EE 462 L Fall 2011 DCDC BuckBoost Converter

EE 462 L, Fall 2011 DC−DC Buck/Boost Converter 1

Boost converter i in + v L 1 I out – V in + V out – L 1 C Buck/Boost converter i in + v L 1 – V in L 1 + v C 1 – I out + C 1 L 2 v L 2 – C + V out – + v C 1 – + C 1 L 2 v L 2 – 2

Buck/Boost converter i in + v L 1 – V in L 1 + v C 1 – I out + C 1 L 2 v L 2 – C + V out – This circuit is more unforgiving than the boost converter, because the MOSFET and diode voltages and currents are higher • Before applying power, make sure that your D is at the minimum, and that a load is solidly connected • Limit your output voltage to 90 V 3

KVL and KCL in the average sense I in + 0 – V in L 1 I in 0 + Vin – C 1 I out L 2 I out + 0 – C 0 + V out – KVL shows that VC 1 = Vin Interestingly, no average current passes from the source side, through C 1, to the load side, and yet this is a “DC - DC” converter 4

Switch closed assume constant i in + Vin – L 1 V in + Vin – + v D– I out + V out – + C 1 L 2 v L 2 – C KVL shows that v. D = −(Vin + Vout), so the diode is open Thus, C is providing the load power when the switch is closed i in V in + Vin – L 1 + Vin – – (Vin + Vout) + C 1 L 2 – Vin + C I out + V out – i. L 1 and i. L 2 are ramping up (charging). C 1 is charging L 2. C is discharging. 5

Switch open (assume the diode is conducting because, otherwise, the circuit cannot work) i in – Vout + V in L 1 + Vin – assume constant I out C 1 L 2 + V out – C 1 and C are charging. L 1 and L 2 are discharging. KVL shows that VL 1 = −Vout The input/output equation comes from recognizing that the average voltage across L 1 is zero 6

Inductor L 1 current rating During the “on” state, L 1 operates under the same conditions as the boost converter L, so the results are the same Use max 7

Inductor L 2 current rating Average values I in + 0 – 0 L 1 V in C 1 I in 2 Iout Iavg = Iout 0 + Vin – I out L 2 I out + 0 – C 0 + V out – i. L 2 ΔI Use max 8

MOSFET and diode currents and current ratings i in + v L 1 – L 1 V in + v C 1 – I out + C 1 L 2 v L 2 – C + V out – i. L 1 + i. L 2 MOSFET 2(Iin + Iout) 0 switch closed 2(Iin + Iout) 0 Take worst case D for each Diode i. L 1 + i. L 2 switch open Use max 9

Output capacitor C current and current rating i. C = (i. D – Iout) 2 Iin + Iout 0 −Iout switch closed As D → 1, Iin >> Iout , so switch open As D → 0, Iin << Iout , so 10

Series capacitor C 1 current and current rating i in V in + Vin – L 1 – (Vin + Vout) + C 1 L 2 i in – Vout + V in + Vin – L 1 – Vin + C + Vin – I out + V out – I out C 1 L 2 + V out – C + V out – Switch closed, IC 1 = −IL 2 Switch open, IC 1 = IL 1 11

Series capacitor C 1 current and current rating Switch closed, IC 1 = −IL 2 i. C 1 2 Iin 0 Switch open, IC 1 = IL 1 switch closed switch open − 2 Iout As D → 1, Iin >> Iout , so As D → 0, Iin << Iout , so 12

Worst-case load ripple voltage i. C = (i. D – Iout) 0 −Iout The worst case is where D → 1, where output capacitor C provides Iout for most of the period. Then, 13

Worst case ripple voltage on series capacitor C 1 i. C 1 switch open 2 Iin 0 − 2 Iout switch closed Then, considering the worst case (i. e. , D = 1) 14

Voltage ratings + Vin – V in L 1 – (Vin + Vout) + C 1 L 2 C + V out – MOSFET and diode see (Vin + Vout) – Vout + V in L 1 + Vin – C 1 L 2 C + V out – • Diode and MOSFET, use 2(Vin + Vout) • Capacitor C 1, use 1. 5 Vin • Capacitor C, use 1. 5 Vout 15

Continuous current in L 1 2 Iin i. L Iavg = Iin 0 (1 − D)T Then, considering the worst case (i. e. , D → 1), use max guarantees continuous conduction use min 16

Continuous current in L 2 2 Iout Iavg = Iout i. L 0 (1 − D)T Then, considering the worst case (i. e. , D → 0), use max guarantees continuous conduction use min 17

Impedance matching Iin + + Source DC−DC Boost Converter Vin − − Iin + Vin Equivalent from source perspective − 18

Impedance matching For any Rload, as D → 0, then Requiv → ∞ (i. e. , an open circuit) For any Rload, as D → 1, then Requiv → 0 (i. e. , a short circuit) Thus, the buck/boost converter can sweep the entire I-V curve of a solar panel 19

Example - connect a 100Ω load resistor D = 0. 80 2Ω equ iv. D = 0. 88 . 4Ω 4. 6 uiv q e D = 0. 50 100Ω equiv. With a 100Ω load resistor attached, raising D from 0 to 1 moves the solar panel load from the open circuit condition to the short circuit condition 20

Example - connect a 5Ω load resistor D = 0. 47 2Ω equ iv. D = 0. 61 . 4Ω 4. 6 uiv q e D = 0. 18 100Ω equiv. 21

BUCK/BOOST DESIGN 9 A 10 A 250 V 5. 66 A p-p Our components 200 V, 250 V 16 A, 20 A 90 V 10 A, 5 A 40 V, 90 V 10 A, 5 A Likely worst-case buck/boost situation L 1. 100µH, 9 A L 2. 100µH, 9 A C. 1500µF, 250 V, 5. 66 A p-p C 1. 33µF, 50 V, 14 A p-p Diode D. 200 V, 16 A MOSFET M. 250 V, 20 A 22

BUCK/BOOST DESIGN 5 A 0. 067 V 1500µF 50 k. Hz L 1. 100µH, 9 A L 2. 100µH, 9 A C. 1500µF, 250 V, 5. 66 A p-p C 1. 33µF, 50 V, 14 A p-p Diode D. 200 V, 16 A MOSFET M. 250 V, 20 A 23

BUCK/BOOST DESIGN 40 V 90 V 200µH 450µH 2 A 50 k. Hz L 1. 100µH, 9 A L 2. 100µH, 9 A C. 1500µF, 250 V, 5. 66 A p-p C 1. 33µF, 50 V, 14 A p-p Diode D. 200 V, 16 A MOSFET M. 250 V, 20 A 24

BUCK/BOOST DESIGN 50 V 14 A p-p Our components 9 A 5 A 40 V 10 A 5 A 3. 0 V 33µF 50 k. Hz 5 A Likely worst-case buck/boost situation L 1. 100µH, 9 A L 2. 100µH, 9 A C. 1500µF, 250 V, 5. 66 A p-p C 1. 33µF, 50 V, 14 A p-p Diode D. 200 V, 16 A MOSFET M. 250 V, 20 A Conclusion - 50 k. Hz may be too low for buck/boost converter 25

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