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author | Vasil Zlatanov <v@skozl.com> | 2018-01-30 10:28:53 +0000 |
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committer | Vasil Zlatanov <v@skozl.com> | 2018-01-30 10:28:53 +0000 |
commit | fe13956bb939bc418dd535bb9a5973d8857b4a04 (patch) | |
tree | 8cda477beb30f14317343a94667ff3c08880f812 | |
parent | b39817280176f5b43bd0ef2dccc456692358ade9 (diff) | |
download | e3-smps-fe13956bb939bc418dd535bb9a5973d8857b4a04.tar.gz e3-smps-fe13956bb939bc418dd535bb9a5973d8857b4a04.tar.bz2 e3-smps-fe13956bb939bc418dd535bb9a5973d8857b4a04.zip |
report intro
-rw-r--r-- | report.md | 20 |
1 files changed, 18 insertions, 2 deletions
@@ -2,5 +2,21 @@ % Vasil Zlatanov, Sheheryar Salman % 99 March 2018 -# Dummy Heading -Dummy text + +# Selection of Components + +## Inductor sizing + +$$ L = \frac{V_{I} \times (V_{O} - V_{I})}{\Delta I_{L} \times f_{S} \times V_{O}} $$ + +We require an input current ripple of 10%. At 10V input 20V output and 90% efficiency this would be 0.22A ripple. + +$$ L = \frac{ 10V \times (20V - 10V)}{0.22A \times 250kHz \times V_O} \approx 100uH $$ + +## Maximum Switch Current + +To find the switch current we will first calculate the duty cycle required to produce the required output voltage, when we are working of the minimum input. + +$$ \delta = 1 - \frac{V_I_M \times \nu}{V_O} = 1 - \frac{8V \times 0.9}{20V} = 0.64 $$ + +Where $\nu$ is the minimum efficienecy as we want to compensate for power loss. |