Finding the Currents and Voltages of a Simple Resonant Inverter The resonant inverter in Figure 7.7a has C = 2 µF, L = 20 µH, R = ∞, and Vs = 220 V. The switching time of the transistor is tsw = 12 µs. The output frequency is fo = 20 kHz. Determine (a) the peak supply current Ip, (b) the average

Question

Finding the Currents and Voltages of a Simple Resonant Inverter

The resonant inverter in Figure 7.7a has [latex]C = 2 \mu F[/latex], [latex]L = 20 \mu H[/latex], [latex]R = \infty[/latex], and [latex]V_s = 220 V[/latex]. The switching time of the transistor is [latex]t_{sw} = 12 µs[/latex]. The output frequency is [latex]f_o = 20 kHz[/latex]. Determine (a) the peak supply current [latex]I_p[/latex], (b) the average device current [latex]I_A[/latex], (c) the rms device current [latex]I_R[/latex], (d) the peak-to-peak capacitor voltage [latex]V_{pp}[/latex], (e) the maximum permissible output frequency [latex]f_{max}[/latex], and (f) the average supply current [latex]I_s[/latex].

Accepted Answer

When device [latex]Q_1[/latex] is turned on, the current is described by

[latex]L\frac{di_0}{dt}+\frac{1}{C}\int{i_0dt+\upsilon _c(t=0)}=V_s[/latex]

with initial conditions [latex]i_0=(t=0)=0 , \upsilon _c(t=0)=V_s=0[/latex]. Solving for the current gives

[latex]i_0(t)=V_s\sqrt{\frac{C}{L} } \sin \omega _rt[/latex] (7.26)

and the capacitor voltage is

[latex]\upsilon _c(t)=V_s(1-\cos \omega_r t)[/latex] (7.27)

where

[latex]\omega _r={1}/{\sqrt{LC} }[/latex]

[latex]\omega _r=\frac{10^{6}}{\sqrt{20 imes 2} }158,114 rad/s[/latex] and [latex]f_r=\frac{158,114}{2\pi }=25,165 Hz[/latex]

[latex]T_r=\frac{1}{f_r} =\frac{1}{25,165} =39.74 \mu s [/latex] [latex]t_1=\frac{T_r}{2}=\frac{39.74}{2} =19.87 \mu s[/latex]

At [latex]\omega _rt=\pi [/latex]

[latex]\upsilon _c(\omega _rt=\pi )=V_{c1}=2V_s=2 imes 220=440 V[/latex]

[latex]\upsilon _c(\omega _rt=0 )=V_c=0[/latex]

a. [latex]I_p=V_s\sqrt{C/L}=220\sqrt{2/20}=9.57 A[/latex].

b. [latex]I_A=f_o\int_{0}^{\pi }{I_p \sin heta d heta }=I_pf_o/(\pi f_r)= 69.57 imes 20,000/(\pi imes 25,165)=17.6 A[/latex]

c. [latex]I_R=I_p\sqrt{f_ot_1/2} =9.57\sqrt{20,000 imes 19.87 imes 10^{-6}/2}= 31.01 A[/latex].

d. The peak-to-peak capacitor voltage [latex]V_{pp} = V_{c1} - V_c = 440 V[/latex].

e. From Eq. (7.24),

[latex]f_{s(max)}=\frac{1}{2t_{sw}}[/latex] (7.24)

[latex]f _{max }= 10^6/(2 × 12) = 41.67 kHz[/latex].

f. Because there is no power loss in the circuit, [latex]I_s = 0[/latex].

Question 7.3: Finding the Currents and Voltages of a Simple Resonant Inver...

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