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Without wasting too much energy in the process? I have something in mind but not sure if it would work....how about have the 12V voltage source, such as a car battery followed by 6 identical capacitors in series....then each cap would be at my desired 2V right? If that works, is there a way I could combine the currents from each cap and sum them?

2007-03-23 12:40:47 · 10 answers · asked by dangerthird 2 in Science & Mathematics Engineering

basically what I want to do is this:

let's say I have a battery charger that puts out 12V @ 12A max = 144W

what I want to do is to bring down the voltage to about 2V while still maintaining a similar amount of power....of course, realistically, I wouldn't want 72A, but maybe 25 to 30A at most

how can I do this without wasting too much energy in the form of heat?

2007-03-23 15:12:06 · update #1

10 answers

you want the capacitors to source the 2 volts for a millisecond or so right? your scheme will never be practical as a means of providing continuous or sustained 2 volts. putting 70 amps of load on a capacitor will draw the voltage down nearly instantaneously. Also capacitors have internal resistance, so that will limit the current draw. you need a capacitor with ESR on the order of a milliohm or less, hard to find. Capacitors put in series never charge up with equal voltage, the ESR and capacitance between "identical" devices is never identical and changes with use. When capacitors have to be put in series, they are paralleled with equalizing resistors which add to the circuit loss to be effective. I don't know of a practical dc voltage reducer that can drive more current to the load than it draws from the source. Theoretical possible but not practical. Circuits like you propose have been successfully done for increasing dc voltages, but at very low current.

2007-03-25 15:20:07 · answer #1 · answered by lare 7 · 0 0

Capacitors, no way. In conjunction with diodes for a voltage multiplier, yes, since that is common practice. By themselves, all that the first one will do is to block the voltage path to the load. You need a good old fashioned voltage regulator. While a 2 volt reference diode, or Zener would do, along with a properly sized resistor to keep the diode from being fried, just get a ready made regulator that will provide you the output voltage that you desire. Such a device, if properly sized would give you the voltage you need, and the current that you want as well. A single lead acid battery cell would also supply you with just 2 volts. Notice, I said cell, not battery. The standard 12 volt battery is 6 cells connected in series to provide that 12 volts. Just one of such cells will give you 2 volts.

2007-03-23 12:58:06 · answer #2 · answered by Anonymous · 1 0

I will assume that using a commercial 2V DC voltage regulator is not permitted for this exercise. So discounting a voltage regulator which is what we use in the real world because of their size, power, efficiency and tolerance of output regulation, what else.
I dont like your idea although theoretically it could work.
Are you familiar with a Zener diode? What does it do when it is back-biased. What is the Vt or threshold.
Try a back-biased zener diode with a turn-on threshold of 2V. Use multiple to handle the current. Watch your current requirements.

2007-03-23 12:51:10 · answer #3 · answered by KingGeorge 5 · 0 0

One way to get the 2V without wasting power, would be to go across one cell only. That at max will give you 2.2V DC
The only problem with that is, that you would run down one cell and none of the others (unless you keep switching from cell to cell as you do your stuff).
Why don't you just get a single acid gel cell and build or buy a small charger for that.

2007-03-23 13:30:19 · answer #4 · answered by Anonymous · 0 0

use a 2 v zener diode in series with a resistor, The voltage across Zener is 2v.

you could use 3 rectifier diodes, each with a voltage drop of .65 v to give you 1.95 v. Just substitute the zener with 3 rectifier diodes. Make sure you include the resistor (10x load resistance)

2007-03-23 12:53:27 · answer #5 · answered by davidosterberg1 6 · 0 1

2v Regulator

2017-02-23 09:37:41 · answer #6 · answered by tougas 4 · 0 0

You can use a buck-boost transformer also. All of the means discussed will have heat waste and you will in the end realize that you can't get something for nothing.

2007-03-23 13:14:00 · answer #7 · answered by Nightstalker1967 4 · 1 0

USE A STEP DOWN TRANSFORMER CONNECTED TO THE 12V INPUT.THEN CONNECT A BRIDGE RECTIFIER COMPRISING OF 4 DIODES .IF YOU WANNA LARGER CURRENT THEN Ns>Np if less then vice versa.THEN CONNECT TWO RESISTORS TO THE OUTPUT YOU CAN ALSO USE A CAPACITOR FOR FILTERING WHICH WILL BE PARALLEL TO THE OUTPUT .

2007-03-24 07:43:30 · answer #8 · answered by kirk b 3 · 0 1

Your idea fails on many levels.

The best way to get 2V is by using a switching regulator like this one. http://www.national.com/pf/LM/LM78S40.html. There are simpler DC/DC converters around.

2007-03-23 12:47:08 · answer #9 · answered by Anonymous · 0 0

Use a resistor out of a busted old piece of electronic equipment.

2007-03-23 12:52:41 · answer #10 · answered by MorgantonNC 4 · 0 1

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