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It must be late in the day or I am just losing it. I can't quite understand what this datasheet is saying for input voltage level. I am running my 74HC245 at 3.3V. I have the OE tied high, but mistakenly tied it to a 10K pull up resistor. It appears that I am not getting high enough to disable the OE with 10K. I understand I should be using something like a 4.7K resistor instead. However, when I read the data sheet, it seems to state that the minimum high voltage at 4.5V VCC is 3.15V but typical is 2.4V. That doesn't seem to make sense to me. If the minimum it recognizes for HIGH is 3.15V how can typical be below this at 2.4?

Here is a link to the full datasheet: https://assets.nexperia.com/documents/data-sheet/74HC_HCT245.pdf

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Here is how the control signal goes to the ULN2803.

enter image description here

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  • \$\begingroup\$ Well after more investigation my OE logic line, which is held HIGH with a 10K pullup is driving an input to a ULN2803 to drive an LED. I'm wondering if having this attached to this NPN transistor array is affecting the ability of a 10K to pull-up to 3.3V. I'll breadboard up in the morning to see if I can duplicate this. st.com/content/ccc/resource/technical/document/datasheet/e4/fa/… \$\endgroup\$
    – Kevin McQuown
    Commented Nov 20, 2023 at 22:01
  • \$\begingroup\$ @KevinMcQuown 10kΩ would easily pull up a 74HC input, since that input's impedance is in the gigohms. Even 1MΩ would work, so the problem is either that the input is damaged (by static discharge, for example) or there's something else preventing the potential there from rising. A ULN2803 input would do exactly that. \$\endgroup\$
    – Simon Fitch
    Commented Nov 21, 2023 at 2:00
  • \$\begingroup\$ ULN2803 inputs take in about 1 mA current, so the pull-up should be less than 1 kohm. \$\endgroup\$
    – jpa
    Commented Nov 21, 2023 at 7:34
  • \$\begingroup\$ Look at fig1 on page 3 of the ULN2803 datasheet. Connecting that to a line with a 10k pullup could make an IMMENSE difference. \$\endgroup\$
    – Russell McMahon ♦
    Commented Nov 21, 2023 at 12:29
  • \$\begingroup\$ The question makes no sense. If the only thing on OE is the pull-up, any pull-up will do as OE input draws no current (OK, fine, maybe 1uA of leakage which we can ignore but still 1 megaohm pull up should work). If you have other things on the OE pin, that will draw current or clamp the voltage, is another thing, but you don't mention you have other stuff on OE. \$\endgroup\$
    – Justme
    Commented Nov 21, 2023 at 16:54

3 Answers 3

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The minimum voltage you must apply to the input for it to be recognized as a '1' with a 4.5V supply is 3.15V (any possible sample of device, over the entire temperature range), but typically anything over 2.4V will do it (at room temperature with many samples of the device).

With a 3.3V supply you can safely assume it's 0.7\$\cdot\$3.3V = 2.31V worst case (anything higher than that is guaranteed to be recognized as a '1'). Similarly anything less than 0.3\$\cdot\$3.3V = 0.99V is guaranteed to be recognized as a '0'. (In both cases within the allowable input range, of course).

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Minimums and maximums are what you use to make sure your design works under every case, and typical is what you use to see what usually happens. The way I read this is:

  • Typically, if VDD = 4.5V, you need to pull the input up to about 2.4V before the device registers the input as a logical '1'.
  • The lowest voltage that will always register a logical '1' with a VDD of 4.5V is 3.15V.

Each input will have a voltage where above that voltage it will register as a logical '1', and below that voltage it will register as a logical '0'. Of course, that middle area is a no-man's land that you usually don't want to be in for two main reasons:

  1. It's a source of variation between parts - aiming to keep your voltage at say 2.7V might usually be a logical '1', but some parts may still register that as a logic '0' and they wouldn't be defective.
  2. Often times digital inputs that aren't close to the rails will draw extra power as both NFET and PFET input stages are partially on. This extra power dissipation may cause issues in the chip aside from just using more power.

Now looking at VDD=4.5V numbers when your VDD is 3.3V, is not super useful, but if the OE pin is above 2.5V it's probably registering as a logic '1'. Both 10kΩ and 4.7kΩ pull-ups sound reasonable, but what is the voltage on the OE pin?

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  • \$\begingroup\$ thank you that is very helpful. I am measuring 1.7V when the pull-up is acting on its own for this control pin. However when my board is disconnect from a main status board that has LED's to show control line states, I measure a full 3.3V even with the 10K resistor. The control board is somehow affecting it. I believe it may be due to the fact that this digital input goes to a ULN2803 and that may be causing the issue as this is connected to a base of an NPN transistor. \$\endgroup\$
    – Kevin McQuown
    Commented Nov 20, 2023 at 22:17
  • \$\begingroup\$ @KevinMcQuown: 1.7 V sounds like the forward voltage drop of a red LED. It could also be approximately the Vbe of a Darlington pair. Are the LEDs/ULN2803 connected directly to the signal lines? They really should be buffered. ULN2803 inputs are NOT TTL inputs. The are "compatible" only if there are no other connections to TTL inputs. \$\endgroup\$
    – Dave Tweed
    Commented Nov 20, 2023 at 22:24
  • \$\begingroup\$ For Vdd values not listed: For typical CMOS, > 70% of Vdd is guaranteed logic 1, < 30% of Vdd is guaranteed logic 0. Typical switching point is near 50%. \$\endgroup\$
    – Mattman944
    Commented Nov 20, 2023 at 22:39
  • \$\begingroup\$ @DaveTweed I updated my original post to show how the control lines are connected to the ULN2803 and drives the LED's \$\endgroup\$
    – Kevin McQuown
    Commented Nov 20, 2023 at 23:25
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    \$\begingroup\$ Yeah, you need an additional resistor in series with the input to the Darlington array - otherwise the double base junction will clamp your output to 1.8V apparently. \$\endgroup\$
    – W5VO
    Commented Nov 20, 2023 at 23:45
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I duplicated my issue. Connecting a 10K pull up resistor to the input on a ULN2803 to drive an LED causes the voltage at drop to 1.7 volts. Exactly what I am seeing on the bench with a breadboard setup and clearly 1.7V on an HC device at 3.3V is seen as a LOW, not what I want. It looks like if I go with a 2K pull-up I get a voltage of 2.5V which should be sufficient for a high reading going into the OE of my 74HC245

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  • \$\begingroup\$ After further testing, I replaced my ULN2803 with a single BS170 MOSFET and got exactly the behavior I want. It is still show 3.3V after the 10K pullup resistor and and it controls the LED properly. I guess I needed to use MOSFETs rather than NPN transistors. \$\endgroup\$
    – Kevin McQuown
    Commented Nov 21, 2023 at 14:32
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    \$\begingroup\$ You never said you have other stuff on OE as well. The 74HC245 has CMOS inputs. The ULN2803 doesn't, it will draw current. \$\endgroup\$
    – Justme
    Commented Nov 21, 2023 at 18:01

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