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I am using this Schmidt trigger IC SN74HCS125 which has 4 channels

I would like to save some board space by hopefully using only one resistor to pull down the enable pins of each channel. I do not plan to connect those enable pins to anything else as I want them to be always enabled.

schematic

simulate this circuit – Schematic created using CircuitLab

I have not done this before since what I have been doing since now was individually giving them their own resistor. Just intuitively speaking it should work, but there might be caveats that I am not aware of. Is doing this perfectly fine?

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    \$\begingroup\$ You do not need a resistor as they are meant to be directly grounded (ON), or tied to Vcc (OFF). If you insist on R1 then keep it under 1 Kohms. \$\endgroup\$
    – user105652
    Jul 25, 2020 at 0:01
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    \$\begingroup\$ @VTNCaGNtdDVNalUy Why should the resistor be less than 1k? What bad thing would happen if the resistor is 2.2k? \$\endgroup\$ Jul 25, 2020 at 0:27
  • \$\begingroup\$ Page 8 Section 8.3.1 last paragraph last sentence did mention of tying it to a pull up/down resistor. Im also curious why about under 1K ohms ? Any reason for that? Is it a general rule or a value specific to this IC ? \$\endgroup\$
    – Jake quin
    Jul 25, 2020 at 1:29
  • \$\begingroup\$ That paragraph is referring to using a pull-up or -down on the output, so that it goes to a defined state when disabled instead of floating. \$\endgroup\$
    – brhans
    Jul 25, 2020 at 1:40
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    \$\begingroup\$ Resistors are not needed to set a static state for CMOS inputs. Bipolar (TTL) inputs need resistors to avoid drawing too much current. Since CMOS inputs are basically open circuit at DC, a resistor is unnecessary. You could use a resistor just so that you can easily leave it out if you are patching in new logic, instead of cutting traces or lifting pins. \$\endgroup\$
    – crj11
    Jul 25, 2020 at 2:48

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The unused inputs can be directly connected to ground (or supply voltage). It even reads in the datasheet.

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