Chapter 2 - Linear Circuit Analysis

Transcription

1 Chapter Linear Circuit Analysis Recommended problems to study: Page Warmups Resistance of a wire Transients,3 Transformers/mutual inductance 4 Linear circuit models 5 Voltage regulator 6 iv diode models 7 Photodetector circuit Temperature dependent diode modeling 3 Hall effect 4 Transformers 5 Concentrates Source/load relationships 5 Matrix analysis of circuits 7 Transformers 3 Reactances 33 Wheatstone bridge 35 Oscillator 38 Transient oscillator 40

2 Chapter Part Warmups. Using the same diameter nichrome wire as in example., determine the length needed to obtain a resistance of ohm at 00 C. Determine the resistance at 0 C. The characteristics of #30 AWG wire are: diameter (mils) diameter (circular mils) in C pounds/000 ft for nichrome C r Wm a / C r 00 r 0 [ a 0 (T0)] [ (000)] r [.36] Wm Ê 3 in d d A Ê p ˆ p Ë p Ë 4 4 R r l A 6 l W ( W m) m 8 mm in ˆ p ( m) m m l ( W) W m 00. m R 0 ( l 00. m) 6 ( W m)( 00. m) 04. W m Note: Circular mils is nothing more than the diameter (expressed in thousandths of an inch) squared.

3 . For the circuit shown in example., assume the switch has been closed for a long time and is opened at t0. Determine (a) the current through the capacitor at the instant that the switch is opened, (b) the time derivative of the capacitor voltage at the instant the switch is opened, and (c) the voltage that will exist across the capacitor after a long time. R ic V S i R C VC The boundary conditions for the capacitor C are no DC current flow v( 0 ) v( 0 ) R By inspection, v R R V C( 0 ) S since there is no DC current through C. (a) When the switch is open i R ic C V C R v v R R V C 0 C 0 i ( 0 ) ( ) ( ) R R R VS Then, ic ( 0 ) i ( 0 ) R R S VS R R (b) Using the derivative relationship i dv t dt (c) v i 0 C ( t ) 0 C ( ) ( ) 0 C VS R R C ( ) C dv for C dt

4 3. For the circuit shown in example.3, assume that the switch has been closed for a long time (the final conditions of example.3 apply), and the switch is opened at t0. Determine at the instant the switch opens (a) the current in the inductance, (b) the time rate of change of current in the inductance, and (c) the voltage across the inductance. R il V S i R L VL The boundary conditions for an inductor are i (a) i L VS ( 0 ) since the inductor is a DC short R Therefore, i L ( 0 ) VS R L ( 0 ) il( 0 ) (b) For an inductor, v il(0 ) L di dt VR S VR il( 0 ) R R di VL VR VR S dt L L LR VR R VL (c) V L VR S R 3

5 Example.4 A transformer consists of a primary winding with 500 turns and two secondary windings of 5 turns and 30 turns. The 5 turn winding has 60 ohms connected to its terminals, and the 30 turn secondary winding has 3 ohms connected to its terminals. If the primary winding is connected to a 0 volt 60 Hz source, determine the current rating for each winding. N 5 60W 0 V N 500 N 336 3W 4. Repeat Example.4 with the 0 volt source connected to winding and the 60 ohm load attached to winding. V V 3 N N V volts ( ) N3 N V volts ( ). 60W V i N 500 N 5 i V i 3 0V i i 3 480V 8amps 60W V 5amps 3W. N 3 36 V 3 3W To find i we use conservation of power vi vi vi i ( )( ) ( )( ) 0i i 35. 3amps 4

6 Example.5 Obtain an expression for the characteristic curve shown over the range from 8 to amps. Obtain linear circuit models from the derived expressions. 0 8 terminal voltage terminal current (amperes) 5. Obtain a linear model of voltage versus current for the characteristic shown in example.5 over the current range of 0 to 6 amps. Obtain the current source and voltage source equivalent circuits. Use the twopoint method to fit a straight line to the data V V0 V V0 I I0 I I0 Read two points from the graph point 0: 0 volts, 0 amps point : 9 volts, 6 amps Evaluating the expression, V I V 0 I 6 0 V /6W 6V 60 I 6V I 60 I V 60 I NOTE: Thevenin is usually easier to visualize I 6V A /6W 5

7 6. A 5 volt, 5 amp d.c. source has voltage regulation of %. Obtain an equivalent circuit for this source. The equivalent circuit is R V NL VR V V NoLoad V FullLoad FullLoad 00% VNoLoad 5V % 00% 5V Solving for V NoLoad 5% V NoLoad 00% 500% V NoLoad ( 00) 505 V NoLoad R 5.05 V R W 5A 5 6

8 7. All silicon diodes have a reverse breakdown voltage essentially the same as the avalanche voltage of a zener diode. This results in three distinct regions of operation: one for the normal forward region, one for the avalanche region, and an intermediate region where the device behaves essentially as an open circuit. Obtain the linear models in these three regions for a amp, watt silicon diode with a reverse breakdown voltage of 5.5 volts and a maximum reverse voltage of 5.6 volts. A I 5.5 V 0.6 V V 5.5 V 0.6V V Theoretical Powerrated vrated irated 5.6 V Linear model.4 W v rated Powerrated watts Volts i Amp rated In region I: V I V. 4I. 4 V 4. I V In region II: nothing, diode is an open circuit In region III: find the rated current using breakdown voltage V I max max watts ( 56. Volts) Imax watts I 036Amps max. ( ) ( ) V I V I V 08. I V 0.8 W 7

9 8. A threeterminal device is described by the following zparameter equations. VIN 50iIN 5iOUT VOUT 00iIN 5iOUT Obtain an equivalent circuit for this device. Rewriting in matrix form. This is the zparameter model. È VIN iin ÎVOUT È 50 5 È Î 00 5 Îi OUT I IN Z Z I OUT V IN Z V OUT Z V IN V OUT I IN 50W 5W I OUT V IN 5I OUT 00I IN V OUT or draw reversed as 5W I OUT 00I IN V OUT 8

10 9. Common base transistor configurations are often described in terms of common base y parameters: y ib, y rb, y fb and y ob. (The b s in the subscripts indicate that the parameters were obtained from a common base configuration with the emitter as the input terminal and the collector as the output terminal.) A common circuit model for the transistor used in this configuration is shown. Obtain the yparameters for this common base circuit. ai e i e e r e r c c i c v eb r b v cb b This is algebraically a very complex problem. Finding twoport parameters is usually algebraically complex. The yparameters are defined by ÈI y y V I Î È Î y y È ÎV to find y and y short V : I yv I yv to find y and y short V : I yv I yv Let s start by shorting V to get: I r e V I r e V r b r c ai Using KCL at the common node: V Ir e V Ir e I ai r r b Ê ˆ I( ) VÁ Ir e Ë r r Ê r ˆ a Á Ë r b Ê I V r r b ˆ Ê c rb r ˆ c ( a ) Á Ir e Á Ë rr Ë rr b c c c I b b c c 9

11 ( )( b c) ( b c) e( b c) I( a )( rr b c) Ir e( rb rc) V( rb rc) I V r r ( b c) y V V [( a )( rr b c) re( rb rc) ] ( rb rc) y ( a )( rr) r( r r) I a rr V r r Ir r r y b c e b c I V Ire ai V V I V r e a ( ) ( rb rc) y y( re a ) ( a )( rr b c) re rb rc ( a )( rr b c) re( rb rc) ( rb rc) ( re a) y ( a )( rr b c) re( rb rc) arr b c rr b c rr e b rr e c rr b e rr c e arc arb y ( a )( rr b c) re( rb rc) arr b c rr b c arc arb y a rr r r r y ( )( b c) e( b c) ( a) rr b c a( rb rc) ( a )( rr) r( r r) b c e b c ( ) ( r a ) e Now short V : ai V I rc Using definitions I yv I y V I y (3) V r e I r b V V' V V' I a I () re rb rc where the first expression is the lower resistances, and the second term is the upper loop V I ' () r e Substituting () into () gives: V V' V' V Ê a ˆ I a V' Á rc re rc Ë r c r e Using (3) 0

12 I I V Ê a ˆ I ( re rb) Á r Ë r r c Ê rr e b Á Ë r r e b r e c e ar ˆ c V rr r c e ( ) Ê re rb rc rr e b arr ˆ b c V I Á Ë ( re rb) r c rc I re rb y V r r r rr arr ( ) e b c e b b c Using (), V ' I re y V V Using () V Ê a ˆ I V' Á r Ë r r I c c e V Ê V r r e a ˆ c ' Á r Ë rr c Solving for V r e ' : c e c re rb a rr r r r ( ) ( ) V' Ê V ˆ Ê ˆ rc Á I Á re Ë rc Ë re arc and substituting into our expression for y Ê V ˆ Ê r ˆ c Ir c V Á I Á Ë rc Ë re arc re arc re arc y V V I rc y yrc V re arc re arc ( re arc) Substituting for y È ( re rb) rc y ( re rc) ( ) rr b c re( rb rc) a Î a ( re rb) rc ( a ) rr b c re( rb rc) y ( re arc) [( a) rr b c re( rb rc) ] rr e c rr b c rr b c arr b c rr e b rr e c y r ar a rr r r r y [( ) ( )] ( ) rb[ arc re] ( ) e c b c e b c [ ] b c e b c rb r ar a rr r r r a rr r r r [( ) ( )] ( ) e c b c e b c ( ) b c e b c

13 0. For a photodiode with characteristics shown in figure.0 and used in the circuit for example. with a 5 volt source and a 0,000 ohm load, determine the load voltage as a function of illumination level. Over what range of illumination levels will this result be valid? Assume the curves extend to 5 volts on the third quadrant of the characteristic curves. I LIGHT 0 fc If (µa) Vf (volts) Fig..0 Typical photodiode characteristics (a) R L I 0 (b) ki LIGHT (a) Biasing circuit for photodiode of figure.0 (b) equivalent circuit I L RL From example, i where the socalled dark current is estimated I L to be about 0.7µA for 0 fc (footcandles). The other point comes from the estimated line for the 500 fc illumination level. i I L (i is in microamperes) () Considering the load electrical circuit using KVL 5 Vf i( 0kW) () Substituting () into () gives 6 5 Vf ( IL ) 0 ( 0kW) 5 Vf IL But 5 V VLOAD is simply the voltage across the load resistor. f The equation VLOAD IL is valid when the illumination I L >0 and breaks down when the diode becomes forward biased, i.e., the voltage across the load resistor becomes greater than the voltage across the photodiode. At this point the voltage across the diode drops to zero and the entire 5 volts appears across the load resistor. At this breakdown point, V I 5Volts LOAD and the corresponding light level is 5 I L ª 6, 000footcandles L

14 Example. The characteristics shown in figure. have voltage divisions of 0. volts and current divisions of amp. Obtain an equivalent circuit for a diode which accounts for the temperature dependence with T 0 C and T 80 C. If (amps) Vs/RL T T V S If V f Vf (volts) Vs R L Figure. Temperature characteristics for a diode. Repeat example. for T 0 C and T 00 C. Using T 0 C and T 00 C. From the graph for T we can estimate the two points on graph to be (0.95V,0Amps) and (.4Volts,.5Amps). Using the two point method Vf if Vf if Vf 03. if 095. There are many ways to find the second curve for T 00 C. The slope remains the same but the yintercept changes. Note that for T 00 C at i f 0, V f 06.. The T curve is then Vf 03. if 060. Assume that the yintercept is a linear function of temperature and again use the two point method. Using the points (0.95,0) and (0.60,00) for the curve we get an equation for the yintercept of C o T C o T The overall equivalent circuit is V i, T 03. i C 03. i T ( ) f f f o f 3

15 Example.3 Copper has an electron density of 0 8 electrons per cubic meter. The charge on each electron is coulombs. For a copper conductor with width 5 mm, height 5 mm, and length 0 cm, determine the Hall voltage for a current of amp when the magnetic field intensity is amp/m, Bµ o H and µ o 4p Use the dimensions given in example.3 but assume the conductor is an ntype semiconductor with 0 9 electrons per cubic meter. The current is 0. amp. Obtain an equivalent circuit for a transducer using the Hall voltage to measure magnetic field intensity. Assume that the voltage pickups are capacitative so there is no connection of the output to the current circuit. y l0cm h5mm z x v w5mm HA/m The fundamental equation for the Hall effect is r r r r F q( E v B) 0 Since there is only a ycomponent of the E field we can rewrite this as r r Ey ( v B) 0 () y The current I x in the xdirection is given by Ix nqvxa where nnumber of carriers per square meter q is the charge on the carriers v x is the xcomponent of the carrier velocity and Awh is the conductor cross sectional area. Solving, Ix Ix vx () nqa nqwh Rewriting () and substituting () into it E v B 0 y x z Vh h Ix Ê ohz Ë Á ˆ m 0 nqwh Solving for V h V Ê 0. Amp ( m) Á Ë ( 9 ) m m ˆ ( 4p 0 ) H h z z ( )( )( ) H 4

16 Example.4 A simplified small signal amplifier circuit is shown. The voltage source Vin is a variable signal source and can be considered to be an ideal voltage source. The transistor has parameters h ie 500, h re 0, h fe 50, and h oe Determine the Thevenin equivalent circuit (as seen from terminals a and b) in terms of the unknown turns N and N on the righthand transformer. kw a V in b Ideal : Ideal N :N 3. For the circuit of example.4, obtain the Norton equivalent circuit. The transistor model is i B 500W i S For the transformer, if the v terminal is at the dot then the current is into the dot. 0 50I B 0kW V V See p. 6 of your reference book The resulting overall circuit is then 000W i C i SC V IN I B 500W 50I B 0kW N :N Note the results of the dot convention. The current flowing into the dot results in current flowing out of the top of the transformer secondary. The voltage source s polarity is reversed by the transformer as shown in the first figure. The output is a short circuit because we are determining the short circuit current. This gives vin vin ib Continuing through the transistor, 5

17 50 ic 50iB vin 500 By the dot conventions for the output transformer, the output current is going into the dot. N The current magnitude is transformed as and the voltage magnitude as N (Power is N N conserved). Since we are looking for a Norton equivalent circuit, the output is short circuited and the load voltage is zero. Consequently, we are only looking at the shortcircuit current. In this context the transformer also appears as an ac short and we can ignore the 0kW output impedance of the transistor. N i i v N 50 N N N N v SC C in in This gives us the short circuit current in the direction shown. To get the Norton equivalent circuit we need to also compute the Thevenin resistance. To get this resistance we short v in, NOT the transistor current source 50i B and look at the resulting circuit. 000W i i 500W 50I B 0kW Z N :N This leaves only the 0kW resistor which gets transformed by the transformer as N R Ê N k Ë Á ˆ 0 W N The final Norton equivalent circuit is then i SC Z or Ê N ˆ Á 30 Ë N V IN Ê Á Ë N ˆ N 0k 6

18 Example.4 A simplified small signal amplifier circuit is shown. The voltage source Vin is a variable signal source and can be considered to be an ideal voltage source. The transistor has parameters h ie 500, h re 0, h fe 50, and h oe Determine the Thevenin equivalent circuit (as seen from terminals a and b) in terms of the unknown turns N and N on the righthand transformer. kw a V in b Ideal : Ideal N :N 4. Consider the circuit of example.4 with an 8 ohm speaker attached to terminals ab. Ê N ˆ Show that the necessary turns ratio Á to provide maximum power transfer is Ë N Using the Norton equivalent circuit from the previous problem 30 Ê N Á Ë N ˆ V IN Ê Á Ë N N ˆ 0k 8W N The maximum power transfer occurs when R Ê N Ë Á N Ê N ˆ Solving for the turns ratio, Á Ë N Ê N Á Ë N ˆ W kW ˆ 0kW 8W 7

19 Example.6 The circuit shown has Z µf and Z 00mH. The voltage source is represented by the partial Fourier series v S 0 cos 000t 5 cos 4000t Determine the voltage across the 00 ohm resistor. 50W Z V S Z 00W 5. The elements Z and Z are interchanged in example.6, and the voltage v S takes on the value of v S 0 5 cos 36t Find the voltage across the 00 ohm resistor. The resulting circuit is then 50W Z 00mH V S Z µf 00W At DC the inductor is a short, the capacitor is an open. 50W 0V VT 0Volts R T 50W At w36 we compute the equivalent impedances to get 8

20 50W Z j447.w 5V Z j447.w Z jw L j( 36)( 0. H) j447. W Z j j C j ( 36 ) ( 0 ). W w The Thevenin voltage for this AC circuit is found using a voltage divider j V 447. T ( j 50 j447 j447 5 ) The Thevenin resistance is that of the two impedances in parallel 50 j447 j447 ZT ( 50 j447. ) ( j447. ( ). )(. ) 3999 j j447 (j447 ).... To find the output voltage we use 50W Using a voltage divider we get V 00 OUT ( 0) 3. 33Volts V rad/sec j44.7v (4000j447.)W 00W Using a voltage divider we get 00 VOUT ( j44. 7) j447. VOUT 0. j The measured voltage across the 00W resistor is then V V V DC AC V 00 TH V TH 3. W cos 36t ( ) 9

21 6. The circuit of example.7 has r b 00 ohms, b50, r c 500 ohms, and r e 0 ohms. Find the voltage gain V OUT when a 000 ohm load is placed across the V OUT terminals. VIN Note that the output current I is V OUT 000 for this load. I I r b bi r c V V V in V V V out R e The circuit is then Figure.6. Combination of port networks with common currents. r b 00 W bi 50i r c500w 000W R e 0W We will convert both the upper and lower circuits to zparameters and then combine them. for zparameters ÈV z z I V Î È Î z z È ÎI For an open output I 0 and V z I 0

22 V z I I I V r b00w bi 50i r c 500W V By inspection V z 00W I V 50I 500 z I I if I 0 V zi V zi and V z 0 I V z 500W I For the 0W resistor ( )( ) I I V R e0w V Z È 0 0 Î 0 0 The total admittance matrix is then È 00 0 Z T Î È 0 0 Î È Î To find the voltage gain we write the network equations resulting from Z T V 0I 0I () V 4990I 50I () For the output, V I R L Using () and (3) V V 4990I 50 Ê ˆ Á Ë R L

23 V 50 V 4990I I. V 8. 0 V 4990 Substituting this result into () and using (3) 5 V V 0(. 8 0 V ) 0 Ê ˆ Ë V V 00. V V V V 6. 9

24 7. Using yparameters, obtain the total circuit yparameters for the circuits indicated by the dashed lines. Hint: first find the yparameters of the two indicated twoport networks, then combine them to obtain the total network yparameters. Rf Vin Vout V rb gmv rc The solution of this problem is similar to that of problem 6 except that yparameters will be used. ÈI ÎI y È Î y y y V È ÎV For V 0: I yv I y V For V 0: I yv I y V I I R f V V R f I I y y I I V V I V R f R f y y I V R f I I V V R f Therefore, Y R È Rf R f Î Rf Rf 3

25 For V 0 ForV 0 I I I I V rb gmv rc V 0 V0 r b gmv r c V y y I V r b I gv m g V V m y y I 0 V I V r c And, therefore, Y transistor È rb g Î m 0 rc È È 0 r R R b f f Ytotal Ytransistor YR g m Î rc Î Rf Rf È Rf rb R f Ytotal gm Î Rf Rf rc 4

26 CONCENTRATES Example.6 The circuit shown has Z µf and Z 00mH. The voltage source is represented by the partial Fourier series v S 0 cos 000t 5 cos 4000t Determine the voltage across the 00 ohm resistor. 50W Z V S Z 00W. In the circuit shown in example.6, the voltage source is operating at a frequency of 000 rad/sec. Specify the impedances Z and Z so that maximum power is transferred to the 00 ohm load resistance. (Hint: Z and Z must be reactive or else they will absorb some of the power.) In specifying Z and Z, give their values in microfarads and/or millihenries. There are two possible sets of answers. When faced with a sourceload problem always Thevenize the source circuit. From the above circuit jx VOC VS jx jx 50 where we have assumed that Z and Z are purely reactive, i.e., Z jx and Z jx Similarly, Z ( jx jx T jx ) jx ( 50 ) jx jx The resulting circuit is then Z T V OC 00W For maximum power transfer we require that Z T ( 00) 00W ( 50 jx) jx 50 jx jx Multiplying out 00 * or 5

27 j50x XX 5000 j00x j00x Equating the real and imaginary parts XX 5000 () 50X 00X 00X () Solving () for X and substituting into () 50X 00X X 00 X X 50 X( X) 5000 X 500 Therefore, X ± 50W The corresponding solutions for X are found by substituting this result into (): XX X / 50W X ± 50 Case : X 50W, X 00W X is inductive so X wl L 0. 05H 000 X is capacitive so X 00W wc Solving for the capacitance gives C 5mf ( 000 )( 00 ) Case : X 50W, X 00W X is capacitive so X 50W wc Solving for the capacitance gives C 0mf ( 000 )( 50 ) X is inductive so X wl L 005. H 000 6

28 . For the circuit shown, obtain the node voltage equations in matrix form. Solve these equations for V and V. (The actual output is V V.) Obtain the voltage gain V OUT for this VS emitter coupled amplifier. 50W V 4 I 500W V3 50I 000W 000W 50I VS 50W 000W 500W I V 000W V 000W This is a complex problem which requires that we know how to construct an admittance matrix representation for the circuit. The standard ( formal ) admittance matrix formulation is: È I È Y Y V Â Â È Â I Y Y V   ΠΠΠwhere  I i sum of current sources feeding into node i (intopositive, outnegative)  Y ii sum of all admittances connected to node i  Y ij sum of all admittances connected between nodes i and j V i node voltages (including dependent sources) The first step in solving the problem is to Nortonize the voltage source, converting it into a current source 50W ISVS/50 50W VS fi Now, convert all resistances into admittances (I will use the unit of millimhos for convenience) and label the nodes. 7

29 node 4 mmhos V3 node 3 I 50I mmho mmho 50I I S 40mmhos* mmho mmhos node node I mmho mmho Note that node 0, the reference node, must always be present. *The 40millimhos is the combination of the two 50W resistors in parallel. node 0 Constructing the admittance matrix representation È 50i È 0 0 ÈV 50i 0 0 V 50i 50i V3 Î I Î S ÎV4 Ground Consider node : A current source of 50i leaves node, therefore the current is 50i The sum of all admittances connected to node is millimhos There are no admittances directly connected between nodes and or 4 and, therefore Y Y4 0 There is millimho connected between nodes and 3 so Y 3 millimho since all offdiagonal terms are negative. Actually each row of the matrix represents a KCL equation for that node and this is the way I prefer to generate the matrix. At node, assuming that currents into the node are positive, 50i ( V 0) ( V V3) 0 which can be rewritten as 50i ( ) V V3 0 which is exactly the first row of the matrix equation. The rest of the problem is simply solving the matrix: È 50i È 0 0 ÈV 50i 0 0 V 50i 50i 7 V3 Î I Î S 0 0 4ÎV4 Expressing i and i in terms of independent variables V 3 and V 4 V4 V3 i ( V4 V3) 500 V3 i V3 500 and converting the source term 8

30 VS 0VS 50 Then, È 00V4 00V3 00V3 50 V V 50 V Î 0VS ( ) ( ) È 0 0 ÈV 0 0 V 7 V3 Î 0 0 4ÎV4 È 00V4 00V3 È 0 0 ÈV 00V V 00V4 00V3 7 V3 Î 0V Î S 0 0 4ÎV4 Rearranging the equations, È 0 È ÈV V V3 Î0V S Î ÎV 4 This equation can now be solved for the variables we are interested in (V and V ) using any method you want. I highly recommend using a calculator which can solve systems of equations since this is where I usually make math errors. However, I will use expansion by minors to solve for V and V since the matrices are only 4 4. È V S V Î È 0 0 È Î 0 4 Î 0 4 The coefficients of all other minors are zero and not shown. Recall that solving for V requires that the column vector of sources be substituted for column. Note also that the signs of the minor coefficients alternate. Solving: 0V V S [( )( 07)( 00) ( 00) (0)( ) ( ) (0)(0)] V [( )( )( ) ( )( )( ) ( )( )( )] ( )( )( ) ( )( )( ) [ ] 0VS 0696 [ ] [ ] [ ] [ 738] [ 8084] 0VS [ ] 9

31 390 V VS. 9V 739 S Similarly, È V S VS V Î ( )( )( ) ( )( )( ) È 0 0 È Î 0 4 Î 0 4 where we saved a little effort by recognizing that the denominator is the same as before. [ ] V [ ]. 0VS V S A V V V VS VS V V S S Expansion by minors: Check sign: 30

32 3. For the circuit shown, obtain the loop current equations. Solve for the current in the 00 ohm resistance connected to the secondary of the ideal transformer. 00W 00W I V 00W 50I 000W 00W ideal 4: As usual we will Thevenize the dependent current source to make the loop analysis easier. 000W 50I 000W 50,000I fi We will also reflect the load resistance through the transformer to make analysis simpler. 00W 300W ideal 4: where ( 4) ( 00) ( 6)( 00) 300W Now, redrawing the circuit for analysis 00W 000W 00W V I Writing the loop equations 00i 00i i 000i 00i 300i 0 00W 50,000I I 300W È Î È Î i È Îi 3

33 We are interested in the transformer primary current i. Solving for i : È i Î ( )( ) 75 76mA È ( 300)( 4400) 0. Î The transformer secondary current is then 75.76mA 4303mA with the current as shown below 00W 303mA 3

34 4. A 60 Hz source has its voltage measured under various loads, with the results shown. Case Load Voltmeter reading open circuit 0 volts rms 60W resistance 7 volts rms 3 60W pure capacitance 360 volts rms 4 60W pure inductance 5.43 volts rms Determine the voltage for a load of 0 ohms pure capacitance. We will assume that the internal impedance of the source is complex and in series with the source. Case : under no load the open circuit voltage is 0 volts rms Case : RjX 7Vrms V Ê 60 OUT Á Ë 60 R ˆ 0 jx 0V 60W ( )( ) ( ) R jx R X ( ) 584 ( ) 60 R X 0000 () Case 3: RjX 360Vrms j V Ê 60 ˆ OUT Á 0 Ë j60 R jx 0V j60w ( j60)( 0) R j X 60 ( ) R X 60 ( ) 9600 R ( ) () X Case 4: 33

35 RjX 5.43Vrms j V Ê 60 ˆ OUT Á 0 Ë j60 R jx 0V j60w ( j60)( 0) R j X 60 ( ) R X 60 ( ) R ( ). (3) X Solving equations () and (3) R X 0X R X 0X R X 0X 300 R X 0X Subtracting 40X X79.995W Substituting this result back into equation () gives R ( ) 400 R ( ) 400 R R W The output voltage for a capacitive j0w load can then be calculated using these results. ( ) j V Ê ˆ j j OUT Á Ë R jx 00 j j j j j4. 0 V OUT V OUT ª 360 volts rms 34

36 5. A bridge network is used to measure the temperature of a chemical bath. R R 4 V m kw kw R and R are identical thermistors with negative temperature coefficients of resistance of 4%/ C. At 5 C they are both 500W. The meter (m) has an internal resistance of 800W. Resistance R is held at 5 C. (a) Determine the meter current when R is at 0 C. (b) Determine the meter current when R is at 30 C. The first step is to calculate the different values of R. R 5 C temperature coefficient 4%/ C R T 5 C R R ( ( )) ( ( )) ( ) ( ) W ( ) C R T 5 C R R ( ( )) ( ( )) ( ) ( ) W ( )

37 R R 500W 4 V V 800W V kw kw We will write the nodal equations as shown below: 4 V V V V V V V V Rewriting 4 V. 5V. 5V. 8V 0 4 V. 875V. 875V. 5V 0 In matrix form, È Î È... V Î È ÎV È Î È.. V Î È ÎV Solving, È V Î. È Î ( ) ( )( ) ( )( ) V È Î È Î ( 505. )( 4) (875. )( 4) 66 ( 505)( 4375) ( 875) ( 5)

38 V V m ( 0 C) mA 800W 800 At 30 C the node equations 4 V V V V 0 R V V V V 0 R V V V V V V V V In matrix form, È 4 È Î 500 Î ÈV Î V Multiplying the first row by 00 and the second row by 500 we get: È4 Î È V 4 Î È ÎV È4 Î È V 4 Î È ÎV V È4. 5 Î È Î V È Î È Î V V m ( 0 C) 067. ma 800W

39 6. An oscillator circuit is shown below. Determine the minimum value of K such that e o E. Also, find the frequency at which e o E. 5000W 09F E 000W KE 5000W 0.00H 09F 400W eo As usual we Thevenize the source so that we can combine resistances. 5000W KE 5000W fi 5000KE Redrawing the circuit and writing the loop equations 0000W 09F E 09F 000W 5000KE 03H i i i3 400W eo ( ) KE 0000i i i jw0 0 3 jw0 i i i i jw0 jw0 ( ) ( ) 3 9 ( i i ) i 400i 0 jw In matrix form, È 3 3 È 5000KE 0000 jw0 jw0 0 Èi jw0 jw0 i 9 9 jw0 jw0 Î 0 Î i Î jw0 jw0 jw0 38

40 Since we are only interested in i 3 : È jw0 jw0 5000KE 3 3 jw0 jw0 0 9 jw j 0 i3 Î w È jw0 jw jw0 jw0 9 9 jw0 jw Î jw0 jw0 9 9 jw0 i3 Ï Ê Ì 0 jw0 Á jw0 Ó Ë ( ) jw0 3 ( ) Á 5000KE jw0 ˆ j 400 j 0 j 0 Ê Á Ë j 0 ˆ ( w ) w w 9 9 Ê Ë 3 3 ˆ 9 jw0 Ê Ë ( ) Á j ˆ Ê ˆ Á 9 jw0 Ë jw0 Ê ˆ Á 0000 jw0 9 Ë jw0 3 ( ) i i 3 3 Ï 0 0 j 0 0 Ì Ó w w KE 0 4j j0 jw w w w KE ÏÊ 0 0 ˆ 5 0 ÌÁ04. 0 j 4000 Ê Á w ÓË w Ë w 5 ˆ 5 5 For o E for oscillation we require i to be real which requires w 0 or w w Solving for w gives w 5. 0 or w. 0 6 rad /sec 4000 Using this value, KE ( 400) E Solving for K gives K( 400) or K ( )

41 The following problem is similar to a problem I encountered on my PE exam (morning session) Sketch the steadystate output of the following circuit. Be sure to indicate values of voltages and time. Assume the opamp operates on ±5 power supplies. R4 R3 R Vo IDEAL C The component values are C 0.µf, R 5kW, R 3 0kW and R 4 0kW. 40

42 4