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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).

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  • \$\begingroup\$ 400ns seems long. TI specs 100ns rise and fall for their LM555 part. What's your supply voltage? What are you driving? Perhaps adding a reasonable load would help speed up the edges... \$\endgroup\$ – BobT Dec 31 '17 at 21:06
  • \$\begingroup\$ I'm using a LMC555 with the 50% duty cycle circuit listed on page 16 with 0.1u for C, 5K for R to get the approx 1.5khz freq (I guess it's supposed to be 1.43khz but my scope is showing 1.5khz). I'm unclear on the proper value for the pull-up on the discharge line as the data sheet says nothing about that. I've tried both a 100 ohm resistor and 1.5k ohm resistor and both seem to work except a lot more current is consumed with the 100ohm pullup on discharge line. 10X scope probe (so 10M ohm load) is placed on the discharge line (alt out in DS). \$\endgroup\$ – acker9 Dec 31 '17 at 21:37
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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.

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  • \$\begingroup\$ I got 3 answers in the first couple of hours that mention a Schmitt Trigger inverter as what I'm looking for, so all three of those could be marked as the answer. But I'm marking this one as the answer and I'm +1'ing the other two. I haven't actually verified a Schmitt Trigger works to lower the rise time because I don't yet have one on hand to test with, but since three different people say it will work, I will assume that it will do so. Thanks for the responses. \$\endgroup\$ – acker9 Jan 1 '18 at 18:31
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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:

enter image description here

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).

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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.

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