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Introduction

I'm having trouble producing an equation that involving my MAX756 boost chip, a load and my 1.5F capacitor.

Basically I'm trying to find out how long it would take for a 1.5F capacitor to discharge from Va to Vb when connected to a loaded RLC boost converter.

Here's what I've got far.

Consider this: the MAX756 boost chip has a feedback loop built in to modify it's frequency so it will always (within reason) output 5V.

There is a load of 62.5 ohms equating to a current demand of 80mA. Now according to the datasheet, the boost chip has approximately 80% efficiency so to produce 400mW (5V 80mA) it would require 500mW to keep outputting 5V at 80mA.

Here's where the capacitor comes into play. I have a 1.5F capacitor supplying the boost chip and the load. Since my power demand is constant, that means \$R=\dfrac{V^2}{P}\$ right? I say this because in order to keep outputting 5 volt power, the boost chip will have to draw more and more current as the voltage across the capacitor falls.

I was trying to balance the equations with these cap draining equations but Wolfram wasn't much help

What am I missing?

Ricardo
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Bruce Lee
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1 Answers1

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Integrating the changing current over the draining of the capacitor is correct and quite fun, but there is an easier way: use the energy formula which already has that integral done for you.

Calculate the energy stored in the capacitor, at \$V_a\$ and \$V_b\$.

$$ Energy = 0.5 \times C \times V^2 $$

Subtract the final from the initial to find the available energy.

Divide this by the power drawn to find the time it will take.

Last thought: the boost converter efficiency will drop as the voltage drops. Most of the energy is delivered while the capacitor is fairly full, so you can assume a constant efficiency near \$V_a\$. To get it more accurate, you might need to try that integral after all.

Ricardo
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tomnexus
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