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bitrex
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This is close enough to a homework question so I've put it in this section, though maybe it should be posted in the EE section. I've been looking at the equations for boost converters today, and I'm interested in the problem of how long it would take a bank of capacitors to charge through an ideal boost converter, assuming there is not any load on the output. I couldn't find an equation for it - so far I've considered the problem like this: the equation for a boost converter operating in continuous mode is:
[tex]\frac{Vs}{L} * DT + \frac{Vs-Vo}{L} * (1-D)T = 0[/tex]
The portion that describes the current through the inductor when the boost converter switch is open, allowing the current stored in the inductor to flow into the capacitor is:
[tex]\Delta I_{L} =\frac{Vs-Vo}{L} * (1-D)T[/tex]
Since the capacitor is in series with the inductor, current will "flow" through the capacitor charging it based on:
[tex]I_{c} = C\frac{dV}{dT}[/tex]
So I do this: [tex]\int \frac{dV}{dT} = \int \frac{Vs - Vo}{CL}(1-D)t[/tex] =
[tex]V(t) = \frac{1}{2}\frac{Vs - Vo}{CL}(1-D)t^{2}[/tex]
Vs is the source voltage and Vo is [tex]\frac{Vs}{1-D}[/tex]
Does this look legit?
[tex]\frac{Vs}{L} * DT + \frac{Vs-Vo}{L} * (1-D)T = 0[/tex]
The portion that describes the current through the inductor when the boost converter switch is open, allowing the current stored in the inductor to flow into the capacitor is:
[tex]\Delta I_{L} =\frac{Vs-Vo}{L} * (1-D)T[/tex]
Since the capacitor is in series with the inductor, current will "flow" through the capacitor charging it based on:
[tex]I_{c} = C\frac{dV}{dT}[/tex]
So I do this: [tex]\int \frac{dV}{dT} = \int \frac{Vs - Vo}{CL}(1-D)t[/tex] =
[tex]V(t) = \frac{1}{2}\frac{Vs - Vo}{CL}(1-D)t^{2}[/tex]
Vs is the source voltage and Vo is [tex]\frac{Vs}{1-D}[/tex]
Does this look legit?
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