I'm trying to create a circuit that will pass a 12V pulse on the rising edge of a push button (it can be triggered by the entire input pulse but rising edge preffered). I've had some experience with microcontrollers but I'm very new to ICs. I believe I can use a monostable multivibrator, constructed with a 555 timer.

Datasheet: http://www.ti.com/lit/ds/symlink/ne555.pdf

The below circuit is taken almost directly from the datasheet (pg. 10) with the addition of the trigger configuration. My understanding of the graph in figure 10 of pg. 10 is that the pulse is generated on a "low" input at the trigger, would my configuration below create this low input when the button is pressed (pull up resistor, I believe)?


simulate this circuit – Schematic created using CircuitLab

The output pulse length is found approx by: \$T=1.1{\times}R_1{\times}C_x\$

I'm having trouble wrapping my head around what the threshold and control inputs do and so in the above equation I'm unsure which capacitor value to use, \$C_1\$ or \$C_2\$?

When triggered, is the output pulse equal to the input source \$V_1\$ or another value?

Not 100% on choosing a value for \$V_2\$, graph would indicate using a value of \$\frac{V_1}{2}\$ works?

Sorry, I know this is a lot of questions, thank you for any and all help.


2 Answers 2


C1 is the capacitor that is used along with R1 to set the pulse length using the formula you gave. So you can substitute R1 for R, and C1 for C in the formula.

The CONTROL lead is used to adjust the interior comparator levels, in this case it is not used. The capacitor C2 just provides some noise immunity to prevent false triggering. It is typically 10 nF to 100 nF.

The output will be equal to V1 when triggered, and ground otherwise.

Instead of using a separate V2 voltage, you can just tie R2 to V1. The TRIGGER voltage just needs to be above V1/3 when not active, but there is no reason it can't be equal to V1. A good value of R2 is 10K.

You should also put a 100 nF capacitor between the Vcc pin and ground.

Here is a simplified view of interior circuit of the 555:

enter image description here

Note the three 5K resistors on the left that create a voltage divider; that's where the name 555 comes from. The resistors set up a voltage of 2/3 V on the - input to the upper comparator C\$_{A}\$, and 1/3 V on the + input of the lower comparator C\$_{B}\$.

When the TRIGGER falls below 1/3 V, the lower comparator C\$_{B}\$ outputs a high and sets the flipflop, and the OUTPUT goes high. The external capacitor C1 also starts to charge. When the external RC network made up of R1 and C1 reaches 2/3 V, the upper comparator C\$_{A}\$ goes high, and resets the flip-flop, and the OUTPUT goes back to 0.

Potential problem: Looking at the interior circuit of the 555, if the TRIGGER input is held low for longer than pulse length, it will keep the lower comparator C\$_{B}\$ high and the OUTPUT will remain high.

You can get around this problem using a differentiating input:

enter image description here

It generates a short negative going pulse regardless of how long you hold the switch down.

  • \$\begingroup\$ Great! thank you, you managed to answer all my questions and simplify my circuit \$\endgroup\$ Commented Jul 19, 2015 at 9:20
  • \$\begingroup\$ For a monostable, Pin 6 should be connected directly to pin 7 and Rb should be gone. \$\endgroup\$
    – EM Fields
    Commented Jul 19, 2015 at 10:56
  • \$\begingroup\$ @EMFields I was just trying to show the interior circuitry, and wasn't really paying attention to how it was hooked up since the OP already had that in his post. I'll see if I can find another picture. Thanks. \$\endgroup\$
    – tcrosley
    Commented Jul 19, 2015 at 11:00
  • \$\begingroup\$ Good catch, it is quite possible for the button to be held down longer than the pulse so this added bit of circuitry will be necessary. rather than redrawing my circuit i will go with the one provided by "EM Fields" as this seems to address all the additions you have brought up, thanks for the explanation \$\endgroup\$ Commented Jul 19, 2015 at 11:25

Im having trouble wrapping my head around what the threshold and control inputs do and so in the above equation im unsure which capacitor value to use, C(1) or C(2)?

The THRESHOLD input determines when the output pulse ends by sensing when the voltage on the timing capacitor (C1 re. your drawing) rises to 2/3Vcc. C2 is used for noise reduction, and is generally not used.

The CONTROL input is used to adjust the threshold voltage sensing point and must float or be cap-coupled to ground when not used.

When triggered, is the output pulse equal to the input source V(1) or another value?

The output pulse amplitude will be equal to Vcc minus Vce(sat) of the output totem-pole's upper transistor.

Not 100% on choosing a value for V(2), graph would indicate using a value of 1/2 V(1) works?

Ignore V2 and use Vcc for everything.

One thing that hasn't been mentioned yet is that the input trigger pulse must be shorter than the output pulse, otherwise the output will just follow the input.

Also, you can run everything from Vcc, as shown below with the TRIGGER input cap-coupled to the switch, ensuring that the input pulse will be short, no matter how long the button's held ON.

RtCt are the timing components, and the output pulse's ON time will be 1.1RtCt.

The diode in the triggering circuit is used to clamp the 555's TRIGGER pin to Vcc plus one diode drop when the switch is released and the coupling cap is acting like a charge pump.

If you use a bipolar (instead of CMOS) 555 you'll need to decouple the chip from the supply wiring with a cap directly across pins 1 and 8 in order to provide a local reservoir for the totem pole's ~ 400mA supply current transient during an output transition and, finally, here's a link to an LTspice schematic you can use to run a simulation if you want to play with the circuit.

enter image description here

  • \$\begingroup\$ All very good info, your last section has me a little confused though, can i tell the difference between a CMOS and bipolar by appearance? and does this mean your diagram is valid only for a bipolar? \$\endgroup\$ Commented Jul 19, 2015 at 11:39
  • \$\begingroup\$ @JordanCartwright: You can tell the difference from the part number on the chip. The bipolar ones will usually say something like "NE 555" or "SE 555" on them and the CMOS ones will be marked with something else peculiar to their manufacturer. Google the part number and something relevant will surely come up. My circuit will work for either flavor. Be aware though, that the bipolars can source and sink a lot more current than the CMOS ones. \$\endgroup\$
    – EM Fields
    Commented Jul 19, 2015 at 12:57
  • \$\begingroup\$ Bipolar then as NE555 was the part i was going for, thankyou for your help ill give it a go tomorrow \$\endgroup\$ Commented Jul 19, 2015 at 13:16
  • \$\begingroup\$ tried constructing the circuit you gave today, no luck. gives a constant ~2.5V with V1=6V. are the open circles on pins 2,4,7 of the 555 representing anything? theyre not "NOT" gates are they? \$\endgroup\$ Commented Jul 21, 2015 at 16:13
  • \$\begingroup\$ @JordanCartwright: The circles mean that those signals are "active low". That is, if you want to trigger the timer you pull pin 2 low, if you want to reset the timer you pull pin 4 low, and when it's time to discharge Ct the chip will pull pin 7 low. Other than a wiring error or a bad chip, I can't imagine why it's not working. In particular, take a look at pin 2 and make sure it's not connected to any other pins. \$\endgroup\$
    – EM Fields
    Commented Jul 22, 2015 at 9:55

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