Question

please answer all parts of the question

LE 211 IU 15 ko -alone X 9 +10v Var = Vol 63 154 = -9 I vow -9.000 What is the approximate output voltage, V.? pts) -be Vin for meding, oksel loop op-amp 52 this is non-inverting and it Saturates - 10 "h 4 km2 5. Rp is the Padbook lesstar Assume that the op-amp is ideal with a saturation voltage of 18 V. Determine the following: (20 pts) 2k12 to IL14K22 0 most be 4V 4k on 4 a) output voltage, Vout they Vor- Um (1.R (Vat-12.00 4/1. 4) = 12 equal 1 I y current through the feedback resistor, In V =4V closed loop 2 Varu 12 - I = 3,00mA) non-inctor Io = I for IL (KCL) c) current through the load resistor, IL ✓ Vav- o - 12-6-I = 3.00 mA] 4K 4K - ( ) output current, I Io = If I - I. - 6.00 mA of output power of the op-amp pavi - (Var) (I) (6nAX 12v) = N P = 22 mW (P= 12.0mW nethado

Answer 1

15kn amou 5KR I 1, + -lov Voul va = -3v § due to virtual ground} Apply kel at node a , we get 11 + z =0 0-(va) + bout - va =0 5k 15k 3 + vout +3 =0 15k gt vout + 3 =0 15k - Vout =- 12v But Saturation voltage ✓ lov vout =- lov ] ps it im A to Do yout ал Arg 4u a) Vow = ? Vazuu Que tovirtual goound } Apply ko at hode a we get I t I fb = 0 o va vout - Va zo k 4 + vod -4 4K = 0 = Voty MK =
1
Vout = 12v \out = lav Ihn - vout va 6 = 12-4 = 2mA uk UK Ifb=2mA I = bout = 14 = 3m ЧЕ (d) - © to a I + Ifl = 32A + 2 (Io = 5mA Output power = vo Xiao toupe Power = 6omw] - 12x 5mA
2