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Answer by Tony Stewart EE75 for Pumping a few amperes for 100 µsec
Consider the effective series resistance (ESR) involved and loss in power transfer.At worst case maximum input levels:Surge forward current, tp = 100 μs IF = 5 A Vf = 3.5 V nominal!!IF = 1 A Vf = 2.0 V...
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Answer by Rocketmagnet for Pumping a few amperes for 100 µsec
This is the most efficient way I can think to do it. There's a MAX1682 charge pump to give you 6.6v at the super capacitor. The voltage doubler is pretty efficient, probably more then 90%, but they...
View ArticleAnswer by Russell McMahon for Pumping a few amperes for 100 µsec
That's an average power of Power = 5 A \$\times\$ 10.5 V \$\times\$ 100 \$\mu\$s / 10 ms = 0.525 W.Average power is easy for almost any battery. You just need a store to accomodate the pulse. A...
View ArticleAnswer by clabacchio for Pumping a few amperes for 100 µsec
A Joule thief may be the answer to your problem: it's a sort of boost converter, where you open a circuit with an inductor in series to create a high voltage. Since the power is delivered by the...
View ArticlePumping a few amperes for 100 µsec
I would like to pump 4-5 A to a high-power LED for 100 µs. My system has only a 3.3 V battery, and this 100 µs high-power event takes place once every 10 seconds.What is the best way of doing this...
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