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.



3 Answers 3


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


I know this post is quite old but the question remains. From TI datasheet SLFS022I :

Rload is used to improved output high voltage level

From TI reference design, they connect a 1kohm between Out and Vcc. My guess is that it fills the same purpose as your problem and it is used to pull some current in order to enhance the output voltage level (as the output High level can be quite low).

  • 1
    \$\begingroup\$ Pulling the output low with 100 ohms like in the question will definitely not improve the output high level, it would just make it worse. \$\endgroup\$
    – Justme
    Jun 14, 2022 at 9:19

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.

  • 5
    \$\begingroup\$ 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! \$\endgroup\$ Jan 24, 2012 at 0:38

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