555 Timer Monostable & Astable operations
Suchendranath Popuri
Internal Diagram of 555 timer
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Functions
The voltage divider has three equal 5K resistors. It divides the input voltage (Vcc) into three equal parts.
The two comparators are op-amps which compare the voltages at their inputs and saturate depending upon which is greater.
The flip-flop is a bi-stable device. It generates two values, a high value equal to Vcc and a low value equal to 0V. The transistor is being used as a switch, it connects pin 7 (discharge) to ground when it is closed
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Astable operation
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Astable Mode
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Astable Mode
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Steps
Assume output(Q) is initially 1 This means,Qbar=0 and transistor Q14 is OFF Capacitor C will charge towards Vcc through Ra and Rb When C voltage crosses Vcc/3,S=0.this however has NO effect on the output since S-0,R=0 maintains the previous state. But when C voltage crosses 2Vcc/3,output of upper comparator becomes 1,resetting the flip flop. Thus output(Q) becomes 0,Qbar =1 which turns ON Q14 Now,C has a path to discharge through Rb (through Q14)(current from Vcc also flows through Ra and Q14 to GND) C voltage decreases exponentially till it becomes just below Vcc/3.At this instant,Lower comparator is triggered and FF is set(S=1) Hence output(Q) becomes high again.Qbar becomes 0 which turns OFF Q14.Hence Capacitor cannot discharge,but starts charing towards Vcc through Ra and Rb. Cycle repeats
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Astable-Time period
Charging Interval: TH Discharging Interval: TL Period of Oscillation: TH RA TL RB C ln 2 RA RA 2RB C ln 2 1 2RB C ln 2 100%
RB C ln 2
1 Frequency of Oscillation: T TH Duty Cycle: d = 100% T
R A RB R A 2RB
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Monostable Operation
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Monostable mode
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Monostable mode
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Steps
Initially 555 is reset,Q=0, Qb=1.Hence output(Q)=0 Pin no.6 is almost at GND since Qb turns ON Transistor Q14 and drives it to saturation Trigger input(Pin no2) is held at some positive value > Vcc/3,say 2 V Hence S=0,R=0 and the circuit is stable at output logic 0(0 volts) When a negative trigger is applied at 2,S becomes logc 1.FF is set,i.e output(pin no3) goes high. Now,Qbar being 0 turns offf Transistor Q14 So,capacitor C,being connected to Vcc through R starts charging towards Vcc The moment capacitor voltage > 2Vcc/3,R becomes 1,resetting the FF.Hence output(Q) becomes 0. Qbar becomes 1 which quickly tutns ON the transistor and discharges the capacitor voltage Circuit stays in 0 untill a trigger comes again
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Monostable mode
The pulse width of time t, which is the time it takes to charge C to 2/3 of the supply voltage, is given by
t
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RC ln(3) 1.1RC
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