4 ms·

Shameless plug: the article pins the minimum latency of your keyboard to 3ms. I recently published the kinX, a keyboard controller for the Kinesis Advantage wit

by secure 8y ago

Shameless plug: the article pins the minimum latency of your keyboard to 3ms. I recently published the kinX, a keyboard controller for the Kinesis Advantage with merely 0.2ms of input latency: https://michael.stapelberg.de/posts/2018-04-17-kinx/ https://michael.stapelberg.de/posts/2018-04-17-kinx/

Sidnicious 8y ago
That article led me to the page on Cherry’s site about their “analog keyboard controller”, which promises no scanning latency. “Analog” suggests to me that each key has its own power-of-two-ish resistor value, and the controller can just look at the total resistance to see which keys are pressed. Does this seem plausible? If so, it could be a fun way to build a DIY keyboard.
- tux1968 8y ago
  Went looking for this after you mentioned it. As far as I can tell, Cherry is not licensing that tech or selling it yet except in the form of a complete keyboard. https://the-gadgeteer.com/2017/08/11/cherry-mx-board-6-0-mechanical-usb-keyboard-review/ https://the-gadgeteer.com/2017/08/11/cherry-mx-board-6-0-mec...
- secure 8y ago
  That does seem plausible to me at first glance. I had read about the analog keyboard controller before, but nobody seemed to _really_ know what they’re doing. It would be interesting to confirm this :)
- duskwuff 8y ago
  > Does this seem plausible? No. 1. The precision of your resistors is a limiting factor. If you use standard 1% resistors, you can't even put 8 switches on a single ADC; the precision of the largest resistor ends up being greater than the value of the smallest one. Increasing the precision of the resistors helps to a degree, but it drives up the price dramatically. 2. Contact resistance and switch bounce becomes a problem -- instead of just making a key show up twice, it could potentially result in a key you didn't press showing up. 3. Using ADCs doesn't get you away from scanning. ADCs have a sampling rate too -- and unless you pay a lot for your ADCs, it'll be slower than you could scan a diode matrix.
  - adrianratnapala 8y ago
    While it sounds difficult to do, I wouldn't underestimate the manufacturers. > 1. The precision of your resistors is a limiting factor. Suppose your resistors and measuring equipment were good enough that you needed 5% between neighbouring values. Then you can squeeze in 141 distinct values between 1 Megohm and 1 kilohm. Seems plenty to me. > 2. Contact resistance and switch bounce becomes a problem -- instead of just making a key show up twice, it could potentially result in a key you didn't press showing up. They definitely become issues, but they strike me as manageable. For good switches, contact resistance should add perhaps 1 ohm of variation , and above we saw that 50Ohm between the nearest switches is good enough. Switch bounce means that you can't afford to accidentally average over a bounce, you need to sample fast (say 200 kHz), and then take say the Nth highest of 20 samples. So you get a result 100 microseconds after the first switch went high. (Obviously that algorithm is just a guess, just to show the kind of options engineers have when they come up against the real system). > 3. Using ADCs doesn't get you away from scanning. ADCs have a sampling rate too -- and unless you pay a lot for your ADCs, ... Yes you need a good ADC, but that ADC will still be an single IC that does nothing vastly fancy. Such a thing is difficult and expensive to buy retail, but if I was Cherry I might well be able to negotiate a deal that brought it to a fraction of the end price of the keyboard (which I would sell at a premium).
    - Stratoscope 8y ago
      > Suppose your resistors and measuring equipment were good enough that you needed 5% between neighbouring values. Then you can squeeze in 141 distinct values between 1 Megohm and 1 kilohm. Seems plenty to me. How would you detect multiple simultaneous keypresses, either for n-key rollover or for shift/ctrl/alt types of keys?
- kevin_thibedeau 8y ago
  More likely they are just using a package with enough pins to discretely input every key without a scan matrix. There are no real cost constraints on doing this nowadays with modern IC packaging. The added hardware inside to track all those inputs is inconsequential with modern process technologies. Then their marketing dept. got a hold of it and came up with some BS to sell it.
  - adrianratnapala 8y ago
    Yes sounds a lot easier than the resistor thing (though even that sounds doable).
- panic 8y ago
  How would a keyboard like this detect simultaneous key presses? It seems like you'd either see only the lower-resistance key (if the resistances are orders of magnitude apart) or end up with a completely nonsense value.
  - Sidnicious 8y ago
    Let's say that Q, W, and E are 1Ω, 2Ω, and 4Ω, respectively. I'll pretend they're wired in series to make the numbers work out nicely, but it should also work in parallel. If you measure 3Ω, then Q and W are pressed. 6Ω: W and E. 7Ω: all three.
ifoundthetao 8y ago
That's the keyboard I use.. it's the most perfect keyboard ever.

by secure 8y ago

Sidnicious 8y ago

That article led me to the page on Cherry’s site about their “analog keyboard controller”, which promises no scanning latency. “Analog” suggests to me that each key has its own power-of-two-ish resistor value, and the controller can just look at the total resistance to see which keys are pressed. Does this seem plausible? If so, it could be a fun way to build a DIY keyboard.

tux1968 8y ago
Went looking for this after you mentioned it. As far as I can tell, Cherry is not licensing that tech or selling it yet except in the form of a complete keyboard. https://the-gadgeteer.com/2017/08/11/cherry-mx-board-6-0-mechanical-usb-keyboard-review/ https://the-gadgeteer.com/2017/08/11/cherry-mx-board-6-0-mec...
secure 8y ago
That does seem plausible to me at first glance. I had read about the analog keyboard controller before, but nobody seemed to _really_ know what they’re doing. It would be interesting to confirm this :)
duskwuff 8y ago
> Does this seem plausible? No. 1. The precision of your resistors is a limiting factor. If you use standard 1% resistors, you can't even put 8 switches on a single ADC; the precision of the largest resistor ends up being greater than the value of the smallest one. Increasing the precision of the resistors helps to a degree, but it drives up the price dramatically. 2. Contact resistance and switch bounce becomes a problem -- instead of just making a key show up twice, it could potentially result in a key you didn't press showing up. 3. Using ADCs doesn't get you away from scanning. ADCs have a sampling rate too -- and unless you pay a lot for your ADCs, it'll be slower than you could scan a diode matrix.
- adrianratnapala 8y ago
  While it sounds difficult to do, I wouldn't underestimate the manufacturers. > 1. The precision of your resistors is a limiting factor. Suppose your resistors and measuring equipment were good enough that you needed 5% between neighbouring values. Then you can squeeze in 141 distinct values between 1 Megohm and 1 kilohm. Seems plenty to me. > 2. Contact resistance and switch bounce becomes a problem -- instead of just making a key show up twice, it could potentially result in a key you didn't press showing up. They definitely become issues, but they strike me as manageable. For good switches, contact resistance should add perhaps 1 ohm of variation , and above we saw that 50Ohm between the nearest switches is good enough. Switch bounce means that you can't afford to accidentally average over a bounce, you need to sample fast (say 200 kHz), and then take say the Nth highest of 20 samples. So you get a result 100 microseconds after the first switch went high. (Obviously that algorithm is just a guess, just to show the kind of options engineers have when they come up against the real system). > 3. Using ADCs doesn't get you away from scanning. ADCs have a sampling rate too -- and unless you pay a lot for your ADCs, ... Yes you need a good ADC, but that ADC will still be an single IC that does nothing vastly fancy. Such a thing is difficult and expensive to buy retail, but if I was Cherry I might well be able to negotiate a deal that brought it to a fraction of the end price of the keyboard (which I would sell at a premium).
  - Stratoscope 8y ago
    > Suppose your resistors and measuring equipment were good enough that you needed 5% between neighbouring values. Then you can squeeze in 141 distinct values between 1 Megohm and 1 kilohm. Seems plenty to me. How would you detect multiple simultaneous keypresses, either for n-key rollover or for shift/ctrl/alt types of keys?
kevin_thibedeau 8y ago
More likely they are just using a package with enough pins to discretely input every key without a scan matrix. There are no real cost constraints on doing this nowadays with modern IC packaging. The added hardware inside to track all those inputs is inconsequential with modern process technologies. Then their marketing dept. got a hold of it and came up with some BS to sell it.
- adrianratnapala 8y ago
  Yes sounds a lot easier than the resistor thing (though even that sounds doable).
panic 8y ago
How would a keyboard like this detect simultaneous key presses? It seems like you'd either see only the lower-resistance key (if the resistances are orders of magnitude apart) or end up with a completely nonsense value.
- Sidnicious 8y ago
  Let's say that Q, W, and E are 1Ω, 2Ω, and 4Ω, respectively. I'll pretend they're wired in series to make the numbers work out nicely, but it should also work in parallel. If you measure 3Ω, then Q and W are pressed. 6Ω: W and E. 7Ω: all three.

ifoundthetao 8y ago

That's the keyboard I use.. it's the most perfect keyboard ever.