On Wed, 08 Jan 2014 17:24:04 -0700, Jim Thompson
wrote:
On Tue, 07 Jan 2014 09:47:23 -0700, Jim Thompson
wrote:
High Side Driver...
http://www.analog-innovations.com/SED/HighSideDriver.pdf
...Jim Thompson
0.5*L*I^2 = Qg*Vg (Qg = Total Gate Charge, from data sheet, at
gate voltage, Vg)
[neglecting losses in diode and resistor, I'll include that RSN,
otherwise someone will have another hissy fit :]
Not "L = R^2 * C"
"Close to it" (pretty close actually) would be...
0.5*L*I^2 = 0.5*C*V^2
It's all about energy transfer from magnetic storage to capacitive
storage... and boost converters rely on it.
...Jim Thompson
...Jim Thompson
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