 myself Sunil Kalshetti, Assistant Professor, Department of Electronics Engineering, Valchan Institute of Technology, Swelapur. Today I am going to explain the step-up chopper, learning outcome. At the end of this session students can analyze step-up chopper. This is the basic arrangement of step-up chopper. The dotted rectangular represents the chopper switch which consists of power device with internal firing circuit plus commutation circuit. It is also called as boost converter. It is a DC to DC power converter with an output voltage greater than its input voltage. It is a class of SMPS, switched mode power supply containing at least two semiconductor switches, diode and transistor and at least one energy storage element, capacitor or inductor. Or the two combination. Boost converter is also used as the voltage increase mechanism in the circuit. For a step-up chopper we can obtain an average output voltage v0 greater than input voltage. Step-up chopper is used to obtain a load voltage higher than input voltage v. The values of L and C are chosen depending upon the requirement of output voltage and current. When the chopper is on the inductor L is connected in series with supply. The inductor current I rises and the inductor stores energy during the on time and this on time is generated as a T on. When the chopper is off the inductor current I forced to flow through the diode D and load for the period T off. The current tends to decrease resulting in reversing the polarity of industry EMF in the inductor. Therefore the voltage across load is given by v0 is equal to vs plus L di by dt where vs is the input voltage and the voltage across inductor is L di by dt that is v0 is greater than vs. That's why the circuitry acts as a step-up chopper and in this chopper output voltage is greater than input voltage. The large capacitor is connected across the load will provide the continuous output voltage. Diode D prevents any current flow from a capacitor to the source operating principle. The key principle that drives the boost converter is the tendency of an inductor to resist changes in current. Mode 1 T on time. When the switch is closed the current flows through the inductor in the clockwise direction and inductor stores the energy. In this duration the current flows through the input voltage source L chopper switch and these are the polarity of inductor. And in this T on the the energy is stored in the L the polarity of the left side of the inductor is positive. This is the circuit diagram for the mode 1 the switch is closed mode 2 off state. When the switch is opened the current will be reduced as the impedance is higher. Therefore the change or reduction in current will be opposed by the inductor. Thus the polarity will be reversed means left side of inductor will be negative and right side will be positive. As a result two sources will be in series causing higher voltage to charge the capacitor through the diode diode. In this duration chopper switch is a non-contacting set therefore current flows through the input input voltage source L D load and the polarity of inductor is like this. Right side is positive left side is negative. Here the switch is open and the only path offered to inductor current is through the flyback diode D and the capacitor C and the load. This is the circuit diagram for the mode 2 when switch is open. When the chopper is on the voltage across inductor is given by V L is equal to L DI by DT. And in this duration the V s is appears across the V L therefore V s is equal to V L is equal to L DI s by DT. And the duration of T is greater than 0 less than T on. Here DI is the change in current I max to I min. Therefore V s is equal to L I max minus I min divided by T on is equal to L delta I upon T on. Therefore delta I is equal to V s upon L into T on. When the chopper is off that is switch s is open. The instantaneous output voltage is V L is equal to V s plus L DI L by DT. For V s plus L DI s by DT. Now delta I is equal to V s upon L into T on. Here V s plus L delta I upon T off. So substitute the value of delta I from the previous equation. Therefore equation becomes V L is equal to V s into bracket 1 plus T on upon T off. Now divide numerator and denominator by capital T. And after solving this we obtain V L is equal to V s into bracket 1 upon 1 minus K. From the equation it is clear that for K is less than 1 load voltage is greater than supply voltage. And the circuit acts as a step up chopper. If K is equal to 0 V L is equal to V s and K is equal to 1 V L is equal to infinity. Equation for minimum and maximum current. Assuming L is large enough and the on time and off time are small enough to maintain linear rise and fall of current. The current waveform is as shown in figure. Now during T on current I rises gradually. The current increases from I into I max and this current is denoted as I 1. And during T off I decreases and this current is denoted as I 2. More on when the chopper is on the equivalent circuit is as shown above. And the voltage equation can be written as V s is equal to L di 1 by dt. Therefore the rate of rise of current di 1 by dt is equal to V s upon L. Therefore di 1 is equal to V s upon L into dt. Referring to waveform the current can be expressed as I 1 of T is equal to V s of V s upon L into T plus I min. Where I min is the initial current at T is equal to 0. At T is equal to KT I 1 of T is equal to I max. Therefore I 1 is equal to V s upon L into KT plus I min. This current is denoted as I max. Therefore I max minus I min is equal to delta I is equal to V s upon L into KT. During mode 1 current must rise and for that di 1 by dt must be greater than 0. And the V s must be greater than 0. More on when the chopper is off the equivalent circuit is as shown above. The voltage equation can be written as V s plus L di 2 by dt is equal to V L. Now during T off assuming linear fall in current and the equation can be expressed as I 2 of T is equal to I max minus delta I upon T off into T. At T is equal to T off I 2 of T is equal to I min. Therefore I 2 of T is equal to I max minus delta I is equal to I min. Why step up chopper is called as a boost converter. In this converter V 0 is equal to V s plus L di by dt. And output is greater than the input. That's why the name is the step up chopper or boost converter. Chopper control techniques. The operation of chopper can be classified in the following two ways. Time ratio control and current limit control. Time ratio control can be divided in two types constant frequency system and variable frequency system. Now in the time ratio control average output voltage of chopper is given by V average is equal to K into V s where K is the T on upon T. Output voltage can be varied either by varying T on with constant F by varying F with constant T on. In the constant frequency system T on is varied either keeping the F or capital T constant and the pulse width changes. That's why it is also called as a pulse width modulation control. And in the variable frequency system T on or T off kept constant vary the F. That's why the name is the variable frequency system. In the current limit control the output current is allowed to vary only between the predetermined maximum and minimum limit. If load current tends to increase beyond the maximum limit thyristor is turned off and if the load current tends to fall below the minimum limit thyristor is turned on. Since chopper operates between Imax and Imin and the current is continuous the switching frequency is determined by difference between Imax and Imin. If the difference is large switching frequency is less and if the difference is small switching frequency is large resulting higher switching losses. These are the waveforms for current limit control. These are the references. Thank you.