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From what I remember, from decades ago, 1980's 5v TTL chips would see a floating input as a zero (because they switched on current, rather than voltage). Update: it seems they were interpreted as ones.

Are there currently any families of discrete logic which interpret a floating inputs as a zero or a one, rather than a non-deterministic value? (AC, ACT, HC, HCT, ALS, AS, AC, CB, CBT, F, S, LV, LC, etc.)

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    \$\begingroup\$ 1980's TTL chips generally saw "floating" inputs as 1 because the outputs were designed to sink much more current to 0V than they supplied at 5V. \$\endgroup\$
    – Finbarr
    Commented May 31, 2017 at 11:14
  • \$\begingroup\$ @Finbarr Thanks, I must have misremembered. The important point is that it was deterministic. \$\endgroup\$
    – fadedbee
    Commented May 31, 2017 at 11:17
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    \$\begingroup\$ The input stage meant that it was pulled up - much more so with LS than HC(T), which has a sensing input stage and doesn't require the driving stage to sink such a large current from it. With HC(T), a high was very likely but in no way would I call it deterministic. A floating input can pick up radiated signals from nearby easily enough. Take no chances, if you're thinking of letting them float, use pull-up/down resistors. \$\endgroup\$
    – TonyM
    Commented May 31, 2017 at 11:29

2 Answers 2

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Unused TTL inputs are high = 1; Fairchild's application note 363: Designing with TLL says:

Unused inputs on TTL devices float at threshold, anywhere from 1.1V to 1.5V, depending upon the device and its family. While this usually simulates a “high”, many application problems can be traced to open inputs. Inputs floating at threshold are very susceptible to induced noise (transmitted from other lines) and can easily switch the state of the device. A good design rule is to tie unused inputs to a solid logic level. Inputs are usually tied to VCC through a 1 kΩ to 5 kΩ resistor, since tying them to ground means supplying the IIL current instead of the IIH current. IIL is several orders of magnitude greater than IIH. The resistor is recommended to protect the input against VCC voltage surges and to protect the system against the possibility of the input shorting directly to ground. A single 1k resistor can handle up to 10 inputs.

While not all TTL-like families based on BJTs ((A)LS, (A)S, F, etc.) use the emitter of a BJT as input, they are designed to behave the same way.

CMOS devices use MOSFET gates as inputs. This implies that floating inputs can lead to an half-on state where the input state is indeterminate, or oscillates (see, for example, TI's application note Implications of Slow or Floating CMOS Inputs).

The only practical way of giving a CMOS input a 'default' value is to use a pull-up or -down resistor, but it's usually not done for discrete logic because it would reduce the input impedance by a great amount.

(Some CMOS devices have bus-hold circuits to prevent their inputs from floating, but the power-up state still is undefined.)

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  • \$\begingroup\$ The other, easier, way to force a CMOS input high or low is to tie it to Vcc or ground. Pullups on TTL were needed to control input current levels, but that is not true on CMOS. \$\endgroup\$ Commented May 31, 2017 at 13:05
  • \$\begingroup\$ @WhatRoughBeast I was speaking of a default level; a direct connection would not allow to override the level. \$\endgroup\$
    – CL.
    Commented May 31, 2017 at 13:14
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This is not a direct answer, but nonetheless relevant information.

While it is true that old TTL would interpret floating inputs as high, this was never something to be relied on in a real design. The inputs floated only a little above the threshold, so the noise margin was low. Because they weren't actively driven, the impedance was high. Between these to affects, floating inputs could, and sometimes did, glitch low.

The usual solution was to pull them to +5 thru 1 kΩ. Sometimes you'd have a unused inverter. You could ground the input of the inverter, then use its output to hold several unused inputs high.

One way or another, you had to make sure that floating inputs that needed to be interpreted as high were actually held high. Just leaving them unconnected was irresponsible design, and really did lead to flaky field behavior.

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  • \$\begingroup\$ Re: "...unused inverter...": Is that the "one generator"? Why not connect the input directly to the (local) Vcc (that should be sufficiently decoupled)? Is there be some extra capacitance inside the chip that would give a PSRR significantly higher than 0 dB? \$\endgroup\$ Commented May 31, 2017 at 21:21
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    \$\begingroup\$ Connecting directly to +5V had the unfortunate effect of killing some TTL chips, because the power supply could have transients that exceeded the voltage tolerance of the inputs. Connection through a current-limiting resistor, often 1kOhm, was recommended. \$\endgroup\$
    – Whit3rd
    Commented Jun 1, 2017 at 9:22

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