What Is The CSC Converter Used In The BLDC Converter

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The CSC converter used in the proposed BLDC drive system operating in DICM. In DICM the inductor current becomes discontinuous in a switching period. The CSC converter have three modes of operation. These are explained as follows: Mode 1: In this mode the CSC converter is in ON position, then the inductor current charges the input current and the capacitor C1 discharges the energy to the dc link capacitor, the DC link capacitor is charging as well as supplies energy to the load. Mode 2: Here the CSC converter is in off position, the inductor L1 discharges the stored energy to the DC link capacitor through diode D. Mode 3: The diode D becomes reverse biased when the inductor is going to be zero, the capacitor C1 is continuous to charging, then …show more content…

DESIGN AND CONTROL OF CSC CONVERTER The CSC converter is similar to the Cuk converter. The CSC converter includes a switch, diode and a capacitor. It is used for the design of various DC -DC converters [4-5]. Fig. 2. shows a CSC converter. In this the voltage at the output if DBR is given as, 〖 V〗_(i )=(2√(2 ))/π V_s where V_s is the the supply voltage. (1) Duty ratio of the converter is given as, D =V_dc/(V_i+ V_dc ) where V_dc is the dc link voltage. (2) Fig. 2. CSC …show more content…

The BLDC motor is driven by a three phase inverter in which the devices are triggered with respect to the rotor position as shown in Fig. 3. The phase A terminal voltage with respect to star point of the stator V_an is given as, V_an =R_(a ) i_a +L_a d_ia/dt +e_an (10) where R_(a ) is the stator resistance, L_a is the phase inductance,e_an is the back emf,i_a is the phase current of the "A" phase. Similar equations can be written for the other two phases as V_bn =R_(b ) i_b +L_b d_ib/dt +e_bn (11) 〖 V〗_cn =R_(c ) i_c +L_c d_ic/dt +e_cn (12) From above equations, the line voltage V_ab can be determined as, V_ab =V_an- V_bn = R(i_a-i_b)+ L (d(i_(a-) i_(b)))/dt +e_an - e_bn Similarly, V_bc =V_bn- V_cn = R(i_b-i_c)+ L (d(i_(b-) i_(c)))/dt +e_bn - e_cn V_ca =V_cn- V_an = R(i_c-i_a)+ L (d(i_(c-) i_(a)))/dt +e_cn - e_an subtracting V_bc from V_ab gives, V_abbc = V_ab- V_bc =R(i_(a )- 2i_(b )+ i_(c )) + L(d(i_(a-) 2i_(b+) i_(c)))/dt

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