33

Possibly your inputs will each pull up with 0.4 mA, giving 0.8 mA total which is safely above the output high-Z leakage of less than 0.25 mA. But it is entirely within datasheet specifications to have only 0.1 mA pull-up current, or even down to 0 mA.

Furthermore, the noise margin is really small. Typically the TTL input only gets pulled only slightly above the threshold voltage, not up to the supply voltage. In this case somewhere around 2.7 V, compared to the 2.0 V minimum guaranteed high input voltage. With a pull-up resistor the input will go up to the supply voltage, typically 5 V.


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