Electronic – How and why is “floating input” a valid state for RF/IR encoder IC’s

floating-pinpulldownpulluptheory

RF/IF encoder/decoder IC's such as this one, will accept the address-pin in one of 3 states:

  • Floating
  • High
  • Low

As per this excellent answer, I believe "Floating" input can assume either High or Low value due to variations in the EM field, if I understood it correctly. If so, then why if floating a valid state ? At the time the IC is reading the value of the pin, how does it determine that the pin is really pulled HIGH or LOW, versus a floating-pin that was inadvertently pulled up/down such as due to ambient EM noise (I'm assuming that such a thing is possible).

Also in line with the question in the context of which the above mentioned answer was given, can someone explain, through non-technical analogy, the difference between weakly pulled up/down versus strongly pulled up/down ?

Best Answer

Weak pullups are overridden by strong pullups. For a non-electronic analogy, imagine the weak pullup as a weak spring and a strong pullup as a strong spring, connected to some lever. You can drive the lever away from the pullup, but doing so requires effort. Letting go of it will return the lever to its resting position. The greater the strength of the spring/pullup, the faster the lever/signal recovers.

Edit: wrote that analogy without reading through to the linked page, which has essentially the same analogy.

I think this particular IC is actually using "floating" to implement a ternary coding system for addresses. So a binary system has notionally one threshold voltage:

  • above threshold: HIGH / 1
  • below threshold: LOW / 0

The ternary has two:

  • above high threshold: HIGH / 2
  • below high threshold, above low threshold: MIDDLE / 1
  • below low threshold: LOW / 0

I suspect that internally it has two large bias resistors connected to each pin, and that if you probe it you'll find them at half the supply voltage (you may need to make sure it's not in standby mode for this).

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