The schematic bellow is for a monostable vibrator (one-shot). Such a circuit is

Question

The schematic bellow is for a monostable vibrator (one-shot). Such a circuit is stable in one output state (monostable).

(a) Analyze the circuit operation and find out which state is stable for Vin = const.

(b) If Vin changes abruptly, is the rise or fall of Vin producing a pulse at the output?

(c) What is the nature of the output pulse? (d) Which components in the circuit determine the pulse duration ?

Use the following component values: R1 = 100k, R2 = 27k, R3 = 47k, R4 = 470k, C1 = 0.01F and C2 = 1000pF.

The schematic bellow is for a monostable vibrator (one-shot). Such a circuit is stable in one output state (monostable). (a) Analyze the circuit operation and find out which state is stable for Vin = const. (b) If Vin changes abruptly, is the rise or fall of Vin producing a pulse at the output? (c) What is the nature of the output pulse? (d) Which components in the circuit determine the pulse duration ? Use the following component values: R1 = 100k, R2 = 27k, R3 = 47k, R4 = 470k, C1 = 0.01?F and C2 = 1000pF.

Explanation / Answer

When the input voltage Vin is constant The Triggering circuit C2 and R4 produces a positive spike, this will reverse bias the diode D2 . The voltage at the non-inverting input V+ = Vout R3 / ( R2 + R3) = KVout Due to virtual short, the voltage at inverting input V- = V+ The diode D1 will be forward biased and conducts through R1 and the clamping voltage will be one diode drop above the ground level VD1 = 0.7V Choosing K such that KVout is greater than VD1 . The steady state output will be maintained HIGH.    When a negative going edge of the trigger pulse is applied, is it coupled to the diode D2 through the capacitor C2 . D2 is now forward biased and conducts heavily. It pulls the voltage at the non-inverting terminal V+ to nearly zero. Now the differential input voltage at the op amp will be negative , thus making the output to go LOW. The output will be in its new state for a predetermined duration
The LOW output voltage , will cause D1 to cutoff and C1 begins to discharge exponentially with w time constant C1R1 . The monostable multivibrator is now in quasi state. This will remain so until the the discharging voltage across C1 that is V- goes below the voltage at the non-inverting input V+ . At this instant the opamp output will go back to HIGH . Capacitor C1 then charges towards HIGH, until the diode D1 turns ON and the circuit remains in this stable state until another NEGATIVE triggering pulse occurs.
The nature of the output is a negative pulse. The pulse duration is given by
W = C1R1 ln [1/ (1-K)] K = 47k / (27k + 47k) = 0.635 W = (0.01*10-6 ) (100000) ln [ 1/(1-0.635)] W = 1.08ms

The voltage Due to virtual short, the voltage at inverting input V- = V+ The diode D1 will be forward biased and conducts through R1 and the clamping voltage will be one diode drop above the ground level VD1 = 0.7V Choosing K such that KVout is greater than VD1 . The steady state output will be maintained HIGH.    When a negative going edge of the trigger pulse is applied, is it coupled to the diode D2 through the capacitor C2 . D2 is now forward biased and conducts heavily. It pulls the voltage at the non-inverting terminal V+ to nearly zero. Now the differential input voltage at the op amp will be negative , thus making the output to go LOW. The output will be in its new state for a predetermined duration
The LOW output voltage , will cause D1 to cutoff and C1 begins to discharge exponentially with w time constant C1R1 . The monostable multivibrator is now in quasi state. This will remain so until the the discharging voltage across C1 that is V- goes below the voltage at the non-inverting input V+ . At this instant the opamp output will go back to HIGH . Capacitor C1 then charges towards HIGH, until the diode D1 turns ON and the circuit remains in this stable state until another NEGATIVE triggering pulse occurs.
The nature of the output is a negative pulse. The pulse duration is given by
W = C1R1 ln [1/ (1-K)] K = 47k / (27k + 47k) = 0.635 W = (0.01*10-6 ) (100000) ln [ 1/(1-0.635)] W = 1.08ms

The voltage The diode D1 will be forward biased and conducts through R1 and the clamping voltage will be one diode drop above the ground level VD1 = 0.7V Choosing K such that KVout is greater than VD1 . The steady state output will be maintained HIGH.    When a negative going edge of the trigger pulse is applied, is it coupled to the diode D2 through the capacitor C2 . D2 is now forward biased and conducts heavily. It pulls the voltage at the non-inverting terminal V+ to nearly zero. Now the differential input voltage at the op amp will be negative , thus making the output to go LOW. The output will be in its new state for a predetermined duration
The LOW output voltage , will cause D1 to cutoff and C1 begins to discharge exponentially with w time constant C1R1 . The monostable multivibrator is now in quasi state. This will remain so until the the discharging voltage across C1 that is V- goes below the voltage at the non-inverting input V+ . At this instant the opamp output will go back to HIGH . Capacitor C1 then charges towards HIGH, until the diode D1 turns ON and the circuit remains in this stable state until another NEGATIVE triggering pulse occurs.
The nature of the output is a negative pulse. The pulse duration is given by
W = C1R1 ln [1/ (1-K)] K = 47k / (27k + 47k) = 0.635 W = (0.01*10-6 ) (100000) ln [ 1/(1-0.635)] W = 1.08ms

The voltage

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---