Question

[20pts Consider the following circuit: R. W R Tout Road 200mA 100 Assume that R, = 10km, Road = 300k1, and V.. = 1.5V. (a) What value of R, will result in Vout = 12V? [2] (b) For what value(s) of the current source is a solution impossible? Explain. [4] (c) What current is flowing out of the output pin of the op-amp? [2] (d) What is the power delivered by the op-amp? Is power balanced? [4] (e) Redraw the network WITHOUT using "electronics" notation as here, i.e., as a single closed circuit, with a single ground on the bottom as a reference point. Label the drawing with the current in each branch (magnitude and direction). [8] Show your work for full credit. NOTES and SUGGESTIONS • Work symbolically, as this will make it easier to solve for R and also easier to see the dependence of R, on the value of the current source. • You may use MATLAB or Python to help solve parts (a), (c), and (d), but you must include your script and output as PDF files for full credit. There must also be a "hand calculation" showing where the equations in your script came from. The hand calculation should agree with your script • The current flowing out of the output pin is not the current through the load. • For part (e), there are 11 branches to consider. Use uA for all branches, including the branch with the current source and the branch with the 100 resistor. • You should be able to verify that KCL is satisfied at every junction.

Answer 1

Solution Ra Assuming M to be ideal vout a + opamp- Virtual Shost Circuit is valid. va = vb) &R load. Vin 5 102 2oomt Iin=0A Giveni IS No Imput Cussent R6 = 10kor Rioad = 300k Voit = 12V + Now, Vj = (200mA) (10) Iva = V₂ = au) Vin= 1.5v. a) Now 20 Apply Nodal Analysis at Node a & vin Va = 2-12 RA Rb at Node Va - Vout 1.5-2. Ra. Rb toos to Ra Rb Ra Rb Ra = (20) (10) Kiz= 200 kis 20 b) Ra Rb 101- 12 Now, Ua=VG = I(10) . 1 - Current Source again Apply Nodal Analysis at Node a vinava a va -vout =) Los - 101 = 101-12 Rb Ra. > 165-10I :- >0.1 Either, Ra 1.5 - 10I 70 I <0.15A Not 10 I-12:20 » I > 102A ponible 11.5-10 I Co Iz 0.15A . 101-12 BTO I < 12A » Ponible for values (I) 1.2AT & I <0.15 • solution Imposible

(0) Ra. -R5 I To 2v IL Vout Rload vin 612) (3ooker) [lisu) 200mAY Que KCL at Noden Apply 2-12 + Io = 12 300ks To 12. 2 7.12 300 ks 2ooks2ook 200kn > Io -0.09 MA Ld) Power-delivered (0.09 (12) = 1.08m watt