I'm creating this circuit (original circuit source here) and it requires a 7555 timer. A 7555 timer is like the next generation 555 since it is CMOS. Now there are no 7555s or CMOS 555 in my area and was wondering what will happen if I replace it with the NE555? Are there any consequences? Will it behave in the same manner? What would be the best thing to do?


Seems like in order to use NE555, adjustments must be made. What are they?

  • \$\begingroup\$ The 555 suffers from terrible disruptions to the supply voltage for the entire circuit when it switches. I will never use a TTL 555 again. CMOS is the way to go. I ordered some ICM7555 and ICM7556 (dual version) from Mouser but they are backordered until August! Hard to understand why; there ought to be lots of demand for these devices! \$\endgroup\$ Commented Apr 30, 2021 at 20:53

1 Answer 1


The NE555, being TTL has a minimum input voltage of 4.5V. It won't work on the 1.5V that your referenced circuit uses. That said, I see no reason why you couldn't use an NE555 at 5V, provided that you adjust the values of R2, R3, and add 5 diodes (3.5V worth) in series with the LED.

  • \$\begingroup\$ Ok so in order to use ne555, resistrs need to be adjusted, hw about the cpacitnce? And other stuff \$\endgroup\$ Commented Sep 8, 2012 at 2:51
  • \$\begingroup\$ Actually, R3 does not need to be adjusted. R2 needs to be adjusted to bias the transistor for maximum sensitivity. The collector voltage needs to sit just above Vcc/3 when there's no signal. And C3 and D2 at the output can be eliminated. Just connect the LED to pin 3 through a suitable resistor. \$\endgroup\$
    – Dave Tweed
    Commented Sep 8, 2012 at 14:29
  • \$\begingroup\$ The NE555 is not a TTL device. In fact, in terms of traditional bipolar digital logic devices, it isn't a logic device. The Trigger and Threshold inputs are analog comparators, the output stage has a huge cross-conduction spike typical of an over-biased class AB circuit, and three of the four output transistors never saturate. The flipflop in the middle of it all is not a saturating-switch latch circuit; it is a linear amplifier with positive feedback (R7). \$\endgroup\$
    – AnalogKid
    Commented Apr 30, 2021 at 22:42

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