Applied Superconductivity - Walther Meißner Institut - Bayerische ...

Section 4.2 APPLIED SUPERCONDUCTIVITY 183I rfsin ω rftI Tsin ω rftLR I c CL TC TampR TMFigure 4.16: The rf-SQUID inductively coupled to the resonant tank circuit.change of the resonance frequency causes a strong change of the rf-current and hence of the rf-voltageof the tank circuit.For β L,rf > 1, the situation is different, since we have to deal with hysteretic Φ(Φ ext ) curves. The totalapplied flux to the SQUID isΦ ext = Φ s + Φ rf sinω rf t . (4.2.9)It is composed of a low frequency (static) signal flux Φ s and a rf-flux Φ rf coupled to the SQUID via thetank circuit. Here,Φ rf = M · I T = M · QI rf (4.2.10)is determined by the applied rf-current I rf , the Q factor of the tank circuit and the mutual inductance M.As soon as Φ s + Φ rf exceeds the critical flux value Φ ext,c , a hysteresis loop is traced out in the Φ(Φ ext )curve. This results in an energy loss proportional to the area of the hysteresis loop and hence in a dampingof the tank circuit. It is obvious from Fig. 4.15 that the damping is minimum, if the signal flux is closeto n · Φ 0 , whereas it is maximum, if the signal flux is close to 2n+12· Φ 0 . This shows that also for the caseβ L,rf > 1 the tank voltage is a periodic function of the applied flux.In order to discuss how the tank voltage V T depends on the signal flux Φ s and the rf-flux Φ rf we discuss thesituations Φ s = nΦ 0 and Φ s = (n+ 1 2 )Φ 0. We first consider the case Φ s = nΦ 0 with n = 0. On increasingI rf the tank voltage V T initially increases linearly with I rf as long as the resulting rf-flux Φ rf = MQI rf doesnot exceed the critical value Φ ext,c . The corresponding critical rf-current is I rf,c = Φ ext,c /M and the tankvoltage isV (0)T = ω rf L T I rf,c = ω rf L TΦ ext,cM . (4.2.11)Here, the superscript 0 indicates Φ s = nΦ 0 with n = 0. If we further increase the rf-current resp. rf-fluxa jump to the k = +1 or k = −1 branch of the Φ(Φ ext ) curve occurs and a hysteresis loop is traced out(L T − αML T /L)I T . That is, we have the effective inductance L T,eff = L T (1 − αM/L). With M = α √ L T L we finally obtainL T,eff = L T (1 − α 2√ L T L/L). For α = 0 we obtain the obvious result L T,eff = L T , for α = 1 the tank circuit inductance isreduced to the effective value L T,eff = L T (1 − √ L T /L).2005