From ef3f89693bc2de9cf3f38ff9878baafec3e8ff12 Mon Sep 17 00:00:00 2001
From: ben-jaynes <1btjaynes@gmail.com>
Date: Wed, 24 Jun 2026 17:21:18 -0700
Subject: [PATCH] vault backup: 2026-06-24 17:21:18

---
 PCB Design/Music Player/Process.md | 10 ++++++++++
 1 file changed, 10 insertions(+)

diff --git a/PCB Design/Music Player/Process.md b/PCB Design/Music Player/Process.md
index 0a7f7b0..8591aaa 100644
--- a/PCB Design/Music Player/Process.md	
+++ b/PCB Design/Music Player/Process.md	
@@ -3,6 +3,8 @@
 
 ## TPS63020
 ### Choosing an Inductor
+Neither of the options given in the datasheet are still available so a different inductor will need to be chosen
+
 The equations needed to find the peak current in steady state operation are 
 Duty cycle boost: $$D=\frac{{V_{OUT} - V_{IN}}}{V_{OUT}}$$
 and peak current: $$I_{PEAK} = \frac{I_{out}}{\eta * (1 - D)} + \frac{{V_{IN} * D}}{2 * f * L}$$
@@ -15,3 +17,11 @@ where
 
 First, to calculate the duty cycle boost we can plug in the output voltage (3.3V) and the "minimum input voltage in boost mode" which for the BQ25985 is listed as 2.304V
 $$D = \frac{{3.3 - 2.304}}{3.3} = 0.302$$
+Now calculating for $I_{PEAK}$ using the values 
+- $D = 0.302$ (calculated previously)
+- $f = 2.5$ (MHz, typical stated in datasheet)
+- $L=1.5$ (uH, value for inductor used)
+- $\eta = 0.9$ (efficiency assumption stated in datasheet)
+$$I_{PEAK} = \frac{3.3}{0.9 * (1 - 0.302)} + \frac{{2.304 * 0.302}}{2 * 2.5 * 1.5} = 5.346$$
+The datasheet states to pick an inductor with a saturation current 20% higher than the calculated value which would mean $$I_{sat} \geq 6.415$$
+The []
\ No newline at end of file