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Is there a nominal dead time for rotary switches? Obviously all switches are different, so this value will not be precise. For making a reasonable situation, let us assume that a detent style switch is independent of the speed and/or vigor of the user.

As I understand it, a non-shorting rotary switch works by completely disconnecting from a pole before connecting to another. Therefore, there must be a non-zero delay time when the throw is not connected to any pole. Has anyone measured this? Is there a good way to estimate it? I had a Clever Solution(tm) to a problem which uses a NOR gate between switch poles to momentarily toggle something else.

Edit for clarity: Assuming the delay time based on how fast someone operates the switch, which we will term vigor (v), goes really high. What is our time delay?

$$ \lim_{v \to \infty} t_{delay} $$

What I really need to know, is how this value compares to the logic delays in my circuit, but I am keeping the question generic to match with the purpose of Stack Exchange.

$$ \mathrm{Test~if:~~} t_{switch~delay} \gg t_{propagation~delay} $$

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  • \$\begingroup\$ what kind of a signal are you switching? ..... what are you using the switch for? \$\endgroup\$ – jsotola Feb 8 at 1:48
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    \$\begingroup\$ It's called Break begore Make in a switch and is not specified as a time value, it is a degree of rotation value between detents. \$\endgroup\$ – Jack Creasey Feb 8 at 1:57
  • \$\begingroup\$ The delay also depends of the operator speed, I don't think it's very repeatable. If you need something dependable, maybe use monostable multivibrator or a microcontroler with a timer? \$\endgroup\$ – pserra Feb 8 at 4:53
  • \$\begingroup\$ @jsotola A selector switch that requires a specific disconnect signal before forming a new connection (for switching USB hub) \$\endgroup\$ – WesH Feb 8 at 22:03
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    \$\begingroup\$ Propogation time for most logic gates is measured in nanoseconds. I'd expect the open circuit time for a manually-operated switch to be at least several milliseconds, (although I've never measured it), so the switch "dead time" will be much, much longer than any likely propogation delay. If you are really concerned, you will have to measure it yourself, as it is not something that the switch manufacturer can measure in any meaningful way. \$\endgroup\$ – Peter Bennett Feb 8 at 23:28
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On the rotary switches that I've used, the dead time between positions depends entirely on the speed with which the operator turns the switch. It may be possible for a careful operator to "park" the switch between positions, so neither contact is made, and the dead time becomes infinite.

(added from a comment)

Propogation time for most logic gates is measured in nanoseconds. I'd expect the open circuit time for a manually-operated switch to be at least several milliseconds, (although I've never measured it), so the switch "dead time" will be much, much longer than any likely propogation delay. If you are really concerned, you will have to measure it yourself, as it is not something that the switch manufacturer can measure in any meaningful way

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  • \$\begingroup\$ This does not fully answer the question. However, my original one may have not been specific enough. If the switch was toggled as fast as possible, there must still be a dead zone--no matter how fast it is toggled. If the dead time is dependent on the operator speed, then so be it. The delay will reach a minimum. What is the minimum time spent not in an ON state? \$\endgroup\$ – WesH Feb 8 at 22:12
  • \$\begingroup\$ @WesH - Well, if you have a 6-position switch, and turn it at 1 million rotations per second, the dead time will be about 167 nsec. Does that help? If not, start thinking more about what you're asking. Stop flailing about and take a deep breath. \$\endgroup\$ – WhatRoughBeast Feb 9 at 3:08
  • \$\begingroup\$ @WhatRoughBeast Your math is problematic. The switch rotating as you describe hits a new area every ~167ns, but that assumes an ideal switch where the connection pad has an area of 0. But I understand what you are trying to say. The time is still long compared to a single logic gate propagation time. \$\endgroup\$ – WesH Feb 11 at 22:34
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The answer to your question is dependent on the mechanical dynamics of the switch design and how the switch is operated by the user. From an electrical perspective, all that matters is that there will be some amount of time between switch states and whatever circuit this switch is connected to needs to gracefully deal with that in-between state. There are numerous methods available for switch "debouncing" to deal with undesired effects of switch mechanics. The final design should be carefully tested with these considerations in mind.

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  • \$\begingroup\$ The notion that electronics are exempt from time is incorrect. Sure, in some cases the small time window can be ignored, but not this one. A real logic gate will take time to reach a threshold level. Although I only mention a single NOR gate in the OP, it would be unsafe to assume that it will be my only logic in the circuit. The assumption I need to confirm is: $$ t_{switch delay} \gg t_{propagation delay} $$ \$\endgroup\$ – WesH Feb 8 at 22:23
  • \$\begingroup\$ Very roughly, "settling" time in typical logic circuits is on the order of a few nanoseconds from reading through various data sheets. For mechanical switches, not so sure, but my guess would be settling times on the order of milliseconds due to movement of mechanical parts involving inertia, acceleration, etc. So, I would estimate your assumption is good in most every practical situation. I was not trying to imply that "electronics are exempt from time" in this answer, but perhaps just making the assumption that electronic timings are much faster than mechanical systems, such as a switch. \$\endgroup\$ – Tim D Feb 9 at 0:59

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