Lunes, Oktubre 6, 2014

Maximum Power Transfer

Maximum Power Transfer Theorem


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.
Taking our Thevenin equivalent example circuit, the Maximum Power Transfer Theorem tells us that the load resistance resulting in greatest power dissipation is equal in value to the Thevenin resistance (in this case, 0.8 Ω):
With this value of load resistance, the dissipated power will be 39.2 watts:
If we were to try a lower value for the load resistance (0.5 Ω instead of 0.8 Ω, for example), our power dissipated by the load resistance would decrease:
Power dissipation increased for both the Thevenin resistance and the total circuit, but it decreased for the load resistor. Likewise, if we increase the load resistance (1.1 Ω instead of 0.8 Ω, for example), power dissipation will also be less than it was at 0.8 Ω exactly:
If you were designing a circuit for maximum power dissipation at the load resistance, this theorem would be very useful. Having reduced a network down to a Thevenin voltage and resistance (or Norton current and resistance), you simply set the load resistance equal to that Thevenin or Norton equivalent (or vice versa) to ensure maximum power dissipation at the load. Practical applications of this might include radio transmitter final amplifier stage design (seeking to maximize power delivered to the antenna or transmission line), a grid tied inverterloading a solar array, or electric vehicle design (seeking to maximize power delivered to drive motor).
The Maximum Power Transfer Theorem is not: Maximum power transfer does not coincide with maximum efficiency. Application of The Maximum Power Transfer theorem to AC power distribution will not result in maximum or even high efficiency. The goal of high efficiency is more important for AC power distribution, which dictates a relatively low generator impedance compared to load impedance.
Similar to AC power distribution, high fidelity audio amplifiers are designed for a relatively low output impedance and a relatively high speaker load impedance. As a ratio, "output impdance" : "load impedance" is known as damping factor, typically in the range of 100 to 1000. [rar] [dfd]
Maximum power transfer does not coincide with the goal of lowest noise. For example, the low-level radio frequency amplifier between the antenna and a radio receiver is often designed for lowest possible noise. This often requires a mismatch of the amplifier input impedance to the antenna as compared with that dictated by the maximum power transfer theorem.

Capacitor Circuits

Next let us consider a single capacitor of capacitance C, here the relationship between the current flow and applied voltage is given by
Current flow for a capacitor
unlike the resistor, the current flow is proportional to the voltage gradient (with respect to time) and consequently this introduces a phase shift between the two. The impedance response for a circuit containing a single capacitor is shown in time, phasor and bode notations below.

Circuit Component Frequency Response 
 Capacitor Frequency response for a capacitor
clearly the current is 90° out of phase with the voltage, the Bode plot shows that this relationship holds for all frequencies although the magnitude of the signal drops as the frequency increases. To explain this behaviour we need to understand how the capacitor resists the passage of current. A measure of this resistance to current flow is given by the capacitative reactance Xc which has units of Ohms. This quantity Xc has both magnitude and phase and calculated using
Reactive capacitance
it may be predicted by invoking Ohms law which tells us that the circuit resistance (in this Xc) is equal to
This shows us that the quantity ‹Xc always has a -90° angle attached to its magnitude and it's usually written as above or in the complex form -j Xc

Inductor Circuits

Finally we consider the response of an inductor with an inductance (L). Here the relationship between the current flow and applied voltage is given by
Response of an inductor circuit
like the capacitor, the current can be seen to be out of phase with the voltage. The impedance response of a circuit containing a single inductor is shown below in time phasor and bode forms.

Circuit Component Frequency Response 
 Inductor Frequency response for an inductor
In an analogous manner to the capacitor the 'resistance' to current flow is given by the inductive reactance Xl which has both a magnitude and phase:
Inductive reactance calculation
it may be predicted by using Ohms law which tells us that the circuit resistance (in this case Xl) is equal to
Ohm's law
This shows us that the inductive reactance always has a 90° angle attached to its magnitude and is usually written in complex form as jXl or in polar form
Inductive resistance
We now have the basic information required to analyse circuits containing combinations of the above components in series or parallel. As the majority of circuits of interest in electrochemical analysis are combinations of resistors and capacitors we will only consider these in the later sections, although the extension to examine inductive circuits requires no further developments.

Thevenins and Norton's Theorem

  Norton's  Theorem


Norton's theorem states that a network consists of  several voltage sources, current sources and resistors with two terminals, is electrically equivalent to an ideal current source " INO" and a single parallel  resistor, RNO. The theorem can be applied to both A.C and D.C cases. The Norton equivalent of a circuit consists of an ideal current source in parallel with an ideal impedance (or resistor for non-reactive circuits).

      

The Norton equivalent circuit is a current source with current "INO" in parallel with a resistance RNO.To find its Norton equivalent circuit,
  1. Find the Norton current "INO". Calculate the output current, "IAB", when a short circuit is the load (meaning 0 resistances between A and B). This is INo.
  2. Find the Norton resistance RNo. When there are no dependent sources (i.e., all current and voltage sources are independent), there are two methods of determining the Norton impedance RNo.
  • Calculate the output voltage, VAB, when in open circuit condition (i.e., no load resistor — meaning infinite load resistance). RNo equals this VAB divided by INo.
    or
  • Replace independent voltage sources with short circuits and independent current sources with open circuits. The total resistance across the output port is the Norton impedance RNo.
    However, when there are dependent sources the more general method must be used. This method is not shown below in the diagrams.
  • Connect a constant current source at the output terminals of the circuit with a value of 1 Ampere and calculate the voltage at its terminals. The quotient of this voltage divided by the 1 A current is the Norton impedance RNo. This method must be used if the circuit contains dependent sources, but it can be used in all cases even when there are no dependent sources.

Example 1:-

Consider this circuit-


To find the Norton’s equivalent of the above circuit we firstly have to remove the centre 40Ω load resistor and short out the terminals A and B to give us the following circuit.


When the terminals A and B are shorted together the two resistors are connected in parallel across their two respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can now be calculated as:

With A-B Shorted :



If we short-out the two voltage sources and open circuit terminals A and B, the two resistors are now effectively connected together in parallel. The value of the internal resistor Rs is found by calculating the total resistance at the terminals A and B giving us the following circuit.


Find the Equivalent Resistance (Rs):

10Ω Resistor in parallel with the 20Ω Resistor

Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the following Nortons equivalent circuit.

Nortons equivalent circuit.

Ok, so far so good, but we now have to solve with the original 40Ω load resistor connected across terminals A and B as shown below.


Again, the two resistors are connected in parallel across the terminals A and B which gives us a total resistance of:


The voltage across the terminals A and B with the load resistor connected is given as:


Then the current flowing in the 40Ω load resistor can be found as:


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.

Learnings:

1.Norton's theorem and Thevenin's theorem are equivalent,and the equivalence leads to source transformation in electrical circuits.

2.For an electric-circuit,the equivalence is given by  VTh= Ith x Rth ie Thevenin's volage=Norton's current x Thevenin's resistance.

3.The applications of Norton theorem is similar to that of Thevenin's theorem.The main application is nothing but the simplification of electrical circuit by introducing source transoformation.

Thevenin Theorem




This theorem is very conceptual. If we think deeply about an electrical circuit, we can visualize the statements made in Thevenin theorem. Suppose we have to calculate theelectric current through any particular branch in a circuit. This branch is connected with rest of the circuits at its two terminal. Due to active sources in the circuit, there is one electric potential difference between the points where the said branch is connected. The currentthrough the said branch is caused by this electric potential difference that appears across the terminals. So rest of the circuit can be considered as a single voltage source, that's voltage is nothing but the open circuit voltage between the terminals where the said branch is connected and the internal resistance of the source is nothing but the equivalent resistanceof the circuit looking back into the terminals where, the branch is connected. So theThevenin theorem can be stated as follows,

  1. When a particular branch is removed from a circuit, the open circuit voltage appears across the terminals of the circuit, is Thevenin equivalent voltage and,
  2. The equivalent resistance of the circuit network looking back into the terminals, isThevenin equivalent resistance.
  3. If we replace the rest of the circuit network by a single voltage source , then thevoltage of the source would be Thevenin equivalent voltage and internal resistanceof the voltage source would be Thevenin equivalent resistance which would be connected in series with the source as shown in the figure below.

To make Thevenin theorem easy to understand, we have shown the circuit below,

Here two resistors R1 and R2 are connected in series and this series combination is connected across one voltage source of emf E with internal resistance Ri as shown. One resistive branch of RL is connected across the resistance R2 as shown. Now we have to calculate the current through RL.
thevenin theorem

First, we have to remove the resistor RL from the terminals A and B.

Second, we have to calculate the open circuit voltage or Thevenin equivalent voltage VTacross the terminals A and B.
Thevenin equivalent voltage

The electric current through resistance R2,


Hence voltage appears across the terminals A and B i.e.


Third, for applying Thevenin theorem, we have to determine the Thevenin equivalentelectrical resistance of the circuit, and for that; first we have to replace the voltage sourcefrom the circuit, leaving behind only its internal resistance Ri. Now view the circuit inwards from the open terminals A and B. It is found the circuits now consist of two parallel paths - one consisting of resistance R2 only and the other consisting of resistance R1 and Ri in series.

Thevenin equivalent resistance


Thus the Thevenin equivalent resistance RT is viewed from the open terminals A and B is given as. As per Thevenin theorem, when resistance RL is connected across terminals A and B, the network behaves as a source of voltage VT and internal resistance RT and this is called Thevenin equivalent circuit. The electric current through RL is given as,

Thevenin Equivalent Circuit


thevenin theorem