--- Sure it is.
If your hookup looks like this:
.Vcc>-----------------+-------+--------+----+ . |R3 8| |R4 | . [910k] +---+---+ [1M] | .__ Rt| 2|_ Vcc _|4 | |C3 .IN>------------------|--O|T R|O---+ [100nF] . | 6| | |C4 | . +---|TH 555| [10nF] | . | 7|_ |3 | | . +--O|D OUT|O---|----|--->OUT . +| | GND | | | . [1µF] +---+---+ | | . Ct|C2 1| U1 | | .GND>-----------------+-------+--------+----+ __ and IN lasts longer than OUT, then the timing will look like this:
t0 t10s __ ____| |_______ IN |__________________________________|
__________________________________ OUT____| |_______
While if the lashup looks like this:
.Vcc>-+---------+-----+-------+--------+----+ . |R1 |R2 |R3 8| |R4 | . [10k] [10k][910k] +---+---+ [1M] | .__ | C1 | Rt| 2|_ Vcc _|4 | |C3 .IN>--+-[100nF]-+-----|--O|T R|O---+ [100nF] . | 6| | |C4 | . +---|TH 555| [10nF] | . | 7|_ |3 | | . +--O|D OUT|O---|----|--->OUT . +| | GND | | | . [1µF] +---+---+ | | . Ct|C2 1| U1 | | .GND>-----------------+-------+--------+----+
then the timing will look like this:
t0s t10s __ ____| |_______ IN |__________________________________|
___ OUT____| |______________________________________ | t1s
---
--- Sounds like something _you'd_ do instead of just differentiating the low going edge of the input, but, if you want to play the semantic edge-triggered VS level-triggered game, an edge is an edge regardless of its slope, and the fact still remains that when the 555 is used as a one-shot, the 555's trigger input must go below Vcc/3 in order to start the output pulse, and must rise to > Vcc/3 before the output pulse terminates, in order for the output time to be as calculated.
--- JF