Question

Please solve by hand

Exercise 1: Consider the circuit below built with an AD 509J operational amplifier. The data sheet indicates the following relevant values: Vos-5 mV; lbias 125 nA; loffset 20 nA 1. 2. 3. What is the AC component of the output waveform you expect to measure on the oscilloscope? What is the DC component of the output waveform you expect to measure on the oscilloscope? What source resistance value would use in place of the current 2.2 k2 to minimize the dc component of the output waveform? What would be the DC component of the output waveform with this new resistor in place? VDD VCC 15V -15V Ext Tri VCC 2.2kΩ 0.01Vpk 100Hz Vin- 0° Vin AD509J 560Ω VDD

Answer 1

I_b1 - i_b2 = 20 nA, i_b1 + i_b2/2 = 125 nA i_b1 = 135 nA, i_b2 = 115 nA First we will calculate output voltage with zero offset voltage(V_os) and source active Then we will calculate output voltage with zero input and offset Final output will be the sum of above two. Circuit with active input and zero input input offset and bias. \r\n V_+ = V_p, KCL at node 'x' (V_o - V_+)/47 = V_+/0.56 = > (V_o - V_p)/47 = V_p/0.56 V_o - V_p = 83.928V_p V_o1 = 84.928 V_p output because of input bias current \r\n V_+ = -i_b1 R_s V_+ = v - (virtual short) V_o - V_+/R_2 = V_+/R_1 + i_b2 V_o - V_+ = R_2/R_1 V_+ + R_2 i_b2 V_o = V_+ (1 + R_2/R_1) + R_2i_b2. V_o = -i_b1 R_s(1 + R_2/R_1) + R_2 i_b2 substituting i_b1 = 135 nA, i_b2 = 115 nA, R_s = 2.2 K R_2 = 47 K, R_1 = 0.56 K V_o = -0.0198 v