A schematic of the converter and how it fits in the system is shown below. The c

Question

A schematic of the converter and how it fits in the system is shown below. The converter switches are realized using current bidirectional switches, these allow both discharging and charging of the battery.

When the vehicle is accelerating, power is drawn out of the battery bank, and the converter boosts the battery voltage Vbatt to the dc bus voltage Vbus. The inductor current iL is positive under these conditions, and transistor Q1 conducts during the first interval for 0 < t < DTs. Diode D2 conducts during the second interval for DTs < t < Ts. Transistor Q2 and diode D1 do not conduct for this case.

When the vehicle is decelerating, the AC motor can act as a generator to supply power extracted from the kinetic energy of the moving vehicle. This causes a reversal of the currents of the system, and the converter extracts power out of the dc bus to charge the battery. The inductor current iL is negative under these conditions. Transistor Q2 or diode D1 conduct, while transistor Q1 and diode D2 do not conduct for this case.

The IGBT transistors can be modeled as a constant voltage source VT in series with a resistance RT. The diodes can be modeled as a constant voltage source Vd. The inductor has dc winding resistance RL.

The vehicle operates with the following conditions and parameter values:

Vbatt = 240 V

Vbus = 500 V

VT = 1 V

RT = 1.6 milli ohms

Vd = 1.2 V

RL = 4 milli ohms

Power input to AC motor drive from dc bus = 150 kW

Derive a dc equivalent circuit model for this converter under the conditions that the vehicle is accelerating and the inductor current is positive and calculate the duty cycle, D.

Explanation / Answer

The grid voltage is rectified to half sine wave by the diode rectifier. By choosing an appropriate DC-link capacitor, this wave is smoothed more and an acceptable stiff DC voltage is provided for buck-boost converter. The diode rectifier can be simplified to the equation circuit in By applying KVL and KCL on this circuit, and will be obtained: W Y = Vb X + WX (3.4) X = X WX + WX _ X (3.5) By rearranging the above equations in the state variable form for a half cycle, the state matrix equation will be obtained: X WX = 0 1 Vp 1 X 1 X_ X X WX + 1 Vp 0 W Y (3.6) This matrix can be implemented in MATLAB. By increasing the size of the capacitor the output voltage will be smoother. But as this capacitor is huge and expensive, there is always a tradeoff between the desired ripple and capital cost. By assuming a ripple around 25~30% and by trial and error, a capacitor value is found to be around X = 1 . The effect of such a dc-link capacitor to build a DC output voltage can be seen in will be obtained. As it can be seen the output voltage average is increased to 311.5V. Without the capacitor, the average voltage of the half sine wave, W with amplitude of Wu was: W = WX = 2Wu = 2 × 2302 206W

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