BOOST REGULATORS

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:

dlL/dt = VL/L

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

CHARACTERISTICS OF ABOOST REGULATORDC-DC switching regulator

OUTPUT voltage is always higher than the INPUT voltage (during normal operation)

OUTPUT cannot be isolated from the INPUT

BOOST REGULATORCIRCUIT DIAGRAM

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.

The inductor L feeds the output capacitor C and load resistor RL through the rectifying diode D.

Regulation of the output voltage is accomplished by varying the duty cycle of the switch with respect to input voltage changes.

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.

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]

lR = lCVR = VC[Eq. 3]tonVoltage V(t) Ipk= (Vin)(ton) LVin0 Time, tCurrent I(t)0 Time,t

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

EQUATION FOR OFF STAGEIdeal case: VD = 0VO = Vin + VL [Eq. 4]

VO = Vin + L IL/tOFF

tOFF = T tON

IL = (VO-Vin) (T tON) / L [Eq. 5]

Voltage V(t)Vo-Vin Ipk=(Vo-Vin)(ton) T ton0Current I(t)Time,t0

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

BOOST REGULATOR APPLICATIONS

Low output power levels for auxiliary supply e.g., to step-up a 5V computer logic level to 15V for use with Op-Amps.

Almost exclusively used to Power Factor Correction (PFC)

EXAMPLEGiven:Ideal Boost ConvertertON = 50 sec (FIXED)Vin = 50VVo = 75VL = 250 HRL = 2.5

Required:f = ?tOFF = ?IINdc = ?IOUTdc = ?IL(t) = ?

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

INDUCTOR CURRENT WAVEFORM[ANSWER to c]

504540 0 50 100 150 200Current IL(A)Time,t(sec)ILdc