Voltage Dividers and Current Dividers | Electronics PPT | web4study

# Voltage Dividers and Current Dividers | Electronics PPT

## Voltage Dividers and Current Dividers

Topics Covered in this ppt

Series Voltage Dividers

Current Dividers with Two Parallel Resistances

Current Division by Parallel Conductances

Series Voltage Divider with Parallel Load Current

Design of a Loaded Voltage Divider

### 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. Fig. 7-2b: Series voltage divider with voltage taps.

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.
• 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.

### 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,
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.
• Current DividerI1 = 4 Ω/(2 Ω + 4 Ω) × 30A = 20AI2= 2 Ω /(2 Ω + 4 Ω) × 30A = 10A

### 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

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 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.

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

GT = G1 + G2 + G3

= 0.1 + 0.5 + 0.2

GT = 0.8 S

I1 = 0.1/0.8 x 40 mA = 5 mA

I2 = 0.5/0.8 x 40 mA = 25 mA

I3 = 0.2/0.8 x 40 mA = 10 mA

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

### 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) R1 and R2 in series without any branch current. (b) A reduced voltage across R2 and its parallel load RL. (c) An equivalent circuit of the loaded voltage divider.

• V1 = 40/60 x 60 V = 40 V
• V2 = 20/60 x 60 V = 20 V
• V1 + V2 = VT = 60 V (Applied Voltage)
• 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.

### Design of a Loaded Voltage Divider Fig. 7-7: the Voltage divider for different voltages and currents from the source VT.

• 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
• 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.