How to amplify current from a source with significant output voltage but low current?

Thread Starter

Rahul Kumar 4

Joined Sep 2, 2016
3
Hi,

I'm working on a LINEAR GENERATOR setup. It has following constructional specifications:

No. of tuns: 3000
Copper wire: 35 AWG
Stroke length: 5 cm
Magnets: Neodymium magnets (Br = 1.43 T)

At a frequency of 5 Hz (Rectilinear motion of magnets inside the copper wire loop) I'm able to produce 20 Volts, 100 mA.

I want to know how I can get significant amount of current instead of voltage?
 

wayneh

Joined Sep 9, 2010
17,152
Fewer turns of a lower gauge (thicker) wire would do it. Or you could use a transformer to step down the voltage, although that may not work well at 5Hz.

The easiest solution is probably a DC-DC converter. Collect your output on an electrolytic capacitor, and feed it to a buck converter. You can dial in the output voltage you want and the conversion might be 90% efficient or so. You can find DC-DC buck converters on E-bay that are very inexpensive. Just be sure the ratings meet your needs with room to spare.

Oh, one more. You could just send the output into a battery. The battery will do a pretty good job of capturing the energy, but probably not as efficiently as the option above.
 

tcmtech

Joined Nov 4, 2013
2,867
Less turns but optionally with heavier gauge (lower resistance) wire as wayneh said. It doesn't get any simpler than that. :rolleyes:

Half the turns will give you half the volts but at twice the current and so on and by optionally #35 wire is rated for around 250 - 400 ma depending on the application and duty cycle of which I doubt you will be shaking this thing vigorously for hours on end into a dead short so the going to heavier wire may not be necessary unless you want to reduce resistance related losses in the coil itself to gain efficiency.
 

wayneh

Joined Sep 9, 2010
17,152
One thing to learn, if you really want to get into this, is that the generator power output depends on the load, and there is some load that will maximize the power output from the coil. Too much load (like a short), the voltage drops and so does the power. All the power gets wasted heating the coil. Too little load (maybe a single LED with a resistor) and you have plenty of voltage but not enough current to maximize the power.

The maximum power point (MPP) depends on the winding resistance and the frequency of the generator.

My point is that if you're trying to efficiently capture energy, there's a lot to consider.
 

Thread Starter

Rahul Kumar 4

Joined Sep 2, 2016
3
One thing to learn, if you really want to get into this, is that the generator power output depends on the load, and there is some load that will maximize the power output from the coil. Too much load (like a short), the voltage drops and so does the power. All the power gets wasted heating the coil. Too little load (maybe a single LED with a resistor) and you have plenty of voltage but not enough current to maximize the power.

The maximum power point (MPP) depends on the winding resistance and the frequency of the generator.

My point is that if you're trying to efficiently capture energy, there's a lot to consider.
I tried using different set of load, but the output power from LG remains the same. I like the idea of using a DC-DC buck converter....just wanted to know whether you've tried it before? Also how well an operational amplifier would work in this scenario?
 
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