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# Elektronika (6)

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### Elektronika (6)

1. 1. Elektronika<br />AgusSetyo Budi, Dr. M.Sc<br />Sesion #06<br />JurusanFisika<br />FakultasMatematikadanIlmuPengetahuanAlam<br />
2. 2. Series Voltage Dividers<br />Current Dividers With Two Parallel Resistances<br /> Current Dividers By Parallel Conductances<br />Series Voltage Divider with Parallel Load Current<br />Series Voltage Divider with Parallel Load Current<br />Design of a Loaded Voltage Divider<br />07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />2<br />
3. 3. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />3<br />Voltage Dividers and Current Dividers<br />
4. 4. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />4<br />Series Voltage Dividers<br />VTis divided into IR voltage drops that are proportional to the series resistance values.<br />Each resistance provides an IR voltage drop equal to its proportional part of the applied voltage:<br /> VR = (R/RT) × VT<br />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.<br />The largest series R has the largest IR voltage drop.<br />
5. 5. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />5<br />R1<br />× VT<br />V1 =<br />RT<br />1 kW<br />=<br />× 1000 V = 1 V<br />1000 kW<br />R2<br />999 kW<br />V2 =<br />× VT<br />RT<br />1000 kW<br />=<br />× 1000 V = 999 V<br />Series Voltage Dividers<br />The Largest Series R Has the Most V.<br />Fig. 7-2a: Example of a very small R1 in series with a large R2; V2 is almost equal to the whole VT.<br />KVL check: 1 V + 999 V = 1000 V<br />
6. 6. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />6<br />Series Voltage Dividers<br />Voltage Taps in a Series Voltage Divider<br />Different voltages are available at voltage taps A, B, and C.<br />The voltage at each tap point is measured with respect to ground.<br />Ground is the reference point.<br />Fig. 7-2b: Series voltage divider with voltage taps.<br />
7. 7. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />7<br />2.5 kW<br />× 24 V = 3 V<br />VBG =<br />20 kW<br />1 kW<br />× 24 V = 1.2 V<br />VCG =<br />20 kW<br />Series Voltage Dividers<br /><ul><li>Voltage Taps in a Series Voltage Divider</li></ul>Note: VAG is the sum of the voltage across R2, R3, and R4.<br />VAG is one-half of the applied voltage VT, because R2+R3+<br /> R4 = 50% of RT.<br />VAG = 12 V<br />
8. 8. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />8<br />Current Dividers with Two Parallel Resistances<br />ITis divided into individual branch currents.<br />Each branch current is inversely proportional to the branch resistance value.<br />For two resistors,R1 and R2, in parallel:<br />Note that this formula can only be used for two branch resistances.<br />The largest current flows in the branch that has the smallest R.<br />
9. 9. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />9<br />Current Dividers with Two Parallel Resistances<br />Current Divider<br />I1 = 4 Ω/(2 Ω + 4 Ω) × 30A = 20A<br />I2= 2 Ω /(2 Ω + 4 Ω) × 30A = 10A<br />Fig. 7-3: Current divider with two branch resistances. Each branch I is inversely proportional to its R. The smaller R has more I.<br />
10. 10. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />10<br />G<br />I =<br />× IT<br />GT<br />Current Division by Parallel Conductances<br />For any number of parallel branches, IT is divided into currents that are proportional to the conductance of the branches.<br />For a branch having conductance G:<br />
11. 11. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />11<br />Current Division by Parallel Conductances<br />G1 = 1/R1 = 1/10 Ω = 0.1 S<br />G2 = 1/R2 = 1/2 Ω = 0.5 S<br />G3 = 1/R3 = 1/5 Ω = 0.2 S<br />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.<br />
12. 12. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />12<br />Current Division by Parallel Conductances<br />The Siemens (S) unit is the reciprocal of the ohm (Ω)<br />GT = G1 + G2 + G3<br /> = 0.1 + 0.5 + 0.2<br />GT = 0.8 S<br />I1 = 0.1/0.8 x 40 mA = 5 mA<br />I2 = 0.5/0.8 x 40 mA = 25 mA<br />I3 = 0.2/0.8 x 40 mA = 10 mA<br />KCL check: 5 mA + 25 mA + 10 mA = 40 mA = IT<br />
13. 13. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />13<br />Series Voltage Divider with Parallel Load Current<br />Voltage dividers are often used to tap off part of the applied voltage for a load that needs less than the total voltage.<br />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.<br />
14. 14. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />14<br />Series Voltage Divider withParallel Load Current<br />V1 = 40/60 x 60 V = 40 V<br />V2 = 20/60 x 60 V = 20 V<br />V1 + V2 = VT = 60 V<br /> (Applied Voltage)<br />Fig 7-6<br />
15. 15. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />15<br />Series Voltage Divider with Parallel Load Current<br />The current that passes through all the resistances in the voltage divider is called the bleeder current, IB.<br />Resistance RL has just its load current IL.<br />Resistance R2has only the bleeder current IB.<br />Resistance R1has<br /> both ILand IB.<br />Fig. 7-6<br />
16. 16. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />16<br />Design of a Loaded Voltage Divider<br />Fig. 7-7: Voltage divider for different voltages and currents from the source VT.<br />
17. 17. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />17<br />Design of a Loaded Voltage Divider<br />I1 through R1 equals 30 mA<br />I2 through R2 is 36 + 30 = 66 mA<br />I3 through R3 is 54 + 36 + 30 = 120 mA<br />V1 is 18 V to ground<br />V2 is 40 − 18 = 22 V<br />V3 is 100 V (Point D) − 40 = 60 V<br />
18. 18. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />18<br />Design of aLoaded Voltage Divider<br />R1 = V1/I1 = 18 V/30 mA = 0.6 kΩ = 600 Ω<br />R2 = V2/I2 = 22 V/66 mA = 0.333 kΩ = 333 Ω<br />R3 = V3/I3 = 60 V/120 mA = 0.5 kΩ = 500 Ω<br />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.<br />
19. 19. 07/01/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />19<br />TerimaKasih<br />