Question

In Figure 3 shown, V1 = 100 millivolts. Further, R1 = 1Kohm, R2 = 10 Kohm, R3 = 5 Kohm, R4= 5 Kohm, R5 = 10 Kohm and RL = 1Kohm.

(i) Determine the amplified output voltage V2 of the op-amp (of open-loop gain A) in circuit shown in Figure 3.

(ii) Suppose this amplified voltage V2 becomes the source voltage to a passive circuit as illustrated in Figure.3. Find Thevenin’s equivalent circuit looking at the terminal XY. Hence, determine the voltage VRL and the power dissipated in the load, RL.

[Solution hints: 1. Gain of a non-inverting amplifier = [1 + R2/R1]2. Thevenin theorem parameters: VTh corresponds to open-circuit voltage at X-Y; and, RThrefers to input resistance deduced at X-Y with voltage-source short-circuited]

FIGURE 3 volt volt Vel

Answer 1

0.1V IK V5 Vo 5K 10K um By Virtual ground concept 5K EIK VRIL jok (o- o. 1) = 0.1-vo 10 - 1 = 0.1-vo Vo=114 (11) Accordurig to question 5K x (ok By Therinen theorm (.1V &5K FIK VRL = For Rsh 5K lok UM Rih 3k Ron= (5/110k = 5x1 = (3.4375k_2 =Rsh) Vih 16 Tok for SK M 1. Vt By voltage divider yn = 11 X 1 = 0.75 6V 1145 i. Vsh = 0.756V

therinin equivalent Rsn=3.4375K V0-7566 for SK M w WRL liv Vx= lex (5/11) (51110 +5 Va=0.448V By Voltage divider VRL= 1 x x = 1 x 0.440 10+1 VR = 0.0407V Power dissipated in load R ted as load R = = VRE VR - (0.0407) RE 1.659 BODO x 103 mW Power durikaten = power dissipated = 1.659 MW/ At RL