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 1e12
l1 2 0 100m
.ac lin 20 100 200
.plot ac i(v1)
.end
freq i(v1) 3.162E04 1.000E03 3.162E03 1.0E02
                                
1.000E+02 9.632E03 . . . . *
1.053E+02 8.506E03 . . . . * .
1.105E+02 7.455E03 . . . . * .
1.158E+02 6.470E03 . . . . * .
1.211E+02 5.542E03 . . . . * .
1.263E+02 4.663E03 . . . . * .
1.316E+02 3.828E03 . . . .* .
1.368E+02 3.033E03 . . . *. .
1.421E+02 2.271E03 . . . * . .
1.474E+02 1.540E03 . . . * . .
1.526E+02 8.373E04 . . * . . .
1.579E+02 1.590E04 . * . . . .
1.632E+02 4.969E04 . . * . . .
1.684E+02 1.132E03 . . . * . .
1.737E+02 1.749E03 . . . * . .
1.789E+02 2.350E03 . . . * . .
1.842E+02 2.934E03 . . . *. .
1.895E+02 3.505E03 . . . .* .
1.947E+02 4.063E03 . . . . * .
2.000E+02 4.609E03 . . . . * .
                                
The 1 picoohm (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
inductorvoltage 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.
