What is a good value for a pull-down resistor, which when switched, connects to 24v? Is there a formula to derive this?

This is effectively how it's drawn in the schematic (the inverter input's impedance is 650kohm):

enter image description here

  • 1
    \$\begingroup\$ That depends entirely on what you are pulling against - but you haven't told us that. Q1. What are you driving? Q2. What is driving it? \$\endgroup\$
    – Transistor
    Commented May 30, 2019 at 17:36
  • \$\begingroup\$ The 24V is switched by an opto-coupler, and fed into a high-impedance pin (650khm) \$\endgroup\$
    – 19172281
    Commented May 30, 2019 at 17:37
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    \$\begingroup\$ "Is that a formula to derive this?" Ohms law is probably applicable but we still need to know more. Show us a diagram. \$\endgroup\$
    – Oldfart
    Commented May 30, 2019 at 17:46
  • \$\begingroup\$ Why 24V swing? This compromises the Opto which has very limited current GBW product. When you amplify from 1.2V in to 24V out you only get 5% of your CTR which may be < 0.5 already \$\endgroup\$ Commented May 30, 2019 at 18:10
  • \$\begingroup\$ @Oldfart, please see updated question. \$\endgroup\$
    – 19172281
    Commented May 30, 2019 at 18:12

1 Answer 1



10K is often a good choice if you don't know what resistor to try.

Simulate or think about the circuit with a 10K resistor. are the currents that will flow apropriate to the devices and the power-supply? will the voltages be good? if not adjust it.

resistor choice for pull-ups and pull-downs is often not critical.

  • \$\begingroup\$ Yes, I think I’ll try it out with 10k. With 24v there’s only 2.4ma flowing which is fine. \$\endgroup\$
    – 19172281
    Commented May 31, 2019 at 12:19

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