Moule 2 DC Circuit
Lesson 9 Analysis of c resistive network in presence of one non-linear element
Objectives To unerstan the volt (V ) ampere ( A ) characteristics of linear an nonlinear elements. Concept of loa-line an analysis of c resistive network having a single nonlinear element using loa-line analysis. L.9.1 Introuction 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, ioes or lamps is not linear as in fig. 3.3 (lesson-3). Temperature effects cause much non-linearity in lamps that are mae of metals. Most materials resistance increases (or ecreases) with rise (or fall) in temperature. On the other han, most nonmetals resistance ecreases 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 issipate an in turn rising the filament temperature. Further note, that the increments of voltage prouce smaller increments of current that causes increase resistance in the filament element. Opposite effects can be observe in case of carbon filament lamp or silicon carbie or thermistor. Aitional increments of voltage prouce large increments of current that causes ecrease resistance in the element. Fig.9.1 shows the characteristics of tungsten an carbon filaments.
Let us consier a simple circuit shown in fig. L.9.2(a) that consists of inepenent sources, combination of linear resistances, an a nonlinear element. It is assume that the nonlinear element characteristics either efine in terms of current ( it ()) (flowing through it) an voltage ( v ( t) ) (across the nonlinear element) relationship or nl Voltage ( vnl ( t )) an current () i relationships of nonlinear element can be expresse as mathematical expression or formula. For example, consier the actual (non-ieal) V I relationship of the typical ioe can be expresse as V a I = I0 e 1, where ' a' is constant ( for germanium ioe a =0.026 an silicon ioe a =0.052). Assume that the network (fig.9.2(a)) at the terminals A an B is replace 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 RTh + VAB or, VTh I AB RTh = VAB Thevenin terminal voltage = loa voltage. If the nonlinear element characteristic is given (note, no any analytical expression is available) then one can aopt graphical metho calle loa-line analysis to etermine the branch variables ( I AB = i nl an V AB = v nl ) of nonlinear element as shown in fig. L.9.2(a). This resulting solution is frequently referre to as the operating point ( Q ) for the nonlinear element characteristic (in the present iscussion, we consier a nonlinear element is a resistor R 2 ). This metho is quite simple an useful to analysis the circuit while the loa has a nonlinear V I characteristic. It is very easy to raw the source characteristic using the intercepts at points v() t = V = V, i = i = 0( open circuit conition) an vt () = V AB = 0, i= iab = AB Th AB
V R Th Th = I (short circuite at A & B ter minals) in two axes. It is obvious that the N values of voltage ( VAB ) an current ( I AB ) at the terminals of the source are exactly same as the voltage across an current in the loa as inicate in fig. 9.2(a). The point of intersection of the loa an the source characteristic represents the only conition where voltage an current are same for both source an loa elements. More-specifically, the intersection of source characteristic an loa characteristic represents the solution of voltage across the nonlinear element an current flowing through it or operating point ( Q )of the circuit as shown in fig.9.2(c). Application of loa-line analysis is explaine with the following examples. L.9.3 Application of loa-line metho Example-L.9.1: The volt-ampere characteristic of a non-linear resistive element connecte in the circuit (as shown Fig.9.3(a)) is given in tabular form. Table: volt-ampere characteristic of non-linear element V nl 0 2 4 6 8 10 12 14 15 I 0 0.05 0.1 0.2 0.6 1.0 1.8 2.0 4.0 nl
(i) Calculate the voltage rop V ab across the non-linear element. (ii) For the same circuit, if the non-linear element is replace by a linear resistance RL, fin the choice of RL that will absorb maximum power elivere by the inepenent sources. Solution: (i) The Thevenin equivalent circuit across the terminals ' a' an ' b' of fig.9.3(b) can be obtain using noe-voltage metho (or one can apply any metho to fin ). V Th Noe voltage at c Vc 10 Vc Vc + 20 + + = 0 Vc = 4.52 volt. 25 10 15 V ( 20) 4.52 20 Current through c-a branch = c + = = 1.032 A 15 15 Voltage across the terminals a an b =Thevenin equivalent voltage ( V Th ) = 4.52 10 1.032= 14.48 volt. (note, point b is higher potential than the point a ).
Thevenin resistance ( R Th ): ( ) ( ) 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 (loa-line), we examine the extreme conitions uner which a given source may operate. If the nonlinear loa is remove (i.e. terminal a an b is open-circuite), the terminal voltage will be equal to the Thevenin voltage ( V Th ). On the other han, if the nonlinear resistance is short-circuite, the current flow VTh through the a an b terminal is Isc = IN =. RTh The operating point of the circuit is foun from the intersection of source characteristic (loa-line) an loa characteristic curves as shown in fig.9.3(). From this graph (see fig.9.3()), one can fin the current flowing through the nonlinear element I = 1.15 A an the voltage across the nonlinear element is V = 10.38 Volt. L ab
(ii) If the nonlinear resistance is replace by a linear resistance, the maximum power transfer conition is achieve when the linear loa resistance RL = RTh = 3.87Ω. Uner such conition, the network will eliver the maximum power to the loa an the 2 2 VTh 14.84 corresponing maximum power is given by Pmax = = = 14.23 W RTh 4 3.87 Remarks: The primary limitation of this metho (loa-line analysis) is accuracy, ue to its graphical nature an thus it provies an approximate solution of the circuit. Example-L.9.2: Fig.9.4(a) shows that a nonlinear element (ioe) is inclue in the circuit.
Calculate the current flowing through the ioe, voltage across the ioe an the power issipate in the ioe using (i) loa-line analysis (graphical) technique (ii) analytical metho. The volt-ampere characteristic of the ioe is given by the expression V V a 6 0.026 I = I0 e 1 = 10 e 1 (9.1) Solution: (i) Loa-line analysis metho The volt-ampere characteristics of the ioe are given in tabular form using the equation (9.1). Table: V = voltage across the ioe in volt, I = current in ioe in ma V 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: Thevenize circuit Remove the nonlinear element (ioe) an 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 aroun the close path of the circuit as shown in fig.9.4(b), we get 2 4 ( 5+ 5) I 2= 0 I = = 0.2A 10 4 5 0.2 3 volt. R = 5 5 + 7.5= 10Ω VTh = = ; ( ) Th The equivalent Thevenin circuit is shown in fig.9.4(c). Step-2: Loa-line analysis Draw the source an loa (ioe) characteristics on a same graph paper as shown in fig.l.9.4(). Note that the Thevenin parameters obtaine in step-1 are use to raw the source characteristic.
The operating point Q in fig.9.4() provies the information of the following quantities: Voltage across the ioe = V = V = 0.33 volt. Current flowing through the ioe = IL = 269mA Power issipate in the ioe = P = V I = 88.77 mw (ii) Analytical metho ab ioe ab L KVL equation aroun the loop of fig.l.9.4(c) is written as V I R V = 0 (9.2) Th L Th VTh IL RTh V 5 38.46 = + or ( V ) 3 = 10 e + V ( note V = 3 V, R = 10 Ω) Th Th The nonlinear algebraic equation can be solve by using any numerical technique. To solve this equation, one can consier the Newton-Raphson metho to the above equation. The above equation is rewritten in the following form 5 V 38.46 ( ) ( ) f V = 10 e + V 3 (9.3) fv (9.4) 3 38.46 V ( ) = 1+ 0.385 10 e
To solve V, Newton s formula is use f( V ) V( new) = V( ol) f ( V ) = V V ( ol) (9.5) Initial guess of V is assume as V ( ) = 0.36V. The final value of is obtaine after ol four iterations an they are shown below. Using equation (9.2), we get Iteration V ( ol) V ( 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 issipate in ioe P = V I = 0.3842 261.58= 100.5mW. ioe ab L I L VTh V 3 0.3842 = = = 261.58mA an power R 10 Th L.9.4 Test your unerstaning (Marks: 30) T.1 The volt-ampere characteristic of a unknown evice (T ) is shown in fig.9.5(a) an it is connecte to a resistive circuit of fig.9.5(b). V
(i) If Vin = 10 V, an Rin = 150Ω, plot the source characteristics on the voltagecurrent axes. [3] (ii) Inicate the operating point on the graph an estimate the values of I an V (ans. I = 43 ma, an V = 3.7V ) [3] T T T T T.2 A Thermistor with the volt-ampere characteristic (see fig.9.6(a)) given below is connecte into one arm of a brige as shown in the circuit of fig.9.6(b).
(i) Determine the value of R so that the thermistor operation point is at VT = 5Volt an IT = 2 ma. Justify the answer using the loa-line analysis technique. (ans. R = 0.5kΩ ) [5] (ii) Fin the new operating point of the thermister when R = 300Ω. Subsequently, compute the power issipate by the thermistor. (Ans. 5.3 V; 2.4 ma an 12.7 mw) [5+2]
T.9.3 The ioe whose volt-ampere characteristic is given by analytical expression V V a 6 0.026 I = I0 e 1 = 10 e 1 is use in the circuit of fig.9.7. Using analytical metho, etermine the following (i) the ioe current I, voltage V an power issipate by ioe. [5] (ii) the ioe voltage for each case while the supply voltage ( Vs ) is change in succession to 30 V, 40 V an 50 V respectively. [7]