Momentary switch to activate 555 timer

Question

I am very new to circuits and I've tried everything online but nothing seems to be working. I am making this circuit for a class.

I am trying to emulate this circuit diagram from this youtube video with Vcc set to 5V.

I am trying to use the 555 timer to send a signal to a small RC servo motor. I have set up the 555 in monostable mode with a momentary switch. The problem is that the switch sends a signal to the output that will not stop unless the switch is depressed.

With the short pulse width requirement of the servo, the output signal from the 555 is corrupted from the output from the momentary switch so long as it is pressed. With the servo connected, it will move sporadically back and forth, but will not complete a 60 degree cycle. I have tried connecting a 10k resistor from 2 to +, but this had no effect. The resistor connecting 7 and 8 is just for this example; I'm actually using a potentiometer to set the timing. Any help would be greatly appreciated.

Answer 1

This sounds like a problem people had when trying to connect a reset switch to a Commodore 64 computer, if the switch was held down it would keep it in reset mode (which wasn't good for the CPU).

The C-64 has a 556 timer that controls the reset. It has a capacitor on the trigger pin that holds it low for a short time after power up, triggering the timer. Then the capacitor charges through a resistor and stops the reset pulse.

To make a reset switch that didn't hold trigger low when held closed, I used a capacitor and resistor in parallel between the switch and the trigger pin. This is the basic circuit:

^{simulate this circuit – Schematic created using CircuitLab}

The capacitor C2 will be held discharged by R2 until the switch is pressed, it will then pull the trigger low and then charge through R1 so that trigger goes to Vcc/2 as long as the switch is held, this should keep the timer from remaining triggered. Releasing the switch allows R2 to drain C2, so that the next time it is pressed it will trigger the timer again. You may have to play around with resistor and capacitor values a bit. You will also probably need to give the switch a quick, sharp press to trigger the timer. The circuit will trigger when first powered up, removing C1 might prevent that if it's a problem.

Answer 2

The standard way of circumventing this common problem is to add an RC differentiator to the pin 2 input. The differentiator consists of C1 & R2.

When S1 is pressed, the left hand side of C1 goes to 0V. The right hand side of C1 must instantly follow. The voltage on the right hand side of C1 then rises exponentially to 5V as C1 charges through R2. When S1 is released, C1 discharges via R1, R2 and D1 ready for the next switch press. The reason for having D1 is to prevent the voltage at the pin 2 input rising to more than about 5.6 V when the switch is released. This protects the pin 2 input from damage.

The result of all this is to generate a short pulse into pin 2 when the switch is pressed.

The pin 3 output will be high for a time period equal to 1.1 * R3 * C2 but for this to be true pin 2 must rise above 1/3 * Vcc before the end of the output pulse otherwise the output pulse will be extended.

And so, 1.1 * R3 * C2 must be greater than 0.4 * C1 * R2

I have specified a value of 10 nF for C3 which is a more typical value than your 10 pF.