This energy consumed by load is supplied by source (V s) and the inductor L. Since, the average current through load is (I 1+I 2)/2, therefore, the energy consumed by load during this OFF time of chopper is given as below. Let the load voltage (output voltage) be Vo. When chopper is switched OFF, this stored energy is transferred to the load. This is evident from the circuit diagram. The voltage drop across L during ON time equal to the source voltage V s. This stored energy in L during the ON period is equal to the multiplication of voltage across the inductor, average current through it and T ON time. From the above analysis of source current and load current waveform, it is clear that, the average value of current flowing through load and inductor are same and equal to (I 1+I 2)/2.Īs discussed earlier in the article, the energy is stored in L during chopper ON time. Let us now find the expression for the output voltage of step-up chopper. Since, load only comes into circuit during the OFF period, it may be said that, load current decreases from I 2 to I 1 during OFF time. Thus, the current through the inductor decreases from I 2 to I 1 during the OFF period. When chopper is switched OFF, the source current starts decreasing from its peak value I 2 to least value I 1. During this time, no current flows through the load as shown in i o versus time (t) graph. Therefore, it may be said that the current through the inductor L rises from I 1 to I 2 during ON period. It may also be noted that, this source current flows through the inductor during ON time. When chopper (CH) is switched ON, the source current increases from its minimum value I 1 to maximum value I 2. Second waveform shows the source current i s. Therefore, it is shown as a straight line parallel to time axis. The first waveform represents the source voltage which is a DC voltage equal to V s. source voltage, source current, load voltage and load current waveform are shown in figure below. It may be noted here that, the voltage across the load increases because the inductor releases its stored energy to the load during the OFF period. Thus, the circuit works as a step-up chopper. The load / output voltage may be written as below. Thus, the output voltage exceeds the source voltage V s. As a result, the voltage across the load becomes equal to the sum of source voltage and emf induced in inductor. Since, the current through the inductor L tends to decrease, the polarity of the emf induced in inductor L is reversed as shown in above figure. Due to this behavior of L, it will force the current through the diode D and load for the entire time period T OFF. Switch OFF period: When the chopper CH is switched OFF, the current through the L can not die instantaneously rather it decays exponentially.
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