Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm using 555 timer in astable mode. I've used Multisim to generate it ant this is what I've got:

555 astable

I'm little confised by the 100 ohms Rl resistor. Do I need it there ? If I do, why ? I've tried with and without it, and it works for me both ways ... I'm using the 555 to drive another CMOS IC input.


share|improve this question
up vote 3 down vote accepted

The 100 ohm is A load - an example - what load you really use is up to you.

If you are driving a CMOS IC you do not need anything else connected to the OUT pin.

The 555 has a "push pull" or bipolar output stage. This means that it drives "actively" both high and low, so you get an output voltage whether there is a load connected or not.

The 555 load is generally connected to the OUT pin when you want a square wave output. There are other places you can connect to if you are doing less usual things.

LM555 / NE555 / xx555 datasheet - Fairchild
LM555 / NE555 / xx555 datasheet - NatSemi

Active low and high drive can be seen here:

enter image description here

More normal block diagram view. Here you cannot tell that the output pin is actively driven high and low.

enter image description here

share|improve this answer

You don't need this 100 ohm resistor because the IC drives the pin high or low as required even if nothing is connected to it.

share|improve this answer
Why not? Your answer should thoroughly explain why it is correct: Draw a schematic, quote the datasheet, explain when it would be needed, and why it's not. This is the schematic produced by Multisim, hundreds or thousands of people might have this same problem, and Google ought to take them here for support. Make your answer count! – Kevin Vermeer Jan 24 '12 at 0:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.