Electronics I: Circuit Analysis - Voltage/Current Division, Mesh/Nodal Analysis, Superposi, Slides of Electronics

Download Electronics I: Circuit Analysis - Voltage/Current Division, Mesh/Nodal Analysis, Superposi and more Slides Electronics in PDF only on Docsity! BME 372 Electronics I – J.Schesser 67 Circuit Analysis Lesson #2 BME 372 Electronics I – J.Schesser 68 Voltage Division • The voltage across impedances in series divides in proportion to the impedances. Z1 Z2 a b c 21 2 2 21 Law sOhm' KVL );( ZZ Z Z ZZ ac bc bc bcabac     V V IV IVVV Z1 Zn a c n i ac i ZZZ Z   21V V BME 372 Electronics I – J.Schesser 71 Nodal Analysis 1. Define a voltage at each node (junction point) of a network. For example, in a 5 node network, define 5 voltage unknowns. 2. Using KCL, write an equation for each node using the unknown voltages. In our 5 node example, you’ll have 5 equations and 5 unknown voltage. 3. Solve for the unknown voltages and now apply these voltages to the network to find the currents for each impedance in the network. BME 372 Electronics I – J.Schesser 72 Mesh Analysis Example 5Vdc 5Vdc 10 5 5 I1 I2 5 1 ;5 1 10155- );(10550 2Mesh 10155 ;5)(1050 1Mesh 21 12122 21211    II IIIII IIIII 5Vdc 5Vdc 10 5 5 .2a .2a- 1v ++ 1v - .4a + 4v - Mesh Analysis I1(15)+I2(-10) = 5 I1(-10)+I2(15) = -5 I1 = 0.2, I2 = -0.2 clear all; R=[15 -10;-10 15]; V=[5;-5]; I=R\V; grid off axis off text(.1,1,'Mesh Analysis') text(.1,.9,['I1(',num2str(R(1,1)),')+I2(',num2str(R(1,2)),') = ',num2str(V(1))]); text(.1,.8,['I1(',num2str(R(2,1)),')+I2(',num2str(R(2,2)),') = ',num2str(V(2))]); text(.1,.7,['I1 = ',num2str(I(1)),', I2 = ',num2str(I(2))]) BME 372 Electronics I – J.Schesser 73 Nodal Analysis Example 5Vdc 5Vdc 10 5 5 V I2 4 22 ;0105 5 5 5 0 1Node 321    V VVVV III 5Vdc 5Vdc 10 5 5 .2a .2a- 1v ++ 1v - .4a + 4v - I1 I3 clear all; R=[1/5+1/5+1/10]; VS=[10/5]; V=R\VS; I1=(5-V(1))/5;I2=(5-V(1))/5;I3=-V(1)/10; grid off axis off text(.1,1,'Nodal Analysis') text(.1,.9,['V = ',num2str(V)]); text(.1,.8,['I1 = ',num2str(I1),', I2 = ',num2str(I2),', I3 = ',num2str(I3)]); Nodal Analysis V = 4 I1 = 0.2, I2 = 0.2, I3 = -0.4 BME 372 Electronics I – J.Schesser 76 Superposition Analysis Example 5Vdc 5Vdc 10 5 5 5Vdc10 55 5 3 25 15 3 105 5 33.33 10 15 50 510 5*105||10 22       s p I R 5Vdc3.33 5 I2s2 5Vdc 10 55 Source 2 I1s2=-.4 I2s2=.6 I3s2=-.2 2.5 1 5 3 15 5 4.5 2 5 3 15 10 23 21   s s I I I1 I2I3 I1s2 I2s2I3s2 BME 372 Electronics I – J.Schesser 77 Superposition Analysis Example 5Vdc 5Vdc 10 5 5 .2a .2a- 1v ++ 1v - .4a + 4v - • Summing the results of each solution: 4.2.2. 2.6.4. 2.4.6. 23133 22122 21111    ss ss ss III III III BME 372 Electronics I – J.Schesser 78 Thevenin and Norton Equivalent Circuits • Thevenin’s Theorem: Any circuit consisting of passive and active components can be represented by a voltage source in series with an equivalent set of passive components – The value of the voltage source equals the voltage seen at the output terminal without any load connected to it, i.e., the open-circuit voltage – The value of the equivalent set of passive components equals the impedance looking back into the terminals with the sources set to zero, i.e., the output impedance. BME 372 Electronics I – J.Schesser 81 Thevenin and Norton Examples 5 10Vdc 15 a b Short Circuit Current at terminals: ab 10 2 5 Output Impedance 5 15 155 ||15 3.75 20 4 abSC o I a R         a b 5 10Vdc 15 a b 2Adc 3.75 BME 372 Electronics I – J.Schesser 82 Input and Output Impedance of a Circuit • Input impedance of a circuit is the impedance looking into the input terminals of the circuit with any load connected to the output and all internal sources are set to zero. • Output impedance of a circuit is the impedance looking into the output terminals of the circuit without any load connected to the output and all internal and input sources are set to zero. BME 372 Electronics I – J.Schesser 83 Examples I2 I1 1 2 1 1 1 1 2 4 2 4 1 2 2 1 4 4 2 8 kBRANCH kSOURCEBRANCH CURRENTSOURCE kBRANCH kSOURCEBRANCH I I a V a V V a V V V V V               - 2V + + 2V - + 2V - - 4V + 4Adc 1 1 11 1 + 2V - - 8V + Find the currents and voltages in these circuits 1 1 1 1 1 4Adc BME 372 Electronics I – J.Schesser 86 Examples R2R1 1k1k V= 1Vac C1 1n C2 1n a b c )(111 11 )()( )( (1) into (2) ngSubstituti 2221112121 2 21 2 1 1 1 2 2 21 21 2121112111 21 2211211 221 22 2 CRCRCRjRRCC VV RR Cj R Cj R Cj R CjCj CjCjV V ZZZZZZZZZZ ZZ V ZZZZZZZ ZZZ ZZ ZV ab RRcRcRRccc cc RccRRcc Rcc cR c ab                 Find the voltage Vab in terms of V ZR1=R1 ZC1= 1/jωC1 a b ZR2=R2 ZC2= 1/jωC2 + -V c BME 372 Electronics I – J.Schesser 87 Homework • Voltage and Current division – How does the voltage divide across two capacitors in series? Show your results. – How does the current divide among two capacitors in parallel? Show your results. • Calculate the Currents and Voltages for the following circuits: 1 10Adc 2 1 25Vdc 2Vdc 1 1 10Adc 2 1 10Adc 2 1 2Vdc BME 372 Electronics I – J.Schesser 88 Homework Calculate the current labeled i and the voltage labeled v in the following circuit R1 = 1Ω, R2 = 2Ω, R3 = 1Ω, R4 = 1Ω, R5 = 2Ω, R6 = 2Ω, R7 = 2Ω, Vcc = 4v R1 R7 R2 R3 R4 R6 R5 + VCC -- i + v -- BME 372 Electronics I – J.Schesser 91 Homework • Repeat the analysis of this circuit using Mesh and Nodal Analysis. That is find and plot Vab as a function of frequency. Use Matlab to perform the plot. a b R2 5k R1 1k 1Vac C1 1n C2 5n BME 372 Electronics I – J.Schesser 92 Homework • Repeat the analysis of this circuit. That is find and plot Vout / Vin as a function of frequency. Assume Va = 0; C=10nF,R1=2.2MΩ, R2 =330kΩ, R3=2.7MΩ. Perform the plot using Matlab Vin VoutR1 R2 R3 C C Va