Finding the Performance Parameters of a Half-Bridge Resonant Inverter The average output voltage of the half-bridge resonant circuit in Figure 13.9a is Vo = 24 V at a resistive load of RL = 0.8 Ω. The inverter operates at the resonant frequency. The circuit parameters are C1 = C2 = C = 1 µF,

Question

Finding the Performance Parameters of a Half-Bridge Resonant Inverter

The average output voltage of the half-bridge resonant circuit in Figure 13.9a is [latex]V_o = 24 V[/latex] at a resistive load of [latex]R_L = 0.8 Ω[/latex]. The inverter operates at the resonant frequency. The circuit parameters are [latex]C_1 = C_2 = C = 1 µF, L = 20 µH[/latex], and [latex]R = 0[/latex]. The dc input voltage is [latex]V_s = 100 V[/latex]. The on-state voltage drops of transistors and diodes are negligible. The turns ratio of the transformer is [latex]a = N_s/N_p = 0.25[/latex]. Determine (a) the average input current [latex]I_s[/latex], (b) the average transistor current [latex]I_A[/latex], (c) the peak transistor current [latex]I_p[/latex], (d) the rms transistor current [latex]I_s[/latex], and (e) the open-circuit transistor voltage V_{oc}. Neglect the losses in the transformer, and the effect of the load on the resonant frequency is negligible.

Accepted Answer

[latex]C_e = C_1 + C_2 = 2 C[/latex]. The resonant frequency [latex]\omega _r=10^6/\sqrt{2 imes 20}= 158,113.8 rad/s[/latex] or [latex]f_r = 25,164.6 Hz[/latex], [latex]a = N_s/N_p = 0.25[/latex] , and [latex]I_o = V_o/R = 24/0.8 = 30 A[/latex].

a. The output power [latex]P_o = V_o I_o = 24 imes 30 = 720[/latex] W. From Eq. (3.11),

[latex]V_{dc}=\frac{2}{T}\int_{0}^{T/2}{V_m \sin \omega t dt}=\frac{2V_m}{\pi }=0.6366 V_m[/latex] (3.11)

the rms secondary voltage [latex]V_2=\pi V_o/(2\sqrt{2} )=1.1107V_o=26.66 V[/latex]. The average input current [latex]I_s = 720/100 = 7.2 A[/latex].

b. The average transistor current [latex]I_A = I_s = 7.2 A.[/latex]

c. For a sinusoidal pulse of current through the transistor, [latex]I_A = I_p/ π[/latex] and the peak transistor current [latex]I_p = 7.2 π= 22.62 A[/latex].

d. With a sinusoidal pulse of current with 180° conduction, the rms transistor current [latex]I_R = I_p/2 = 11.31 A[/latex].

e. [latex]V_{oc} = V_s = 100 V[/latex].

Question 13.4: Finding the Performance Parameters of a Half-Bridge Resonant...

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