Series Resonance
In the RLC series circuit, when the circuit current is in phase with the applied voltage, the circuit is said to be in Series Resonance. The resonance condition arises in the series RLC Circuit when the inductive reactance is equal to the capacitive reactance XL = XC or (XL – XC = 0). A series resonant circuit has the capability to draw heavy current and power from the mains; it is also called as Acceptor Circuit.
Simple Series Resonance
A similar effect happens in series inductive/capacitive circuits. When a state of resonance is reached (capacitive and inductive reactances equal), the two impedances cancel each other out and the total impedance drops to zero!
With the total series impedance equal to 0 Ω at the resonant frequency of 159.155 Hz, the result is a short circuit across the AC power source at resonance. In the circuit drawn above, this would not be good. I’ll add a small resistor (Figure below) in series along with the capacitor and the inductor to keep the maximum circuit current somewhat limited, and perform another SPICE analysis over the same range of frequencies:
As before, circuit current amplitude increases from bottom to top, while frequency increases from left to right. (Figure above) The peak is still seen to be at the plotted frequency point of 157.9 Hz, the closest analyzed point to our predicted resonance point of 159.155 Hz. This would suggest that our resonant frequency formula holds as true for simple series LC circuits as it does for simple parallel LC circuits, which is the case:
A word of caution is in order with series LC resonant circuits: because of the
high currents which may be present in a series LC circuit at resonance, it is
possible to produce dangerously high voltage drops across the capacitor and the
inductor, as each component possesses significant impedance. We can edit the
SPICE netlist in the above example to include a plot of voltage across the
capacitor and inductor to demonstrate what happens:
According to SPICE, voltage across the capacitor and inductor reach a peak somewhere around 70 volts! This is quite impressive for a power supply that only generates 1 volt. Needless to say, caution is in order when experimenting with circuits such as this. This SPICE voltage is lower than the expected value due to the small (20) number of steps in the AC analysis statement (.ac lin 20 100 200). What is the expected value?
Given: fr = 159.155 Hz, L = 100mH, R = 1
XL = 2πfL = 2π(159.155)(100mH)=j100Ω
XC = 1/(2πfC) = 1/(2π(159.155)(10µF)) = -j100Ω
Z = 1 +j100 -j100 = 1 Ω
I = V/Z = (1 V)/(1 Ω) = 1 A
VL = IZ = (1 A)(j100) = j100 V
VC = IZ = (1 A)(-j100) = -j100 V
VR = IR = (1 A)(1)= 1 V
Vtotal = VL + VC + VR
Vtotal = j100 -j100 +1 = 1 V
The expected values for capacitor and inductor voltage are 100 V. This voltage will stress these components to that level and they must be rated accordingly. However, these voltages are out of phase and cancel yielding a total voltage across all three components of only 1 V, the applied voltage. The ratio of the capacitor (or inductor) voltage to the applied voltage is the “Q” factor.
Q = VL/VR = VC/VR