Analysis of the Basic Resonant Inverter The series resonant inverter in Figure 7.2 has L1 = L2 = L = 50 H, C = 6 F, and R = 2 . The dc input voltage is Vs = 220 V and the frequency of output voltage is fo = 7 kHz. The turnoff time of transistors is toff = 10 µs. Determine (a) the available

Question

nalysis of the Basic Resonant Inverter
The series resonant inverter in Figure 7.2 has [latex]L_1 = L_2 = L = 50 \mu H[/latex], [latex]C = 6 \mu F[/latex], and [latex]R = 2 Ω[/latex] . The dc input voltage is [latex]V_s = 220 V[/latex] and the frequency of output voltage is [latex]f_o = 7 kHz[/latex]. The turnoff time of transistors is [latex]t_{off} = 10 \mu s[/latex]. Determine (a) the available (or circuit) turn-off time [latex]t_{off}[/latex], (b) the maximum permissible frequency [latex]f_{max}[/latex], (c) the peak-to-peak capacitor voltage [latex]V_{pp}[/latex], and (d) the peak load current I_p. (e) Sketch the instantaneous load current io(t), capacitor voltage [latex]v_c(t)[/latex], and dc supply current [latex]i_s(t)[/latex]. Calculate (f) the rms load current [latex]I_o[/latex]; (g) the output power [latex]P_o[/latex]; (h) the average supply current [latex]I_s[/latex]; and (i) the average, peak, and rms transistor currents.

Accepted Answer

[latex]V_s = 220 V, C = 6 \mu F, L = 50 \mu H, R = 2 \Omega , f_o = 7 kHz, t_q = 10 \mu s, and ω_o = 2 \pi imes 7000 =43,982 rad/s[/latex] From Eq. (7.4),

[latex]\omega _r=\left(\frac{1}{LC}-\frac{R^2}{4L^2} ight)^{1/2}=\left(\frac{10^{12}}{50 imes 6}-\frac{2^2 imes 10^{12}}{4 imes 50^2} ight)^{1/2}= 54,160 rad/s[/latex]

The resonant frequency is [latex]f_r=\omega _r/2 \pi =8619.8 Hz, T_r = 1/f_r = 116 \mu s[/latex]. From Eq. (7.6),

[latex]\alpha =\frac{R}{2L}[/latex] (7.6)

[latex] α= 2/(2 × 50 × 10^{-6}) = 20,000[/latex].

a. From Eq. (7.17),

[latex]\frac{\pi }{\omega _0}-\frac{\pi }{\omega _r}=t \gt t_{off}[/latex] (7.17)

[latex]t_{off}=\frac{\pi }{43.982}-\frac{\pi }{54.160}=13.42 \mu s[/latex]

b. From Eq. (7.18),

[latex]f_o\leq f_{max}=\frac{1}{2(t_{off}+\pi /\omega _r)}[/latex] (7.18)

the maximum possible frequency is

[latex]f_{max}=\frac{1}{2(10 imes 10^{-6}+\pi /54.160)}=7352 Hz[/latex]

c. From Eq. (7.14),

[latex]V_c=\frac{V_s}{e^z-1}[/latex] (7.14)

[latex]V_c=\frac{V_s}{e^{\alpha \alpha \pi /\omega _r}-1}=\frac{220}{e^{20\pi /54.16}-1} =100.4 V [/latex]

From Eq. (7.16),

[latex]V_s + V_c = V_{c1}[/latex] (7.16)

[latex]V_{c1} = 220 + 100.4 = 320.4 V[/latex]. The peak-to-peak capacitor voltage is [latex]V_{pp} = 100.4 + 320.4 = 420.8 V[/latex].

d. From Eq. (7.7), the peak load current, which is the same as the peak supply current, occurs at

[latex]t_m=\frac{1}{ \omega _r}tan^{-1}\frac{\omega _r}{\alpha }=\frac{1}{54,160}tan^{-1}\frac{54.16}{20}=22.47 \mu s[/latex]

and Eq. (7.5)

[latex]i_1(t)=\frac{V_s +V_c}{\omega _rL}e^{-\alpha t}sin \omega _rt[/latex] (7.5)

gives the peak load current as

[latex]i_1(t=t_m)=I_p=\frac{320.4}{0.05416 imes 50}e^{-0.02 imes 22.47}sin(54,160 imes 22.47 imes 10^{-6})=70.82 A[/latex]

e. The sketches for i(t), [latex]v_c(t)[/latex], and [latex]i_s(t)[/latex] are shown in Figure 7.5.

f. The rms load current is found from Eqs. (7.5) and (7.11)

[latex]i_3(t)=\frac{V_{c1}}{\omega _rL}e^{-\alpha t}sin\omega _rt[/latex] (7.11)

by a numerical method and the result is

[latex]I_o=\left[2f_o\int_{0}^{T_r/2}{i^2_0(t)dt} ight]^{1/2}=44.1 A[/latex]

g. The output power [latex]P_o = 44.1^2 × 2 = 3889 W[/latex].

h. The average supply current [latex]I_s = 3889/220 = 17.68 A[/latex].

i. The average transistor current

[latex]I_A=f_o\int_{0}^{T_r/2}{i_0(t) dt}=17.68 A[/latex]

The peak transistor current [latex]I_{pk }= I_p = 70.82 A[/latex], and the root-mean-square (rms) current transistor [latex]I_R=I_o/\sqrt{2}=44.1/\sqrt{2} =31.18 A[/latex].

Question 7.1: Analysis of the Basic Resonant Inverter The series resonant ...

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