From fe13956bb939bc418dd535bb9a5973d8857b4a04 Mon Sep 17 00:00:00 2001
From: Vasil Zlatanov <v@skozl.com>
Date: Tue, 30 Jan 2018 10:28:53 +0000
Subject: report intro

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 report.md | 20 ++++++++++++++++++--
 1 file changed, 18 insertions(+), 2 deletions(-)

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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.
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