Chapter 6. Series-Parallel Circuits ISU EE. C.Y. Lee

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Chapter 6 Series-Parallel Circuits

Objectives Identify series-parallel relationships Analyze series-parallel circuits Determine the loading effect of a voltmeter on a circuit Analyze a Wheatstone bridge circuit Apply Thevenin s theorem to simplify a circuit for analysis Apply the maximum power transfer theorem Apply the superposition theorem to circuit analysis 2

Identify Series-Parallel Relationships A series-parallel circuit consists of combinations of both series and parallel current paths 3

Total Resistance Identify the series and parallel relationships, then apply what you have previously learned Example: Determine R T between terminals A and B R 23 = (100 100)/(100+100) = 50 Ω R T = R 1 + R 2 3 = 10 + 50 = 60 Ω 4

Total Current Using the total resistance and the source voltage, find the total current by applying Ohm s law Example: Determine total current drawn from a 1.5V source R 23 = (100 100)/(100+100) = 50 Ω I T R T = R 1 + R 2 3 + R 4 = 56 + 50 + 22 = 128 Ω I T = V S /R T = 1.5/128 = 11.7 ma 5

Branch Currents Using the current-divider formula, KCL, Ohm s law, or combinations of these, you can find the current in any branch of a series-parallel circuit Example: Determine the branch currents, I 2 and I 3 I T = V S /R T = 1.5/128 = 11.7 ma I 2 I 3 I 2 = I 3 = (50/100) 11.7 = 5.85 ma 6

Loading Effect of a Voltmeter Being in parallel, the internal resistance of the voltmeter will have a loading effect on the circuit that is being measured V R L 7

Loading Effect of a Voltmeter The higher the resistance across which a voltage is measured, the more the loading effect Example: Calculate the voltage across R 2, if the meter resistance (R M ) is 10 MΩ R 2M = (100 10M)/(10.0001M) = 99.999 Ω V R2 = 15 (99.999)/(50+99.999) = 9.99997 V (a) 9.99997V 9.67742V R 2M = (1M 10M)/(11M) = 909.09 kω V R2 = 15 (909.09)/(500+909.09) = 9.67742 V (b) 8 (a) (b)

9 Balanced Wheatstone Bridge The Wheatstone bridge is in the balanced bridge condition when the output voltage between terminals A and B is equal to zero 4 4 2 2 3 3 1 1 R I R I R I R I = 4 2 3 1 R R R R = = 4 2 3 1 R R R R

Balanced Wheatstone Bridge Example: Determine the value of R X in the balanced bridge 150Ω 1200Ω 100Ω R X = R V (R 2 /R 4 ) = (1200)(150/100) = 1800 Ω 10

Unbalanced Wheatstone Bridge The unbalanced bridge, when V OUT is not equal to zero, is used to measure some transducer quantities, such as strain, temperature, or pressure 900 Ω @ 50 C T 50 C = 0V = 0.32V ΔV OUT = 12((1/1.9)-(1/2)) = 0.32 V 11

比色計 比色計 (Colorimeters) 為臨床常見的檢驗設備, 其原理是利用光線對樣品的穿透度或吸收度來作為檢驗分析的參考 + - V in V out = V in R PD1 RPD2 + R PD2 R2 R + R 1 2 12

血液細胞計數儀 由於血球細胞的導電性遠比離子溶液要來的低, 因此當血球細胞通過兩片極板中間時, 極板間的阻抗值就會增加 白血球體積最大, 阻抗值增加最多, 紅血球次之, 血小板最小 (Coulter counter method) PBS: Phosphate Buffered Saline 13

應變計 應變計 (Strain Gauge) 為一種電阻性的換能器, 常作為力學訊號與電訊號之間轉換的工具 ( R + ΔR) ( R ΔR) ( R ΔR) ( R + ΔR) ( R ΔR) L ΔL = ρ A + ΔA L + ΔL + = ρ A ΔA ( R ΔR) 14

應變計 下圖之應變計, 受上下或左右外力時, 皆有兩個電阻元件受壓縮及拉伸應變 這樣的設計可以增加輸出電壓, 提高電路的靈敏度 ( R + ΔR) ( R ΔR) ( R ΔR) + - ( R + ΔR) V out = V = V ΔR R R + ΔR R ΔR ( ) ( ) ( ) ( ) R + ΔR + R ΔR R ΔR + R + ΔR 15

應變計 Example: 如下圖, 應變計輸入電壓為 12V, 而每一個電阻元件為 300Ω 若將此應變計黏貼於股骨頭表面, 且股骨受外力變形, 使得應變計上的每一個電阻改變量為 20Ω, 則輸出電壓為多少伏特? + - V out = V ΔR R 20 = 12 = 0. 8 300 V 16

Thevenin s Theorem Thevenin s theorem provides a method for simplifying a circuit to a standard equivalent form The Thevenin equivalent voltage (V TH ) is the open circuit (no-load) voltage between two terminals in a circuit The Thevenin equivalent resistance (R TH ) is the total resistance appearing between two terminals in a given circuit with all sources replaced by their internal resistances 17

Thevenin Equivalent of a Circuit Example: Simplify the circuit by Thevenin theorem V TH = 10((69)/(100+69)) = 4.08 V R TH = 100+(100)(69)/(100+69) = 141 Ω 18

Thevenin Equivalent of a Circuit Example: Simplify the circuit by Thevenin theorem 19

Thevenin Equivalent of a Circuit 1. Open the two terminals (remove any load) between which you want to find the Thevenin equivalent circuit 20

Thevenin Equivalent of a Circuit 2. Determine the voltage (V TH ) across the two open terminals V TH V TH = (5)(47/(11+47+33)) = 2.58 V 21

Thevenin Equivalent of a Circuit 3. Determine the resistance (R TH ) between the two open terminals with all sources replaced with their internal resistances (short voltage sources and open current sources) R TH R TH = (44)(47)/(44+47) = 22.7 Ω 22

Thevenin Equivalent of a Circuit 4. Connect V TH and R TH in series to produce the complete Thevenin equivalent for the original circuit 22.7 Ω 2.58 V 23

Thevenin Equivalent of a Circuit 5. Place the load resistor removed in Step 1 across the terminals of the Thevenin equivalent circuit. The load current and load voltage can now be calculated using only Ohm s law. They have the same value as the load current and load voltage in the original circuit 22.7 Ω 2.58 V 24

Maximum Power Transfer Maximum power is transferred from a source to a load when the load resistance is equal to the internal source resistance 75 Ω 10 V 25

Maximum Power Transfer A typical application of the maximum power transfer theorem is in audio systems, where the speaker resistance must be matched to the audio power amplifier in order to obtain maximum output 26

Superposition Theorem The superposition theorem is a way to determine currents and voltages in a circuit that has multiple sources by considering one source at a time Example: Find the current through R 2 and the voltage across 27

Applying Superposition Theorem 1. Take one voltage (or current) source at a time and replace the remaining voltage sources with shorts (and remaining current sources with opens) 2. Determine the particular current or voltage that you want I 2(S1) R T(S1) = 100 + 50 = 150 Ω I T(S1) = 10/(150) = 66.7 ma I 2(S1) = 66.7/2 = 33.3 ma 28

Applying Superposition Theorem 3. Take the next source in the circuit and repeat Steps 1 and 2 for each source I 2(S2) R T(S2) = 100 + 50 = 150 Ω I T(S2) = 5/(150) = 33.3 ma I 2(S2) = 33.3/2 = 16.7 ma 29

Applying Superposition Theorem 4. To find the actual current in a given branch (algebraically sum the currents due to each individual source) I 2 = I 2(S1) + I 2(S2) = 33.3 + 16.7 = 50 ma I 2(S1) I 2(S2) V R2 = I 2 R 2 = (50mA)(100Ω) = 5 V 30

Summary To determine total resistance in a series-parallel circuit, identify the series and parallel relationships, and then apply the formulas for series resistance and parallel resistance To find the total current, apply Ohm s law and divide the total voltage by the total resistance To determine branch currents, apply the currentdivider formula, KCL, or Ohm s law To determine voltage drops, use the voltagedivider formula, KVL, or Ohm s law 31

Summary A load resistor should be large compared to the resistance across which it is connected, in order that the loading effect may be minimized A bridge is balanced when the output voltage is zero, balanced Wheatstone bridge can be used to measure an unknown resistance An unbalanced Wheatstone bridge can be used to measure physical quantities using transducers 32

Summary Any two-terminal resistive circuit, no matter how complex, can be replaced by its Thevenin equivalent, made up of an equivalent resistance (R TH ) in series with an equivalent voltage source (V TH ) The maximum power transfer theorem states that the maximum power is transferred from a source to a load when Load Resistance equals Source Resistance The superposition theorem is a way to determine currents and voltages in a circuit that has multiple sources by considering one source at a time 33