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## 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.

## 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 $C = 1.875 μF$