The 555 timer chip in monostable mode in an electronic circuit works like an egg timer. When you start it, the timer turns on the output, waits for the time interval to elapse, and then turns the output off and stops. This mode is called monostable because when wired this way, the 555 has just one stable mode, with the output at pin 3 off.
When the 555 is sent a trigger pulse, this stable state is temporarily interrupted for an interval that’s determined by the value of a resistor and a capacitor. During this interval, the output at pin 3 goes high, but once the time interval has passed, the 555 returns to its stable state, with pin 3 going low.
Monostable mode is sometimes called one-shot mode, which seems a little more descriptive. One-shot mode conveys the idea that when triggered, the 555 gives one and only one output pulse. When the time interval is reached, the output pulse stops, and the circuit goes quiet until another trigger pulse is detected. Each trigger pulse results in a single output pulse.
Typical 555 monostable circuit
To understand how this circuit works, first look at the way the 10 kΩ resistor and the switch are wired to pin 2, the trigger input. The switch is a normally open pushbutton. When the button isn't depressed, the 10 kΩ resistor provides a voltage input to pin 2, which keeps the trigger input high. With the trigger input high, the output voltage at pin 3 is near zero.
When the pushbutton switch is depressed, the supply voltage is short-circuited to ground. This causes the voltage at pin 2 to drop to zero, and the timer is triggered. Once the timer is triggered, the output voltage at pin 3 goes high and the timing interval begins.
Resistor-capacitor circuit in a monostable timer
Now that you understand how the trigger circuit works, look at how the RC circuit (R1 and C1) works. The resistor and capacitor work together to determine how long the output will remain high. In a nutshell, once the circuit is triggered, C1 begins to charge.
Pins 6 and 7 — the threshold and discharge pins — are tied together in a monostable 555 circuit. Pin 6 watches the voltage across the capacitor. As the capacitor charges, this voltage increases. When the capacitor voltage reaches two thirds of the Vcc supply voltage, the timing cycle ends, and the output at pin 3 goes low.
The discharge pin (pin 7), charges and discharges the capacitor. To understand how pin 7 works, it may be helpful to visualize the internal workings of pin 7.
Here, pin 7 is connected to a switch that's controlled by the status of the output at pin 3. When the output is high, the switch is open; when the output is low, the switch is closed. When the switch is closed, a small 10 Ω resistor within the 555 connects pin 7 to ground.
When the output on pin 3 is low, the imaginary switch inside the 555 is closed, and pin 7 is connected to ground through the 10 Ω resistor. This allows the voltage on C1 to discharge through the 555.
But when the output on pin 3 goes high, the imaginary switch inside the 555 is opened. This forces the current flowing through R1 to go through C1, which in turn causes the capacitor to charge at a rate that depends on the values of R1 and the capacitor.
While the capacitor is charging, pin 6 monitors the voltage that builds up across the capacitor. Once this voltage reaches two-thirds of the supply voltage, pin 6 signals the 555 that the timing interval is ended, and the output goes low. This, in turn, closes the imaginary switch inside the 555, which allows the capacitor to discharge.
