Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING LECTURE 2

Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING

LECTURE 2

Circuit Analysis Techniques

• Voltage Division Principle

• Current Division Principle

• Nodal Analysis with KCL

• Mesh Analysis with KVL

• Superposition

• Thévenin Equivalence• Norton’s Equivalence

Circuit Analysis using Series/Parallel Equivalents1. Begin by locating a combination of resistances that are in series or

parallel. Often the place to start is farthest from the source.

2. Redraw the circuit with the equivalent resistance for the combination found in step 1.

3. Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with a single source and a single resistance.

4. Solve for the currents and voltages in the final equivalent circuit.

Working Backward

Voltage Division

total321

Application of the Voltage-Division

Principle

156000200010001000

total4321

Current Division

total21

206030

32eq RR

A10152010

Application of the Current-Division Principle

•Voltage division •Voltage division and •current division

•Current division

Although they are veryimportant concepts,series/parallel equivalents andthe current/voltage divisionprinciples are not sufficient to solve all circuits.

Node Voltage (Nodal) Analysis

Definitions of node and Supernode

(-1.4)- 5

At node 1

At node 2

Obtain values for the unknown voltages across the elements in the circuit below.

Writing KCL Equations in Terms of the

Node Voltages for Figure 2.16

node 1

node 2

node 3

node 1

node 2

node 3

No. of unknown: v1, v2, v3

No. of linear equation : 3

Setting up nodal equation withKCL at Node 1, Node 2, Node 3

No. of unknown: v1, v2, v3

No. of linear equation : 3

Setting up nodal equation withKCL at Node 1, Node 2, Node 3

Problem with node 3, it is rather hard to set the nodal equation at node 3 but still solvable.

Problem with node 3, it is rather hard to set the nodal equation at node 3, but still solvable. Why? As there is no way to determinethe current through the voltage source, but v3=Vs

Same as before.

May Not Be That Simple

Circuits with Voltage SourcesWe obtain dependent equations if we use

all of the nodes in a network to write KCL equations.Any branch with a voltage source:

• define SUPERNODE, sum all current either in or out at the supernode with KCL

• use KVL to set up dependent equation involving the voltage source.

(a) The circuit of Example 4.2 with a 22-V source in place of the 7- resistor. (b) Expanded view of the region defined as a supernode; KCL requires that all currents flowing into the region must sum to zero, or we would pile up or run out of electrons.

3 38 3121 vvvv

231312

vvvvvv

At node 1:

At the “supernode:”

0)15()15(

For supernode A, EXCLUDE THE SOURCE

Why? As the current via the 10V source is equal to the current via R4 plus the current via R3

Summing all the current outfrom the supernode A

For supernode B, EXCLUDE THE SOURCE

Summing all the current into the supernode B

0)15()15(

-v1 -10 + v2 = 0

Any branch with a voltage source:• define supernode, sum all current either in or out at the supernode with KCL• use KVL to set up dependent equation involving the voltage source.

010 21 vv

Node-Voltage Analysis with a Dependent Source•First, we write KCL equations at each node, including the current of the controlled source just as if it were an ordinary current source.

•Next, we find an expression for the controlling variable ix in terms of the node voltages.

xs iiR

Substitution yields

Node-Voltage Analysis1. Select a reference node and assign variables for the unknown node voltages. If the reference node is chosen at one end of an independent voltage source, one node voltage is known at the start, and fewer need to be computed.

2. Write network equations. First, use KCL to write current equations for nodes and supernodes. Write as many current equations as you can without using all of the nodes. Then if you do not have enough equations because of voltage sources connected between nodes, use KVL to write additional equations.3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the node voltages. Substitute into the network equations, and obtain equations having only the node voltages as unknowns.4. Put the equations into standard form and solve for the node voltages.

5. Use the values found for the node voltages to calculate any other currents or voltages of interest.

Step 1.Reference node

Step 1. v1

Step 1 v2

Step 2.

vvvvvv

055210

331221

refStep 1

Step 1 Step 1

supernode

node 1Step 2

supernode

Step 3

Mesh Current Analysis

Definition of a loop

Definition of a mesh

Choosing the Mesh Currents

When several mesh currents flow through one element, we consider the current in that element to be the algebraic sum of the mesh currents.

Writing Equations to Solve for Mesh

CurrentsIf a network contains only resistors and independent voltage sources, we can write the required equations by following each current around its mesh and applying KVL.

For mesh 1, we have

For mesh 2, we obtain

024123 BviRiiR

For mesh 3, we have

031132 BviRiiR

0213312 AviiRiiR

Determine the two mesh currents, i1 and i2, in the circuit below.

For the left-hand mesh,

-42 + 6 i1 + 3 ( i1 - i2 ) = 0

For the right-hand mesh,

3 ( i2 - i1 ) + 4 i2 - 10 = 0

Solving, we find that i1 = 6 A and i2 = 4 A.

(The current flowing downward through the 3- resistor is therefore i1 - i2 = 2 A. )

Mesh Currents in Circuits Containing

Current Sources*A common mistake is to assume the voltages across current sources are zero. Therefore, loop equation cannot be set up at mesh one due to the voltage across the current source is unknown

0105)(10 212 iii

Anyway, the problem isstill solvable.

As the current source common to two mesh, combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined.

01042 32311 iiiii

Mesh 3: 0243 13233 iiiii

512 ii

It is the supermesh.

Three linear equations and three unknown

Find the three mesh currents in the circuit below.

Creating a “supermesh” from meshes 1 and 3:

-7 + 1 ( i1 - i2 ) + 3 ( i3 - i2 ) + 1 i3 = 0 [1]

Around mesh 2:

1 ( i2 - i1 ) + 2 i2 + 3 ( i2 - i3 ) = 0 [2]

Rearranging,

i1 - 4 i2 + 4 i3 = 7 [1]

-i1 + 6 i2 - 3 i3 = 0 [2]

i1 - i3 = 7 [3]

Solving,

i1 = 9 A, i2 = 2.5 A, and i3 = 2 A.

Finally, we relate the currents in meshes 1 and 3:

i1 - i3 = 7 [3]

026420 221 iii

supermesh of mesh1 and mesh2

current source

branch current

026420 221 iii

Three equations and three unknown.

Mesh-Current Analysis1. If necessary, redraw the network without crossing conductors or elements. Then define the mesh currents flowing around each of the open areas defined by the network. For consistency, we usually select a clockwise direction for each of the mesh currents, but this is not a requirement.

2. Write network equations, stopping after the number of equations is equal to the number of mesh currents. First, use KVL to write voltage equations for meshes that do not contain current sources. Next, if any current sources are present, write expressions for their currents in terms of the mesh currents. Finally, if a current source is common to two meshes, write a KVL equation for the supermesh.3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the mesh currents. Substitute into the network equations, and obtain equations having only the mesh currents as unknowns.4. Put the equations into standard form. Solve for the mesh currents by use of determinants or other means.

5. Use the values found for the mesh currents to calculate any other currents or voltages of interest.

Superposition• Superposition Theorem – the response of a

circuit to more than one source can be determined by analyzing the circuit’s response to each source (alone) and then combining the results

Insert Figure 7.2

Superposition

Insert Figure 7.3

Superposition• Analyze Separately, then Combine Results

Use superposition to find the current ix.

Current source is zero – open circuit as I = 0 and solve iXv

Voltage source is zero – short circuit as V= 0 and solve iXv

XcXvX iii

Use superposition to find the current ix.

The controlled voltage source is included in all cases asit is controlled by the current ix.

Voltage and Current Sources

Insert Figure 7.7

Insert Figure 7.8

Insert Figure 7.9

Source Transformation

RP RLIS

Under what condition, the voltage and current of the load is the same whenoperating at the two practical sources?For voltage source

For current source

We have,

SSSP R

ViRR ,

• Equivalent Voltage and Current Sources – for every voltage source, there exists an equivalent current source, and vice versa

Thevenin’s Theorem• Thevenin’s Theorem – any resistive circuit

or network, no matter how complex, can be represented as a voltage source in series with a source resistance

Thevenin’s Theorem• Thevenin Voltage (VTH) – the voltage present at the

output terminals of the circuit when the load is removed

Insert Figure 7.18

Thevenin’s Theorem• Thevenin Resistance (RTH) – the resistance

measured across the output terminals with the load removed

Thévenin Equivalent Circuits

Finding the Thévenin Resistance Directly

When zeroing a voltage source, it becomes a short circuit. When zeroing a current source, it becomes an open circuit.

We can find the Thévenin resistance by zeroing the sources in the original network and then computing the resistance between the terminals.

Computation of Thévenin resistance

Equivalence of open-circuit and Thévenin voltage

A circuit and its Thévenin equivalent

Superposition

As the voltage source does not contribute any output voltage,Only the current source has the effect.

Determine the Thévenin and Norton Equivalents of Network A in (a).

Source transformation

Find the Thévenin equivalent of the circuit shown in (a).

As i = -1, therefore, the controlled voltage source is -1.5V.Use nodal analysis at node v,

6.0,123

vv Thus, Rth =v/I = 0.6/1 = 0.6 ohms

Applications of Thevenin’s Theorem

• Load Voltage Ranges – Thevenin’s theorem is most commonly used to predict the change in load voltage that will result from a change in load resistance

• Maximum Power Transfer– Maximum power transfer from a circuit to a

variable load occurs when the load resistance equals the source resistance

– For a series-parallel circuit, maximum power occurs when RL = RTH

• Multiload Circuits

Insert Figure 7.30

Norton’s Theorem• Norton’s Theorem – any resistive circuit or network,

no matter how complex, can be represented as a current source in parallel with a source resistance

Norton’s Theorem• Norton Current (IN) – the current through

the shorted load terminals

Insert Figure 7.35

Computation of Norton current

Norton’s Theorem

• Norton Resistance (RN) – the resistance measured across the open load terminals (measured and calculated exactly like RTH)

Norton’s Theorem• Norton-to-Thevenin and Thevenin-to-Norton

Conversions

Insert Figure 7.39

Step-by-step Thévenin/Norton-Equivalent-

Circuit Analysis

1. Perform two of these: a. Determine the open-circuit voltage Vt = voc.

b. Determine the short-circuit current In = isc.

c. Zero the sources and find the Thévenin resistance Rt looking back into the terminals.

2. Use the equation Vt = Rt In to compute the remaining value.

3. The Thévenin equivalent consists of a voltage source Vt in series with Rt .

4. The Norton equivalent consists of a current source In in parallel with Rt .

Maximum Power Transfer

The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance.

Graphical representation of maximum power transfer

Power transfer between source and load

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING LECTURE 2

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