Calculate Vout for Fig. A.4, with Vin=4Vp-p, R=20kohm, Vcc = 5V, C = 10,000pF Th

Question

Calculate Vout for Fig. A.4, with Vin=4Vp-p, R=20kohm, Vcc = 5V, C = 10,000pF

This integrator, shown in Figure A.4, is not very practical because there is no method of discharging the capacitor; hence any current leakage eventually charges the capacitor until the circuit becomes saturated. The positive input of the integrator is biased at Vcc/2 to center the output voltage at Vcc/2. thus allowing for positive and negative voltage swings. The bias resistors are selected as 2R so that the parallel combination equals R. This offsets the input current drawn through R. Functionally, the circuit in Figure A.5 is the same as that shown in Figure A.4, but a current compensation network has been added to offset the input current. Vcc, R1, and R2 bias the positive input at Vcc/2 to center the output voltage at Vcc/2, thus allowing for positive and negative voltage swings.

Explanation / Answer

Cant see the picture can you make it a bit smaller,

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