Homework #2 Solutions (2)

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University of California, Los Angeles *

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121B

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Electrical Engineering

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Nov 24, 2024

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EE121B/Winter 2015/Marko Sokolich/UCLA 12 10 ; I =50 pA Yec g [ & l3=40‘pA R, f(E: 6 | o,\9;° 30uA L Vit = 4 b k'D\ _ &L‘A VBET ‘-/CE ,F E ?,9. 1S pA T Wl N 0 0.5 1 1.5 2 2.5 Vce (Volts) 6) Referring to the common-emitter Ic-Vce characteristic above. The transistor with the above characteristic is biased from a 1.5V supply {Vcc = 1.5V) with a 150 Q load. Vg is chosen such that the base current is 30 YA (the traces represent even steps in base current). a) Draw the load line on the transistor plot. b) What is the approximate collector-emitter voltage (V) across the transistor? c) We discussed 4 different regimes of operation of a BJIT depending on the biases on the two junctions. In what regime is the transistor operating for the above conditions? d) If the base current is reduced to 15 pA what I¢ will result? What will Ve be? What regime of operation is this? e} If the base current is increased to 50 pA, in what regime will the transistor be operating? G—) Yee Gxi's l.n‘l-ercep‘l' S Vcc = ISV = ,‘S’-y = IOmA T. axis intecrcepT 1S Vcc/R‘_ = S0 (Se.e_ ')'Lp_ ‘oaJ ’:V\Q o +LQ f’a‘f) b 0.6V /' [ 2 06‘/;0 = 0.9V Vge = d> Fprt...c.(-a( ‘ic-l\'vt VBQ Eeven More reverse L“‘tJta‘ T.=3mA Vee = .06V

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I Homework 6b 03182022.docx.pdf - Adobe Reader File Edit View Window Help 2 / 4 Tools Fill & Sign Comment Оpen 125% IT Sign In • Export PDF • Create PDF EP: RRE (Generalijed) Liiear Phase Filieus Euan Mart hin) - hcH-n) Gpes I I hin) --h(M.n) Types I&Z 2. Meven; (Ma6) M + maltipliro M adders X(n) Mesen be M odd ; (M=5) MA multijpleis M adders Madld Modd + For an ideal silicon p-n abrupt junction with NA = 5 x1016 cm and Np = See also cascade raalizationo in b.5.2 of OxS. 2021-10-20 19:02:14 -3 ст 5 x 10'5/cm³, 3 • Send Files • Store Files Calculate Vbi at 300K and 600K. Find the depletion layer width and maximum field at zero bias at T=300K. Hints: For Vbi calculation at T=600K, use Eg versus T formula (HW 1). 600 kT (600K)= kT(300K)x 300 3/2 600 N(600K)=N (300K)× 300 3/2 600 N, (600K)=N, (300K)× 300 3:36 AM Type here to search 日 27°C W 24-Mar-22
ZI Q.2: For the circuit shown, if Z1=10445°Q & Z2=Z3=102-45°0 1. Find the Z-Parameters? 2. Find the Y-Parameters? 001 0: V2 Z11=Z22=6.32-26.6°Q, (Ans.: 71.6°Q, y1=Y22=0.1414Z-0°u & y12=Y21=0.12135°u) Z12=Z21=4.462-
I ned answers for 2 a) and 2 b)
Ex. 592. Refer to Figure 592. [R1,C1,S1,P1]=[17, 8,1,0]. [R2,C2,S2,P2]=[12,10,1,0]. Determine A,B,C,D,E,F.
SOLVE IN DIGITAL FORMAT, DON'T USE AI Simplify the following Karnaugh map and obtain the simplest possible expression: d0 e 0 d0 U 1 d 1 e 1 d 1 e 0 b cabc a b 00 0 0 0 101 a X 0 1 0 1 X 0 0 1 1 0 X ca c b c a b cabcabcabc 1 0 1 0 1 1 0 1 1 1 10 1 100 1 1 1 X 1 1 0 X 0 X 0 0 0 X 0 1 1 1 1
Please solved it in 30 min q
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