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authorVasil Zlatanov <v@skozl.com>2018-01-30 10:28:53 +0000
committerVasil Zlatanov <v@skozl.com>2018-01-30 10:28:53 +0000
commitfe13956bb939bc418dd535bb9a5973d8857b4a04 (patch)
tree8cda477beb30f14317343a94667ff3c08880f812
parentb39817280176f5b43bd0ef2dccc456692358ade9 (diff)
downloade3-smps-fe13956bb939bc418dd535bb9a5973d8857b4a04.tar.gz
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report intro
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% 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.