555 Monostable Kit Instructions

Before the 555 monostable circuit can be explained in detail, it is important to see what is actually inside the 555 timer!

The image (figure 2) shows the circuit diagram of what is inside a 555 timer IC. Looking at the internal circuit diagram of the 555 timer, the following can be seen:

• 3 resistors (highlighted in green. 5kΩ each, hence the name 555)
• A flip-flop (highlighted in blue. Has two resets)
• Two comparators (one red (call this comparator 1), and one orange (comparator 2))
• One inverter (highlighted in pink. Connects to the output)
• One NPN transistor (highlighted in turquoise)

In the starting state of the circuit, the following is true

• The voltage across the capacitor C1 is 0V
• Both comparators inside the 555 chip have 0V on their outputs to the flip flop
• The flip-flop is in reset mode
• The flip-flop output voltage is therefore 9V
• The 555 output voltage is 0V
• The trigger input is at 9V (due to the pull-up resistor R1)
• VCC is 9V

To start the monostable the trigger voltage needs to quickly be dropped to 0V. This is done by pressing the switch (when the switch contacts are shorted, the trigger input is connected directly to the ground and, therefore, will have a voltage of 0V). COMP1  inverting input is connected to the trigger pin, and the non-inverting input is connected to 1/3VCC. This means that normally the output of COMP1 is off but when the switch is pressed, the inverting pin (for a short time), is at a lower voltage than the non-inverting pin. Therefore the output of COMP1 is on for a short
amount of time.

This short pules from COMP1 sets the flip flop and therefore, the output of the flip flop will be 0V. The flip-flop output is connected to the base of the discharge transistor, so when the switch is pressed, the transistor stops discharging the capacitor. Even when the switch is depressed, the transistor stays off because the flip flop remembers the last input (i.e., it stays latched). As the flip-flop output is 0V, the output stage of the 555 inverts this, and therefore, the output switches to 9V. 

With the discharge pin no longer discharging, the capacitor C1 begins to charge through R3 and the potentiometer. Note that the capacitor is connected to the threshold input on the 555, which is internally connected to the non-inverting input to COMP2.

Eventually, the voltage across C1 will become larger than 2/3 VCC. Since the inverting input of COMP2 is connected to 2/3 VCC and the non-inverting input to the capacitor, the non-inverting input voltage will become greater than the inverting input. When this happens, COMP2’s output will turn on, which will, in turn, reset the flip-flop. This results in the flip-flop output being on and, therefore, turning on the discharge transistor. C1 very quickly discharges through the discharge transistor, making the voltage across C1 0V. Therefore, the output switches back to 0V, and the circuit is back in the starting state.

This circuit is called monostable because it is only stable in one state, the off state. While the 555 is in the on state, it will only be a matter of time (depending on the resistance of the potentiometer and the size of C1), before the system is reset back into the off state.