EECS40, Midterm 1 Solutions, Spring 2000, Exams of Microelectronic Circuits

Download EECS40, Midterm 1 Solutions, Spring 2000 and more Exams Microelectronic Circuits in PDF only on Docsity! EECS40, Spring 2000 Midterm 1 solutions Prof King Problem #1 a) A B G 0 0 1 0 1 0 1 0 0 1 1 1 b) G = (not A)(not B) + AB c) 3 unit gate delays The longest path between the input variables and the output variable is 3 logic gates. Therefore, we need to wait for a period of 3 unit gate delays after an input variable is changed, before we can trust the value of G to be valid. d) EECS40, Midterm 1 solutions, Spring 2000 EECS40, Spring 2000 Midterm 1 solutions Prof King 1 Problem #2 a) Rab = 13 ohms EECS40, Midterm 1 solutions, Spring 2000 d) 2 Problem #4 a) V_Th = 2 V R_Th = 4 kohm (x is the node between the 3 kohm and 2 kohm resistors) The open-circuit voltage, Voc, is equal to Vab, which is equal to Vxb since no current is flowing through the 2 kohm resistor. Applying KCL to node x (defining node b as the reference node) -> (5-Vx)/3 + (-4-Vx)/6 = 0 => 6 = 3 Vx => Vx = 2 V therefore Voc = V_Th = 2 V To find R_Th, set all the independent sources to zero: b) I_N = 0.5 mA R_N = 4 kohm R_N = R_Th = 4 kohm I_N = V_Th/R_Th = (2 V)/(4 kohm) = 0.5 mA c) When I = 0, V = -6V When V = 0 (i.e. terminals a and b shorted together), I = (0 - (-6 V))/200 => I = 30 mA d) P1k = 25 mW EECS40, Midterm 1 solutions, Spring 2000 nodal equations:(V_AA - Va)/R1 + I_BB - (V_CC + Vb)/R3 + I_CC = 0(Vb - Vc)/R4 + I_CC = 0Vb - Va = V_BB5 Using voltage-divider formula, V = (1000/(1000+200))*(-6) = -5 V P = IV = (V/R)*V = V^2/R = ((-5)^2)/1000 = 25 mW Posted by HKN (Electrical Engineering and Computer Science Honor Society) University of California at Berkeley If you have any questions about these online exams please contact examfile@hkn.eecs.berkeley.edu. EECS40, Midterm 1 solutions, Spring 2000 P1k = 25 mW 6