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):
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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\$– TransistorMay 30, 2019 at 17:36
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\$\begingroup\$ The 24V is switched by an opto-coupler, and fed into a high-impedance pin (650khm) \$\endgroup\$– 19172281May 30, 2019 at 17:37
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1\$\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\$– OldfartMay 30, 2019 at 17:46
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\$\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\$– Tony Stewart EE75May 30, 2019 at 18:10
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\$\begingroup\$ @Oldfart, please see updated question. \$\endgroup\$– 19172281May 30, 2019 at 18:12
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1 Answer
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10K
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.
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\$\begingroup\$ Yes, I think I’ll try it out with 10k. With 24v there’s only 2.4ma flowing which is fine. \$\endgroup\$– 19172281May 31, 2019 at 12:19