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Dec. 26, 2010

Electrical Engineer in (Maintenance Dep.) at . AkzoNobel India Ltd (ICI Dulux PAINTS),

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- 1. EE362L, Fall 2008 DC − DC Buck/Boost Converter
- 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. 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. + 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Impedance matching DC − DC Boost Converter + V in − + − I in + V in − I in Equivalent from source perspective Source
- 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. 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. 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. 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. 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. 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. 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

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