Question

how did he get this ohms value for R1

In other words, we need to shift these voltage levels down by -21.9V. Using the known value of R we'set the equation R, REV 12k 15V -21.9 R Va R. Ry Letting VR equal one of the op-amp DC supplies, we need to take V15V. This results in having R,=8219.17O Watch the units!

Answer 1

Assume op amp ideal that voltage infinite G = V_s/V_id = infinity here V_id = 0 V_id = V_1 - V_2 = 0 v_1 = v_2 input impedance ideal op amp infinite R_in = infinity that means input converts I_1 - I_2 = 0 Now the (2) KCL I_s + I_R = I_2 + I_4 v_1 = 0 v_1 - v_2 = 0 v_2 = 0 v_s - v_2/R_s + v_R - v_s/R_1 = 0 + v_2 - v_s/R_f v_s/R_s + v_R/R_1 = -V_0/R_f V_0 = -R_f/R_s V_s - R_f/R_1 V_R this output consists of AC voltage and DC voltage Now separate AC voltage output and DC voltage output V_0 AC = -R_f/R_s V_s V_0 DC = -R_f/R_1 V_R (1) Given data - 21.9 = V_) DC 12 k/R_1 15 v = R_f/R_1 V_R V_R = 15 v R_f = 12 k ohm