Finding the Values of L and C for a Series-Loaded Resonant Inverter to Yield Specific Output Power A series resonance inverter in Figure 7.8a with series loaded delivers a load power of PL = 1 kW at resonance. The load resistance is R = 10 Ω. The resonant frequency is f0 = 20 kHz. Determine (a) the

Question

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

A series resonance inverter in Figure 7.8a with series loaded delivers a load power of [latex]P_L = 1 kW[/latex] 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 quality factor [latex]Q_s[/latex] if it is required to reduce the load power to [latex]250 W[/latex] by frequency control so that [latex]u = 0.8[/latex], (c) the inductor L, and (d) the capacitor C.

Accepted Answer

a. Because at resonance [latex]u = 1[/latex] and [latex]|G(j\omega )|_{max}=1[/latex], the peak fundamental load voltage is [latex]V_p=V_{i(pk)}=4V_s/\pi [/latex].

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

which gives [latex]V_s = 110 V[/latex].

b. To reduce the load power by [latex](1000/250 = ) 4[/latex], the voltage gain must be reduced by 2 at [latex]u = 0.8[/latex]. That is, from Eq. (7.35),

[latex]|G(j\omega )|=\frac{1}{[1+Q_s^2(u-1/u)^2]^{1/2}}[/latex] (7.35)

we get [latex]1 + Q_s^2(u - 1/u)^2 = 2^2[/latex], which gives [latex]Q_s = 3.85[/latex].

c. [latex]Q_s[/latex] is defined by

[latex]Q_s=\frac{\omega _0L}{R}[/latex]

or

[latex]3.85=\frac{2\pi imes 20 kHz imes L}{10}[/latex]

which gives [latex]L = 306.37 \mu H[/latex].

d. [latex]f_0=1/2\pi \sqrt{LC}or 20 kHz -1/[2\pi \sqrt{306.37 \mu H imes C} ][/latex] which gives [latex]C = 0.2067 μF[/latex].

Question 7.6: Finding the Values of L and C for a Series-Loaded Resonant I...

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