Consider the Oscillator below tran 0 3m 0 C1 +VS tV2 22n +VS COMP +VS 12 U1 R1 U

Question

Consider the Oscillator below tran 0 3m 0 C1 +VS tV2 22n +VS COMP +VS 12 U1 R1 U2 22K OUT LT1007 LT1007 VS R2 56K VS VS V3 R3 12 18K The Lispice simulation is shown below .tran 0 3m 0 C1 +VS V2 22n +VS COMP 12 U1 R1 U2 LT1007 22K OUT LT1007 vs R2 56K .VS VS V3 R3 12 18K The Relaxation oscillator simulations above gives a frequency value of 1.6KHz square wave signal which is equivalent to a time period of 6228 (micro seconds = 10-6 sec The square wave ouput is less than the upply voltages of ±12V at nearly ±11V If the supply voltage varied from +10V to +15V answer the two questions below 1. What would be the approximate frequency in KHz? 2. What would be the minimum and maximum output voltage of the square wave?

Explanation / Answer

1) Frequency of relaxation oscillator is given by formula

fr = R2 /4R1C1R3

R2 = 56 K,R1 = 22 K , R3 = 18 K & C1 =22 nF

fr = 56 K / 4*22K*18K*22nF

fr = 0.0016*106

fr = 1.6 KHz                          ...............1

This is the required frequency in KHz

2) supply voltage is varied from +/- 10 V to +/- 15 V

Hence minimum output voltage would be 1 V less than +/- 10 V i.e. +/- 9 V

Hence maximum output voltage would be 1 V more than +/- 10 V i.e. +/- 11 V

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