TRANSIENT RESPONSE OF A SINGLE-PHASE FULL-BRIDGE INVERTER WITH VOLTAGE-CONTROLLED SWITCHES A single-phase bridge-resonant inverter is shown in Figure 15.8. The transistors and diodes can be considered as switches whose on-state resistance is 10 mΩ and whose on-state voltage is 0.2 V. Plot the

Question

TRANSIENT RESPONSE OF A SINGLE-PHASE FULL-BRIDGE INVERTER WITH VOLTAGE-CONTROLLED SWITCHES

A single-phase bridge-resonant inverter is shown in Figure 15.8. The transistors and diodes can be considered as switches whose on-state resistance is 10 mΩ and whose on-state voltage is 0.2 V. Plot the transient response of the capacitor voltage and the current through the load from 0 to 2 msec in steps of 10 μsec. The output frequency of the inverter is [latex]f_{o}[/latex] = 4 kHz.

Accepted Answer

When transistors [latex]Q_{1}[/latex] and [latex]Q_{2}[/latex] are turned on, the voltage applied to the load will be [latex]V_{s}[/latex], and the resonant oscillation will continue for the entire resonant period, first through [latex]Q_{1}[/latex] and [latex]Q_{2}[/latex] and then through diodes [latex]D_{1}[/latex] and [latex]D_{2}[/latex]. When transistors [latex]Q_{3}[/latex] and [latex]Q_{4}[/latex] are turned on, the load voltage will be −[latex]V_{s}[/latex], and the oscillation will continue for another entire period, first through [latex]Q_{3}[/latex] and [latex]Q_{4}[/latex] and then through diodes [latex]D_{3}[/latex] and [latex]D_{4}[/latex]. The resonant period of the circuit is calculated approximately as

[latex]\omega _{r}=\left\lgroup \frac{1}{LC}-\frac{R^{2} }{4L^{2} } \right\rgroup ^{1/2}[/latex]

For L = [latex]L_{1}[/latex] = 50 μH, C = [latex]C_{1}[/latex] = 6 μF, and R = [latex]R_{1}[/latex] + [latex]R_{1(sat)}[/latex] + [latex]R_{2(sat)}[/latex] = 0.5 + 0.2 +0.2 = 0.54 Ω,[latex]ω_{r}[/latex] = 57572.2 rad/sec, and [latex]f_{r}[/latex] = [latex]ω_{r}[/latex]/2π = 9162.9 Hz. The resonant period is [latex]T_{r}[/latex]= 1/[latex]f_{r}[/latex] = 1/9162.9 = 109.1 μsec. The period of the output voltage is [latex]T_{o}[/latex] = 1/[latex]f_{o}[/latex] = 1/4000 = 250 μsec.

The switching action of the inverter can be represented by two voltage-controlled switches as shown in Figure 15.9(a). The switches are controlled by voltages as shown in Figure 15.9(b). The on time of switches, which should be approximately equal to the resonant period of the output voltage, is assumed to be 112 μsec. Switch [latex]S_{2}[/latex] is delayed by 115 μsec to take account of overlap. The model parameters of the switches are RON = 0.01, ROFF = 10E+6, VON=0.001, and VOFF = 0.0.

The PSpice schematic with voltage-controlled switches is shown in Figure 15.10. The listing of the circuit file is as follows:

Example 15.3 Full-bridge resonant inverter

SOURCE * The controlling voltage for switch S1:

V1 1 0 PULSE(0220V01US1US110US250US)
* The controlling voltage for switch S2 with a delay time
of 115 μs:
V2 3 0 PULSE(0−220V115US1US1US110US250US)

CIRCUIT S1 1 2 1 0 SMOD ; Voltage-controlled switches

S2 2 3 0 3 SMOD
* Switch model parameters for SMOD:
.MODEL SMOD VSWITCH (RON=0.01 ROFF=10E+10E+6 VON=0.001
VOFF=0.0)
RSAT1 2 4 10M
VSAT1 4 5 DC 0.2V
RSAT2 9 0 10M
VSAT2 8 9 DC 0.2V
* Assuming an initial capacitor voltage of −250V to
reduce settling time:
C1 5 6 6UF IC=−250V
L1 6 7 50UH
R1 7 8 0.5

ANALYSIS *Transient analysis with UIC condition:

.TRAN 1US 500US UIC ; Transient analysis with UIC
condition
.PROBE ; Graphics post-processor

.END

The transient response for Example 15.3 is shown in Figure 15.11.

Question 15.3: TRANSIENT RESPONSE OF A SINGLE-PHASE FULL-BRIDGE INVERTER WI...

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