EE462L, Fall 2011 DC−DC Buck/Boost Converter

Boost converter Buck/Boost converter + v L1 – i I in out L1 C + V V in out – Buck/Boost converter V in i L1 + v L1 – + v L2 – C1 + v C1 – L2 + V out – I C + v L2 – C1 + v C1 – L2

Buck/Boost converter V in i L1 + v L1 – + v L2 – C1 + v C1 – L2 + V out – I C 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 90V

KVL and KCL in the average sense + Vin – + 0 – I I I in out out L1 + – + C1 I V out V in out I C L2 in – KVL shows that VC1 = Vin Interestingly, no average current passes from the source side, through C1, to the load side, and yet this is a “DC - DC” converter

Switch closed i I + L1 C1 V V C L2 – i I + L1 C1 V V C I L2 – assume constant + Vin – + Vin – + v D – i + V out – I C in L1 + v L2 – C1 V in L2 KVL shows that vD = −(Vin + Vout), so the diode is open Thus, C is providing the load power when the switch is closed + Vin – – (Vin + Vout) + + Vin – i I in out L1 – Vin + + C1 V V in out C L2 I out – iL1 and iL2 are ramping up (charging). C1 is charging L2. C is discharging.

Switch open (assume the diode is conducting because, otherwise, the circuit cannot work) assume constant + Vin – – Vout + i I in out L1 + + C1 V V V in out out C L2 – – C1 and C are charging. L1 and L2 are discharging. KVL shows that VL1 = −Vout The input/output equation comes from recognizing that the average voltage across L1 is zero

Inductor L1 current rating During the “on” state, L1 operates under the same conditions as the boost converter L, so the results are the same Use max

Inductor L2 current rating Average values + Vin – + 0 – I I I in out out L1 + – + C1 I V out V in out I C in L2 – iL2 2Iout Iavg = Iout ΔI Use max

MOSFET and diode currents and current ratings + v L2 – C1 + v C1 – L2 + v L1 – i I in out L1 + V V in out C – iL1 + iL2 MOSFET Diode iL1 + iL2 2(Iin + Iout) 2(Iin + Iout) switch closed switch open Take worst case D for each Use max

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

Series capacitor C1 current and current rating + Vin – – (Vin + Vout) + + Vin – i I in out L1 – Vin + + C1 V V in out C L2 I out – + Vin – – Vout + i I in out L1 + + C1 V V V in out out C L2 – – Switch closed, IC1 = −IL2 Switch open, IC1 = IL1

Series capacitor C1 current and current rating iC1 Switch closed, IC1 = −IL2 Switch open, IC1 = IL1 2Iin switch closed switch open −2Iout As D → 1, Iin >> Iout , so As D → 0, Iin << Iout , so

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

Worst case ripple voltage on series capacitor C1 iC1 switch open 2Iin −2Iout switch closed Then, considering the worst case (i.e., D = 1)

Voltage ratings + L1 C1 V V C L2 – + L1 C1 V V C L2 – + Vin – – (Vin + Vout) + L1 + C1 V V in out C L2 – MOSFET and diode see (Vin + Vout) + Vin – – Vout + L1 + C1 V V in out C L2 – Diode and MOSFET, use 2(Vin + Vout) Capacitor C1, use 1.5Vin Capacitor C, use 1.5Vout

Continuous current in L1 iL 2Iin Iavg = Iin (1 − D)T Then, considering the worst case (i.e., D → 1), use max guarantees continuous conduction use min

Continuous current in L2 iL 2Iout Iavg = Iout (1 − D)T Then, considering the worst case (i.e., D → 0), use max guarantees continuous conduction use min

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

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

Example - connect a 100Ω load resistor 2Ω equiv. 6.44Ω equiv. 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

Example - connect a 5Ω load resistor 2Ω equiv. 6.44Ω equiv. D = 0.18 100Ω equiv.

Likely worst-case buck/boost situation BUCK/BOOST DESIGN 5.66A p-p 200V, 250V 16A, 20A Our components 9A 250V 10A, 5A 10A 90V 40V, 90V Likely worst-case buck/boost situation 10A, 5A MOSFET M. 250V, 20A L1. 100µH, 9A C. 1500µF, 250V, 5.66A p-p Diode D. 200V, 16A L2. 100µH, 9A C1. 33µF, 50V, 14A p-p

BUCK/BOOST DESIGN 5A 0.067V 1500µF 50kHz L1. 100µH, 9A L2. 100µH, 9A C. 1500µF, 250V, 5.66A p-p C1. 33µF, 50V, 14A p-p Diode D. 200V, 16A MOSFET M. 250V, 20A

BUCK/BOOST DESIGN 40V 90V 200µH 450µH 2A 50kHz 2A 50kHz L1. 100µH, 9A C. 1500µF, 250V, 5.66A p-p C1. 33µF, 50V, 14A p-p Diode D. 200V, 16A MOSFET M. 250V, 20A

Likely worst-case buck/boost situation BUCK/BOOST DESIGN Our components 9A 14A p-p 50V 10A 5A 40V Likely worst-case buck/boost situation 5A 33µF 50kHz 3.0V L1. 100µH, 9A L2. 100µH, 9A C. 1500µF, 250V, 5.66A p-p C1. 33µF, 50V, 14A p-p Diode D. 200V, 16A Conclusion - 50kHz may be too low for buck/boost converter MOSFET M. 250V, 20A