Designing Information Devices and Systems I Discussion 8B

Transcription

1 Lst Updted: :40 1 EECS 16A Fll 2018 Designing Informtion Devices nd Systems I Discussion 8B 1. Why Bother With Thévenin Anywy? () Find Thévenin eqiuvlent for the circuit shown elow. 2kΩ 5V 2kΩ i R V R V T h = 2kΩ 5V = 2.5V 2kΩ 2kΩ R T h = 2kΩ 2kΩ = 1kΩ 1kΩ 2.5V () Wht hppens to the output voltge V if we ttch lod of 8 kω to the output s depicted in the circuit elow? Use your Thévenin equivlent from prt (). UCB EECS 16A, Fll 2018, Discussion 8B, All Rights Reserved. This my not e pulicly shred without explicit permission. 1

2 Lst Updted: :40 2 2kΩ 5V 2kΩ V R 8kΩ i R We just ttch the 8kΩ resistor to our Thévenin equivlent circuit nd clculte the voltge cross it. 1kΩ 2.5V 8kΩ V R = (c) Wht if the lod is 8 3kΩ? Wht if the lod is 80 kω? R = 8 3 kω: 8kΩ 2.5V = 2.22V 1kΩ 8kΩ 1kΩ 2.5V 8 3 kω R = 80kΩ: 8 3 kω V R = 1kΩ 8 2.5V = 1.82V 3kΩ UCB EECS 16A, Fll 2018, Discussion 8B, All Rights Reserved. This my not e pulicly shred without explicit permission. 2

3 Lst Updted: :40 3 1kΩ 2.5V 80kΩ V R = 80kΩ 2.5V = 2.46V 1kΩ 80kΩ (d) Sy tht we wnt to support lods in the rnge of 8kΩ to 10kΩ. We would like to mintin 4V cross these lods. How cn we pproximtely chieve this y setting R 1 nd R 2 in the following circuit? R 1 5V R 2 i R 8 10kΩ V R 4V V T h = R 2 R 1 R 2 5V R T h = R 1 R 2 = R 1R 2 R 1 R 2 R 1 R 2 R 1 R 2 R 2 R 1 R 2 5V 8 10kΩ V R = R R 2 R R 1R 2 5V 4V R R 1 R 1 R 2 2 RR 2 R(R 1 R 2 ) R 1 R 2 = 4 5 UCB EECS 16A, Fll 2018, Discussion 8B, All Rights Reserved. This my not e pulicly shred without explicit permission. 3

4 Lst Updted: :40 4 RR 2 R(R 1 R 2 )R 1 R 2 RR 2 R(R 1 R 2 = R 2 If we set R 1,R 2 R, then resistors R 1,R 2 8kΩ, such tht R 2 = 4R 1. R 1 R 2. Therefore, we cn just choose two smll (e) For prt (), how much power does ech element dissipte? Clculte the power using your Thévenin equivlent nd using the originl circuit. Are the vlues the sme? We will ignore the power dissipted y R T h initilly nd just explore V s vs. V T h nd R lod in either cse. This could e done for the specific exmple ove, ut it s more useful to go through this exercise generlly. Thus, we will use the circuit shown elow: R 1 V s R 2 R lod Recll tht the Thévenin equivlent for the circuit ove looks s follows: R T h V T h where R T h = R 1R 2 R 1 R 2 nd V T h = R 2 R 1 R 2 V s. Becuse we re going to end up writing few expressions multiple times, we re going to define new vrile: β = R 1 R 2 R lod R 1 R lod R 2 Let s strt with our equivlent circuit. In the equivlent circuit, the current through the lod resistor nd equivlently every other element in the circuit is: I = V R lod = V T h R lod R T h With this current, we find the power disspted cross the source nd the lod resistor. V 2 T h P VT h = IV = = V T 2 h (R 1 R 2 ) R lod R T h β = V 2 s R 2 2 β(r 1 R 2 ) P Rlod = I 2 VT 2 h R = (R lod R T h ) 2 R lod = V T 2 h (R 1 R 2 ) 2 β 2 R lod = V s 2 R 2 2 β 2 R lod Let s try to find the nswer from the originl circuit. We will egin y clculting the current through the source. UCB EECS 16A, Fll 2018, Discussion 8B, All Rights Reserved. This my not e pulicly shred without explicit permission. 4

5 Lst Updted: :40 5 I s = V s V s = = V s(r 1 R 2 ) R eq R 1 R 2 R lod β Now, we cn clulte the power through the source. P Vs = I s V s = V 2 s (R 2 R lod ) β The power dissipted y the source in the originl circuit is not the sme s the power dissipted in the new circuit. Wht out the lod resistor? We will first clculte the voltge cross the lod resistor. V lod = R R 2 R lod 2 R lod R V s = 2 R lod R 1 R 2 R lod R 1 R 2R lod R 2 R lod P lod = V lod 2 = V 2 R lod s R 2 2 β 2 R lod V s = R 2R lod β The power through the lod is the sme! Thévenin equivlents cn e used to clculte the power through elements tht re not prt of the circuit tht ws trnsformed. V s 2. Series And Prllel Cpcitors Derive C eq for the following circuits. () () (c) C 2 C 2 C 4 C 2 C 3 () C eq = C 2 Notice these cpcitors re in prllel. We cn derive their equivlent cpcitnce y connecting them to voltge source with constnt derivtive, s shown y the circuit elow: i test dv test i 1 C 2 i 1 UCB EECS 16A, Fll 2018, Discussion 8B, All Rights Reserved. This my not e pulicly shred without explicit permission. 5

6 Lst Updted: :40 6 Since oth cpcitors hve the sme voltge cross them: Since we know i test = C eq dv out, dv C1 = dv C 2 i 1 = dv test i 2 = C 2 dv test = dv test i t = i 1 i 2 = ( C 2 ) dv test C eq = C 2 () In order to find the equivlence cpcitnce of the circuit, we plug in test current source, nd mesure the rte of chnge of voltge cross it. u 2 C 2 i C2 u 1 V test I test i C1 From KCL, we know tht ll of the currents re equl. i C1 = i C2 = I test For ech cpcitor, we plug in our I dv reltionship: i C1 = I test = du 1 i C2 = I test = C 2 d(u 2 u 1 ) = C 2 ( du2 Next, we eliminte u 1 from the equtions ove nd rerrnge. du ) 1 du 1 = I test du 2 I test = C 2 C 2 I test I test = C 2 du 2 1 C 2 UCB EECS 16A, Fll 2018, Discussion 8B, All Rights Reserved. This my not e pulicly shred without explicit permission. 6

7 Lst Updted: :40 7 Finlly, we plug in tht u 2 = V test nd solve for the equivlent cpcitnce with C eq = I test / dv test I test = C 2 dv test 1 C 2 C eq = C 2 1 C 2 = C 2 C 2 Note tht this is the sme s sying C eq = C 2. Rememer tht the opertor is mthemticl nottion; in this cse, the cpcitors re ctully in series, ut mthemticlly their equivlent circuit is found vi the prllel resistor opertion. (c) Given tht we know wht the reltionship for cpcitors in series nd prllel re from the lst two prts, we cn just simply the cpcitors step y step: 3. Cpcitnce Equivlence C eq = (C 4 ( (C 2 C 3 ))) = C 4( C 2 C 3 ) (C 2 C 3 ) C 2 C 3 For the structures shown elow, ssume tht the pltes hve depth L into the pge nd wih W nd re lwys distnce d prt. () Wht is the cpctnce of the structure shown elow? The cpcitnce of two prllel plte conductors is given y C = ε A d. The cross-sectionl re A is WL, so the cpcitnce is C = ε WL d. () Suppose tht we tke two such structures nd put them next to ech other s shown elow. Wht is the cpcitnce of this new structure? Here, we hve just douled the wih of the cpcitor pltes. The new cpcitnce is C = ε 2WL d. Notice tht this is just doule the cpcitnce from the first prt. (c) Now suppose tht rther thn connecting the together s shown ove, we connect them with n idel wire s shown elow. Wht is the cpcitnce of this structure? UCB EECS 16A, Fll 2018, Discussion 8B, All Rights Reserved. This my not e pulicly shred without explicit permission. 7

8 Lst Updted: :40 8 Intuitively, nothing hs chnged here since we hve just dded n idel wire etween two cpcitors. Thus, the nswer remins C = ε 2WL d. (d) Suppose tht we now tke two cpcitors nd connect them s shown elow. Wht is the cpcitnce of the structure? We know tht cpcitors plced in series follow the prllel rule. Thus, the overll cpcitnce is hlf the individul cpcitnce. C eq = C C = C C C C = C 2 (e) Wht is the cpcitnce of the structure shown elow? Notice here tht we re ignoring the mteril in the middle. Thus, from modeling perspective, we cn think of this s the originl cpcitor with the distnce etween the pltes douled. C eq = ε WL 2d = 1 2 ε WL 2d = C 2 UCB EECS 16A, Fll 2018, Discussion 8B, All Rights Reserved. This my not e pulicly shred without explicit permission. 8