Thus spake jwisnia@aol.com (JWISNIA) in sci.electronics.misc: >Today I was trying to show someone that the total energy of the I^2*R >losses when resistively charging a capacitor from a dc source is exactly >equal to the energy stored in the capacitor after it has charged for an [snip] >I'd appreciate someone helping me out here, my calculus is rusty.....and >never was very good anyway! / + Vr - ___/ ___________/\/\/\/\____ _|_ -----> I R | / + \ ___|___ | | V C _______ \_-_/ | |_____________________________| <---- I When the switch is closed (time t=0), the capacitor acts as a short circuit. Current is V/R. Once the cap charges up, current flow is 0. The charging process is exponential, so instantaneous current is: V -t/RC I=--- * e R (This means I'm waving my hand, instead of actually doing the integration!) Instantaneous voltage across resistor is: -t/RC Vr= V * e Instantaneous power dissipated in resistor is voltage * current: V * V -2t/RC Pr= ----- * e R Integrate this to get total energy dissipated in resistor: oo / | V * V -2t/RC Er= | ----- * e dt | R 0 / _ _ | | V * V * R * C | -oo 0 | = ------------- * | e - e | R * (-2) | | |_ _| _ _ V * V * C | | = ---------- * | 0 - 1 | (-2) |_ _| V * V * C = --------- 2 The capacitor charges to voltage V, so the energy it stores is also: V * V * C Ec = --------- 2 >BTW, we got into this while discussing "charge pump" voltage multiplying >power supplies. I'd read an article in a recent mag which claimed they Here is one charge pump circuit. It works best if the square-wave oscillator has a low output impedance, and if Schottky diodes are used: VCC O | _|_ \ / __V__) | | + ( | / VCC | | | |\ | O .----+ +------+---| >|------+---------> 2 * VCC | | | | |/ | | | | | | + / + | + ______|_____ | ___|___ | | | _______ | square- | | | | wave +---+ ___|___ | oscillator | | _____ | | | ___ |____________| | + | | / . | | | | | /| | '---+ +-----+------|< |------+---------> -2 * VCC | | | | | \| | ___|___ + | | | / | _____ _|_ ___|___ ___ \ / _______ . __V__) + | + ( | ___|___ | _____ ___|___ ___ _____ . ___ . (I love ASCII art!) More capacitor-diode stages can be added to get higher output voltages. I don't know how efficient it is. Hope this helps. >Thanks guys... >Jeff Wisnia W1BSV >Winchester, MA. > "Common sense isn't very common..." -- _ _ _ _ _ _ | Sales rushes in |_| |_| |_| / |_| | | where Engineering _| |_ |_ /_ |_ | @ nconnect.net | fears to tread.

**Date:** Fri, 23 Aug 1996 05:22:49 GMT

**Original Subject:** Re: Help my rusty math please

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