TITLE Prople Q1 [60 marks] The problem with ripple in the output current from a single-phase full bridge inverter shown in Figure Q1(a) is to be studied. The first harmonic of the output voltage is given by Vor at f = 47 Hz. The load is given as L = 100 mH in series with an ideal voltage source e.(1). It is assumed that the inverter works in square wave mode. (1) Calculate the value of Va which gives Vo1 = 220 V. [8 marks] (ii) Calculate the peak value of the ripple current based on Figure Q1(b). [15 marks] The conditions are the same as in (1), but the inverter operates in sinusoidal PWM-mode, bipolar modulation. The frequency and amplitude modulation index are m=21 and m, 0.8., respectively. (iii) Which value of Va gives Vor = 220 V? (6 marks) (iv) Explain why the ripple current has its peak value at the zero crossing of the first harmonic voltage, and find this value. (8 marks) (v) Compare the values found in sections (il) and (iv). [5 marks) The conditions are the same as in (I), but the output voltage is given by voltage cancellation and Va=389 V (see Figure Q1(c)). (vi) Calculate the "overlap angle" a and peak value of the ripple current. [8 marks) (vii) Discus and justify your results. [10 marks] im Animation att (6) Figure Q1 Ripple in inverter output (left) square-wave, (right) PWM. VAN Induction-motor load E. VL VLI + Uripple Single-phase inverter 180 V + M. E, sin , PA DR. ITA Ts. [ 180 V Figure Q1(a) Full bridge inverter supply induction-motor. B (180-a) IDA- D- ITA- N Ed -V (180-a) .. (180 (90 - ) Figure Q1(c) TITLE Prople Q1 [60 marks] The problem with ripple in the output current from a single-phase full bridge inverter shown in Figure Q1(a) is to be studied. The first harmonic of the output voltage is given by Vor at f = 47 Hz. The load is given as L = 100 mH in series with an ideal voltage source e.(1). It is assumed that the inverter works in square wave mode. (1) Calculate the value of Va which gives Vo1 = 220 V. [8 marks] (ii) Calculate the peak value of the ripple current based on Figure Q1(b). [15 marks] The conditions are the same as in (1), but the inverter operates in sinusoidal PWM-mode, bipolar modulation. The frequency and amplitude modulation index are m=21 and m, 0.8., respectively. (iii) Which value of Va gives Vor = 220 V? (6 marks) (iv) Explain why the ripple current has its peak value at the zero crossing of the first harmonic voltage, and find this value. (8 marks) (v) Compare the values found in sections (il) and (iv). [5 marks) The conditions are the same as in (I), but the output voltage is given by voltage cancellation and Va=389 V (see Figure Q1(c)). (vi) Calculate the "overlap angle" a and peak value of the ripple current. [8 marks) (vii) Discus and justify your results. [10 marks] im Animation att (6) Figure Q1 Ripple in inverter output (left) square-wave, (right) PWM. VAN Induction-motor load E. VL VLI + Uripple Single-phase inverter 180 V + M. E, sin , PA DR. ITA Ts. [ 180 V Figure Q1(a) Full bridge inverter supply induction-motor. B (180-a) IDA- D- ITA- N Ed -V (180-a) .. (180 (90 - ) Figure Q1(c)
