quarz simulation
Dieses Beipiel wurde durch ein VNA-Seminar des DARC G01/G02 in Aachen im Juni 2026 angeregt.
Vielen Dank dafür!
Eine einfache Ersatzschaltung eines Schwingquarzes kann so aussehen:
+---L---C---R---+
| |
p1 ---+ +--- p2
| |
+------C0-------+
L = 18.19716 e-3 mH
C = 14.47256 e-15 fF
C0 = 4.5559235307 e-12 pF
R = 28.80912679 Ohm
F = 9.8072119097 MHz
R*Q = 1121.3190689 ==> Q = 38922.3
import matplotlib.pyplot as plt
import skrf as rf
from skrf import circuit, media, Network, Frequency
rf.stylely()
#freq = Frequency(9.5, 10.5, 1001, 'MHz')
#freq = Frequency(9.805, 10.81, 1001, 'MHz')
freq = Frequency(9.805, 9.81, 1001, 'MHz')
line = media.DefinedGammaZ0(frequency=freq, z0=50)
L = line.inductor(18.19716e-3)
L.name = "Inductor"
C = line.capacitor(14.47256e-15)
C.name = "Cap"
C0 = line.capacitor(4.55592e-12)
C0.name = "Cap_par"
R = line.resistor(28.80912679)
R.name = "Resistor"
plt.rcParams['figure.figsize'] = [4, 4]
LCR = L ** C ** R
LCR.name="LCR"
LCR.plot_s_db()
port1 = circuit.Circuit.Port(freq, name='P_1', z0=50)
port2 = circuit.Circuit.Port(freq, name='P_2', z0=50)
connections = [
[(port1, 0), (C0, 0), (L, 0),],
[(L, 1), (C, 0)],
[(C, 1), (R, 0)],
[(port2, 0), (C0, 1), (R, 1), ],
]
Q_circuit = circuit.Circuit(connections)
Q = Q_circuit.network
Q.name = "quarz"
Q.plot_s_db()
Free software enabled me to learn, many thanks to many creators:
