7 Voltage Dividers and Current Dividers

Transcription

1 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 McGraw-Hill Companies, Inc. All rights reserved.

2 7-1: Series Voltage Dividers V T 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: V R = (R/R T ) V T 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. McGraw-Hill 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 7-1: Series Voltage Dividers The Largest Series R Has the Most V. V 1 = = R 1 R T V T 1 kω 1000 kω 1000 V = 1 V V 2 = R 2 R T V T 999 kω = 1000 V = 999 V 1000 kω KVL check: 1 V V = 1000 V Fig. 7-2a: Example of a very small R 1 in series with a large R 2 ; V 2 is almost equal to the whole V T. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

4 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. 7-2b: Series voltage divider with voltage taps. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

5 7-1: Series Voltage Dividers Voltage Taps in a Series Voltage Divider Note: V AG is the sum of the voltage across R 2, R 3, and R 4. V AG is one-half of the applied voltage V T, because R 2 +R 3 + R 4 = 50% of R T. V AG = 12 V V CG = 1 kω 20 kω V BG = 2.5 kω 20 kω 24 V = 1.2 V 24 V = 3 V Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

6 7-2: Current Dividers with Two Parallel Resistances I T 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. I = R R R 1 2 I T

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

8 7-3: Current Division by Parallel Conductances For any number of parallel branches, I T is divided into currents that are proportional to the conductance of the branches. For a branch having conductance G: I = G G T I T

9 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 McGraw-Hill Companies, Inc. Permission required for reproduction or display.

10 7-3: Current Division by Parallel Conductances The Siemens (S) unit is the reciprocal of the ohm (Ω) G T = G 1 + G 2 + G 3 = G T = 0.8 S I 1 = 0.1/0.8 x 40 ma = 5 ma I 2 = 0.5/0.8 x 40 ma = 25 ma I 3 = 0.2/0.8 x 40 ma = 10 ma KCL check: 5 ma + 25 ma + 10 ma = 40 ma = I T

11 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 R L. (c) Equivalent circuit of the loaded voltage divider. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

12 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 = V T = 60 V (Applied Voltage) Fig 7-6

13 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, I B. Resistance R L has just its load current I L. Resistance R 2 has only the bleeder current I B. Resistance R 1 has both I L and I B. Fig. 7-6

14 7-5: Design of a Loaded Voltage Divider Fig. 7-7: Voltage divider for different voltages and currents from the source V T. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

15 7-5: Design of a Loaded Voltage Divider I 1 through R 1 equals 30 ma I 2 through R 2 is = 66 ma I 3 through R 3 is = 120 ma V 1 is 18 V to ground V 2 is = 22 V V 3 is 100 V (Point D) 40 = 60 V

16 7-5: Design of a Loaded Voltage Divider R 1 = V 1 /I 1 = 18 V/30 ma = 0.6 kω = 600 Ω R 2 = V 2 /I 2 = 22 V/66 ma = kω = 333 Ω R 3 = V 3 /I 3 = 60 V/120 ma = 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.