Question

Véase el circuito oscilador mostrado de la figura VOD RFC Cc2 Сс w RG Rs (Z) Cs (2) th C C L El inductor es de 1 uH y los capacitores Ciy Cz son de 100 pF y 200 pF respectivamente. Asumiendo que los capacitores Ccı, Cc2, y Cs se comportan como cortocircuitos, encuentre la frecuencia de resonancia del circuito en Hertz

See the oscillator circuit shown in the figure

The inductor is 1 uH and the capacitors C1 and C2 are 100 pF and 200 pF respectively. Assuming that the
Capacitors CC1, CC2, and CS behave like short circuits, find the resonant frequency of the circuit in
Hertz

Answer 1

The inductor is 1 uH and the capacitors C1 and C2 are 100 pF and 200 pF respectively.

Assuming that the Capacitors CC1, CC2, and CS behave like short circuits

By This information we can form modified circuit,we can idenify that it is a collpits oscialltor,

The solution is given below In a systematic manner:

Solution stepl: Given ci 100 pF C2 = 200 pF ТАН deo given CCI, CC2 and co behaves like short circuits. Stip2 : There Modified Circuit will be VDO REC (212 (2-2) 카 = C = 10opf C2 -200pF 0070 L=1411

The marked section is called "Tank Cireset" Tank. Circuit is the 2eacn for the generation of Oscillations. And the oscillations can be deteanned frequency of by uning the resonant resonant frequency frequency of the 1. Tank Circuit (fo) Stép3 : Calculation of Repait for meryl he can alsowe that the coscillator is known as Calpitty Oscillato?". The resonant is given by frequency fr 1 2 TLIC Here for ý the resonant frequenry c is the equivalent Capacitauco. of the series Combination of cr and C2 of the tank circuit.

C = (C1X02) cltc2 (100x10-12 x 200x10-12) (100x10-12 + 200x10-12) 66.667 PF 11 J Resonant frequency for 2T (J1X104) (66-667x1013) 11 2 TT (16.66X10-17 - = 2T1 (8.164 X10-9) 1 5.130 X10-8 = 19,492,371 HZ or 19.49 M H₂ Announ: The Resonanant frequency In is 19.49 MHz 87 19,492., 371. Hz