As a part of my school project I have a timer astable with NE555 and I need to generate a square signal. It has 12V supply and 65% duty cycle normally.

In my simulation on LTSpice I get exactly what I want but when I do the setup on my breadboard, although I have the same duty cycle I don't have 12V-0 as the peaks for my output, it is 10.4V.

It's not a very important problem but since it's a school project I have to explain the cause of a such difference in my report.

I can't read the manufacturer's name on the component and I have tried with different NE555 we have in the lab but there is always an offset of about 1,5V. Probably it's because something inside is not ideal but I just need to know what.

I put the setup just in case.

My Setup

Thank you!


2 Answers 2


If you look at a bipolar 555's data sheet you'll see that there's a Darlington-connected NPN emitter follower on the top of the totem pole output, so it can't be saturated.

What that means is that even if its base voltage rises to the positive rail its emitter will always be about a couple of base-to-emitter drops less than the base, so the 555's output must always be less than Vcc. Sometimes substantially less, depending on the output load.


If you have a load attached to the NE555 (perhaps something like 1K from the output to ground) that is about right.

If you have no load, the output should be closer to 11.5V.

If the 555 is the CMOS type, the output will swing to almost 12V with no load.

The bipolar type output section looks like this:

enter image description here

For a very light load and output 'high', it looks like 6.2K to Vcc in series with the base-emitter junction of Q27, in series with 3.9K. In other words, there is a diode drop in there.

If you remove whatever load is on there (!), and add a pullup resistor such as 10K to Vcc to the output you should get very close to a 0-12V swing.


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