Voltage Dividers Voltage Dividers and Current Dividersand Current Dividers

Topics Covered in Chapter 77-1: Series Voltage Dividers

7-2: Current Dividers with Two Parallel Resistances7-3: Current Division by Parallel Conductances

7-4: Series Voltage Divider with Parallel Load Current7-5: Design of a Loaded Voltage Divider

Chapter 7

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7-1: Series Voltage Dividers7-1: 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.

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7-1: Series Voltage Dividers7-1: Series Voltage Dividers The Largest Series R Has the Most V.

V2 =R2

RT× VT

999 k1000 k

= × 1000 V = 999 V

V1 =R1

RT

× VT

=1 k

1000 k× 1000 V = 1 V

KVL check: 1 V + 999 V = 1000 VFig. 7-2a: Example of a very small R1 in series with a large R2; V2 is almost equal to the whole VT.

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7-1: Series Voltage Dividers7-1: 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.

7-1: Series Voltage Dividers7-1: Series Voltage Dividers

Note: VAG is the sum of the voltage across R2, R3, and R4.

VAG is one-half of the applied voltage VT, because R2+R3+

R4 = 50% of RT.

VCG =1 k

20 k× 24 V = 1.2 V

VBG =2.5 k

20 k× 24 V = 3 VVAG = 12 V

Voltage Taps in a Series Voltage Divider

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7-2: Current Dividers with 7-2: Current Dividers with Two Parallel ResistancesTwo Parallel Resistances

IT is divided into individual branch currents. Each branch current is inversely proportional to the

branch resistance value. For two resistors,

R1 and R2, 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.

IR

R RIT1

2

1 2

7-2: Current Dividers with 7-2: Current Dividers with Two Parallel ResistancesTwo Parallel Resistances

Current Divider

Fig. 7-3: Current divider with two branch resistances. Each branch I is inversely proportional to its R. The smaller R has more I.

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I1 = 4 Ω/(2 Ω + 4 Ω) × 30A = 20A

I2= 2 Ω /(2 Ω + 4 Ω) × 30A = 10A

7-3: Current Division by 7-3: Current Division by Parallel ConductancesParallel 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 = × IT

G

GT

7-37-3:: Current Division by Current Division by Parallel ConductancesParallel Conductances

Fig. 7-5: Current divider with branch conductances G1, G2, and G3, 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.

G1 = 1/R1 = 1/10 Ω = 0.1 S

G2 = 1/R2 = 1/2 Ω = 0.5 S

G3 = 1/R3 = 1/5 Ω = 0.2 S

7-3: Current Division by 7-3: Current Division by Parallel ConductancesParallel Conductances

KCL check: 5 mA + 25 mA + 10 mA = 40 mA = IT

The Siemens (S) unit is the reciprocal of the ohm (Ω)

GT = G1 + G2 + G3

= 0.1 + 0.5 + 0.2GT = 0.8 S

I1 = 0.1/0.8 x 40 mA = 5 mAI2 = 0.5/0.8 x 40 mA = 25 mAI3 = 0.2/0.8 x 40 mA = 10 mA

7-4: Series Voltage Divider with 7-4: Series Voltage Divider with Parallel Load CurrentParallel 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) R1 and R2 in series without any branch current. (b) Reduced voltage across R2 and its parallel load RL. (c) Equivalent circuit of the loaded voltage divider.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7-4: Series Voltage Divider with7-4: Series Voltage Divider withParallel Load CurrentParallel Load Current

V1 = 40/60 x 60 V = 40 V

V2 = 20/60 x 60 V = 20 V

V1 + V2 = VT = 60 V

(Applied Voltage)

Fig 7-6

7-4: Series Voltage Divider with 7-4: Series Voltage Divider with Parallel Load CurrentParallel 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 R2 has only the bleeder current IB. Resistance R1 has

both IL and IB.

Fig. 7-6

7-5: Design of a 7-5: Design of a Loaded Voltage DividerLoaded Voltage Divider

Fig. 7-7: Voltage divider for different voltages and currents from the source VT.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7-5: Design of a 7-5: Design of a Loaded Voltage DividerLoaded Voltage Divider

I1 through R1 equals 30 mA I2 through R2 is 36 + 30 = 66 mA I3 through R3 is 54 + 36 + 30 = 120 mA

V1 is 18 V to ground V2 is 40 − 18 = 22 V V3 is 100 V (Point D) − 40 = 60 V

7-5: Design of a7-5: Design of aLoaded Voltage DividerLoaded Voltage Divider

R1 = V1/I1 = 18 V/30 mA = 0.6 kΩ = 600 Ω

R2 = V2/I2 = 22 V/66 mA = 0.333 kΩ = 333 Ω

R3 = V3/I3 = 60 V/120 mA = 0.5 kΩ = 500 Ω

NOTE: When these values are used for R1, R2, and R3 and connected in a voltage divider across a source of 100 V, each load will have the specified voltage at its rated current.