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bcwilson_oz

Voltage Multiplier

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Yes, thank you, that was very much more helpful and I will consider finding diodes with a lower voltage drop, or just looking for a different method - or maybe giving up altogether, after all, I've already got rechargeable batts.

But for interest, before I savaged the motor, it was a 'fancy' one which lit up when switched on - via 4 LEDs inserted in its casing. When used in reverse as a dynamo, with no other load attached, these 4 LEDs lit up as I cycled down the street and the fan turned. Yes, the LEDs were lower power output than the high luminosity ones in my 'traffic light' (as you have christened it), and there were less of them, but clearly there is SOME usable power being produced.

The motor/dynamo has 4 windings, and 3 terminals. 2 points are common and I'm taking the output from the non-common terminals, after finding that this gave the greatest (AC) p.d. My half-way understanding of the subject leads me to conclude that the common terminal is the 0V rail, and the non-common terminals are producing a simple AC wave. However, with no oscilloscope, these are educated guesses based on investigations with my multimeter, but either way, with no load, I can blow on the dynamo and register upwards of 8V AC across these terminals. Not for very long though - I start to go dizzy!!

I'm using this particular one because it was in the bargain bin and cost me all of 98p (GBP0.98). Being a brushless motor(dynamo), however, (and uses bearings, not bushes) I feel that it will be quite efficient at its job, plus it has the added benefit of already having a rather good fan attached, and a casing with points I can use to attach it either to my bike or the 'traffic light'.

I will return (maybe) when/if I can update you with my achievements or failures. Though I suppose it's more likely if I fall on the side of 'achievement' rather than 'failure'......

Many thanks again,
Cheers,
Clark.

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Hi Clark,
Thanks for the link of the guy producing 64,000V! Where's Steven?
That guy doesn't care about the voltage drop across his diodes.

You might not need a voltage multiplier and its losses if your dynamo produces 8V when you blow it enough to get dizzy. Your meter probably measures AC RMS which means the peak voltage is 11.3V. After rectification, you'll get 10.5VDC. It will probably sag with a load and the fan will slow down but it is worth trying.

Why not test the dynamo with a load and voltmeter while you are holding it out the window of a moving car (let someone else drive)? Record different loads and speeds. Record the loads and speeds when its unloaded output voltage drops to half. Please let us know the results.

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There is an update to my numbers diagram. I forgot that the negative cycle does indeed show up because the first capacitor is clamped at 10 volts. The fist diode will remain reverse biased and you get the complete cycle which is passed through the capacitors.

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Hi Clark,
Welcome to the forum.
The little power plant from a computer fan sounds interesting. I hope that you will continue with the project. Is it possible for you to display a drawing of what you have made? I am sure others will take an interest. Please do not be discouraged from comments posted by others. Wind power is a good way to get a trickle charge. You might find that these computer fans all give different outputs. Have you looked at the signal on a scope? You might not have a sine wave. It is very likely that you have a pulse or square wave, which will give you different results as well.
Good luck!

MP

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First things first. audioguru was right to have reservations about how much power the fan/dynamo can produce.

When blowing on it, taking the output directly into a bridge rectifier and not into the multiplier, I can keep a single high luminosity LED lit. On the other hand, a fully charged 2200 micro-F cap does not keep the LED lit for a significant amount of time (i.e. not long enough to sit at traffic lights) [if I remembered all the equations from school I could have worked that out!]

So - regardless of whether my dynamo can create enough power to light up the LED(s), I'd need some stonkingly large caps to keep them going while I'm stopped. Back to my rechargeable batts then.**

My crude calculations suggest I can get about 10-40mW from the dynamo. Let's say 1mA @ 20V. Doesn't really seem worth the hassle trickle charging the batts either, it's gonna take a long time to put any significant charge into 750mAH AA's.

I do have a slightly different question though, if anyone's still reading (!).

My battery charger registers, for the sake of argument, 1.4V when charging a batt that when isolated registers 1.3V. Now I've convinced myself of this, but could someone confirm?

With the battery in the charger, the charger MUST have a driving potential of 2.7V mustn't it? Because the driving potential from the battery is fighting against it.

Back to a possible trickle charging dynamo. To charge a 10V battery pack, I guess I need to generate a driving potential of 20V (or so).

Although the v-multiplier worked with no load, as soon as I attached a load, the voltage dropped drastically (almost to zero!) Am I right in thinking the lower the capacitance of the caps in the v-mulitiplier, the less will be the voltage drop (if at the expense of a higher ripple)?

Anyway, in conclusion, I think the output from a wind-powered dynamo is too small to be really worth utilising.

**One could maybe attach a (well balanced) weight to the dynamo so that when stopped the angular momentum kept it going... But that's mechanics not electronics!

MP, sorry, no diag, there's not really one to talk of. I simply rectified the dynamo output with an off-the-shelf bridge rectifier. The v-mulitiplier circuit is already on this site. In both instances I just connected the AC out from the dynamo to whichever circuit.

Also, no, I have no oscilloscope and am not likely to have access to one so I can't investigate the exact waveform of the AC.

Thanks everyone.
Cheers,
Cloink.

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Hi Clark,
I am sorry that your generator project didn't work.
But I am glad you realise that if the fan was large enough to create the same wind-resistance to your cycling effort as a mechanical dynamo, then you wouldn't gain anything.

I am also glad that you realise the huge size required for a "supercap" to provide LED current for any length of time.
Small capacitor size is the same problem with the voltage multiplier.

Answers to your new questions:
1) You are correct, a Ni-Cad or Ni-MH rechargeable cell's voltage peaks at about 1.4V when fully-charged and is still charging at 1/10th the current of its A/hr rating. The charger's voltage must be higher than the battery's voltage so that current is continuing to flow into the battery. Each cell's voltage will quickly drop to about 1.25V to 1.3V after being removed from the charger without a load.

2) Since each cell needs only about 1.4V to fully charge, a higher voltage would be required if you used a normal current-limiting resistor, so the extra voltage is lost across the resistor. Ohm's Law says the voltage across the resistor determines the charging current. I don't know where you got a voltage of 2.7V. Maybe your charger produces 2.7V (without a battery cell) and loses the extra 1.3V across its current-limiting resistor.

3) If the generator doesn't produce too much current, it can be connected directly to the battery, but use a series diode to keep the battery from discharging into the generator when it is slow or stopped. The extra generator voltage will be lost across the generator's internal resistance, and 0.7V to 0.9V will be lost across the diode.

4) Ni-Cad and Ni-MH battery cells are usually rated at 1.2V with a reasonable load. So your "10V" battery will have 8 cells and be rated at 9.6V. To provide 1.4V per cell, its charger must have at least 11.2V plus extra voltage for its current-limiting resistor.

5) Some battery manufacturers (excellent info at www.energizer.com ) recommend a maximum continuous "trickle-charge" current that is only 1/25th the battery's A/hr rating. They assume the trickle is not for charging but for keeping a fully-charged battery above self-discharge.

6) With small caps in the voltage multiplier and a high current load, the ripple voltage is maximum. With much larger caps or lower load current the ripple voltage will be much less because the caps will be big enough to provide the current between input half-cycles without the voltage dropping too much. The input current will be multiplied by the multiplier amount.

Have you considered using Ni-MH cells for powering your LEDs? AA cells are available with 2200A/hr ratings.

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Hi audioguru

Thanks for all your time - you've still missed some of my points though. I already have 8AA's in use for the circuit. My hope was to reduce the AAs to AAAs (to save weight and also 'cos the AAAs would fit in the casement in which the LED array is housed whereas 8 AAs don't). Of course I can do this but I'll only get around 2hours usage instead of 8 or so as the charger won't make a significant difference.

I'm fortunate enough to have a dad who sells rechargeable batts so I'm ok on that score!!

The batts are indeed rated at 1.2V (some at 1.25V), but even when in use, the 8 together still registered over 10V using my multimeter (even as high as 10.5V), this was at c.350mA.

I still however believe that, given a clean design sheet, one could manufacture a wind-powered dynamo-and-LED light set. As I mentioned, the crude arrangement I had was enough to power one high luminosity LED - not much I know but this was from cobbled together bits and bobs from someone who doesn't really know what they're doing. Obviously I'll have to wait 'til I'm a millionaire and I can commmission someone to make one for me!

Finally, I'm sure you've picked up on this, but with a diode in place to prevent the batts draining through the dynamo, the small current that the diode allows in reverse would surely be enough to negate any output from the dynamo in the first place, especially if left connected when not in use.

Anyway, I've fallen out with the whole project, 3 of the LEDs have stopped working (presumably blown) and I've lost the battery pack which had my 8 AAs........... Oh dear.

Cheers,
Clark.

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Hi Clark,
Now is the time to replace the old low-capacity 750mA/hr AA Ni-Cad cells that you lost with tiny but modern high-capacity 850mA/hr AAA Ni-MH cells (Energizer have them). You are lucky that your dad can give them to you!!

The reverse leakage current of a silicon diode at your low voltages is so low that you can hardly measure it with extremely sensitive equipment. The 10 femto-amperes of leakage current from a cheap 1N4002 diode will discharge your new 850mA/hr battery in 9,703 years! A slightly more expensive but higher voltage rated 1N4007 diode will leak even less.
D ;D ;D

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