Simple parallel (tank circuit) resonance
A condition of resonance will be
experienced in a tank circuit when the reactances of the capacitor and
inductor are equal to each other. Because inductive reactance increases
with increasing frequency and capacitive reactance decreases with
increasing frequency, there will only be one frequency where these two
reactances will be equal.
In the above circuit, we have a 10 µF
capacitor and a 100 mH inductor. Since we know the equations for
determining the reactance of each at a given frequency, and we're
looking for that point where the two reactances are equal to each other,
we can set the two reactance formulae equal to each other and solve for
frequency algebraically:
So there we have it: a formula to tell us
the resonant frequency of a tank circuit, given the values of inductance
(L) in Henrys and capacitance (C) in Farads. Plugging in the values of L
and C in our example circuit, we arrive at a resonant frequency of
159.155 Hz.
What happens at resonance is quite
interesting. With capacitive and inductive reactances equal to each
other, the total impedance increases to infinity, meaning that the tank
circuit draws no current from the AC power source! We can calculate the
individual impedances of the 10 µF capacitor and the 100 mH inductor and
work through the parallel impedance formula to demonstrate this
mathematically:
As you might have guessed, I chose these
component values to give resonance impedances that were easy to work
with (100 Ω even). Now, we use the parallel impedance formula to see
what happens to total Z:
We can't divide any number by zero and
arrive at a meaningful result, but we can say that the result approaches
a value of infinity as the two parallel impedances get closer to
each other. What this means in practical terms is that, the total
impedance of a tank circuit is infinite (behaving as an open circuit)
at resonance. We can plot the consequences of this over a wide power
supply frequency range with a short SPICE simulation:
tank circuit frequency sweep
v1 1 0 ac 1 sin
c1 1 0 10u
* rbogus is necessary to eliminate a direct loop
* between v1 and l1, which SPICE can't handle
rbogus 1 2 1e-12
l1 2 0 100m
.ac lin 20 100 200
.plot ac i(v1)
.end
freq i(v1) 3.162E-04 1.000E-03 3.162E-03 1.0E-02
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
1.000E+02 9.632E-03 . . . . *
1.053E+02 8.506E-03 . . . . * .
1.105E+02 7.455E-03 . . . . * .
1.158E+02 6.470E-03 . . . . * .
1.211E+02 5.542E-03 . . . . * .
1.263E+02 4.663E-03 . . . . * .
1.316E+02 3.828E-03 . . . .* .
1.368E+02 3.033E-03 . . . *. .
1.421E+02 2.271E-03 . . . * . .
1.474E+02 1.540E-03 . . . * . .
1.526E+02 8.373E-04 . . * . . .
1.579E+02 1.590E-04 . * . . . .
1.632E+02 4.969E-04 . . * . . .
1.684E+02 1.132E-03 . . . * . .
1.737E+02 1.749E-03 . . . * . .
1.789E+02 2.350E-03 . . . * . .
1.842E+02 2.934E-03 . . . *. .
1.895E+02 3.505E-03 . . . .* .
1.947E+02 4.063E-03 . . . . * .
2.000E+02 4.609E-03 . . . . * .
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The 1 pico-ohm (1 pΩ) resistor is placed
in this SPICE analysis to overcome a limitation of SPICE: namely, that
it cannot analyze a circuit containing a direct inductor-voltage source
loop. A very low resistance value was chosen so as to have minimal
effect on circuit behavior.
This SPICE simulation plots circuit
current over a frequency range of 100 to 200 Hz in twenty even steps
(100 and 200 Hz inclusive). Current magnitude on the graph increases
from left to right, while frequency increases from top to bottom. The
current in this circuit takes a sharp dip around the analysis point of
157.9 Hz, which is the closest analysis point to our predicted resonance
frequency of 159.155 Hz. It is at this point that total current from the
power source falls to zero.
Incidentally, the graph output produced
by this SPICE computer analysis is more generally known as a Bode
plot. Such graphs plot amplitude or phase shift on one axis and
frequency on the other. The steepness of a Bode plot curve characterizes
a circuit's "frequency response," or how sensitive it is to changes in
frequency.
- REVIEW:
- Resonance occurs when capacitive and
inductive reactances are equal to each other.
- For a tank circuit with no resistance
(R), resonant frequency can be calculated with the following formula:
-
- The total impedance of a parallel LC
circuit approaches infinity as the power supply frequency approaches
resonance.
- A Bode plot is a graph plotting
waveform amplitude or phase on one axis and frequency on the other.
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