Chapter 7 Voltage Dividers and Current Dividers Topics

Chapter 7 Voltage Dividers and Current Dividers Topics Covered in Chapter 7 7 -1: Series Voltage Dividers 7 -2: Current Dividers with Two Parallel Resistances 7 -3: Current Division by Parallel Conductances 7 -4: Series Voltage Divider with Parallel Load Current 7 -5: Design of a Loaded Voltage Divider © 2007 The Mc. Graw-Hill Companies, Inc. All rights reserved.

7 -1: Series Voltage Dividers § VT is divided into IR voltage drops that are proportional to the series resistance values. § Each resistance provides an IR voltage drop equal to its proportional part of the applied voltage: VR = (R/RT) × VT § This formula can be used for any number of series resistances because of the direct proportion between each voltage drop V and its resistance R. § The largest series R has the largest IR voltage drop. Mc. Graw-Hill © 2007 The Mc. Graw-Hill Companies, Inc. All rights reserved.

7 -1: Series Voltage Dividers § The Largest Series R Has the Most V. V 1 = R 1 RT × VT 1 k. W × 1000 V = 1000 k. W R 2 × VT RT 999 k. W = × 1000 V = 999 V 1000 k. W V 2 = KVL check: 1 V + 999 V = 1000 V Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 7 -2 a: Example of a very small R 1 in series with a large R 2; V 2 is almost equal to the whole VT.

7 -1: Series Voltage Dividers § Voltage Taps in a Series Voltage Divider § Different voltages are available at voltage taps A, B, and C. § The voltage at each tap point is measured with respect to ground. § Ground is the reference point. Fig. 7 -2 b: Series voltage divider with voltage taps. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

7 -1: Series Voltage Dividers §Voltage Taps in a Series Voltage Divider § Note: VAG is the sum of the voltage across R 2, R 3, and R 4. § VAG is one-half of the applied voltage VT, because R 2+R 3+ R 4 = 50% of RT. VAG = 12 V 2. 5 k. W VBG = × 24 V = 3 V 20 k. W 1 k. W VCG = × 24 V = 1. 2 V 20 k. W Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

7 -2: Current Dividers with Two Parallel Resistances § IT is divided into individual branch currents. § Each branch current is inversely proportional to the branch resistance value. § For two resistors, R 1 and R 2, in parallel: § Note that this formula can only be used for two branch resistances. § The largest current flows in the branch that has the smallest R.

7 -2: Current Dividers with Two Parallel Resistances § Current Divider I 1 = 4 Ω/(2 Ω + 4 Ω) × 30 A = 20 A I 2= 2 Ω /(2 Ω + 4 Ω) × 30 A = 10 A Fig. 7 -3: Current divider with two branch resistances. Each branch I is inversely proportional to its R. The smaller R has more I. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

7 -3: Current Division by Parallel Conductances § For any number of parallel branches, IT is divided into currents that are proportional to the conductance of the branches. § For a branch having conductance G: I= G GT × IT

7 -3: Current Division by Parallel Conductances G 1 = 1/R 1 = 1/10 Ω = 0. 1 S G 2 = 1/R 2 = 1/2 Ω = 0. 5 S G 3 = 1/R 3 = 1/5 Ω = 0. 2 S Fig. 7 -5: Current divider with branch conductances G 1, G 2, and G 3, each equal to 1/R. Note that S is the siemens unit for conductance. With conductance values, each branch I is directly proportional to the branch G. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

7 -3: Current Division by Parallel Conductances The Siemens (S) unit is the reciprocal of the ohm (Ω) GT = G 1 + G 2 + G 3 = 0. 1 + 0. 5 + 0. 2 GT = 0. 8 S I 1 = 0. 1/0. 8 x 40 m. A = 5 m. A I 2 = 0. 5/0. 8 x 40 m. A = 25 m. A I 3 = 0. 2/0. 8 x 40 m. A = 10 m. A KCL check: 5 m. A + 25 m. A + 10 m. A = 40 m. A = IT

7 -4: Series Voltage Divider with Parallel Load Current § Voltage dividers are often used to tap off part of the applied voltage for a load that needs less than the total voltage. Fig. 7 -6: Effect of a parallel load in part of a series voltage divider. (a) R 1 and R 2 in series without any branch current. (b) Reduced voltage across R 2 and its parallel load RL. (c) Equivalent circuit of the loaded voltage divider. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

7 -4: Series Voltage Divider with Parallel Load Current § V 1 = 40/60 x 60 V = 40 V § V 2 = 20/60 x 60 V = 20 V § V 1 + V 2 = VT = 60 V (Applied Voltage) Fig 7 -6

7 -4: Series Voltage Divider with Parallel Load Current § The current that passes through all the resistances in the voltage divider is called the bleeder current, IB. § Resistance RL has just its load current IL. § Resistance R 2 has only the bleeder current IB. § Resistance R 1 has both IL and IB. Fig. 7 -6

7 -5: Design of a Loaded Voltage Divider Fig. 7 -7: Voltage divider for different voltages and currents from the source VT. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

7 -5: Design of a Loaded Voltage Divider § I 1 through R 1 equals 30 m. A § I 2 through R 2 is 36 + 30 = 66 m. A § I 3 through R 3 is 54 + 36 + 30 = 120 m. A § V 1 is 18 V to ground § V 2 is 40 − 18 = 22 V § V 3 is 100 V (Point D) − 40 = 60 V

7 -5: Design of a Loaded Voltage Divider § R 1 = V 1/I 1 = 18 V/30 m. A = 0. 6 kΩ = 600 Ω § R 2 = V 2/I 2 = 22 V/66 m. A = 0. 333 kΩ = 333 Ω § R 3 = V 3/I 3 = 60 V/120 m. A = 0. 5 kΩ = 500 Ω NOTE: When these values are used for R 1, R 2, and R 3 and connected in a voltage divider across a source of 100 V, each load will have the specified voltage at its rated current.