I designed the following circuit, using a 555 timer, to supply a single trigger pulse with the same width to digital circuits, no matter how long the button is pressed. The actual pulse output works perfectly. I was just wondering if there was a special way to connect a 555 timer output to CMOS and/or TTL inputs.

Here's the schematic:

Original Schematic

[Edit by OP]

In my original schematic I used a 7555 timer, but when I built the circuit I used a standard 555. Also, when I built it, I did use a base resistor!

Here is what I really built:

Fixed Shematic

[Another Edit By OP]

So, I went back to my bench and tried out the circuit again and and realized that the switch was bouncing! I then added a switch debouncer(See below schematic). I was wondering if there is a simpler way to get rid of the bouncing.

Here is the schematic with the switch debouncer:

enter image description here


2 Answers 2


If you run the CMOS from the same supply voltage there should be no problem in directly connecting the 555 to a CMOS input.

The TTL chips are pretty much limited to a supply of around 5V +/- 0.5V. This may be an issue. If the 555 is running from a 5V supply they can be directly connected as the output of the 555 is both source and sink (sink being required for TTL input)

  • \$\begingroup\$ Thanks! I use CMOS MUCH more than TTL so that should be a non issue! \$\endgroup\$
    – Zack Frost
    Commented Mar 2, 2014 at 16:44

You need a series base resistor on that 2N2222, to limit the current. 10K will do.

If you want to run your 7555 from 9V, one way to translate that to 3.3V or 5V CMOS/TTL levels is to connect it to a transistor or MOSFET and use the open-drain output to drive the TTL or CMOS input (pull up to the power supply of the logic).

Another method is to use a CMOS buffer such as a CD4049 (inverting) or CD4050 (non-inverting). The buffer is powered from the same supply voltage as the logic. This particular type of chip allows the input voltage to exceed the supply voltage.

Another method, applicable for CMOS inputs without pullups or pulldowns, and assuming the 9V power supply is regulated, is to use a voltage divider. Since the 7555 swings from 0 to 9V at the output, a ratio of a bit under 4:5 is required for the resistors, such as 20K+24K.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ I forgot to include the base resistor in the schematic, but when I built the circuit I did use one! \$\endgroup\$
    – Zack Frost
    Commented Mar 2, 2014 at 16:48
  • \$\begingroup\$ That's good. Otherwise it might only work once, which has limited (but not zero) application. \$\endgroup\$ Commented Mar 2, 2014 at 16:49
  • \$\begingroup\$ I'm just curious, what does "0V/9V" and "Node 2" mean in you schematic? \$\endgroup\$
    – Zack Frost
    Commented Mar 2, 2014 at 17:01
  • \$\begingroup\$ The 0V/9V is the output of your 7555 when operated from a 9V supply. Node 2 should say 5V, and I'll fix it. \$\endgroup\$ Commented Mar 2, 2014 at 17:44

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