To investigate further the series LCR circuit, especially around the point of minimum impedance. 1 Electricity & Electronics Constructor EEC470

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1 Series esonance OBJECTIE To investigate further the series LC circuit, especially around the point of minimum impedance. EQUIPMENT EQUIED Qty Apparatus Electricity & Electronics Constructor EEC470 Basic Electricity and Electronics Kit EEC Multimeters or Milliammeter 0 5 ma ac oltmeter 0 0 ac Function generator 00 Hz khz 20 pk pk sine (eg, Feedback FG60) PEEQUISITE ASSIGNMENTS Assignment 23 KNOWLEDGE LEEL See prerequisite assignment. EEC

2 Series esonance Fig 26.3 EEC

3 Series esonance EXPEIMENTAL POCEDUE We have seen from Assignment 23 that the impedance of a series LC circuit varies with frequency, and that the form of variation is given in fig 26.. Fig 26. Connect up the circuit as shown in the patching diagram of fig 26.3 corresponding to the circuit diagram of fig Fig 26.2 Connect the voltmeter across the input to the circuit (points marked P and S on fig 26.2) and adjust the generator output to give 4 rms at 200 Hz. ary the frequency of the generator from about 00 Hz to 300 Hz and note the variation in current and voltage. EEC

4 Series esonance Find the frequency at which the circuit impedance is a minimum. This is where the current is a maximum and the voltage is a minimum. Question. What is this frequency? This frequency is termed the ESONANT FEQUENCY of the circuit. At the resonant frequency of the circuit the capacitive reactance is equal to the inductive reactance, and they cancel each other out. The circuit impedance at the resonant frequency is thus just the resistance of the circuit. Measure the current and voltage at resonance and calculate the impedance at resonance. Question 2. Does the calculated value agree with the value expected? The formula for XC is: X C = 2πfC and for X L : X L = 2πfL at resonance X C = X L. Thus 2πfC = 2πfL at resonance Let us denote the resonant frequency by the symbol fo. Then 2πf o C = 2πfoL From this derive an expression for fo in terms of L and C. Question 3. fo =...? Substitute L = 00 x 0 3 H (00 mh) C = 2.2 x 0 6 F (2.2 µf) in your expression and work out fo. EEC

5 Series esonance Question 4. Does your calculated value agree with the frequency found previously for minimum impedance? Set the generator frequency to 20 Hz and the output voltage to 4 rms. The generator frequency may be taken from the dial if the calibration is sufficiently accurate, or a digital frequency meter used for more accuracy. Take readings of the circuit current I, the inductor voltage L, and the capacitor voltage C. Copy the results table as shown in fig 26.3, reproduced at the end of this assignment, and record your readings. Increase the frequency to 40 Hz, and re-adjust the generator output to give 4 rms again. Take readings of, I, L and C as before. epeat the procedure for the frequencies given in the table. Ensure that the input voltage is 4 rms for each frequency setting. Find the resonant frequency again (the frequency at which the current is a maximum) and take a set of readings at this frequency, fo. On a sheet of 2 cycles logarithmic graph paper, draw curves of I, L and C against frequency, using the axes shown in fig EEC

6 Series esonance Fig 26.4 emove the resistor from the circuit and connect points P and Q together. Set the generator frequency back to 20 Hz, but now set the output amplitude to.0 rms as shown on the meter. Take measurements as before, and copy another table as in fig 26.3, using the same frequencies, but always ensuring that the generator output is.0 rms. On another sheet of 2-cycles log graph paper plot the second set of I, L and C curves, on axes as used before for fig Notice the differing shapes of the two sets of curves. Questions 5. From the curves plotted with = kω in circuit, what is the resonant frequency of the circuit (where I is a maximum)? 6. What is the value of I at this frequency? 7. What is the value of L at fo? 8. What is the value of C at fo? 9. Do they have any relationship to each other, if so why? EEC

7 Series esonance 0. What is the ratio of L or c at fo to the input voltage? The ratio of Q, of the circuit. L or C in in at f o is termed the Quality Factor, or. What is the Q of your circuit? Now, at fo, the phasor diagram for the circuit voltages is given in fig Fig 26.5 L = C thus = in at resonance Now I = in Z = in at resonance But also L = I X L Therefore, at resonance L = in. X L thus L in = X L EEC

8 Series esonance But L in = Q Q = X L Q = ω o L ω L Calculate o for your circuit and compare it with the value found for Q from the graphs. Now similarly, C = I X C C = in X C at resonance ie, C X = C in Q = Calculate ω C o ω C o and compare it with the graphical Q. Now examine your graphical results for the circuit with removed. Questions 2. What is the resonant frequency? 3. Does it differ from fo when = kω? 4. What are the value of I, C, and L, at f o? 5. What is the Q of this circuit, calculated from the graphs? EEC

9 Series esonance 6. Using Q = ω o L or what is the calculated Q of ω o C the circuit? 7. Is the Q of the circuit when = 0 higher or lower than that when = kω? With = 0 the calculated value of Q is infinite. Obviously this is unreasonable, as there would then be an infinite voltage across the inductor and the capacitor. 8. Where do you think resistance is present to limit the Q to the value obtained experimentally? 9. What would the value of this resistance have to be? Let us now examine the shapes of the curves. At resonance I is a maximum, and as I = Z in then Z must be a minimum. We have seen this from Assignment 23. Thus we sometimes call a series LC circuit an 'acceptor circuit' because its impedance to ac signals about its resonant frequency is low, whereas it presents a high impedance to other frequencies. The LC circuit will thus pass signals near to its resonant frequency much more readily than those of other frequencies. It is said to have a 'bandpass' effect. By examining the two sets of curves it can be seen that the current curve when = 0 is much sharper in its peaking than when = kω. It follows that the 'band' or range of frequencies accepted by the circuit with the sharper peaked current curve will be narrower than for the flatter one. We say that the 'bandwidth' of the circuit with = 0 is smaller than that with = kω. Bandwidth is defined as the change in frequency between the points on the current curve which are of the current at resonance. This is shown in fig EEC

10 Series esonance Fig 26.6 Here Bandwidth ( BW ) = ( f2 - f ) Hz Copy the table given in fig 26,7, reproduced at the end of this assignment, and from your two sets of graphs, complete the table and calculate the bandwidths. We see then that a circuit with a low value of Q has a large bandwidth, and one with a higher Q has a smaller bandwidth. It would seem reasonable that Q and BW are inversely proportional in some way. Let us investigate this. We know I max = I at resonance = in By definition: at f 2 I f2 = I max I f2 = in EEC

11 Series esonance But also Z f2 = in = I f2 in in = Z f2 = 2. X + X 2 But Z = + ( ) = + ( ) 2 L X + X 2 2 = 2 + (X L + X C ) 2 L C C = X L X C at f 2 However, we know that, at resonance (f 0 ), X L = X C. Because X L is directly proportional to frequency, and X C is inversely proportional to frequency, X L and X C must change by approximately the same amount for small changes in frequency. Ie, for a frequency change of f 0 to f 2 : X L changes to X L + x X C changes to X C - x where x is a small reactance change. At resonance X L X C = 0 at f 2 ( X L + x) ( X C x) = X L ( X C +2x = from which x = 2 we can say at f 2 : X L = 2πf2L = 2πf 0 L + 2 f 2 = f L π EEC

12 Series esonance Similarly at f : X L = 2πf L = 2πf0L 2 f = f 0 4 L π But Bandwidth = f 2 f = (f L π ) (f 0 4π L ) BW = 2πL Dividing by f0 we get: BW = f 0 2πf 0 L = ω 0 L BW = f Q 0 or f Q = 0 BW From your experimental figures for f 0 and bandwidth, calculate Q using the above expression. Question 20. How do these values compare with those previously calculated? The series LC circuit is thus a frequency selective circuit, and is often used as such in electronics. In the majority of cases a high Q circuit is desirable, giving as narrow a bandwidth as possible, thus there is normally no resistor in circuit. However, the inductance will never be pure, but will have some internal resistance due to the resistance of the wire, thus the Q of the circuit will be limited by this. EEC

13 Series esonance PACTICAL CONSIDEATIONS AND APPLICATIONS The phenomenon of resonance is found in many branches of physics when the physical properties of a system allow oscillations to occur much more severely at one particular frequency, so that when the system is excited by some outside source at this frequency the vibrations build up to a large amplitude. No doubt you all have seen pictures of the bridge at Taccoma apids, USA, as it shattered itself. This was due to the resonant effect excited by the wind. The bridge started to vibrate due to the wind, which was of such a velocity to reinforce the vibration, and the whole bridge became resonant, and literally whipped itself into fragments. In electrical circuits resonance occurs when there are both inductance and capacitance in circuit at the frequency at which the inductive reactance is equal to the capacitive reactance. At resonance the response of the circuit is only limited by the losses of the circuit due to resistance, etc. There are several sources of loss in a resonant circuit. Power losses in an inductor are: copper loss, due to the resistance of wire; losses due to hysteresis and eddy current losses in the core; losses due to the induction of currents in screening cans and objects in close proximity, causing eddy currents to be set up in these, and dissipating power. At high frequencies the coil former may show appreciable dielectric loss also. Losses due to the capacitor are: dielectric losses, and resistance due to to the capacitor plates. EEC

14 Series esonance The full equivalent circuit of a series tuned circuit, representing the loss mechanisms by an equivalent resistor, is given in fig Fig 26.8 These losses are frequently represented in terms of a single equivalent series or parallel resistance for each component. The tuned circuit (resonant circuit) is used extensively in electronics for its frequency selective properties. It can be seen from your graphs that the impedance of the series tuned circuit varies widely with frequency, and that the impedance is a minimum at the resonant frequency. The series resonant circuit will thus form an acceptor circuit which will pass frequencies near to the resonant frequency and attenuate other frequencies. EEC

15 Series esonance If the series resonant circuit is connected across a signal line, as in fig 26.9, it will shunt to earth signals of frequency near its resonant frequency. This facility is useful if it is wished to cut out a particular interfering signal, and a series tuned circuit could be used. Fig 26.9 The degree to which a resonant circuit responds to one frequency rather than another is termed its selectivity. From your results it should be seen that circuits with a high Q have also high selectivity and vice-versa. Normally, high selectivity is desired, and thus resonant circuits should be designed to have the highest Q (ie, lowest losses) possible. EEC

16 Typical esults and Answers ESULTS FO ASSIGNMENT 26 frequency (Hz) input voltage in () current I (ma) C () L () , f 0 = Fig 26.3 EEC

17 Typical esults and Answers esults graph for Fig 26.3( in = 4 ) EEC

18 Typical esults and Answers frequency (Hz) input voltage in () current I (ma) C () L () f0 = Fig 26.3 EEC

19 Typical esults and Answers esults graph for Fig 26.3( in = ) EEC

20 Typical esults and Answers From the given component values at f 0 ω 0 L 2π = 0 3 Q = 0.5 Similarly ω0c = 2π Q = 0.3 From the measured value Q = 0.27 The differences may be accounted for by the tolerance of the components which produce an actual resonant frequency different to that predicted by calculation using given component values. (See Q4) I max ma I max ma f Hz f2 Hz f2 - f Hz for = kω for = 0 Ω Fig 26.7 EEC

21 Typical esults and Answers ANSWES TO ASSIGNMENT Hz 2. From the measured values at resonance: Z = I = Z = 064 Ω From the calculated value at resonance: Z = 2 + X 2 where X = X L X C and = 000 Ω X L = 2πfL = 6.28 x 240 x 00 x 0 3 = 5 Ω X C = 2 fc π = = 30 Ω Z = ( 000) 2 + ( 50) 2 Z = 0Ω The difference in value is accounted for by the tolerances of components. EEC

22 Typical esults and Answers 3. As 2πf0C = 2πf 0 L 4π 2 f 0 2 LC = 2 f 0 = 4π 2 LC f 0 = f 0 = 4π 2 LC 2π LC 4. From the given values of components: f 0 = 2π 00 x 0 3 x 2.2 x 0 6 =339 Hz Again the difference in value may be accounted for by the tolerances of the reactive components Hz ma L = C at f 0 0. L 09. = 4 in = Q = Hz 3. Yes EEC

23 Typical esults and Answers 4. At f 0 = 230 Hz: I C L = 9.88 ma = 3.0 = Q = L in or Q = C in = 30. = 293. Q = 3.0 Q = Q = ω o L and when = 0, Q is infinitely large. However, because the inductance possesses some small resistance, practically Q will have a high value. 7. Q is higher when = 0 than when = kω. 8 esistance is present in the inductor, the connecting wires and the meters, but mainly in the inductor. 9 To give a value of Q = 3.0 with a 00 mh inductor: Q = ω o L = ω o L Q = 2π = 50 Ω EEC

24 Typical esults and Answers 20. From the experimental results: For = kω: f 0 = 240 Hz BW = 888 Hz Q = f 0 = Q = 0.27 Comparing Q Q = 0.27 Similarly for = 0: f 0 = 230 Hz BW = 80 Hz Q = Q = 2.88 Comparing Q5 Q = 3.0 or 2.97 EEC

25 Typical esults and Answers frequency (Hz) input voltage in () current I (ma) C () L () f 0 =240 4 Fig 26.3 EEC

26 Typical esults and Answers I max ma I max ma f Hz f2 Hz f2 - f Hz for = kω for = 0 Ω Fig 26.7 EEC

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