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#31
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Bill Bowden wrote in news:1177821138.653191.285430
@u30g2000hsc.googlegroups.com: Does anyone know why the distributed winding capacitance of a loop antenna, or any inductor, degrades the efficiency? It would seem that a loop antenna with 100pF of winding capacitance in parallel with a external capacitor of 200pF would resonate at the same frequency as a antenna with no winding capacitance and a external capacitor of 300pF,but apparently that's not the case. The best explanation I got was that winding capacitance represents 'low Q' and a external tuning capacitor represents ' High Q' What is the difference between high and low Q, and why should a loop antenna with no winding capacitance perform any better than one with 50% of the total capacitance in the windings? Where is the energy loss? Bill, Some thoughts about inductor loss and self capacitance: Consider and ideal coil (ie lossless, no distributed capacitance) in series with a small ideal resistor to represent its loss, the combination having high Q. Connect it to a constant voltage source at some frequency and observe that the current lags the voltage by almost 90 deg. Now shunt that combination coil+resistor with a small lossless capacitor, and note that the current in the capacitor will be small in magnitude, and leading the applied voltage by 90 degrees. The effect of the capacitor is to reduce the total current, and not change its phase slightly. So the combination of coil & series resistance, & shunt capacitance draws less current and at slightly lower (lagging) phase, so it appears to be a smaller but lossier inductor. The discussion above is about conditions below the self resonance of the total combination. Now, real inductors might be represented by a simple circuit as dealt with above, but it is an approximation only. A better representation of real inductors is more complex and highly dependent on the frequency, geometry and materials. An example of the influence of these factors is that a ferrite cored inductor usually needs less turns (and less capacitance) than an air cored inductor of the same inductance; a bifilar split transformer winding on a toroid increases the self capacitance compared to a normal winding, albeit with higher flux leakage; close spaced windings reduce the number of turns needed, and resistance due to decreased wire length however proximity effect increases the resistance per turn. Design is about finding an optimal solution to these effects for the intended usage. Distributed capacitance is not of itself necessarily lossless, the materials in which the electric field alternates might not be ideal dielectrics, and so a further loss is contributed by dielectric losses. Operation of coils approaching their self resonance increases the loss due to this effect. Owen |
#32
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Owen Duffy wrote in
: .... Some thoughts about inductor loss and self capacitance: Consider and ideal coil (ie lossless, no distributed capacitance) in series with a small ideal resistor to represent its loss, the combination having high Q. Connect it to a constant voltage source at some frequency and observe that the current lags the voltage by almost 90 deg. Now shunt that combination coil+resistor with a small lossless capacitor, and note that the current in the capacitor will be small in magnitude, and leading the applied voltage by 90 degrees. The effect of the capacitor is to reduce the total current, and not change its phase slightly. So the combination of coil & series resistance, & shunt capacitance draws less current and at slightly lower (lagging) phase, so it appears to be a smaller but lossier inductor. A workup at 10MHz of some numbers for a 10uH inductance in series with 10 ohms loss resistance gives Z=10+j628, Q is 62.8. When this is shunted by a 2pf ideal capacitor, the impedance is now 11.8 +j682, Q is 58, apparent inductance is 10.9uH in series with 11.8 ohms of resistance. The small shunt capacitor has increased the apparent inductance, and decreased the Q. Where has this newfound loss come from? The current in the coil's loss resistance is higher than the current from the source, so whilst the two terminal equivalent has a higher impedance, the higher internal current is generating larger loss from the smaller resistance. This is the "circulating current" people are talking about. Owen Note |
#33
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In message , Owen Duffy
writes Owen Duffy wrote in : ... Some thoughts about inductor loss and self capacitance: Consider and ideal coil (ie lossless, no distributed capacitance) in series with a small ideal resistor to represent its loss, the combination having high Q. Connect it to a constant voltage source at some frequency and observe that the current lags the voltage by almost 90 deg. Now shunt that combination coil+resistor with a small lossless capacitor, and note that the current in the capacitor will be small in magnitude, and leading the applied voltage by 90 degrees. The effect of the capacitor is to reduce the total current, and not change its phase slightly. So the combination of coil & series resistance, & shunt capacitance draws less current and at slightly lower (lagging) phase, so it appears to be a smaller but lossier inductor. A workup at 10MHz of some numbers for a 10uH inductance in series with 10 ohms loss resistance gives Z=10+j628, Q is 62.8. When this is shunted by a 2pf ideal capacitor, the impedance is now 11.8 +j682, Q is 58, apparent inductance is 10.9uH in series with 11.8 ohms of resistance. The small shunt capacitor has increased the apparent inductance, and decreased the Q. Where has this newfound loss come from? The current in the coil's loss resistance is higher than the current from the source, so whilst the two terminal equivalent has a higher impedance, the higher internal current is generating larger loss from the smaller resistance. This is the "circulating current" people are talking about. Owen Note Just out of interest, if you increased the inductance to 10.9uH by increasing the number of turns, what effect would it have on the Q? Ian. -- |
#34
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Ian Jackson wrote in
: .... Just out of interest, if you increased the inductance to 10.9uH by increasing the number of turns, what effect would it have on the Q? Ian. Ian, that depends on the type of coil. A very simple view (eg if a toroidal core was used) would be that it would take a (10.9/10)^0.5 increase in turns (4.4%), inductive reactance would increase by 9% and R would increase by 4.4%, Q would increase by 4.4%. I don't really understand the relevance of the questions. Owen |
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