“inverse” voltage divider - Electrical and Computer Engineering

ECE 3183 – EE Systems Chapter 2 – Part A Parallel, Series and General Resistive Circuits

Chapter 2 Resistive Circuits 1. Solve circuits (i.e., find currents and voltages of interest) by combining resistances in series and parallel. 2. Apply the voltage-division and current-division principles. 3. Solve circuits by the node-voltage technique. 4. Solve circuits by the mesh-current technique. 5. Find Thévenin and Norton equivalents. 6. Apply the superposition principle. ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-2

FIRST GENERALIZATION: MULTIPLE SOURCES  v R1 

 v2  + -

 v5

 v3 

i(t) + -



+ -

R1

+ -

 v1



R2

Voltage sources in series can be algebraically added to form an equivalent source.

vR2 

 + -

KVL

 v4 

We select the reference direction to move along the path. Voltage drops are subtracted from rises ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-3

FIRST GENERALIZATION: MULTIPLE SOURCES

veq = ?

R1

veq ECE 3183 – Chapter 2 – Part A

+ -

R2

CHAPTER 2 A-4

SECOND GENERALIZATION: MULTIPLE RESISTORS

APPLY KVL TO THIS LOOP

v R  Ri i  i

VOLTAGE DIVISION FOR MULTIPLE RESISTORS ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-5

SECOND GENERALIZATION: MULTIPLE RESISTORS FIND I ,Vbd , P (30k )

APPLY KVL TO THIS LOOP

LOOP FOR Vbd

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-6

SECOND GENERALIZATION: MULTIPLE RESISTORS

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-7

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-8

THE CONCEPT OF EQUIVALENT CIRCUITS THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT HERE WITH A VERY SIMPLE VOLTAGE DIVIDER

i vS

R1

i vS

+ -

R2

i

R1  R2

+ -

vS R1  R2

AS FAR AS THE CURRENT IS CONCERNED BOTH CIRCUITS ARE EQUIVALENT. THE ONE ON THE RIGHT HAS ONLY ONE RESISTOR SERIES COMBINATION OF RESISTORS

R1

ECE 3183 – Chapter 2 – Part A

R2



R1  R2

CHAPTER 2 A-9

THE DIFFERENCE BETWEEN ELECTRIC CONNECTION AND PHYSICAL LAYOUT

SOMETIMES, FOR PRACTICAL CONSTRUCTION REASONS, COMPONENTS THAT ARE ELECTRICALLY CONNECTED MAY BE PHYSICALLY FAR APART

IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-10

SUMMARY OF BASIC VOLTAGE DIVIDER

VR1 = ? VR2 = ?

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-11

SUMMARY OF BASIC VOLTAGE DIVIDER

EXAMPLE: VS  9V , R1  90k, R2  30k

VOLUME CONTROL?

R1  15k 

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-12

SUMMARY OF BASIC VOLTAGE DIVIDER

A “PRACTICAL” POWER APPLICATION

HOW CAN ONE REDUCE THE LOSSES?

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-13

THE “INVERSE” VOLTAGE DIVIDER

R1 

VS

+ -

R2

VO 

VOLTAGE DIVIDER

VO 

R2 VS R1  R2

ECE 3183 – Chapter 2 – Part A

"INVERSE" DIVIDER

VS 

R1  R2 VO R2

CHAPTER 2 A-14

THE “INVERSE” VOLTAGE DIVIDER

COMPUTE VS

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-15

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-16

Current Division R2 v i1   itotal R1 R1  R2

R1 v i2   itotal R2 R1  R2

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-17

FIND Vx, i3

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-18

Find i1, i2, and i3

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-19

SERIES AND PARALLEL RESISTOR COMBINATIONS UP TO NOW WE HAVE STUDIED CIRCUITS THAT CAN BE ANALYZED WITH ONE APPLICATION OF KVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR)

WE HAVE ALSO SEEN THAT IN SOME SITUATIONS IT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT

NOW WE EXAMINE SOME MORE COMPLEX CIRCUITS WHERE WE CAN SIMPLIFY THE ANALYSIS USING THE TECHNIQUE OF COMBINING RESISTORS… … PLUS THE USE OF OHM’S LAW

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-20

SERIES AND PARALLEL RESISTOR COMBINATIONS

SERIES COMBINATIONS

1 1 1 1    ... Gs G1 G2 GN

PARALLEL COMBINATION

G p  G1  G2  ... GN

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-21

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-22

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-23

FIRST WE PRACTICE COMBINING RESISTORS

FIND RAB

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-24

FIRST WE PRACTICE COMBINING RESISTORS (10K,2K)SERIES = 12K

3k SERIES 6k||3k

(12K||6K) = 4K (4K,2K)SERIES = 6K

We need to re-draw!

5k 12k

3k RAB = 5K ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-25

EXAMPLES COMBINATION SERIES-PARALLEL

9k AN EXAMPLE WITHOUT REDRAWING

12 k

6k || (4k  2k )

12k || 12k  6k 3k || 6k  2k 18k || 9k  6k

RAB  3k

6k  6k  10k

ECE 3183 – Chapter 2 – Part A

RAB  22k

CHAPTER 2 A-26

EXAMPLES COMBINATION SERIES-PARALLEL

RESISTORS ARE IN SERIES IF THEY CARRY EXACTLY THE SAME CURRENT (SHARE ONE COMMON NODE)

RESISTORS ARE IN PARALLEL IF THEY HAVE THE SAME VOLTAGE ACROSS THEM AND ARE CONNECTED EXACTLY BETWEEN THE SAME TWO NODES

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-27

Strategy for analyzing circuits with series and parallel combinations of resistors: • Systematically reduce the resistive network so that the resistance seen by the source is represented by a single resistor. • Determine the source current for a voltage source or the source voltage for a current source. • Expand the network, apply Ohm’s law, KVL, KCL, voltage division, and current division to determine all currents and voltages in the network

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-28

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-29

Circuit analysis example Find Io in the circuit shown. 6 k 6 k 3 k 12 V

2 k

4 k Io

ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-30

FIND VO

60k

V1 6V

FIND VS  2V

30k || 60k  20k

STRATEGY : FIND V1 USE VOLTAGE DIVIDER

9V

V1  60k * 0.1mA

0.15mA  6V 

0.05mA

I1 

THIS IS AN INVERSE PROBLEM WHAT CAN BE COMPUTED?

VS  20k * 0.15mA  6V

20k  + -

20k

V1 

12V



20k (12)  6V 20k  20k

VOLTAGE DIVIDER 20k VO  V1 20k  40k ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-31

6V 120k

Circuits with dependent sources • When writing the KVL and/or KCL equations for the network, treat the dependent source as though it were an independent source. • Write the equations that specify the relationship of the dependent source to the controlling parameters. • Solve the equations for the unknowns. Be sure that the number of linearly independent equations matches the number of the unknowns. ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-32

KCL TO THIS NODE. THE DEPENDENT SOURCE IS JUST ANOTHER SOURCE

FIND VO

A PLAN: IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER. TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION ALGEBRAICALLY, THERE ARE TWO UNKNOWNS AND JUST ONE EQUATION SUBSTITUTION OF I_0 YIELDS

VOLTAGE DIVIDER

* / 6k  5VS  60 VO 

ECE 3183 – Chapter 2 – Part A

4k 2 VS  (12)V 4k  2k 3

CHAPTER 2 A-33

Problem solving strategy: Circuit with dependent sources For the network shown, what is the resulting ratio , Vo /Vs ? RS

Ro +

VS

ECE 3183 – Chapter 2 – Part A

Rin

Vin _

+

 Vin

RL

Vo _

CHAPTER 2 A-34

Problem solving strategy: Circuit with dependent sources For the network shown, what is the resulting ratio , Vo /Vs ? RS

Ro +

VS

 Rin Vin    RS  Rin

  VS 

ECE 3183 – Chapter 2 – Part A

Rin

Vin _

+

 Vin

RL

Vo _

(voltage divider for resistors in series)

CHAPTER 2 A-35

Problem solving strategy: Circuit with dependent sources For the network shown, what is the resulting ratio , Vo /Vs ? RS

Ro +

VS

 Rin Vin    RS  Rin

  VS 

Rin

Vin _

 Vin

RL

Vo _

(voltage divider for resistors in series)

 R   R  R  Vo   L    Vin     in  L  VS  Ro  RL   RS  Rin  Ro  RL 

ECE 3183 – Chapter 2 – Part A

+

(voltage divider for resistors in series)

CHAPTER 2 A-36

Problem solving strategy: Circuit with dependent sources For the network shown, what is the resulting ratio , Vo /Vs ? RS

Ro +

VS

 Rin Vin    RS  Rin

  VS 

Rin

Vin _

+

 Vin

RL

Vo _

(voltage divider for resistors in series)

 R   R  R  Vo   L    Vin     in  L  VS  Ro  RL   RS  Rin  Ro  RL 

(voltage divider for resistors in series)

 Rin  RL  Vo     VS R  R R  R in  o L   S ECE 3183 – Chapter 2 – Part A

CHAPTER 2 A-37