Objectives
Introduction
The volt-ampere characteristic of a linear resistance is a straight line through the origin as in fig. 3.2 (see Lesson-3) but the characteristic for non-linear element for example, diodes or lamps is not linear as in fig. 3.3 (lesson-3). Temperature effects cause much non-linearity in lamps that are made of metals. Most materials resistance increases (or decreases) with rise (or fall) in temperature. On the other hand, most nonmetals resistance decreases or increases with the rise or fall in temperature. The typical tungsten lamp resistance rises with temperature. Note, as the voltage across the lamp increases, more power is dissipated and in turn rising the filament temperature. Further note, that the increments of voltage produce smaller increments of current that causes increase resistance in the filament element. Opposite effects can be observed in case of carbon filament lamp or silicon carbide or thermistor. Additional increments of voltage produce large increments of current that causes decrease resistance in the element. Fig.9.1 shows the characteristics of tungsten and carbon filaments.
Let us consider a simple circuit shown in fig. L.9.2(a) that consists of independent sources, combination of linear resistances, and a nonlinear element. It is assumed that the nonlinear element characteristics either defined in terms of current (i(t)) (flowing through it) and voltage (vnl(t)) (across the nonlinear element) relationship or Voltage (vnl (t)) and current (i) relationships of nonlinear element can be expressed as mathematical expression or formula. For example, consider the actual (non-ideal) relationship of the typical diode can be expressed as where ‘a’ is constant ( for germanium diode ‘a’=0.026 and silicon diode ‘a’=0.052). Assume that the network (fig.9.2(a)) at the terminals ‘A and B’ is replaced by an equivalent Thevenin network as shown in fig. 9.2(b). From an examination of the figure one can write the following expression: VTh = IAB X RTh + VAB 0r, VTh IAB X RTh = VAB ⇒ Thevenin terminal voltage = load voltage. If the nonlinear element characteristic is given (note, no any analytical expression is available) then one can adopt graphical method called load-line analysis to determine the branch variables (IAB = inl and VAB = vnl) of nonlinear element as shown in fig. L.9.2(a). This resulting solution is frequently referred to as the operating point (Q) for the nonlinear element characteristic (in the present discussion, we consider a nonlinear element is a resistor R2) . This method is quite simple and useful to analysis the circuit while the load has a nonlinear V − I characteristic. It is very easy to draw the source characteristic using the intercepts at points v(t) = VAB = VTh , i = iAB = 0 (open circuit condition) and v(t) = VAB = 0, i = iAB = (short − in two axes. It is obvious that the values of voltage (VAB) and current (IAB) at the terminals of the source are exactly same as the voltage across and current in the load as indicated in fig. 9.2(a). The point of intersection of the load and the source characteristic represents the only condition where voltage and current are same for both source and load elements. More-specifically, the intersection of source characteristic and load characteristic represents the solution of voltage across the nonlinear element and current flowing through it or operating point (Q) of the circuit as shown in fig.9.2(c). Application of load-line analysis is explained with the following examples.
Application of load-line method
Example- The volt-ampere characteristic of a non-linear resistive element connected in the circuit (as shown Fig.9.3(a)) is given in tabular form.
Table: volt-ampere characteristic of non-linear element
Vnl | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 15 |
lnl | 0 | 0.05 | 0.1 | 0.2 | 0.6 | 1.0 | 1.8 | 2.0 | 4.0 |
(i) Calculate the voltage drop Vab across the non-linear element.
(ii) For the same circuit, if the non-linear element is replaced by a linear resistance RL, find the choice RL of that will absorb maximum power delivered by the independent sources.
Solution:
(i) The Thevenin equivalent circuit across the terminals “ 'a' of 'b' fig.9.3(b) can be obtain using ‘node-voltage’ method (or one can apply any method to find VTh).
Node voltage at ‘c’
Voltage across the terminals ‘a’ and ‘b’ = Thevenin equivalent voltage (VTh) = -4.52-10 x 1.032 = -14.48volt. (note, point ‘b’ is higher potential than the point ‘a’).
Thevenin resistance (RTh):
RTh =[(25|| 10) +10] || 5 = (7.14 +10)|| 5 = 3.87Ω
Thevenin equivalent circuit for the network as shown in fig.9.3(b) is given below:
To construct the source characteristic (load-line), we examine the extreme conditions under which a given source may operate. If the nonlinear load is removed (i.e. terminal ‘a’ and ‘b’ is open-circuited), the terminal voltage will be equal to the Thevenin voltage (VTh). On the other hand, if the nonlinear resistance is short-circuited, the current flow through the ‘a’ and ‘b’ terminal is
The operating point of the circuit is found from the intersection of source characteristic (load-line) and load characteristic curves as shown in fig.9.3(d). From this graph (see fig.9.3(d)), one can find the current flowing through the nonlinear element IL = 1.15A and the voltage across the nonlinear element is Vab = 10.38 Volt.
(ii) If the nonlinear resistance is replaced by a linear resistance, the maximum power transfer condition is achieved when the linear load resistance RL = RTh = = 3.87Ω . Under such condition, the network will deliver the maximum power to the load and the corresponding maximum power is given by
Remarks: The primary limitation of this method (load-line analysis) is accuracy, due to its graphical nature and thus it provides an approximate solution of the circuit.
Example- Fig.9.4(a) shows that a nonlinear element (diode) is included in the circuit.
Calculate the current flowing through the diode, voltage across the diode and the power dissipated in the diode using (i) load-line analysis (graphical) technique (ii) analytical method. The volt-ampere characteristic of the diode is given by the expression
Solution:
(i) Load-line analysis method The volt-ampere characteristics of the diode are given in tabular form using the equation (9.1).
Table: Vd voltage across the diode in volt, I = current in diode in mA
Vd | 0 v | 0.1 v | 0.2 v | 0.25 v | 0.3 v | 0.32 v | 0.35 v |
I | 0 mA | 0.046mA | 2.19mA | 15mA | 102.6mA | 268mA | 702mA |
Step-1: Thevenized circuit Remove the nonlinear element (diode) and replace the rest of the circuit (as shown in fig.9.4(b)) by a thevenin equivalent circuit (see fig.9.4(c)).
Applying KVL around the closed path of the circuit as shown in fig.9.4(b), we get
The equivalent Thevenin circuit is shown in fig.9.4(c).
Step-2: Load-line analysis Draw the source and load (diode) characteristics on a same graph paper as shown in fig.L.9.4(d). Note that the Thevenin parameters obtained in step-1 are used to draw the source characteristic.
The operating point Q in fig.9.4(d) provides the information of the following quantities:
Voltage across the diode = Vd =Vab= 0.33 volt
Current flowing through the diode = IL = 269 mA
Power dissipated in the diode = 88.77 Pdiode =Vab×IL = 88.77mW
(ii) Analytical method
KVL equation around the loop of fig.L.9.4(c) is written as\
The nonlinear algebraic equation can be solved by using any numerical technique. To solve this equation, one can consider the Newton-Raphson method to the above equation. The above equation is rewritten in the following form
To solve Vd, Newton’s formula is used
Initial guess of is assumed as Vd (oid) = 0.36V . The final value of is obtained after four iterations and they are shown below.
Iteration | Vd (old) | Vd (new) |
1. | 0.36 v | 0.3996 v |
2. | 0.3996 v | 0.3886 v |
3. | 0.3886 v | 0.3845 v |
4. | 0.3845 v | 0.3842 v |
Using equation (9.2), we get and power dissipated in diode Pdiode=Vab xIL = 0.3842X261.58 = 100.5mW .
57 docs|62 tests
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1. What is a DC resistive network? |
2. What does it mean to have a non-linear element in a resistive network? |
3. How does the presence of a non-linear element affect the analysis of a DC resistive network? |
4. What are some techniques used to analyze a DC resistive network in the presence of a non-linear element? |
5. What are some practical applications of analyzing DC resistive networks with non-linear elements? |
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