Maximum Power Transfer

           The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. If the load resistance is lower or higher than the Thevenin/Norton resistance of the source network, its dissipated power will be less than maximum.

            This is essentially what is aimed for in radio transmitter design , where the antenna or transmission line “impedance” is matched to final power amplifier “impedance” for maximum radio frequency power output. Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance.

            The Maximum Power Transfer Theorem states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power.

            The Maximum Power Transfer Theorem does not satisfy the goal of maximum efficiency.

Maximum Power Transfer Example No1.

Where:   RS = 25Ω   RL is variable between 0 – 100Ω   VS = 100v

 

Then by using the following Ohm’s Law equations:

 

We can now complete the following table to determine the current and power in the circuit for different values of load resistance.

Table of Current against Power

RL (Ω) I (amps) P (watts) 0 4.0 0 5 3.3 55 10 2.8 78 15 2.5 93 20 2.2 97

RL (Ω) I (amps) P (watts) 25 2.0 100 30 1.8 97 40 1.5 94 60 1.2 83 100 0.8 64

Using the data from the table above, we can plot a graph of load resistance, RL against power, P for different values of load resistance. Also notice that power is zero for an open-circuit (zero current condition) and also for a short-circuit (zero voltage condition).

Graph of Power against Load Resistance

Norton's Theorem

                Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Just as with Thevenin's Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots).

Contrasting our original example circuit against the Norton equivalent: it looks something like this:

               Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.

               As with Thevenin's Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin's Theorem are the steps used in Norton's Theorem to calculate the Norton source current (INorton) and Norton resistance (RNorton).

              Norton's Theorem is a way to reduce a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load.

Steps to follow for Norton's Theorem:

• (1) Find the Norton source current by removing the load resistor from the original circuit and calculating current through a short (wire) jumping across the open connection points where the load resistor used to be.
• (2) Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for parallel circuits.

Example:

Find R_N, I_N, the current flowing through and Load Voltage across the load resistor in fig (1) by using Norton’s Theorem.

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Solution:-

Step 1

Short the 1.5Ω load resistor as shown in (Fig 2)

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Step 2

Calculate / measure the Short Circuit Current. This is the Norton Current (I_N).

We have shorted the AB terminals to determine the Norton current, I_N. The 6Ω and 3Ω are then in parallel and this parallel combination of 6Ω and 3Ω are then in series with 2Ω.

So the Total Resistance of the circuit to the Source is:-

2Ω + (6Ω || 3Ω) ….. (|| = in parallel with)

R_T = 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)] → I_T= 2Ω + 2Ω = 4Ω

R_T = 4Ω

I_T = V / R_T

I_T = 12V / 4Ω

I_T = 3A..

Now we have to find I_SC = I_N… Apply CDR… (Current Divider Rule)…

I_SC = I_N = 3A x [(6Ω / (3Ω + 6Ω)] = 2A.

I_SC= I_N = 2A

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Step 3

Open Current Sources, Short Voltage Sources and Open Load Resistor. Fig (4)

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Step 4

Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (R_N)

We have Reduced the 12V DC source to zero is equivalent to replace it with a short in step (3), as shown in figure (4)  We can see that 3Ω resistor is in series with a parallel combination of 6Ω resistor and  2Ω resistor. i.e.:

3Ω + (6Ω || 2Ω) ….. (|| = in parallel with)

R_N = 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]

R_N = 3Ω + 1.5Ω

R_N = 4.5Ω

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Step 5

Connect the R_N in Parallel with Current Source I_Nand re-connect the load resistor. This is shown in fig (6) i.e. Norton Equivalent circuit with load resistor.

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Step 6

Now apply the last step i.e. calculate the load current through and Load voltage across load resistor by Ohm’s Law as shown in fig 7.

Load Current through Load Resistor…

I_L = I_N x [R_N / (R_N+ R_L)]

= 2A x (4.5Ω /4.5Ω +1.5kΩ) → = 1.5A

I_L = 1. 5A

And

Load Voltage across Load Resistor…

V_L = I_L x R_L

V_L = 1.5A x 1.5Ω

V_L= 2.25V

Click image to enlarge 

Thevenin's Theorem

                  Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The qualification of “linear” is identical to that found in the Superposition Theorem, where all the underlying equations must be linear (no exponents or roots). If we're dealing with passive components (such as resistors, and later, inductors and capacitors), this is true. However, there are some components (especially certain gas-discharge and semiconductor components) which are nonlinear: that is, their opposition to current changes with voltage and/or current. As such, we would call circuits containing these types of components, nonlinear circuits.

                Thevenin's Theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source, series resistance, and series load.


 Steps to follow for Thevenin's Theorem:

• (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for series circuits.

Example:

             Find V_TH, R_THand the load current flowing through and load voltage across the load resistor in fig (1) by using Thevenin’s Theorem.

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1.Open the 5kΩ load resistor

2.Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (V_TH). Fig (3)

We have already removed the load resistor from figure 1, so the circuit became an open circuit as shown in fig 2. Now we have to calculate the Thevenin’s Voltage. Since 3mA Current flows in both 12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor as it is open.

So 12V (3mA  x 4kΩ) will appear across the 4kΩ resistor. We also know that current is not flowing through the 8kΩ resistor as it is open circuit, but the 8kΩ resistor is in parallel with 4k resistor. So the same voltage (i.e. 12V) will appear across the 8kΩ resistor as 4kΩ resistor. Therefore 12V will appear across the AB terminals. So,

V_TH = 12V 

3. Open Current Sources and Short Voltage Sources. Fig (4)

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4.Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (R_TH)

We have Reduced the 48V DC source to zero is equivalent to replace it with a short in step (3), as shown in figure ()  We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and  12k Ω resistor. i.e.:

8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)

R_TH = 8kΩ +  [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]

R_TH = 8kΩ + 3kΩ

R_TH = 11kΩ

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5. Connect the R_THin series with Voltage Source V_TH and re-connect the load resistor. This is shown in fig (6) i.e. Thevenin circuit with load resistor.

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6. Now apply the last step i.e. calculate the total load current & load voltage as shown in fig 6.

I_L = V_TH/ (R_TH + R_L)

= 12V / (11kΩ + 5kΩ) → = 12/16kΩ

I_L= 0.75mA

And

V_L = I_Lx R_L

V_L = 0.75mA x 5kΩ

V_L= 3.75V

Click image to enlarge  

Superposition Theorem

The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (currents) within the modified network for each power source separately. Let's look at our example circuit again and apply Superposition Theorem to it:

Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in effect. . .

. . . and one for the circuit with only the 7 volt battery in effect:

When re-drawing the circuit for series/parallel analysis with one source, all other voltage sources are replaced by wires (shorts), and all current sources with open circuits (breaks). Since we only have voltage sources (batteries) in our example circuit, we will replace every inactive source during analysis with a wire.

Source Transformation

Source Transformation of Circuits- Explained

The source transformation of a circuit is the transformation of a power source from a voltage source to a current source, or a current source to a voltage source.

In other words, we transform the power source from either voltage to current, or current to voltage. 

Voltage Source Transformation

We will first go over voltage source transformation, the transformation of a circuit with a voltage source to the equivalent circuit with a current source.

In order to get a visual example of this, let's take the circuit below which has a voltage source as its power source:

Using source transformation, we can change or transform this above circuit with a voltage power source and a resistor, R, in series, into the equivalent circuit with a current source with a resistor, R, in parallel, as shown below:

We transform a voltage source into a current source by using ohm's law. A voltage source can be changed into a current source by using ohm's formula,I=V/R. 

Mesh Current, conventional method

 

The first step in the Mesh Current method is to identify “loops” within the circuit encompassing all components. In our example circuit, the loop formed by B₁, R₁, and R₂ will be the first while the loop formed by B₂, R₂, and R₃ will be the second. The strangest part of the Mesh Current method is envisioning circulating currents in each of the loops.

 

The choice of each current's direction is entirely arbitrary, just as in the Branch Current method, but the resulting equations are easier to solve if the currents are going the same direction through intersecting components (note how currents I₁ and I₂ are both going “up” through resistor R₂, where they “mesh,” or intersect). If the assumed direction of a mesh current is wrong, the answer for that current will have a negative value.
The next step is to label all voltage drop polarities across resistors according to the assumed directions of the mesh currents.

Using Kirchhoff's Voltage Law, we can now step around each of these loops, generating equations representative of the component voltage drops and polarities. As with the Branch Current method, we will denote a resistor's voltage drop as the product of the resistance (in ohms) and its respective mesh current (that quantity being unknown at this point). Where two currents mesh together, we will write that term in the equation with resistor current being the sum of the two meshing currents. 


  
            

 

Wye Delta

Delta to wye conversion

It is easy to work with wye network. If we get delta network, we convert it to wye to work easily. To obtain the equivalent resistance in the wye network from delta network we compare the two networks and we confirm that they are same. Now we will convert figure 3 (a) delta network to figure 2 (a) wye network.

From figure 3 (a) for terminals 1 and 2 we get,

R₁₂ (∆) = R_b || (R_a + R_b)

From figure 2 (a) for terminals 1 and 2 we get,

R₁₂(Y) = R₁ + R₃

Setting wye and delta equal,

R₁₂(Y) = R₁₂ (∆) we get,

Equations (v), (vi) and (vii) are the equivalent resistances for transforming delta to wye conversion. We do not need to memorize these equations. Now we create an extra node shown in figure 4 and follow the conversion rule,

Each resistor in the Y network is the product of the resistors in the two adjacent Del branches, divided by the sum of the three Del resistors.

 

Wye to delta conversion

For conversion to wye network to delta network adding equations (v), (vi) and (vii) we get,

 R₁R₂ + R₂R₃ + R₃R₁ = R_aR_bR_c(R_a +R_b + R_c)/(R_a + R_b + R_c)²

                                = R_aR_bR_c/(R_a + R_b + R_c) —————— (ix)

Dividing equation (ix) by each of the equations (v), (vi) and (vii) we get,

R_a = R₁R₂ + R₂R₃ + R₃R₁/ R₁

R_b = R₁R₂ + R₂R₃ + R₃R₁/ R₂

R_c = R₁R₂ + R₂R₃ + R₃R₁/ R₃

For Y to delta conversion the rule is followed below,

Each resistor in the delta network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor.

see link http://waleedeid.tripod.com/Lecture3_cir_analysis.pdf

for more information... :)

Nodal Analysis

Nodal Analysis with Voltage Sources

Case 1

            If a voltage source is connected between the reference node and non-reference node, we simply set the voltage at the non-reference node equal to the voltage of the voltage source.

Case 2

            If a voltage source (dependent or independent) is connected between two non-reference nodes, the two non-reference nodes form a generalized node or super node; we apply both KCL and KVL to determine the node voltages.

Note:

            A super node is formed by enclosing a dependent or independent voltage source connected between two non-reference nodes and any elements connected in parallel with it.

Principles of super node:

a.    The voltage source inside the super node provides a constraint equation needed to solve for the node voltages.

b.    A super node has no voltage of its own.

c.    A super node requires the application of both KCL and KVL.