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 PL = 1 kW at a peak sinusoidal load voltage of Vp = 330 V and at resonance. The load resistance is R = 10 Ω. The resonant

Question

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

Accepted Answer

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

[latex]P_L=\frac{V_p^2}{2R}=\frac{4^2V_s^2}{2\pi ^2R}[/latex] or [latex]1000=\frac{4^2V_s^2}{2\pi ^2 imes 10}[/latex]

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

b. From Eq. (7.37),

[latex]|G(jω) |_{ max} = Q[/latex] (7.37)

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

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

we get

[latex](1-u^2)^2+(u/2.356)^2=2^2[/latex]

which gives [latex]u = 1.693[/latex].

c. Q is defined by

[latex]Q=\frac{R}{\omega _0L}[/latex] or [latex]2.356=\frac{R}{2\pi imes 20 kHzL}[/latex]

which gives [latex]L = 33.78 μH[/latex].

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

Question 7.7: Finding the Values of L and C for a Parallel-Loaded Resonant...

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