Therefore the size of the resistor and inductor that makes the equivalent circuit is,
And
7. Now that the left hand side of the equivalent circuit has been calculated it is possible to re assembly the circuit to calculate the voltage at the initial break point. This is done my simple mathematics and takes into account all previous calculations. The final answer presents both a voltage and a phase angle.
8. To enable a comparison of results the original circuit was assembled in the lab using the bread-board and discrete components. An oscilloscope was used to determine voltage and phase angle at the break-point. The results obtained from the oscilloscope were. Obviously these values fall short of the calculated values, however this is most likely due to the tolerances of the discrete components.
9. Having completed the calculations and compared results by building the circuit I decided to attempt Nodal analysis on the circuit. Hopefully to obtain results identical to those gained using Thevenin’s theorem. To assist in Nodal analysis it was first necessary to re arrange the circuit for ease of analysis.
10. Having re-arranged the circuit for ease it was now possible to calculate the voltage and phase angle using Nodal analysis. The analysis was carried out as shown below.
11. This formula was then factorized to give;
This then equates to;
Re-arranging this gives,
12. The final outcome of all these calculations is a voltage and phase angle as shown below.
13. It can therefore be seen that the final figures using both Thevenin’s theorem and Nodal analysis are the exact same, thus proving the initial results.
TASK 2
14. Nodal analysis was utilized to find the value of voltage and phase angle of the voltage at the centre of the circuit. Firstly in was necessary to obtain the values in format of both the inductor and capacitor within the circuit.
15. This was achieved by using the formula below;
16. Having defined the values of these two components nodal analysis could now begin. Using the formula previously used in task 1 it was possible to obtain the correct voltage as seen below;
Transposing the formula to give;
17. Having calculated the voltage using Nodal analysis the circuit was constructed using the bread board and discrete components. Once built the voltage, phase angle and peak to peak voltages were measured using an oscilloscope. The measured voltage and phase angle was. This figure is very close to the calculated figure, any slight difference is most likely due to component tolerances and or the oscilloscope measurements taken. The peak to peak values for channel one was, and for channel two. It is interesting to note that throughout this experiment the output frequency remained unchanged at.
TASK 3
18. This task involves finding certain values as described below given that an ideal air cored transformer is fed with mains supply. The primary winding, has 2000 turns and a secondary winding, has 400 turns. Resistor draws from the secondary winding which has an inductance of. The mutual inductance between and is. Given that the maximum flux in is, calculate the following.
-
The maximum flux, in the primary winding.
-
The coupling coefficient, , between the windings.
-
The value of maximum flux in
-
The voltage, , induced in the secondary winding.
-
The value of resistor, .
-
The value of the primary current,
Mutual inductance between windings of
Maximum flux in,
19. By utilizing and manipulating the formula’s below it was possible to obtain all the necessary figures.
-
Coefficient of coupling
20. Firstly to find the maximum flux in the primary winding.
Transposed to find =
21. Coupling coefficient, this equates to,
22. The value of inductance in the primary winding. This would involve the manipulation of the formula; .
23. Voltage induced into,
Therefore,
24. Value of resistor, this required none of the formula’s stated it was a simple case of using Ohm’s law, therefore;
25. Finally to find the current in the primary winding it was necessary to initially find the power output of the secondary winding.
Utilizing the fact that we know it is possible to calculate the current in the primary winding by manipulation of the power formula;
TASK 4
Series R-L-C tuned circuit
26. This task involves the prediction of resonant frequency, bandwidth, impedance at resonance and Q-factor of an R-L-C tuned circuit. The first to be considered will be an R-L-C series circuit as shown below.
27. The resonant frequency for the circuit can be found by using the formula,
28. The circuit current can be found by using Ohms law,.
29. The circuit Q-factor can be found,
30. The circuit bandwidth can be found by dividing the by the Q-factor,
31. If the resistor within the circuit was increased in size too the effects upon the circuit characteristics would change. The current flow would decrease as would the Q-factor, however the bandwidth would increase. The re-calculated figures can be seen below.
32. The current within a series R-L-C circuit will always be at its maximum when the circuit is operating at resonant frequency. However the amount of current flowing in the circuit and its phase angle can be found by using the universal resonance curve (URC) (Enclosure 1). Providing is known the current flowing in the circuit can be found. In the next example the frequency given to look at was, we already are aware thatfor this circuit is. By using the equation below we are able to find a value for this is then applied to the URC to find both current and.
The is known to be 21.679, is 7.3412 , is the difference in frequency between and the frequency you wish to look at, in this case -341 , minus because it is less than the value.
33. With a value of -1 the value of current and its can be found. Plotting the -1 figure against both current curve and phase angle curve will provide both values, the current value has to be multiplied by the amount of circuit current prior to resonance in the circuit, thus the I value is 0.45 multiplied by 10 this will give a current at of . The phase angle is simply read from the curve, in this instance it is.
Parallel R-L-C tuned circuit
34. A parallel R-L-C circuit as shown below will operate in a different way, several calculations focus upon different values as can be seen.
35. Firstly to calculate the circuit resonant frequency () the equation used is different to that of the series R-L-C circuit.
36. The circuit resistance is found using the formula,
37. The current flow at can be found once again using Ohms law,
38. The Q-factor can be calculated using the exact same equation as used for the series circuit therefore,
39. Bandwidth is calculated as previously, therefore,
40. Perhaps the most noticeable difference between the two circuits is the construction. However at resonance, the circuit current will be maximum in a series circuit and minimum in the parallel. In order to show this clearly the circuits were built and tested using Multisim.
For the series circuit it can clearly be seen by the screen print that at resonance the circuit current is at a maximum. Also at resonance the phase change takes place. A closer look at the results and you can see the bandwidth as shown by the cursors to be .
AC analysis results for series R-L-C circuit.
41. Replacing the resistor with a larger resistor has the effect of increasing the bandwidth to. It does however reduce the circuit current and the Q-factor. This can be seen in the second screen print.
AC analysis of series R-L-C circuit with larger resistor.
42. As previously stated the parallel circuit has different characteristics to that of the series circuit. The screen prints that follow illustrate this.
The resistor is included in the circuit to allow better measurement of the circuit characteristics, it effect upon the circuit is unnoticeable.
AC analysis for parallel R-L-C circuit.
43. Once again Multisim gives results for bandwidth via cursors and current values at these points. Most noticeable is the change in the shape of the response curve, from convex to concave.
44. For further study an additional resistor was added to the circuit. Its addition gives an increase in bandwidth to and increase in minimum current to. By utilizing the equation,we can find the value of the bandwidth and the value of current at these two points, . This can be seen in the screen print that follows.
Parallel circuit with resistor added
AC analysis of parallel circuit with larger resistor added.
TASK 5
45. There are two main types of attenuator, the -type and the T-type. Task five involves the design of an attenuator of the T-type with a characteristic resistanceof and an attenuation factorof 10.
Figure 8.-Type Attenuator Figure 9.-Type attenuator
46. The equations used to calculate the value of resistors to be used can be seen below and are dependant upon the type of attenuator.
For the -Type attenuator the value for the value for .
For the -Type attenuator the value for the value for .
47. This part of the assignment involved the design of a -Type attenuator, therefore the equations for this type of attenuator were used. The characteristics of the attenuator needed to be such that it’s would be equal to with a value of 10. With this in mind the equations below were utilized to give the values for and .
48. With the values for and calculated the circuit could be constructed to confirm its attenuation and characteristic resistance. By re-arranging the circuit it’s total resistance could be calculated prior to circuit assembly. With a resistor of value placed across the circuit output the total circuit resistance felt at the input should equal , the value of .
Figure 10. Circuit resistance calculation.
49. By using the equations to calculate resistance in series and parallel circuits the following was calculated.
50. To enable a clearer understanding Multisim was utilized to show the circuit functionality. Thus showing clearly the input resistance felt from the attenuator.
51. With the input proven the next task was to prove that the attenuator would fulfill its role and attenuate the input voltage by a factor of ten. Once again the versatility of Multisim was used to show this in the clearest manner for an input voltage of for the attenuator to function correctly an output voltage of would be expected. The result can be seen in figure 12 .
52. With both the attenuator proven and the value the task was completed. For further study I carried out the calculations for a -Type attenuator which gave values for and respectively as and . Once again Multisim was utilized and proved all calculations to be correct.
Figure 11. Input resistance
Figure 12. Output Voltage
TASK 6
53. This task involves finding the first nine components of the Fourier series for the waveform shown below.
Figure 13. Waveform for Fourier analysis
54. The first task is to define the limits, taking into consideration the pulse duration of and knowing that a full cycle is equal to it can be derived that the pulse is equal to .
By using the basic equations for Fourier analysis as shown below, the first nine components can be found.
55. In the last formula is the DC component of the waveform and are the fundamental frequencies with the preceding etc know as the Fourier co-efficient.
56. With the basic formula now described the formula for this particular waveform was derived as below.
The second portion of the equation is equal to zero and will be ignored,
Therefore, and so, and finally,
.
This final equation can now be used to calculate the components of the waveform.
57. The component is the value of DC for the period of the pulse, this is calculated to be which is equal to of.
58. To derive the components of the waveform the basic formula was derived as below,
And finally
59. This final equation was then used to derive the co-efficient as seen below.
60. With known and the first four and co-efficient calculated the final equation can be completed.
61. The equations themselves could be used to continue to calculate the co-efficient to infinity if you chose to do this, however the task only asked for the first nine which have been derived and placed within the final formula.
CONCLUSION
62. Once again this assignment has provided a chance to attain knowledge and understanding of electrical theory’s and circuits. The knowledge gained will no doubt be of considerable use in future studies. Perhaps the most notable matter is how important Multisim can be in proving the theory. The circuits that have been built during this assignment could have been constructed on ‘bread-boards’ using discrete components. The tolerances would however have been questionable, Multisim gives the modern circuit designer and student alike the ability to design, test and improve on the circuit with the up most speed, accuracy and ease.
BIBLIOGRAPHY
Mike Tooley & Lloyd Dingle, Higher National Engineering, Edexcel, Newnes, Oxford, 2000.
