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The Q of a Tuned Circuit

Posted: 15 May 2020 11:56 AM PDT
http://www.verulam-arc.org.uk/posts/...tuned-circuit/






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In “radio science�, ‘Q’ stands for Quality factor. In a simple resonant or
tuned circuit, (and indeed also in a single inductor or capacitor), Q is
conveniently summarised as “the component’s reactance divided by the
component’s resistance�. The higher the Q, the lower the loss in the
component’s resistance. (In most tuned circuits it is usually assumed that
capacitors are relatively perfect compared to inductors). Thus, the Q of a
tuned circuit is virtually the Q of the inductor. By some mathematical
trickery, Q also determines the magnification factor between the voltage
you feed in, (in series with the inductor and capacitor of a tuned circuit
at resonance), and the voltage you get out across either the inductor or
the capacitor alone. How does this voltage magnification come about without
the aid of a transformer?




Basically it is a matter of persuading a modest current to flow through a
high impedance. To understand this we have to go back to Ohms law, where
V=IR, or in this case V=I X, where X is the impedance of either the
inductor or the capacitor in the tuned circuit at the resonant frequency.
The secret lies in two facts: Firstly, the impedance of a reactive
component, (inductor or capacitor), is “at right angles to� ordinary
resistive impedance, (even though they are both measured in Ohms). Thus,
although reactive ohms can be added or subtracted directly to or from other
reactive ohms, they cannot be added or subtracted directly to or from
resistive ohms. They can only be added or subtracted by using the
“parallelogram of vectors� method. Secondly, the impedance of an inductor
is conventionally positive, whereas that of a capacitor is negative. So
capacitive impedance can subtract from inductive impedance and, at the
resonant frequency, produce zero impedance. The end result is that at the
resonant frequency of the tuned circuit, the current through the resistor,
inductor and capacitor in series is just that limited by the resistor, even
though the impedances of the inductor and the capacitor separately can be
very high. Thus, for a modest current, the volts across either the inductor
or capacitor separately can also be very high. As an example, figure 1
shows the RF voltage across either the inductor or the capacitor of a
resonant circuit with a Q of 12, as the frequency of an input of 1Volt in
series with the components is changed.








Although we have looked at ‘Q’ from the impedance point of view,
fundamentally, Q is defined as; “The number of radians required for the
stored energy in an isolated resonant circuit to decay to 1/e, (about 37%),
of its initial value�. However, explaining “radians� and “e� is beyond the
scope of this article except to say that each cycle of oscillation is equal
to 2 Pie radians, Pie is about 3.14159, and e is the base of Natural
Logarithms and its value is about 2.718282.












It is worth noting that after Q cycles, (as opposed to Q radians), the
amplitude has fallen to less than 5% of its initial value, which in
practice means that it can usually be taken to have died out. A little more
maths reveals that the Q of a resonant circuit also determines its relative
“bandwidth�, i.e. how it behaves when it is continuously energised at
frequencies away from its resonant frequency. For Qs significantly greater
than unity, the difference in frequency between the two “half power�
points, (3dB points), of an “amplitude versus frequency plot�, divided by
the peak or resonant frequency is equal to Q.




Only single resonant circuits can be defined by their Q. Two or more
coupled resonant circuits have a more complicated response to an input
frequency. They may have a relatively flat response over a restricted
frequency band, or even a double or multi-humped response. Much use is made
of coupled resonant circuits when designing various types of bandpass
filter.


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John, G0NVZ

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