Electronics 1: Midterm Review - Laws, Circuits, Amplifiers & Feedback, Study notes of Physics

Download Electronics 1: Midterm Review - Laws, Circuits, Amplifiers & Feedback and more Study notes Physics in PDF only on Docsity! Electronics 1: Lecture 8 Midterm I Review • Ohm’s and Kirchoff’s Laws • Thevenin and Norton Equivalent Circuits • Node, Mesh, Superposition Analysis • Operational Amplifiers and Feedback • Homework Solutions • http://www.allaboutcircuits.com/vol_1/chpt_1 0/1.html First Midterm and Homework: Tuesday, September 23, in class: Contents: • Ohm & Kirchhoff • Thevenin & Norton • Node, Mesh, Superposition Analysis • Operational Amplifiers and Feedback Homework Due Sept 30 Read Bobrow Chapters 11 and 12 problems: 11.53, 11.55 , 11.58 12.3, 12.9, 12.30 One sheet of notes. Bring your calculator Voltage Divider 2) Vou = . Vin Ly + 2 Proof: Vin = I . (21 + Z2) Vout =I. Lo 1 [= Vin . Zy + Zo Vout = Vin . ; Z + 29 Nodal Analysis: determine node voltages: Select reference node Apply KCL to each non-reference node Solve the resulting equations… R1 I1 R3 R2 I2 v1v2 GND Let ii be the current through Ri node 1 v1 currents: I1 = I2 + i1 + i2 node 2 v2 currents: I2 + i2 = i3 i2 i1i3 Keep current directions clear by labeling!!! 2.2 Determinants and Cramer’s Kule For the analysis of circuits previously encountered, we had situations in which we © were required to solve two equations in two.unknowns. Although the approaches taken were slightly different, there is yet another technique, known as Cramer’s _rule, that can be employed. To see how this is done, first consider the pair of simultaneous equations, yx, + GX. = d, bx + bom = where the coefficients a, a, bi, b2 are given, as-are dy, do, and the unknowns X, Xp, are to be determined. The determinant D of the coefficients a), a, b,, bo, is defined by a | P= obs = ayby — bia | = ibe — 9142 | To use Cramer’s rule, two more determinants D, and Dz must be calculated. To find Dy, replace coefficients a) and b, by d, and dp, respectively, that is, qd, a% 2 2 D, = = dbp — doar To find D», replace the coefficients a, and by by d; and d, respectively. That is, ja al _ bra |t ol aude — bs By Cramer’s rule, _D and = p 7 % xy ole Cramer’s Rule [254] Seep. “fac mesh, to by - 282.54 TER a (2826: FHC AFD ~ Fe 1a 4e,, 228 eee Fav mesh, te, b KVL,. : DF lead BIR+ 40€2-4,9=0 Uytscyt 4lz- 4,20 ~40)4+8l, =O .. —t) + 2t, 70. ———————— q -4) — . GI aIB-AT14 .. 21. ee : 28 “4p Dd; Le <56 a t= se 4A De — O.: 21. Se a me ee Mesh Analysis Node or Mesh? non-planar node… else: #nodes < #meshes node #meshes < #nodes mesh also: voltage sources mesh current sources node Voltage and Current Sources zero current source = open zero voltage source = short Combining sources: Voltage sources in series OK Current sources in parallel OK NO Voltage sources in parallel No Current sources in series 2S7T(@> mee 7 vo — to 3. <a $$ —_-_-_-__- —— te Mv av(t 30 2a( 4 R 5 . Zvi 32 bn. - j Ms _. The. resis. Penen, Lemdinney, tte voltage <. ta P2.39 x00. E243 CFs) = 24S ifie = 24 See Superposition eth: eo ~ — ChMBY - “ails rer ak 22a. | - . by cunrnsit DeAsion 5. - ith2 en, _ eee gg e oy ey - mS tate) nee WE Ts ie? in f° oe = O.5984 Thevenin/Norton Equivalent Norton: INeq = short-circuit current RNeq = independent sources 0 RNeq = Req = RTeq UTeq = INeqReq Thevenin: UTeq = open-circuit voltage RTeq = independent sources 0 Thevenin’s Theorem Thevenin's Theorem Any combination of batteries and resistances with two terminals can be replaced by a single voltage source e and a single series resistor r. The value of e is the open circuit voltage at the terminals, and the value of r is e divided by the current with the terminals short circuited. t< VaB open circuit = The Thevenin voltage @ isthe open circuit voltage at terminals. Aand B. The Thevenin resistance Fis the resistance seen at AB with all voltage sources replaced by short circuits and all current sources replaced by open circuits. wv Fr Thevenin equivalent circuit 12.29 ew. 22 a, (Qa, BQ wee Ye = - 2v - 2A db : OT ee n kX __ Thevenin Ch atweda ty Usbet mda bes” . Sao te igen teem Equivalent ee “ete =oee. - fees 226 (as) e 2Ye- 227A OF aig ae - Sa 78 p 2a 252 (By 151W  oe eel ; 3a y 1S @ Wy Wy . gy © en vie Norton Equivalent “By KEL at mode. 4 Kel), mH -EN so. EFT « Sf - : i= 3 Ca e-b . au te +4 teuee : ie are 4ye1B log = - 3.5 A we-2V - — . &Sh. = =[lays)+ i] G> (PO +e = =3f/6 (oy . 18 ge SO 2S 20 carveck division y he te a aa ‘1A Ideal Op-Amp • infinite amplification • infinite input impedance • negligible output impedance output inputs: inverting non-inverting also: unlimited frequency response !!! (Voltage controlled voltage source…) Negative Feedback V+=V- (virtual equality) Inverting Amplifier Non-Inverting Amplifier: v2 Vo V1 R2 R1 amplification determined by resistor values !!! virtual equality: infinite impedance: V1/R2 = V0/(R1+R2) V1 = v2 i1 = i2 V o = V 1 1+ R 1 R 2 R1 R2 R3 R4 Rf V1 V2 V4V3 Vo Inverting Adder: virtual GND ii = Vi/Ri I = i1 + i2 + i3 + i4 V o R f = V 1 R 1 + V 2 R 2 + V 3 R 3 + V 4 R 4 V o = R f V 1 R 1 + V 2 R 2 + V 3 R 3 + V 4 R 4 Binary Weighted Ladder: A simple Digital-to-Analog Converter (DAC): R1 R2 R3 R4 Rf V1 V2 V4V3 Vo MSB LSB MSB smallest R LSB biggest R in between: factors of 2 e.g. R1=10k , R2=20k , R3=40k , R4=80k Vo = V1 + 0.5 V2 + 0.25 V3 + 0.125 V4 typical Vi: 0V O 5V I (binary) A Simple Comparator RT Uth Uo R Uo (through a diode…) drives a heater: desired: T>Tth Uo = -Vo T<Tth Uo = +Vo responds to tenths of μV unstable !!! Wouldn’t it be nice… Stay where you are… responds to tenths of μV unstable !!! 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