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Chapter 7

Q. 7.7

Finding the Values of L and C for a Parallel-Loaded Resonant Inverter to Yield Specific Output Power

A series resonance inverter with parallel-loaded delivers a load power of P_L = 1  kW at a peak sinusoidal load voltage of V_p = 330  V and at resonance. The load resistance is R = 10  Ω. The resonant frequency is f_0 = 20  kHz. Determine (a) the dc input voltage V_s, (b) the frequency ratio u if it is required to reduce the load power to 250  W by frequency control, (c) the inductor L, and (d) the capacitor C.

Step-by-Step

Verified Solution

a. The peak fundamental component of a square voltage is V_p = 4V_s/π.

P_L=\frac{V_p^2}{2R}=\frac{4^2V_s^2}{2\pi ^2R}              or               1000=\frac{4^2V_s^2}{2\pi ^2\times 10}

which gives V_s = 110 V. V_{i(pk)} = 4V_s/π = 4 × 110/π= 140.06  V.

b. From Eq. (7.37),

|G(jω) |_{ max} = Q                              (7.37)

the quality factor is Q = V_p/V_{i(pk)} = 330/140.06 = 2.356. To reduce the load power by (1000/250 =) 4, the voltage gain must be reduced by 2. That is,from Eq. (7.36),

|G(j\omega )|=\frac{1}{[(1-u^2)^2+(u/Q)^2]^{1/2}}                                                     (7.36)

we get

(1-u^2)^2+(u/2.356)^2=2^2

which gives u = 1.693.

c. Q is defined by

Q=\frac{R}{\omega _0L}      or          2.356=\frac{R}{2\pi \times 20 kHzL}

which gives L = 33.78  μH.

d. f_0=1/2\pi \sqrt{LC}  or 20 kHz = 1/2\pi \sqrt{\left(33.78  \mu H\times C\right) },which gives C = 1.875   μF