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In a design schematic I have seen series resistors being used in WIRED OR logic.

All 4 input signals are open-drain signals and hence being represented as switches. Can anyone explain why we need R1-R4? Is it for any fault tolerance?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What does 'output' connect to? \$\endgroup\$
    – Tom Carpenter
    Commented Jun 21, 2020 at 18:07
  • \$\begingroup\$ You'd have to ask the person who designed it. Calling on others to speculate, especially in the absence of any information about the sort of situation where it was proposed is just too open-ended to meet that Stack Exchange requirement that questions be specifically answerable. And that's not open drain - maybe it's a conceptual simplification of such, but those are physical switches not semiconductor ones and changing details further eliminates any hints about why something might be done. \$\endgroup\$
    – Chris Stratton
    Commented Jun 21, 2020 at 19:37
  • \$\begingroup\$ depending on loading if it's capacitive, and you expect these switches to last for a million cycles, the resistors will help get you there. \$\endgroup\$
    – MadHatter
    Commented Jun 22, 2020 at 2:02
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    \$\begingroup\$ I wonder why one might need four resistors wired this way instead of one serial with the output. \$\endgroup\$
    – Eugene Ryabtsev
    Commented Jun 22, 2020 at 5:03
  • \$\begingroup\$ Maybe you could check the voltage and see how many buttons were on at the same time \$\endgroup\$
    – Voltage Spike ♦
    Commented Jun 24, 2020 at 20:14

1 Answer 1

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Yes, it is a wired OR logic "gate". The 10 k resistors R1-R4 protect the output from being short connected to ground when a button is pressed. They form voltage dividers with R5 but their resistance is negligible relative to R5 (only 1/10).

At first glance, this seems pointless, but probably the output is connected to a microcontroller port pin that can be programmed as both input and output. If it is accidentally programmed as an output, when a button is pressed, it will be short-circuited to the ground.

In the explanation above, I assumed these were buttons. If they are open drains, the reason is similar but now they can be damaged.

There is always a risk of an open drain (collector) being damaged. Many years ago, a friend of mine (designer) told me, "If you leave a collector open, there will always be a fool to connect it to the supply voltage."

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    \$\begingroup\$ There could be one resistor in series with the microcontroller port. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Jun 22, 2020 at 8:50
  • \$\begingroup\$ I agree... 1 k will do the job, I think... but only if there is not a fool to connect a drain to Vdd:) \$\endgroup\$
    – Circuit fantasist
    Commented Jun 22, 2020 at 8:56
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    \$\begingroup\$ then the 1k will protect them \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Jun 22, 2020 at 9:06
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    \$\begingroup\$ You often see a variant on this in automotive where you use an analogue input pin and add an extra 100k to ground.... This then lets you detect open and short circuits in the wiring as well as switch pressed or not (And if you play with the resistor values and switch wiring, which switch was pressed). \$\endgroup\$
    – Dan Mills
    Commented Jun 22, 2020 at 10:30
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    \$\begingroup\$ It is also used in home alarm systems to detect these states... \$\endgroup\$
    – Circuit fantasist
    Commented Jun 22, 2020 at 12:42

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