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## Electrical guru's continued

Hey you guys --- im going to go down and play with my bike trainer a little tonight, I had to work today but got to thinking --- back in the day when I was trying to learn simple electronics they were teaching that a good way to look at it was kinda like useing pressureized water in pipes, pressure was volts, size of the pipe was amps, batteries and capasitors were like water storage tanks and resistors were orfices for how fast the electricity could drain from these tanks and such and of course diodes were check valves,,,

Any how i got to thinking about what i wired with you guyzes help and how good it worked --- but now im thinking about this --- i have these resistors before the capasitor and then i pull my reading for my volt meter, what if i ran resistors before and after the capasitor and then got my reading? would this not be like filling the water tank at a more steady flow (what im doing right now) but then taking it a step further and allowing the water to drain at a more steady drain to get my reading, If this sounds insane then i shoulda just kept my yap shut

If you catz forgot what neede to be wired the old electrical guru topic is back on page two or three now --- just didnt want to dredge all that info back up,,, thanks..

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Not crazy, but not really necessary if you leave out the diode.

What you are describing is the two time constants I referred to. The "filling up" one, and the "draining" one. I called them "atttack" and "decay", but those are just names, although common ones in electronics for such a circuit.

With the diode, those functions are separated. The series resistor sets the 'filling" rate, and the one across the capacitor (includes the meter resistance) sets the "draining" rate. If the filling resistor is considerable smaller value than the draining one, the two rates are pretty independent. The closer in value they are, the more they interact.

If you don't use the diode, the same resistor will do both "filling up" and "draining", depending on whether the input voltage is lower or higher than what is on the capacitor right then. That will achieve what you described as an even effect for both filling and draining. The result will be a time average, which is what you want as I understand it.

As far as the meter, it has a small current/large resistance, and is probably pretty much neglectable, as most digital meters have at least 1 megohm input resistance, and usually more. So it needn't be considered in all this at the level of precision you want, nor will it affect the "draining" effect significantly.
Last edited by J Tiers; 11-28-2006 at 11:22 PM.

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If I assume that the second resistor you're talking about is in series with a lead to the meter, then it's not needed at all, and won't make any difference at all. If you're talking about the resistor across the capacitor leads, then that's a different story. That one is needed if you use the diode because without it all you'll be reading is the highest voltage peak attained. You could stop pedalling entirely and the reading won't change except verrrry slowly. Without the diode, this second resistor wouldn't normally be required as long as you have at least some load on the generator. I'll have to assume that will always be the case, since when would you be training without any resistance to your efforts- even the fan will be a fairly light load.

As JTiers has suggested, all you really need is one resistor in series between the generator's output and the capacitor. That might give you an entirely satisfactory reading. Personally I think you would want a faster charge time and a slower discharge time to give a suitable meter reading, which requires one more resistor and possibly the diode, but that's still a very simple circuit. You are the experimenter here, and if the single resistor solves the meter reading problem, then that's that.

Interesting that if you do use a fan motor, it will tend to help create a shorter charge time and longer discharge time, since it is also a generator and will give some back voltage anytime its turning. Your capacitors discharge rate will be a little slower that its charge rate, even with the single resistor. This simplest circuit might just work perfectly for you.

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Thanks -- im keeping it the way it is then unless the fan changes things drastically --------- on an even more off topic post ------ I just modified my electronic fly swatter the other day, You used to have to hold down the charge button and chase the fly's down, this was a big waste of battery's and i knew they had some kind of "bleed down" resistor across the capasitor to make the thing safer for kids and stuff -- so i tore into it and found the resistor that was mounted across the two Cap. leads and removed it --- works awesome ,, one three second charge and you can hunt down a fly for a half hour if you want --- and a little more power to boot, Im thinking of going back into it and getting an identical Capasitor from radio shack and soldering it in parralell with the other for those stubborn horseflys, I can see why some of you catzs like electronics, thanks for your help...

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Don't lick the swatter grid.

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Yes it can be a real charge playing with electricity, one can easily get wrapped up in all the wires and parts, however the amount of things you can do with it is just shocking. Currently as long as you do not have any resistance and are conductive to trying something new your capacity to generate new ideas is potentially unlimited.

OK I will stop now.

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You sound like "data" or R2D2

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Sheesh..... don't get Aviemoron... aka Norm, in on this..... if he starts I'll have to put in some, and I'm tired tonight

9. Norman Atkinson Guest

## Electrical guru's continued

J!
Nope, dunno nothin' bout the stuff.
Retired 21 years ago from supplying electricity to 1.3 million households and businesses.
My pension check has been coming in- and that's fine by me.

Money, money is a rich man's friend!
And that's me , folks.

Be my jest.( Well, that's one way to spell it- Mine!)

Norm

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