What is the name of a parallel resonance circuit
Parallel resonant circuit
The parallel resonant circuit is an arrangement of two electrical energy stores of different nature, it is of great importance in electrical engineering, especially in communications engineering. A resonant circuit is characterized by the fact that energy is periodically exchanged between different forms of energy. In electrical engineering, these are the electric and magnetic fields.
1.1 Analogy of mechanics to electrical engineering
In mechanics, the mass corresponds to the inductance and the spring force to the capacitance. If you calculate both systems, the electrical or the mechanical, you get the same type of differential equation. Their approach can be found in countless works of physics and mathematics.
At the end of the 19th century, attention was drawn to electrical oscillating circuits, including the interaction of electrical sparks, Leyden bottles (early capacitors) and wound wire coils. At that time, mechanically driven spark inductors could be used to generate high electrical voltages, which in this way store separate charges in glass containers on metallic coverings. In investigations into the measurement of the duration of electrical spark discharges, it was also found that the spark discharge does not take place as a direct current, but as an alternating current. During this measurement, very rapidly rotating mirrors were placed near the spark gap and the reflected light from the spark was photographed on a film strip. The duration of the spark could be calculated from the speed of rotation, the distance and the exposed length. In the process, unexposed periods were also discovered and a periodicity of the spark light was assumed from this. In this case, the spark has excited an oscillating circuit. Parasitic oscillating circuits are basically always present, even without consciously arranged components, it is only a question of the order of magnitude of the quality and the area where they are effective. Spark inductors were the first transmission systems to be of technical importance. The incentive was also to improve transmission systems in order to simplify and improve the controllability of constant frequencies. It happened around 1914 as an important discovery by Mr. Meißner (then Telefunken) to develop a feedback oscillator that enabled a constant, easily controllable frequency, around this time the first electron tubes were in series production. This combination of amplifying tube and LC resonant circuit with inductive positive feedback got the name of the inventor, Meißner oscillator. A large number of other circuit variants of coupled LC oscillation were developed in the following years, but they are all based on the same principle of the three elements: positive feedback - amplification - LC resonant circuit. Electrical engineering is unthinkable without oscillating circuits.
1.3 Historical literature
Those interested in electrical engineering are particularly advised to study electrical engineering literature from the turn of the century and the time of the world wars. These books are usually very vivid and narrative, they explain the discovery, the why and how something happened. Incredibly helpful bridges to the past, many of the most modern methods used today can be understood more quickly in their basis.
2. Loss-free parallel resonant circuit
Starting with the lossless oscillating circuit, it is a theoretical special case. The ideal parallel resonant circuit consists of two energy stores, a capacitance C and an inductance L. The assumed special case is the assumption that the resonant circuit would be loss-free, in reality this does not occur, but it is a very helpful thought model for understanding, as in many applications poverty loss is one of the prime objectives.
Figure 1 shows the two energy stores C1 and L1.
in a capacitance C is the electric charge in the form of the electric field stored, the amount of energy is: W = 0.5 * U² * C 
in an inductance L is the electrical charge in the form of the magnetic field stored, the amount of energy is: W = 0.5 * I² * L 
 is derived from the integral of work = integral from C * u * du, for  it is similar.
The stored amounts of energy (here in the units VAs) are proportional to the capacitance, inductance and even quadratic to the voltage and current. Both energy storage devices, electric and magnetic fields, are inextricably linked. The breakdown and build-up of an electric field always causes a magnetic field and vice versa, it can also be said that the loss of one field increases the other, the forms of energy exchange periodically.
2.1 lossless parallel resonant circuit at rest
The observation here is based on a dormant state, which means that none of the energy stores carries its energy form. No current flows and there is no charge in the capacitor. In addition, the ideal oscillating circuit is located in an electrically energized environment (- otherwise the ideal undamped oscillating circuit would not remain at rest - thought model).
2.2 Excitation of the lossless oscillating circuit
Now the resting oscillating circuit is stimulated by an energy surge from the outside, which is carried to the oscillating circuit. For this purpose, a short current surge into the resonant circuit would be sufficient, for example. The free numerical circuit simulation program LTspice / Switcher CADIII from Linear Technology, available at www.linear.com, is a very welcome aid for representation
Figure 2 shows the circuit for the excitation process of the lossless resonant circuit
on the left a DC voltage with 2 volts and an internal resistance of 0.001 ohms.
Duration of the simulation 0 to 5 ms, the smallest maximum calculation step size is limited to 10 ns.
S2 is a voltage-controlled switch with a forward resistance of 0.001 ohms, blocking resistance 100 megohms, Vt is the switch-on voltage of 0.5 volts. The hysteresis is -0.4 volts, i.e. the switch switches off again at 0.1 volts. S1 and S2 are identical.
the voltage sources V2 and V3 are used to control S1 and S2.
V2 and V3 switch 1 volt rectangles at the specified times, these are from left to right. listed as x-y value pairs.
C1 and L1 form the resonant circuit.
all parameter settings are quasi-ideal and describe an almost loss-free circuit. The theoretically ideal values cannot be set for all parameters, as the program requires a minimum of reality.
Figure 3 shows the control voltages V (eina) and V (einb) and the voltage V (out) on the resonant circuit
At the time 0 ms of switching on, there is no energy whatsoever in C1 and L1.
At the point in time 0.5 ms, V2 suddenly rises to 1 volt. When the threshold voltage Vt = 0.5 volts is exceeded, switch S2 switches through, the switch resistance drops suddenly from 100 megohms to 0.001 ohms, as a result of which the capacitor C1 charges through the low-impedance voltage source to the nominal value of the DC voltage source of 2 volts in the shortest possible time, a very large charging current flows briefly (not shown). After a millisecond, S2 switches off again and is high-resistance again. The capacitor C1 is now charged to 2 volts and is, so to speak, disconnected from the rest of the circuit (both switches are open), it has stored its energy as an electric field, measurable as a voltage. (In terms of mechanics, the C is comparable to a stretched spring).
At 1.5 ms, switch S1 closes and has a low resistance, which suddenly connects the capacitor to the inductance. The charged capacitor now acts as a voltage source for the inductance connected to it. A current flow from the capacitor into the uncharged inductance begins immediately, the current flow from the capacitor means a loss of its charge, the amount of energy in the capacitor decreases, measurable as a voltage reduction.
At the time of the swing process; now this energy that has flown out of the capacitor has to go somewhere? Yes, since energy can neither be generated nor destroyed according to the rules of classical physics, the energy loss of the electrical field in the capacitor is absorbed by the inductance. C and L are connected in series here (yes in series, otherwise nothing is connected to them!), I.e. the same current flows in C and also in L according to the knot rule, it can be called an oscillating current. The beginning of the current flow from the C out into the L means at the same time the build-up of a magnetic field in the inductance, everything happens at the expense of the electric field. At the point in time when the capacity is almost discharged, the change in voltage of the capacity over time reaches its highest value; at this point in time the magnetic field reaches its maximum, and the highest current then flows. A high current flow in an inductor means a high level of stored energy. After half a period, the inductance plays the ball back and builds up the electric field again, this time at the expense of its own magnetic field. A magnetic field with the associated directed current is clearly comparable to a piece of moving mass; it moves once if it wants to move on in its old direction (current direction). The energy of the moving mass lies in its kinetic energy; in the case of inductance, the built-up magnetic field corresponds to the kinetic energy. With a moving mass you need force to stop its movement, only when the movement has been stopped, the kinetic energy is also zero. In the case of inductance, it is comparable that the magnetic field is only zero when the flow of current has been stopped. The moving mass is stopped by a spring that is tensioned by the movement and the tensioning of the spring lasts as long as the kinetic energy can still exert a movement of the spring. The spring thus corresponds to the capacity of electrical engineering. The spring can store energy in its tensioned state; if it is released, the released movement can accelerate masses through its force and thereby supply kinetic energy to them again.
The essence of a loss-free circle, both fields periodically exchange their energy with each other, no energy once supplied leaves the oscillating circuit.
As soon as the capacitor is connected, the oscillating circuit is excited, but the circuit does not absorb more energy than has been supplied to it. In this case, according to , these were W = 0.5 * 2V * 2V * 4.7As / V * exp-6 = 9.4 * exp-6 * VAs with (VAs = watt seconds = joules). The voltage amplitude in the resonant circuit is the excitation voltage of 2 volts, since the law of conservation of energy naturally also applies to a lossless resonant circuit, this knowledge is important. Only as much energy can oscillate in the loss-free oscillating circuit as has been fed in once.
If the simulation is extended over time to long values, the once excited oscillation lasts infinitely long; this simulation is of course pointless. (Correctly, one would have to say that this applies to 100% loss-free switches and sources, but this has been omitted here for reasons of simulability).
One must realize that the circle is oscillating because the oscillation process is the only possible form of energy conservation, the only possible state of equilibrium of this system. Nature constantly tries to keep an existing state and not to disturb it, every change of a basic state requires either the supply of energy or removal from a system. In this case, energy was supplied, the answer of the circle is the oscillation, since there is no loss, the oscillation lasts indefinitely. One speaks here of a free undamped oscillation at the resonance frequency, in this free, unaffected case also of the natural frequency.
Figure 4 with oscillating current I (L1)
In Figure 4, the oscillating current I (L1) was also drawn. It is the oscillating current, phase-shifted by 90 degrees, which flows back and forth between C and L in a sinusoidal manner. The phase shift can be explained by considering the differential equation for the behavior of capacitance and inductance.
UL = L * di / dt, the voltage UL across an inductance is proportional to the rate of change of the sign-evaluating current through it. In Figure 4, the current I (L1) shows its maximum change over time during the highest voltage V (out).
IL = C * du / dt, the oscillating current I (L1), which also flows through the capacitance, is then highest during the maximum change in voltage V (out) over time.
The current amplitude read is about 135mA. In order to satisfy equation , it should be possible to calculate this:
IL = sqrt (W / 0.5 * L), recalculated results in: 137.11mA, this oscillating current flows for an infinitely long time as long as the energy in the circuit remains unchanged.
3. Loss-prone parallel resonant circuit
In the ideal oscillating circuit, no more current flowed out of the oscillating circuit after excitation, its energy remained constant. That was a bare thought model of how it would be lossless and is not compatible with reality.
3.1 Forms of loss
Nature is lossy, with electrical oscillating circuits the losses are of various kinds in order to lose the absorbed oscillating circuit energy again:
Ohmic heat losses on cables and components; on wires, coil and the capacitor (ESR, active components in the ESB)
Losses due to the active power component of the emitted electromagnetic wave (despite a closed resonant circuit, a low level of radiation cannot be prevented)
Hysteresis losses in inductance, especially when using core materials
The electrical losses in capacitance
The energy given off is ultimately converted into thermal energy.
Figure 5 Lossy parallel resonant circuit
Figure 5 shows a lossy parallel resonant circuit; a resistor R connected in parallel to the resonant circuit is selected as a possible equivalent circuit ESB. Of course, this ESB does not exist in nature as a real circuit, as always, everything is much more complicated. The ESB shows an arrangement which is mathematically easy to handle and which deliberately leads the losses parallel to the resonant circuit. From a circuit point of view, it is extremely helpful to include all losses from C and L in a parallel resistor; there are conversion methods for this purpose. Another advantage of "parallelizing" the losses is that the connected external circuit can be mathematically linked to it. For example, in terms of circuitry, each connected load is nothing more than a further parallel resistance to the actual loss resistance R of the resonant circuit. The internal generator resistance of the driving signal jellyfish, for example, can also be connected in parallel easily and conveniently, making the whole thing computationally simpler and based on the circular resistance Rk.
In many cases it is perfectly legitimate to regard the real lossy oscillating circuit as lossless and to only consider external damping; Of course, this is especially the case if one does not even know the distribution of internal losses, which often happens. To anticipate, the calculation with the ESB close to reality would be complicated. The calculation by means of differential equation (especially if additional parasitic energy storage or non-linearities are added) - have fun - it quickly reaches the impossibility of analytical solvability, even transformations in image areas are no longer of any use because the reverse transformation will probably no longer succeed. The only possible solution to highly complicated networks known to me that is still meaningful is numerical methods, such as simulation, but also numerical and analytical mathematics both have one problem in common: the model has to be correct - at least the approach, in terms of approach nature is gracious and gives the calculator Even with an imperfect model, a lot is already on the way. Computational accuracy fanatics are mercilessly punished by nature, in these points experience, knowledge, experimentation, measurement and, last but not least, the spark of luck of having taken the right direction count.
3.2 single excitation of the lossy oscillating circuit
To do this, the experiment from Figure 4 is repeated with a parallel resistor.
Figure 5 Circuit for the excitation of the lossy parallel resonant circuit. A 200 ohm resistor R1 was added to the circuit, the previous parameters remained unchanged.
Figure 6 Excitation of a lossy circle.
The energy supplied once to the resonant circuit is converted into heat by the resistor R1. R1 continuously draws energy from the circuit, reducing the amplitude of the voltage V (out) and the oscillating current I (L1). Due to the decreasing circular energy, the current amplitude I (R1) also continuously decreases. The current in the resistor is strictly in phase with V (out). The amplitudes in the circle are reduced following an exponential function, they practically reach the value zero at some point (purely mathematically, however, never).The amplitudes are said to fade away, depending on a damping constant. The smaller the parallel resistance, the faster the circuit loses its energy. In the present case, one speaks of a free damped oscillation. The exponential function obeys: u = Ures * e to the power of -R / 2L * t * sin (wt + phi), the derivation of which comes from the solution of the differential equation, that would go too far on the Internet, I have to dig out and read books myself , everything else is freely written down so far.
The reader is advised to study independently with a simulation program and to look at the effects of the variation of R1 step by step. Especially with decreasing resistance, i.e. strong damping, no whole oscillations occur anymore, the result is more and more similar to the transient processes that can also be observed in practice, e.g. with square responses, an example for this, of course the same ESB does not apply, but the processes are based on also on LC oscillation processes. For example, the vector diagram of the evanescent circle would give a spinning spiral.
3.3 intermittent excitation of a lossy circle
up to now the circle has only been excited once with energy; the losses cause the amplitudes to subside. In order to maintain the oscillation permanently, it makes sense to occasionally pump up energy again.
Fig. 7 Switching of intermittent excitation
The switch S3 also switches a current pulse to the resonant circuit in order to supply the circuit, which has almost died down, with fresh energy and to stimulate renewed vibrations.
Figure 8 Intermittent excitation
The upper coordinate system shows voltages, the lower only currents. Up to 4 ms the processes are unchanged from the switching of the single excitation, the amplitudes have almost subsided. Now the DC voltage source is switched on via V (einc) and switch S3 for a pulse duration of 200 µs. During this short period of time, fresh current I (R2) flows into the circuit and stimulates the amplitudes to higher values again. The whole process could now be continuously stimulated anew, the result would be this oscillation. The analytical mathematical handling of this circuit is becoming more and more difficult, at the latest here simulation programs show their advantages, as long as the models are still so simple.
Similar processes take place, for example, when operating historical spark inductors and transmitters; this method can be used to set up transmitter systems for a Morse code. One example from today is the ignition system in motor vehicles. Here, however, the oscillation processes are dampened by a resistance, often in the spark plug connector, as they are undesirable at this point. The spectral distribution of such signals is relatively broadband and easily disturbs the environment (radio reception, etc.) Further examples are the brush fire on an electric motor, switched-mode power supplies or, even more simply, even fast "playing" on the light switch in the house is in principle a similar process. Depending on the application, the oscillation processes are either desired or undesirable and are either stimulated or dampened.
3.4 continuous excitation of the lossy circle
What could be more obvious than to stimulate the circle continuously, rather than intermittently. The continuous excitation should lead to a desired constant amplitude. The idea is now to permanently postpone losses through damping in well-dosed, very small portions. How about the idea of feeding in a constant current all the time? - to anticipate it, it doesn't work. The problem is that the coil is low-resistance for a DC current, the fed-in DC current would be converted into heat at the ESR of the coil, as this is low-resistance. The DC sizes would be short-circuited. The only solution is to feed in a current that is identical or at least very similar to the natural frequency of the circuit. If one now takes part of the oscillating circuit voltage itself as an aid for increased excitation, we have a simple active oscillator.
Figure 8 Circuit of the forced constant excitation by an external alternating voltage
In the previous diagrams, a resonance frequency of the order of 2300 Hz could be recognized. If you now feed an alternating current from the source V4 via V (einc) and switch S3, the circuit is continuously recharged again.
Figure 9 Forced constant excitation by an external alternating voltage
Up to the point in time 4 ms the circle runs as a freely oscillating damped circle, the amplitudes have strongly decayed. At 4ms, the alternating voltage V4 is added, the circuit begins to charge again approximately to the original energy value. As time progresses, for example, the original amplitudes are set again. However, the circle now does not vibrate at the original, as yet unknown, natural frequency but most likely a few Hertz next to it at the forced resonance frequency. The 4 volt AC supply voltage and the 200 Ohm series resistor were dimensioned in such a way that the original amplitudes are set approximately. Note the current I (R2), it corresponds exactly to the current that is converted into heat in R1. This means that here too the law of conservation of energy has been confirmed again. For a constant amplitude, the resonant circuit needs to be supplied with the same amount of energy as is converted into losses in it.
An important finding in this experiment is the fact that the 4 volts alternating supply voltage at node V (out) is halved - there is a voltage drop of apparently 2 volts at R2 and the remaining 2 volts at V (out). One could get to the view that the two resistances R1 and R2 are connected in series and form a voltage halver - it is the same. The important thing is, if the node V (out) appears to the outside like a 200 ohm resistor, what role do L and C have? The parallel connection of ideal L and C results in an infinitely high impedance at the natural frequency, or, in other words, the addition of the complex conductance values results in 0 Siemens.
Figure 10 Forced oscillation to 1500 Hertz, the circuit is unchanged from Figure 8 except for the exciting frequency
Figure 10 shows the forced oscillation with an arbitrarily changed frequency to 1500 Hz. The circuit oscillates as usual at its resonance frequency, at 4 ms 4 volts, 1500 Hz switch on via the 200 ohm R2. The circle begins to recharge, whereby it has now been impressed with a frequency that is below its favorite frequency, the free resonance frequency (natural frequency). As a result, V (out) extends here until it reaches 1500 Hz. The original 2 volt amplitude at V (out) remains unmatched, which is due to the fact that at 1500 Hz the parallel connection of the ideal C and L no longer results in an infinitely high impedance, but a smaller value. The parallel connection of C, L and R1 now again forms the voltage divider with R2, whereby R2 receives the most voltage drop here.
In the steady state (> 8ms), the circuit carries out a forced, undamped oscillation at a constant amplitude. Forced because it is forced to vibrate outside of its natural frequency. Undamped because it receives just as much energy as it loses in the form of losses.
4. Oscillation resistance Zs, natural frequency fo, quality Q and circular resistance Rk of the parallel resonance circuit
4.1 Oscillation resistance Zs
It was noticeable from Fig. 9 that in the case of resonance the parallel connection of C and L results in an infinitely high impedance. This is to be examined in more detail.
Figure 11 shows the magnitude and phase of the impedances of C1 and L1
The natural frequency lies exactly at the intersection of the two impedance curves of C1 and L1, in the present case at 2328 Hertz. In the case of resonance, the capacitor and the inductance have the same amount of impedance of approx. 14.6 Ohm, only with different complex signs. These 14.6 ohms are a significant size of the circle; it is called the oscillation resistance. In practice, it plays a not insignificant role in the dimensioning of resonant circuits in communications technology, but it will not be discussed in detail here; it is important, for example, in the areas of power adjustment. Take another look at Figure 4, the oscillating current I (L1) calculated and read from  and knowledge of the amount of energy was around 137mA. It will soon be clear how the oscillating current can be calculated without knowing the current or voltage.
|Oscillation resistance Zs = sqrt (L / C) |
The oscillation resistance Zs is based on the consideration that there are moments in time during the steady state of the circuit in which the energy in the capacitance and in the inductance are the same, this always happens twice per period, one single time when the current flows out of the capacity and once when the current flows back into the capacity. In these moments of energy equilibrium from the electric and magnetic field, the two equations  and  can of course be equated, the often unknown amount of energy in the circle can be easily eliminated and the result for the oscillation resistance can be found in . The oscillation resistance can also be calculated as the quotient of maximum voltage and maximum current of the steady, free undamped circuit.
The following applies to our previous examples: Zs = sqrt (0.001H / 4.7µF) = 14.5865 Ohm
It should be emphasized that if it is necessary to require a certain oscillation resistance at a certain resonance frequency, then there is only one possible combination of a value pair of C and L that fulfills this condition; one must not then build any circuit that contains the Frequency, but only those from this single CL pair.
4.2 Natural frequency fo of the lossless circle
It is the frequency at which a free, lossless oscillating circuit oscillates. It can be determined from the parallel connection of the two impedances of C and L. The parallel connection of two impedances is somewhat difficult to perform graphically. It would be much easier with the series resonant circuit, where it is easy to graphically add both impedances. But don't worry, there is a little trick to do the same with the parallel circle. The parallel connection of two resistors / impedances is equivalent to the addition of two conductance values. For example, the G or B can be selected as a symbol for the conductance, and the X for the impedances.
Figure 12 Derivation of the equation for calculating the natural frequency
Figure 12 shows the derivation of the natural angular frequency of the loss-free circuit. Derived from the finding, the addition of the tail units is Siemens at zero resonance.
Natural angular frequency where lossless circle wo = 1 / sqrt (L * C)  where wo = 2 * pi * fo
Natural frequency fo lossless circle fo = 1 / sqrt (L * C) * 2 * pi 
The following applies to our previous examples: Natural frequency fo = 1 / sqrt (0.001mH * 4.7µF) * 2 * pi = 2321.5134 Hertz
Both equations apply to the lossless circuit and only to a limited extent for circuits with high damping, since the loss resistance has an influence on the natural frequency. For circles with small losses, however, it is also possible to count with  or  without any problems. In circles with higher damping, the actual natural frequency is increasingly lower than predicted . For this, reference must be made to the damping constant.
4.3 Quality Q of the resonant circuit
the quality of the parallel resonant circuit decreases with increasing load. In the experiments with Fig. 9 and 10 it could already be observed that the voltage divider consisting of the source and the resonant circuit can deteriorate considerably to the detriment of the resonant circuit voltage with increasing load. The following simulation shows the amplitude response of a fixed LC combination with variable loss resistance.
Figure 13 Circuit for simulating the effect of different loss resistances.
Fig. 14 right associated amplitude response
Figure 13 is a parametric simulation. Note the program line .step param Rvar 40 1k 100. The loss resistance R1 receives a starting value of 40 ohms (the green bottom curve), in the next simulation process the resistance increases by the step width 100 ohms to 140 ohms, then to 240 ohms up to 940 ohms, the top curve represents the final value of 1 kOhm.
The parallel resonant circuit is fed from a constant alternating current of 1 mA, the frequency of which is simulated from 1.8 kHz to 3.1 kHz and the amplitude responses of V (out) have been shown. The program line .ac dec 100000 1kHz 5kHz describes the sweep width of the source I1, internally simulated from 1kHz to 5 kHz with 100000 calculations per frequency decade. The high number of calculations is necessary in order to represent the peak of the resonance curve in an unadulterated manner.
The knowledge so far was that in the case of resonance, the parallel connection of C1 and L1 has an extremely high resistance. If the loss resistance is 1 kohm, a current of 1 mA flows through it; according to Ohm's law, this results in an AC voltage drop of 1 volt, shown in the diagram as the top curve. It is now obvious that all further losses dampen the oscillating circuit more and more; it is also said that it loses its quality. If you were to start the simulation with a very high resistance R1 of e.g. 100k, then a V (out) of 100 volts would already result in the case of resonance. High resistance values of R1 were dispensed with in order to be able to be represented in a common diagram.
A very high R1 causes the resonant circuit to come closer and closer to the loss-free circuit, in practice this means that the usable voltage drop across the circuit increases significantly and the circuit gains in quality. In the transmission of messages, parallel resonant circuits are very often interconnected in such a way that the resulting loss resistance in the circuit is as high as possible and a maximum of feasible voltage resonance of a certain frequency arises. The reception circuits of radio and television sets are an example. The task of the resonant circuits in it is to selectively filter out the desired transmission frequency and to raise the very small antenna voltages received as much as possible from the multitude of frequencies in the room in order to enable further amplification in the first place. A lot of practical problems arise here, the art of developing transmitting and receiving systems is rooted in the knowledge of resonant circuits. In electrical engineering, for reasons of component properties, it is usually advantageous to work with voltages and not currents as signals, which is why a common, easily generated current was used as the source in the simulation in order to obtain an easily processed voltage as a signal.
For the sake of completeness, Figure 15 shows the associated phase response. The green line belongs to the loss resistance 40 Ohm, the line with the steepest slope to the 1k resistance. As the quality of the oscillation circle increases, the phase in the area of resonance also steepens.
Fig. 16 Derivation of equation  for calculating the amplitude response when power is fed in. Once as a complex equation for the frequency response (phase and amplitude response) and as an absolute value equation for the amplitude response. The phase response is obtained by calculating the arctangent of the complex equation.
Simulation example for the quality in Figure 17
Figure 18 Simulation example of the quality
In Figure 17, the circuit was excited with a resonance frequency of 2321 Hz, the excitation is provided by an AC current source with 5 mA. As a result, 5mA flow in the ohmic loss resistance 1k. Since the circuit oscillates on resonance, the parallel connection of C and L has an extremely high resistance, i.e. no current flows into L or C from the outside. Compare I (R1) with the constant feed current of 1 mA, they are both identical.
The high oscillating currents with which the electrical and magnetic energy exchange in a circle are astonishing. Approx. 350 mA can be read in the diagram. In memory of the oscillation resistance , Zs = sqrt (L / C) = 14.58 Ohm. 5 volts alternating voltage develop on the circle, calculate 5 volts / 14.58 ohms = 343 mA.
We now simply form the ratio of the measured or calculated oscillating current Is to the current that has to be supplied to the circuit by the generator:
Case A: 343 mA / 5 mA = factor 68.6 the oscillating current is increased by a factor of 68.6 compared to the active current of the source.
Case B: In theory the same game with a loss resistance of 100 kOhm, this would result in a circuit voltage of 5mA * 100k = 500V, which would generate an oscillating current of 500V / 14.58Ohm = 34 amperes! The formation of the ratio Is / Iact = 6858 times the current increase.
The quality describes the increase in the oscillating currents compared to the feed current of the source in the case of resonance. The circle in case A has a quality Q of 68.6, case B a quality of 6858.
Therefore, in the case of resonance, the resonant circuit current Is = ICres = ILres = Q * Iactive, the equation is changed Q = Is / Iact 
The quality is also the reciprocal value of the loss factor d, so the following applies d = 1 / Q 
The derivation of the quality applies in the case of resonance when the currents in the capacitor and the coil show the same amplitude. The circulating currents can be observed with a clamp meter, but note that the energy absorbed by the clamp meter already represents a damping.
If the circuit is operated outside of resonance, e.g.below fo, the amplitude in the coil is greater than that in the capacitor - the matter can then be complicated to determine - in principle, the mean value of both qualities (determined from Ic and IL) could be about the right one, which also applies to resonance ? This subject is perhaps more illuminating.
In practice, qualities of several hundred are already very good values for conventional LC oscillating circuits made of classic individual components. In mechanics, resonant circuits can achieve very high quality levels, quality levels of several thousand are not uncommon (examples are quartz, even modern micromechanics are one of them).
4.4 Consideration of the resonant circuit quality over the bandwidth
The bandwidth in connection with resonant circuits can be determined from those frequencies in the amplitude response at which the amplitude has already decreased by -3 dB, V (out) res / sqrt (2). A band is obtained to the right and left of the resonance point, the width of this band is called the bandwidth. High quality resonant circuits have a smaller bandwidth and vice versa.
Fig. 19 Bandwidths of a circle with high and low Q
In practical measurement technology, the bandwidths can easily be read off on an oscilloscope, for example, but they can also be calculated using equation .
We use i = 5mA as in the example from Fig. 18, fres = 2321Hz, R = 1k, C = 4.7µF and L = 0.001H for this a circular amplitude of: 4.9977 volts is calculated
The bandwidth was B = Uo * 0.707106 ... i.e. the amplitude for the bandwidth is 3.533909 volts. These 3.53..V are to be inserted in  and resolved according to f. The result is: a disgusting beast of an equation, forget it. At first it is easier to determine it from the diagram:
Fig. 20 Bandwidth displayed with the cursor
The frequencies of the bandwidth were approached with the cursor, they are 2.30473 kHz on the left side of the resonance and 2.3384 kHz on the right side of the resonance. The difference between the two results in: 33.67 Hertz, exactly the same thing that the program automatically determines.
The program is programmed really intelligently, also very easy to use, my big praise to Linear Technology - these were electrical technicians who told the software engineers exactly what they wanted. Often times, the work of a real software technician looks significantly different on average - cryptic - read help fifteen times online - google after two hours and rummage through forums - only then understand how to put the tick in the program where you are looking for.
The quality Q and the bandwidth B are related as follows: Quality Q = fo / B .
So you can easily calculate Q = 2321 Hz / 33.67 Hz = 68.93 hurray - that is the same result as it was calculated from Fig. 18 with the oscillating current. (hooray is not a new SI unit).
4.5 Circular resistance Rk
The circular resistance Rk is an important variable in the resonant circuit, not to be confused with the resonant resistance Zs . In the case of our examples, the circuit resistance is the loss resistance, the circuit resistance Rk is thus parallel to the C L, which is of interest because an external load (internal generator resistance || connected load) corresponds exactly to the circuit resistance Rk.
The following relationships apply to the circular resistance Rk: Rk = wo * L * Q = Q * sqrt (L / C) .
Recalculated, the result is Rk = 2 * pi * 2321.5134Hz * 0.001H * 68.6 = 1000 Ohm, see there identical to our loss resistance 1k.
Of course, the game can be reversed:
one knows the load resistance (internal generator resistance @ fres || load resistance @ fres) this parallel connection results in the circuit resistance Rk. If this is known, the quality of the circuit can be calculated very easily with its knowledge.
Q = 1k / (2 * pi * 2321Hz * 0.001H).
Together with the knowledge of C, L and the circuit resistance Rk (Rk includes all losses for fres, generator, load, coils and capacitor losses), a complete characterization of the circuit is possible. There is no need for differential equations, image area transformations or anything else - that of course only applies to sinusoidal signals in the steady state. Nowadays, in addition to direct measurement, simulation has clearly become a valuable aid for complex signals. What kind of poor dogs the electrical engineers of the old school were without these methods - to say it in advance - thanks to their practiced mathematical knowledge, dexterity and habit of using their intellect - they still managed to develop great radios and television systems, for example.
Unfortunately, in practice there are also dominant losses, which unfortunately are in series with the inductance, for example, and therefore do not fit into the ESB being treated at all. One would now have to convert this series resistance into an equivalent circular parallel resistance.
If the coil is wound very well and there is little loss or the load resistance is already very low, e.g. when drawing significant power from the circuit, the Rk is dominantly low, so that the coil losses are negligible with a closed eye. If you wanted to calculate this correctly, the correct ESB of the circuit would have to be set and everything recalculated for this. This more correct ESB is, however, more complicated to handle than the parallel model.
4.5 Conversion of the series resistance of an inductance into a parallel resistance.
This point will be continued at the same point at a later point in time. I need a break.
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