Hi,

I'm working on a physics project for which I have to charge a 1.0 F capacitor "as completely as possible" by dropping a magnet through a solenoid coil. I have tried this with a voltmeter in place of the capacitor and found that the graph of induced voltage vs. time looks something like this.

If I were to hook up the capacitor directly to this voltage source, it would simply charge and then immediately discharge. Obviously, I will need some sort of (preferably full-wave) rectifier. Since I'm dealing with pretty small voltages (about 6.3 V max), minimizing the forward voltage drop across the rectifier is going to be key. I was reading here about center-tapped rectifiers, but I know very little about different sorts of transformers. So, my questions for you all:

1. Would a center-tapped rectifier be an appropriate solution, given the small input voltage I'll be working with?

2. If so, what properties should I be looking for in a center-tapped transformer?

3. What's to stop me from using a step-up transformer in order to increase the voltage from the primary to the secondary coils? Wouldn't this cause the capacitor to charge faster/more completely?

4. Is there an advantage to my using a diode bridge over a center-tapped rectifier?

Thanks very much!
xenolalia

 

A single germanium diode would be appropriate(if you want to keep this thing super simple) to block any discharge of the cap.

Are you free to design the solenoid coil or must you use one provided for you?

Stepping up the voltage will only step DOWN the current. A 1F cap needs the current to be high. The voltage must simply achieve a higher level than the cap charge reaches to maintain current input.

 

Thanks for your speedy reply!

1. If I were to use only one diode, wouldn't I only be utilizing half of the input waveform?

2. I am free to design the coil (lots and lots of turns ).

3. Could you please explain to me why a 1.0 F cap needs a high current input in order to charge?

Thanks again!

 

The waveform pictured by you would require a full wave rectifier to capture the energy of both the negative and positive peaks.

Creative winding and placement of the coil, combined with a magnet that presents both poles simultaneously could reduce the need for the extra diode and increase the energy available for capture.

 

The physics of capacitor charging and why current levels are important should be researched on Google. To much info to convey in a one minute reply.

A coil of many turns connected to another coil of many turns but wound in the opposite direction and place sandwich style on the outside of tube, would combine the voltage induced in each coil additively. The magnet would need to be other than a bar magnet. Non axial polarized with one 180 degree side as a North and the other 180 degree side as the South. Such magnets are not as common but still available with a little searching

 

http://www.magnet4less.com/index.php?cPath=133

 

Though you are absolutely correct that by varying the design of the solenoid coil and the magnet itself I may be able to eliminate the need for the diode(s) altogether, for now, my primary design consideration is the rectifying circuit. There are a few other logistical aspects to this project (such as the required mass of the magnet) which I did not address in my initial post, as they are not strictly relevant to the electronics. For now, I must work from the assumption that my magnet will be a cube magnet polarized through its thickness (for example, this).

This in mind, I still do not quite understand the whole "current level" issue, as the vast majority of internet articles about the physics of charging a capacitor (such as this one, and the relevant Wikipedia article for that matter) assume a constant voltage source.

Here is what I do understand about charging a capacitor:

where the time constant τ0 = R*C and assuming a constant source V0.

Also, more generally,

It is not obvious to me how, from these equations, it follows that charging a 1.0 F cap requires a large current. If you could point me in the right direction (maybe a link or two) that would be much appreciated.

Thanks!

 

3. Could you please explain to me why a 1.0 F cap needs a high current input in order to charge?

Here's the simple form, to use it we have to assume that the current is a constant. (To solve for any current waveform needs calculus.)

Volts = Current * Time / Capacitance

Example: We put 1 amp of current into a 1 F cap for .02 second (which is about your pulse's time)

Volts = 1 * .02 / 1 F = .02 / 1 F = .02 Volts

A smaller cap would charge to a higher voltage (hint: don't do that cause you might get the highest voltage but you'll get caught for cheating)

 

Ah, I see.

So (Kermit2, I think this is what you were saying), the applied voltage (i.e., the Vout from the center-tapped transformer) needs to be bigger than the initial voltage across the cap (in order to cause charge to flow onto it), while simultaneously maximizing the current, so as to shove as much charge onto the cap as possible.

Correct?

Here's a related question (pertaining to that induced voltage vs. time graph I linked to above): during what sections of the graph would the capacitor discharge--when the voltage is negative or when it has a negative slope (i.e. dV/dt < 0), assuming a non-polarized cap?

 

Disregard that last question; I believe I found an answer here.

 

Disregard that last question; I believe I found an answer here.

Those are not the droids you are looking for.

Move along.