Finding the Values of L and C for a Parallel Resonant Inverter to Yield Specific Output Power The parallel resonant inverter of Figure 7.16a delivers a load power of PL = 1 kW at a peak sinusoidal load voltage of Vp = 170 V and at resonance. The load resistance is R = 10 Ω. The resonant frequency

Question

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

The parallel resonant inverter of Figure 7.16a delivers a load power of [latex]P_L = 1 kW [/latex]at a peak sinusoidal load voltage of [latex]V_p = 170 V[/latex] and at resonance. The load resistance is [latex]R = 10[/latex] Ω. The resonant frequency is [latex]f_0 = 20 kHz[/latex]. Determine (a) the dc supply input current [latex]I_s[/latex], (b) the quality factor [latex]Q_p[/latex] if it is required to reduce the load power to [latex]250 W[/latex] by frequency control so that [latex]u = 1.25[/latex], (c) the inductor L, and (d) the capacitor C.

Accepted Answer

a. Because at resonance [latex]u = 1 and |Z(jω)|_{ max} = 1[/latex], the peak fundamental load current is [latex]I_p = 4I_s/π[/latex].

[latex]P_L=\frac{I^2_pR}{2}=\frac{4^2I^2_sR}{2\pi ^2}[/latex] or [latex]1000=\frac{4^2I^2_s10}{2\pi ^2}[/latex]

which gives [latex]I_s = 11.1 A[/latex].

b. To reduce the load power by [latex](1000/250 = ) 4[/latex], the impedance must be reduced by 2 at [latex]u = 1.25[/latex]. That is, from Eq. (7.43),

[latex]|Z(j\omega )|=\frac{1}{[1 +Q^2_p(u-1/u)^2]^{1/2}}[/latex] (7.43)

we get [latex]1 + Q^2_p (u - 1/u2) = 2^2[/latex], which gives [latex]Q_p = 3.85[/latex].

c. [latex]Q_p[/latex] is defined by [latex]Q_p = ω_0CR[/latex] or [latex]3.85 = 2π × 20 kHz × C × 10[/latex], which gives [latex]C = 3.06 μF[/latex].

[latex]f_o=1/2\pi \sqrt{LC}[/latex] or [latex]20 kHz =1/[2\pi\sqrt{3.06 \mu F imes L} ][/latex], which gives [latex]L = 20.67 µH[/latex]

Question 7.8: Finding the Values of L and C for a Parallel Resonant Invert...

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