Is the rise time of the output of a logic IC or optocoupler independent of the rise time of the input?

Question

Is the rise time of the output of a logic IC (e.g. flip-flop or inverter) or optocoupler independent or dependent on the rise time of the input? If the answer is dependent, are there any other devices where the rise time of the output is independent of the rise time of the input?

For example, I have a 555 timer output running in astable mode with a 50% duty cycle producing a 1.5kHz square wave where my scope says the rise time is approximately 400ns (and this changes if I change the frequency of the 555 output). I'd like to hook something up to the output of the 555, like a flip-flop or inverter or optocoupler (or anything that does the trick) to get a square wave with a rise time that's: 1) less than 400ns and 2) preferably, independent of the 555 frequency (i.e., same whether I configure the 555 to output at 1.5kHz or 10kHz).

Answer 1

For really slow input transitions there may be some jitter on the output. Try looking at schmitt triggers- They have built-in hysteresis that takes a slow input and creates a fast jitterless output. A more technical discussion is here. Adding one would also make the rise time frequency independent. An example part might be the CD40106B CMOS Hex Schmitt-Trigger Inverter. It has a similar voltage range to the 555, but double check to make sure it works for your application.

Answer 2

It's usually pretty much independent of the input rise time. With the old 4000-series CMOS they were originally unbuffered and didn't have much voltage gain even at the transition (on the plus side you could bias them into stable amplifiers, which you can't do with more modern chips). Later parts have the 'B' suffix, meaning buffered (the linked TI application note SCHA004 goes into more detail).

Here is an unbuffered inverter 74HCU04:

An input change from about 2V to 2.5V will change the output from about 4V to about 1V (gain of 6) so a 400ns rise time could be reduced to about 70ns. You would not generally use this part as an ordinary inverter- it's more aimed at crystal oscillators and that sort of linear application.

An ordinary 74HC04 has another two inverters in series inside, so the time (with a 400ns input transition) will not be determined by the input rise and fall time, but rather by the characteristics of the transistors and loading of the output.

For very slow input rise and fall times, noise immunity becomes a concern and you might prefer to use a 74HC14 which has a Schmitt trigger action. By adding some hysteresis the output rise and fall times are completely independent of even the slowest input rise and fall times (however the actual switching points are quite loosely specified).

Answer 3

Think of most systems with an input that drives an output as having both a gain, and a maximum output edge rate. Both will be affected by the supply voltage, and the output loading, as well as the circuit.

The actual output edge rate will be limited by whichever is slower of the maximum rate and the gain times the input rate. Gain is usually very large, and if the system if being driven by a similar gate, all the edges will be at the maximum slew rate.

The exception to the above is something like a schmidt trigger, where the output rate is independent of the input rate.