BOOSt SEMINAR.ppt

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<ul><li><p>BOOST REGULATORS</p></li><li><p>BASIC CONCEPTS: INDUCTOR CURRENT SLOPE VS VOLTAGEIn any given inductor with inductance L, the current slope dlL /dt is proportional to the applied voltage VL , with the proportionality constant equal to 1/L:</p><p>dlL/dt = VL/L</p></li><li><p>BASIC CONCEPTS:INDUCTOR POWER VS CURRENTIn any given inductor with inductance, L, the power absorbed and delivered by the inductor in one cycle is equal to the ff:PCYCLE = PABSORBED = PDELIVERED = 1/2L(I22-I12)fDuring steady state, the net power in the inductor per cycle is zero:PLnet = PABSORBED + PDELIVERED = 0</p></li><li><p>CHARACTERISTICS OF ABOOST REGULATORDC-DC switching regulator</p><p>OUTPUT voltage is always higher than the INPUT voltage (during normal operation)</p><p>OUTPUT cannot be isolated from the INPUT</p></li><li><p>BOOST REGULATORCIRCUIT DIAGRAM</p></li><li><p>BASIC OPERATION OF A BOOST REGULATORDC input voltage is chopped by the switch Q to produce a rectangular voltage with the respect to ground at the other end of the inductor L.</p><p>The inductor L feeds the output capacitor C and load resistor RL through the rectifying diode D.</p><p>Regulation of the output voltage is accomplished by varying the duty cycle of the switch with respect to input voltage changes.</p></li><li><p>DETAILED OPERATION:BOOST REGULATOR ON STAGETransistor Q is ON, Vin causes lL to ramp up linearly as energy is stored in inductor L.Load current lR is supplied by capacitor C.</p></li><li><p>ON STAGE: WHEN THE SWITCH IS CLOSEDdlL / dt = VL/L[Eq. 1]Ideal case: VSW = 0,lL /tON = VIN/L lL = (Vin) (tON) / L[Eq. 2]</p><p>lR = lCVR = VC[Eq. 3]tonVoltage V(t) Ipk= (Vin)(ton) LVin0 Time, tCurrent I(t)0 Time,t</p></li><li><p>OFF STAGE: WHEN THE SWITCH IS OPENWhen SW is opened, the voltage across Vsw will fly high. This is clamped by the voltage across capacitor C (assume C is very high)In trying to maintain its current, the voltage across L reverses.Energy stored in L is delivered to the load and excess inductor current IC recharges capacitor C, smoothing out load current IR</p></li><li><p>EQUATION FOR OFF STAGEIdeal case: VD = 0VO = Vin + VL [Eq. 4]</p><p>VO = Vin + L IL/tOFF</p><p>tOFF = T tON</p><p>IL = (VO-Vin) (T tON) / L [Eq. 5]</p><p>Voltage V(t)Vo-Vin Ipk=(Vo-Vin)(ton) T ton0Current I(t)Time,t0 </p></li><li><p>ADVANTAGE AND DISADVANTAGESADVANTAGESTransistor is referred to GND making it simpler to driveLow ripple current reflected to the inputProvides least-costly PFC solutionDISADVANTAGESNo isolation bet. Input and output Series diode contributes extra voltage drop.Does not provide either in-rush current or short circuit protectionLarge output voltage spikesLarge output capacitor required High output ripple current</p></li><li><p> BOOST REGULATOR APPLICATIONS</p><p>Low output power levels for auxiliary supply e.g., to step-up a 5V computer logic level to 15V for use with Op-Amps.</p><p>Almost exclusively used to Power Factor Correction (PFC) </p></li><li><p>EXAMPLEGiven:Ideal Boost ConvertertON = 50 sec (FIXED)Vin = 50VVo = 75VL = 250 HRL = 2.5</p><p>Required:f = ?tOFF = ?IINdc = ?IOUTdc = ?IL(t) = ?</p></li><li><p>SOLUTION Assuming continuous mode of operation, from [Eq. 6]:D = tON/T = (Vo-Vin)/Vo,(50 sec )/T = (75-50)/75 = 1/3T = 150 sec, f = 6.67kHz[ANSWER to b]tOFF = T tON = 100 sec [ANSWER to a]Knowing that ideally, Pin = Pout, Vin(IINdc) = Vout(IOUTdc)IOUTdc = Vo/RL = (75V)(2.5)=30A[ANSWER to b]IINdc = Vout(IOUTdc)/Vin = (75V)(30A)/(50V)=45A [ANSWER to b] From [Eq. 1], dlL / dt =VL/LIL = Vin(tON)/L = 50V(50 sec)250 H) = 10A IINmax = IINdc + 1/2 IL = 45 + (10/2) = 50AIINmin = IINdc = 45 (5) = 40A</p></li><li><p>INDUCTOR CURRENT WAVEFORM[ANSWER to c]</p><p>504540 0 50 100 150 200Current IL(A)Time,t(sec)ILdc</p></li></ul>