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# 8 ee362_l_dc_dc_buckboost_ppt (1)

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### 8 ee362_l_dc_dc_buckboost_ppt (1)

1. 1. EE362L, Fall 2008 DC − DC Buck/Boost Converter
2. 2. Boost converter + V out – I out V in i in L1 + v L1 – Buck/Boost converter C + v L2 – C1 + v C1 – L2 V in i in L1 + v L1 – + v L2 – C1 + v C1 – L2 + V out – I out C
3. 3. Buck/Boost converter This circuit is more unforgiving than the boost converter, because the MOSFET and diode voltages and currents are higher <ul><li>Before applying power, make sure that your D is at the minimum, and that a load is solidly connected </li></ul><ul><li>Limit your output voltage to 90V </li></ul>V in i in L1 + v L1 – + v L2 – C1 + v C1 – L2 + V out – I out C
4. 4. + V out – I out C V in I in L1 + 0 – + 0 – KVL and KCL in the average sense 0 0 I out I in C1 L2 I out + V in – KVL shows that V C1 = V in Interestingly, no average current passes from the source side, through C1, to the load side, and yet this is a “DC - DC” converter
5. 5. Switch closed V in i in L1 + V in – + v L2 – C1 + V in – L2 assume constant + v D – KVL shows that v D = − (V in + V out ), so the diode is open Thus, C is providing the load power when the switch is closed V in i in L1 – V in + C1 + V in – L2 + V out – I out C – (V in + V out ) + I out i L1 and i L2 are ramping up (charging). C1 is charging L2. C is discharging. + V in – + V out – I out C
6. 6. Switch open (assume the diode is conducting because, otherwise, the circuit cannot work) V in i in L1 – V out + C1 + V in – L2 + V out – I out C C1 and C are charging. L1 and L2 are discharging. + V out – KVL shows that V L1 = − V out The input/output equation comes from recognizing that the average voltage across L1 is zero assume constant
7. 7. Inductor L1 current rating Use max During the “on” state, L1 operates under the same conditions as the boost converter L, so the results are the same
8. 8. Inductor L2 current rating 2I out 0 I avg = I out Δ I i L2 Use max + V out – I out C V in I in L1 + 0 – + 0 – 0 0 I out I in C1 L2 I out + V in – Average values
9. 9. MOSFET and diode currents and current ratings 0 2(I in + I out ) 0 Take worst case D for each V in i in L1 + v L1 – + V out – I out C MOSFET Diode i L1 + i L2 Use max switch closed switch open 2(I in + I out ) i L1 + i L2 + v L2 – C1 + v C1 – L2
10. 10. Output capacitor C current and current rating 2I in + I out − I out 0 As D -> 1, I in >> I out , so i C = (i D – I out ) As D -> 0, I in << I out , so switch closed switch open
11. 11. Series capacitor C1 current and current rating Switch closed, I C1 = − I L2 V in i in L1 – V in + C1 + V in – L2 + V out – I out C – (V in + V out ) + I out + V in – V in i in L1 – V out + C1 + V in – L2 + V out – I out C + V out – Switch open, I C1 = I L1
12. 12. Series capacitor C1 current and current rating 2I in − 2 I out 0 As D -> 1, I in >> I out , so i C1 As D -> 0, I in << I out , so switch closed switch open Switch closed, I C1 = − I L2 Switch open, I C1 = I L1
13. 13. Worst-case load ripple voltage The worst case is where D -> 1, where output capacitor C provides I out for most of the period. Then, − I out 0 i C = (i D – I out )
14. 14. Worst case ripple voltage on series capacitor C1 2I in − 2 I out 0 i C1 switch closed switch open Then, considering the worst case (i.e., D = 1)
15. 15. Voltage ratings MOSFET and diode see (V in + V out ) <ul><li>Diode and MOSFET, use 2(V in + V out ) </li></ul><ul><li>Capacitor C1, use 1.5V in </li></ul><ul><li>Capacitor C, use 1.5V out </li></ul>V in L1 C1 + V in – L2 + V out – C – ( V in + V out ) + V in L1 – V out + C1 + V in – L2 + V out – C
16. 16. Continuous current in L1 2I in 0 I avg = I in i L (1 − D)T guarantees continuous conduction Then, considering the worst case (i.e., D -> 1), use max use min
17. 17. Continuous current in L2 2I out 0 I avg = I out i L (1 − D)T guarantees continuous conduction Then, considering the worst case (i.e., D -> 0), use max use min
18. 18. Impedance matching DC − DC Boost Converter + V in − + − I in + V in − I in Equivalent from source perspective Source
19. 19. Impedance matching For any R load , as D -> 0, then R equiv -> ∞ (i.e., an open circuit) For any R load , as D -> 1, then R equiv -> 0 (i.e., a short circuit) Thus, the buck/boost converter can sweep the entire I-V curve of a solar panel
20. 20. Example - connect a 100 Ω load resistor D = 0.80 6.44 Ω equiv. 100 Ω equiv. D = 0.50 D = 0.88 2 Ω equiv. With a 100 Ω load resistor attached, raising D from 0 to 1 moves the solar panel load from the open circuit condition to the short circuit condition
21. 21. Example - connect a 5 Ω load resistor D = 0.47 6.44 Ω equiv. 100 Ω equiv. D = 0.18 D = 0.61 2 Ω equiv.
22. 22. 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. 100 µ H, 9A C. 1500 µ F, 250V, 5.66A p-p Diode D. 200V, 16A L2. 100 µ H, 9A C1. 33 µ F, 50V, 14A p-p
23. 23. 5A 1500µF 50kHz 0.067V MOSFET M. 250V, 20A L1. 100 µ H, 9A C. 1500 µ F, 250V, 5.66A p-p Diode D. 200V, 16A L2. 100 µ H, 9A C1. 33 µ F, 50V, 14A p-p BUCK/BOOST DESIGN
24. 24. 40V 2A 50kHz 200µH 90V 2A 50kHz 450µH MOSFET M. 250V, 20A L1. 100 µ H, 9A C. 1500 µ F, 250V, 5.66A p-p Diode D. 200V, 16A L2. 100 µ H, 9A C1. 33 µ F, 50V, 14A p-p BUCK/BOOST DESIGN
25. 25. MOSFET M. 250V, 20A L1. 100 µ H, 9A C. 1500 µ F, 250V, 5.66A p-p Diode D. 200V, 16A L2. 100 µ H, 9A C1. 33 µ F, 50V, 14A p-p BUCK/BOOST DESIGN Conclusion - 50kHz may be too low for buck/boost converter 10A 5A 40V Likely worst-case buck/boost situation 5A 5A 33µF 50kHz 3.0V Our components 9A 14A p-p 50V