BOOST CONVERTER DESIGN

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By Jackson Taylor

PROBLEM DESIGN AN BOOST CONVERTER WITH FOLLOWING REQUIREMENTS
  • VIN=4V 
  •  IIN=6A
  •  VOUT=16V ,
  • IOUT =1A
ANSWER ASSUMPTIONS FOR DESIGN OF THE CIRCUIT WE NEED FOLLOWING ADDITIONAL  PARAMETERS AND WE WILL ASSUME THESE VALUES  CONSIDERING THE FACT THAT OUTPUT WILL NOT IDEAL BUT CLOSE TO THE IDEAL STATE
  • OUTPUT RIPPLE CURRENT = 30% OF OUTPUT CURRENT = 0.3A
  • OPERATING FREQUENCY OF THE SWITCH (MOSFET HERE) =30KHz
  • OUTPUT RIPPLE VOLTAGE = 0.5% OF OUTPUT VOLTAGE
DUTY CYCLE CALCULATION DUTY CYCLE = (1-   VIN / VOUT                           =(1-   4 / 16)                            =.75  INDUCTOR VALUE CALCULATION WHEN THE MOSFET IS ON ,EQUIVALENT CIRCUIT IS GIVEN BELOW IN ABOVE STATE ,VOLTAGE ACROSS THE INDUCTOR WILL BE EQUAL THE VOLTAGE OF THE BATTERY VIN =  V FOR AN INDUCTOR VOLTAGE CURRENT BASIC RELATION IS VL = L * dI / dt         THEN,  L = VL  *  dt  / dI L =  VIN  *  dt   / dI  HERE dt = DUTY CYCLE / FREQUENCY AND dI IS THE RIPPLE CURRENT  L =  VIN * DUTY CYCLE /( FREQUENCY *dI ) L =  4 * .75 /(30KHz * 0.3) L =  333uH CAPACITOR VALUE CALCULATION FOR A CAPACITOR VOLTAGE CURRENT BASIC RELATION IS I  =  C *  dV  /  dt   I  *  dt   =  C* dV = Q (CHARGE) IN BELOW FIGURE , FIRST WAVE FORM SHOWS THE CURRENT WAVEFORM ACROSS IT. WE CAN SEE CURRENT RIPPLE IN WAVE FORM .THIS OSCILLATING WAVE IS CORRECTED WITH HELP OF CAPACITOR TO GET PURE DC VOLTAGE. CAPACITOR WILL ABSORB THIS OVER CURRENT AND STORES IN IT. IF CURRENT GOES BELOW THE AVERAGE VALUE CAPACITOR WILL DISCHARGE TO MAINTAIN CONSTANT DC CURRENT.  SHADED AREA IN THE SECOND SHOWS CHARGE STORED IN THE CAPACITOR . INDUCTOR AND CAPACITOR CURRENTS SHADED AREA SHOWS THE TRIANGLE. WE DIVIDE THIS INTO TWO RIGHT TRIANGLES. THEN CHARGE Q = AREA  OF TRIANGLE = (1/2) * (TON/2) (∆I/2)  +  (1/2) * (TOFF/2) (∆I/2)
See also
IRON BOX ENERGY SAVING SYSTEM USING ACCELEROMETER AND ARDUINO
UNDER WORK…….