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Ee2 chapter17 monstable_operation

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• 1. IT2001PAEngineering Essentials (2/2)Chapter 17 - Monostable Operation of the555 IC Timer Lecturer Namelecturer_email@ite.edu.sg Nov 20, 2012Contact Number
• 2. Chapter 17 - Monostable Operation of the 555 IC TimerLesson ObjectivesUpon completion of this topic, you should be able to: Students should be able to describe and verify the circuit operation and applications of a monostable multivibrator. Calculate the pulse width of a monostable multivibrator circuit. IT2001PA Engineering Essentials (2/2) 2
• 3. Chapter 17 - Monostable Operation of the 555 IC TimerSpecific Objectives Students should be able to :  Draw the circuit of a monostable multivibrator using IC 555.  Explain the operation of the monostable multivibrator.  State the applications of monostable multivibrator. IT2001PA Engineering Essentials (2/2)
• 4. Chapter 17 - Monostable Operation of the 555 IC TimerMonostable Multivibrator  It is also known as one-shot multivibrator.  Is stable in only one of two voltage levels.  When a trigger is received it switches to the other state for a period of time (t ) and then returns to the stable state.  Application as a timer. Input Circuit Output Monostable Multivibrator t IT2001PA Engineering Essentials (2/2) 4
• 5. Chapter 17 - Monostable Operation of the 555 IC Timer 555 as Monostable Multivibrator +V Trigger input R Vcc 8 Output 1 Vcc 7 3 3C 6 555 Vo tp Vcc OutputTrigger 5 Voutinput 2 1 0.01µF 0 t t = 1.1RC Duration of output pulse at HIGH state, t = 1.1 RC IT2001PA Engineering Essentials (2/2) 5
• 6. Chapter 17 - Monostable Operation of the 555 IC TimerExample : To calculate the duration of the quasi-stable state of themonostable multivibrator. Given that C = 0.001µf and R = 10KΩ.Solution : Input Circuit Output Monostable Multivibrator t T = 1.1 RC = 1.1(10 KΩ x 0.001µf ) = 1.1(10 x 103Ω ) ( 0.001 x 10-6f ) = 0.000 011s = 11µs IT2001PA Engineering Essentials (2/2) 6
• 7. Chapter 17 - Monostable Operation of the 555 IC TimerInternal Circuitry of a 555 in the Monostable Mode Vcc R C OUTPUT 1 + 3 _ 1 S Q 0 2 3 INPUT 1 3 1 + Vcc 2 R 0 2 _ Vcc When power is applied initially Comparator 2 output is driven Low by positive High at the trigger input. Comparator 1 output is driven High by the voltage charged across the external capacitor. S = 1 , R = 0, Q = 1 and output is Low. IT2001PA Engineering Essentials (2/2) 7
• 8. Chapter 17 - Monostable Operation of the 555 IC Timer Vcc R C OUTPUT 1 + _ 1 S Q 0 2 3 INPUT 1 3 + 1 Vcc 2 R 0 2 _ When the transistor is ON Vcc This set the flip-flop, Q=1 and the transistor is ON. The capacitor then discharges to the ground through the transistor and causes the output of comparator 1 to go Low. The flip-flop will now remain set because both inputs are Low (unchanged state). •The output will remain at this stable state until a Low trigger inputs is received. IT2001PA Engineering Essentials (2/2) 8
• 9. Chapter 17 - Monostable Operation of the 555 IC Timer Vcc R C OUTPUT 1 + 3 _ 1 S Q 0 2 3 INPUT 1 3 1 Vcc + 0 2 _ 2 R Vcc When a Low trigger input is applied at pin 2 Inverting input of comparator 2 from pin 2 will go Low. This is lower than the non inverting input which is 1/3Vcc. The output of comparator 2 is driven High. S = 0, R = 1, Q = 0, and it turns OFF the transistor. The circuit is now in its unstable state, output is logic ‘1’. IT2001PA Engineering Essentials (2/2) 9
• 10. Chapter 17 - Monostable Operation of the 555 IC Timer Vcc R C OUTPUT 1 + 3 _ 1 S Q 0 2 3 INPUT 1 3 1 + 0 2 Vcc _ 2 R Vcc When the transistor is OFF Input has returned to High, Logic “1” . The external capacitor will charge through resistor R. When the voltage across the capacitor exceeds 2/3Vcc, the output of comparator 1 is driven High. Therefore S = 1, R = 0 and this sets the flip-flop. Q = 1. The circuit returns to its stable state, output is logic ‘0’. IT2001PA Engineering Essentials (2/2) 10