jbm@eos.UUCP (Jeffrey Mulligan) (06/03/89)
From article <5170036@hplsla.HP.COM>, by tomb@hplsla.HP.COM (Tom Bruhns): > Interesting question: Can you design a simple circuit to connect the > two caps and end up with 7.07 volts on each? (no other energy source) > (I'll take a ckt that even manages to get 7 volts :-) Obviously this can't be done if charge is conserved, but if we can use a little energy to get some charge from somewhere... How about this: switch S1 big inductor |------/ .--------|----OOOOOOOOOOO--------| | | | --- - --- cap C1 --- ^ diode D1 --- cap C2 | | | | | | |-----------------|-----------------------| Initially, C1 is charged to 10 volts, switch is open, no charge on C2, no current flowing. Switch is closed. Current flows through the inductor onto C2. When the voltage on C1 drops to 7.07 volts, the switch is opened. Current "inertia" through the inductor causes current to be pulled through the diode. If we ignore resistive losses, diode drop, etc., charge on C2 should be 7.07 by conservation of energy. I have used a similar circuit for voltage doubling: replace S1 by another diode (pointing to the right), C1 is a power supply, C2 is an electronic flash discharge cap (C2<<C1). C2 is momentarily shorted by the flashtube, The inductor prevents the tube from shorting out the power supply, and subsequently charges C2 to twice the voltage on C1. -- Jeff Mulligan (jbm@aurora.arc.nasa.gov) NASA/Ames Research Ctr., Mail Stop 239-3, Moffet Field CA, 94035 (415) 694-6290
jeffw@midas.STS.TEK.COM (Jeff Winslow) (06/05/89)
From: jbm@eos.UUCP (Jeffrey Mulligan) >>(Tom Bruhns): >> Interesting question: Can you design a simple circuit to connect the >> two caps and end up with 7.07 volts on each? (no other energy source) >How about this: > > switch S1 big inductor > > |------/ .--------|----OOOOOOOOOOO--------| > | | | > --- - --- > cap C1 --- ^ diode D1 --- cap C2 > | | | > | | | > |-----------------|-----------------------| > >Initially, C1 is charged to 10 volts, switch is open, no charge on >C2, no current flowing. Switch is closed. Current flows through the >inductor onto C2. When the voltage on C1 drops to 7.07 volts, >the switch is opened. Current "inertia" through the inductor causes >current to be pulled through the diode. If we ignore resistive losses, >diode drop, etc., charge on C2 should be 7.07 by conservation of energy. I was skeptical of this one at first - then I worked out the math, and, sure enough it works. I guess I got so wrapped up with my "where's the missing charge" questions that I forgot we're no longer in the realm of electrostatics with this circuit. And, the inductor doesn't have to be particularly big. If C1 = C2, I don't think too many people would dispute that, if the switch were closed and the diode is ideal, the voltage on C1 will be a cosine wave starting at peak voltage V and oscillating about V/2. The voltage on C2 will have identical amplitude but be 180 degrees out of phase. The current in the inductor will be maximum when its voltage passes through zero, and the energy stored in the inductor at that time will be the famous "missing energy", or (C/4)*V^2. Thus the inductor current is a sine wave with an amplitude given by (L/2)*I^2 = (C/4)*V^2, or I = V*sqrt(C/(2*L)). So, when the switch is closed, the voltage on C1 is (V/2)*(1 + cos wt), and the current in the inductor is V*sqrt(C/(2*L))*sin wt. When the voltage on C1 has decreased to V/sqrt(2), all of the current flowing from C1 has been flowing through C2, so the voltage increase on C2 is the same as the decrease on C1. Thus the voltage on C2 is V - V/sqrt(2). I'd rather write this (V/sqrt(2))*(sqrt(2) - 1), for reasons which may become clear later (don't you wish all your math profs talked like this? :-)). The energy stored in C2 at this point is (C/4)*V^2 * (sqrt(2) - 1)^2. Now the switch is opened. We can find the wt when this occurs by solving (V/2)*(1 + cos wt) = V/sqrt(2), then substituting the result into the expression for inductor current to find out how much energy will be dumped into C2. Specifically, wt = arccos(sqrt(2) - 1). Thus, the amount of energy stored in the inductor at this time, and the amount that will eventually be dumped into C2, is found through (L/2)*I^2: (C/4) * V^2 * sin^2(arccos(sqrt(2)-1)). Ah, but C2 already had some energy stored in it. See the equation back there? So after the inductor finishes dumping, the energy stored in C2 will be the sum of the two energies, or (C/4) * V^2 * ( (sqrt(2) - 1)^2 + sin^2(arccos(sqrt(2) - 1)) ). Urgh - a mess. But wait - make up an angle phi such that cos(phi) = sqrt(2)-1. Now the mess in the outer brackets becomes cos^2(phi) + sin^2(phi), which is one. So the final energy stored in C2 will be, no surprise, (C/4) * V^2. And the voltage necessary for C2 to store that energy is sqrt(2*E/C), or V/sqrt(2) --- 7.07 volts if we started with 10v on C1. With 7.07 volts left on C1. Are we all bored yet? Jeff Winslow