7 ee462_l_dc_dc_boost_ppt

776 views
678 views

Published on

boost converter ppt

Published in: Education, Business, Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
776
On SlideShare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
112
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

7 ee462_l_dc_dc_boost_ppt

  1. 1. 1 EE462L, Spring 2013 DC−DC Boost Converter
  2. 2. 2 Vin + Vout – C iC Ioutiin Buck converter iL L + vL – Boost converter Vin + Vout – C iC Iout iin iL L + vL –
  3. 3. 3 Boost converter This is a much more unforgiving circuit than the buck converter Vin + Vout – C iC Iout iin iL L + vL – iD • If the MOSFET gate driver sticks in the “on” position, then there is a short circuit through the MOSFET – blow MOSFET! • If the load is disconnected during operation, so that Iout = 0, then L continues to push power to the right and very quickly charges C up to a high value (≈250V) – blow diode and MOSFET! • Before applying power, make sure that your D is at the minimum, and that a load is solidly connected !
  4. 4. 4 Boost converter Vin + Vout – C iC Iout iin iL L + vL – iD • Modify your MOSFET firing circuit for Boost Converter operation (see the MOSFET Firing Circuit document) • Limit your output voltage to 120V
  5. 5. 5 Boost converter Using KVL and KCL in the average sense, the average values are + 0 V – Iout Vin + Vout – C Iout L 0 A Iin Vin + Vout – C iC Iout iin iL L + vL – iD Find the input/output equation by examining the voltage across the inductor
  6. 6. 6 Switch closed for DT seconds Reverse biased, thus the diode is openL V dt di inL = for DT seconds Vin + Vout – C Iout iin iL L Iout Note – if the switch stays closed, the input is short circuited! + Vin −
  7. 7. 7 Switch open for (1 − D)T seconds Diode closed. Assume continuous conduction.L VV dt di outinL − = Vin + Vout – C Iout iin iL L for (1−D)T seconds (iL – Iout) + (Vin − Vout ) −
  8. 8. 8 Since the average voltage across L is zero ( ) ( ) 01 =−•−+•= outininLavg VVDVDV inininout VDVDVDV •−•+=−• )1( D V V in out − = 1 The input/output equation becomes A realistic upper limit on boost is 5 times !
  9. 9. 9 Examine the inductor current Switch closed, Switch open, L V dt di Vv inL inL == , L VV dt di VVv outinL outinL − =−= , sec/A L Vin DT (1 − D)T T Imax Imin Iavg = Iin Iavg = Iin is half way between Imax and Imin sec/A L VV outin − ΔI iL
  10. 10. 10 Inductor current rating ( )22222 12 1 12 1 IIIII inppavgLrms ∆+=+= ( ) 2222 3 4 2 12 1 inininLrms IIII =+= ∆ Max impact of ΔI on the rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iin inLrms II 3 2 = 2Iin 0 Iavg = Iin ΔI iL Use max
  11. 11. 11 MOSFET and diode currents and current ratings inrms II 3 2 = Use max 2Iin 0 2Iin 0 Take worst case D for each Vin + Vout – C iC Iout iin iL L + vL – iD
  12. 12. 12 Capacitor current and current rating 2Iin −Iout −Iout 0 Max rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iout outCrms II = Use max iC = (iD – Iout) Vin + Vout – C iC Iout iin iL L iD See the lab document for the derivation
  13. 13. 13 Worst-case load ripple voltage Cf I C TI C Q V outout = • = ∆ =∆ The worst case is where C provides Iout for most of the period. Then, −Iout 0 iC = (iD – Iout)
  14. 14. 14 Voltage ratings Diode sees Vout MOSFET sees Vout C sees Vout • Diode and MOSFET, use 2Vout • Capacitor, use 1.5Vout Vin + Vout – C Iout iin iL L Vin + Vout – C Iout iin iL L
  15. 15. 15 Continuous current in L sec/A L VV outin − ( ) ( ) ( ) fL D D V TD L V D V TD L VV I boundary in boundary in in boundary inout in −      − − =−• − −=−• − = 11 1 1 1112 fI DV L in in boundary 2 = 2Iin 0 Iavg = Iin iL (1 − D)T fI V L in in 2 > guarantees continuous conduction Then, considering the worst case (i.e., D → 1), use max use min ,2 fL DV I boundary in in =
  16. 16. 16 Impedance matching out out load I V R = equivR ( ) ( ) ( ) load out out out out in in equiv RD I V D D I VD I V R 22 11 1 1 −=−= − − == DC−DC Boost Converter + Vin − + − Iin + Vin − Iin Equivalent from source perspective Source D V V in out − = 1 ( ) inout IDI −= 1
  17. 17. 17 Example of drawing maximum power from solar panel Isc Voc Pmax is approx. 130W (occurs at 29V, 4.5A) Ω== 44.6 5.4 29 A V Rload For max power from panels, attach I-V characteristic of 6.44Ω resistor But as the sun conditions change, the “max power resistance” must also change
  18. 18. 18 Connect a 100Ω resistor directly, extract only 14W 130W 6.44Ω resistor 100Ω resistor 14W ( ) 75.0 100 44.6 11,1 2 =−=−=−= load equiv loadequiv R R DRDR To extract maximum power (130W), connect a boost converter between the panel and the load resistor, and use D to modify the equivalent load resistance seen by the source so that maximum power is transferred So, the boost converter reflects a high load resistance to a low resistance on the source side
  19. 19. 19 Worst-Case Component Ratings Comparisons for DC-DC Converters Converter Type Input Inductor Current (Arms) Output Capacitor Voltage Output Capacitor Current (Arms) Diode and MOSFET Voltage Diode and MOSFET Current (Arms) Boost 1.5 2 5A 10A10A 120V 120V Likely worst-case boost situation 5.66A 200V, 250V 16A, 20A Our components 9A 250V MOSFET. 250V, 20A L. 100µH, 9A C. 1500µF, 250V, 5.66A p-p Diode. 200V, 16A BOOST DESIGN
  20. 20. 20 Comparisons of Output Capacitor Ripple Voltage Converter Type Volts (peak-to-peak) Boost 5A 1500µF 50kHz 0.067V BOOST DESIGN MOSFET. 250V, 20A L. 100µH, 9A C. 1500µF, 250V, 5.66A p-p Diode. 200V, 16A
  21. 21. 21 Minimum Inductance Values Needed to Guarantee Continuous Current Converter Type For Continuous Current in the Input Inductor For Continuous Current in L2 Boost – 40V 2A 50kHz 200µH BOOST DESIGN MOSFET. 250V, 20A L. 100µH, 9A C. 1500µF, 250V, 5.66A p-p Diode. 200V, 16A

×