The 150 ohms is the load (your battery) matched to the resistance of your coil for maximum power transfer. To accomplish the same thing with the boost converter you would limit the current from the coil so that the impedance of the load would look like 150 ohms.
If you like I can attach the simulation and you can see if you can get more power to the load. It's kind of like having your scope in the basket.
You can google maximum power transfer.
PS. The 150 ohms is not attached to the capacitor all the time only when the input voltage is above the threshold. That's why you see the little stair steps.

 

Can you increase the number of magnets?

Probably easier and/or cheaper to just get a dynamo.

The bottle type ("rubwheel") usually produce the highest output, but tend to be difficult to insulate from the frame (can't use a bridge rectifier otherwise)'

Obviously a hub dynamo involves some expense (replacing the whole wheel).

In either case there would be sufficient current to overcharge a lithium cell - without a full charge safety cut off, they can be quite entertaining.

 

Thanks Ian field for your advice.
However, my post is not just about getting something to work well by replacing hubs and gear boxes, and all other expensive ways to "fix" the problem.
My post is about understanding energy harvesting from an incompetent 150-ohm source impedance to charge a large capacity battery, using a pulse outputting bubba generator that was only made to flash an axel mounted safety LED flasher for bikes at night(see pic above).

 

The 150 ohms is the load (your battery) matched to the resistance of your coil for maximum power transfer. To accomplish the same thing with the boost converter you would limit the current from the coil so that the impedance of the load would look like 150 ohms.
If you like I can attach the simulation and you can see if you can get more power to the load. It's kind of like having your scope in the basket.
You can google maximum power transfer.
PS. The 150 ohms is not attached to the capacitor all the time only when the input voltage is above the threshold. That's why you see the little stair steps.

The idea of making the source impedance match the load works well with transmission lines but maybe it is not always best for operation of a boost converter for energy harvesting.

Adding 150 ohms in series to the boost converter (and removing 4500uf (value corrected) capacitor as well??) makes the whole circuit and boost converter worthless as a energy harvesting device.

I would really like to see your simulation, especially interested in seeing the stored power versus startup voltage and capacitor size, that is the stored power in the capacitor(it is actually 4500uf) v. time and startup voltage with/without the 150 ohm in parallel. Of course, you must consider that the boost converter is less than an ohm when active and reaches current limit, otherwise essentially an open circuit.

 

I took my trusty fluke DVM out with me and looked at the output of the generator with 4500uf of capacitance directly connected to the output of the generator through a schottky bridge rectifier. All my readings are with the 4500uf capacitors in circuit.
I was unable to see waveforms, the DVM only measures current and voltage but has a peak hold feature.

At low speeds (approx. 5 mph) I was able to get a peak reading of approx 4 mA short circuit, at approx 10 mph I was able to see about 10mA peak.
Open circuit the capacitors(4500uf) charge to .9V at 5mph in about 5-seconds, at 10mph the capacitors charge to .9v in approx 3-seconds.
Open circuit the capacitors charge to 2.7V in about 8-secs at 5mph in with 6-secs at 10mph.
Peak open circuit voltage was >10.6V and this voltage might have been higher but likely limited by use of three 1500uf/6.3V capacitors.

I measured charge time this by short circuiting the output at the capacitors and watching the wheel rotation I waited until the voltage reached the goal voltage and recorded the time.
A better test of course, would be to record the start and end voltages of the battery after several hours of biking during the day, and the best test would be recording the weekly voltage flashed out on my bike light to see how fast things are charging/discharging with my usual style of riding and using flashlight momentarily and mostly just using flashing modes of operation.

Having perfectly aligned spokes on the wheel is also of great importance, because the magnets will rub on some rotational positions due to slop in the ball bearings and rim "trueing" errors. I had to position the pickup to magnet spacing so that only one magnet passed within a tenth of a inch while others would pass the pickup coil up to about .3 in to prevent any of the magnets from rubbing.

I was quite in error about the distance of the magnets on the spokes from the center. They are about 3-in from the axle of the 28-in rim. That still gives 8 PPS for 10mph. I clearly understand that the closer the magnets are to the axle, the less distance exists between the magnets. There is no more room on the spokes for any more of magnets of the type used. My guess is approx 25% of the time the magnets are interacting with the coil pickup.

 

Madsi said "simulate", not stimulate.

I need to get my eyes checked again.

"

 

I took my trusty fluke DVM out with me and looked at the output of the generator ....

A very positive development. You got 10.16V open-circuit voltage and I guessed 9V. But your short circuit current was lower at 10mA than my guess of 30mA into a 150Ω load. Unfortunately that shortfall works against you in terms of total power available.

One thing not clear to me at all is the behavior of the converters, since I have not read the data sheets and I'm ignorant of these devices. I understand the cutoff voltage, but do they draw no significant current after they shut off? Once they turn on, is there a way to know how much current they will draw? You'll need this sort of information if you hope to simulate various scenarios.

If your converter was ideal, and very fast, then the capacitor would not serve any purpose and would actually introduce a loss. With no capacitor the converter would turn on "instantly" and efficiently process every portion of the pulse profile that it can, then shut off and draw no current until the next pulse arrives. On the other hand, if the converter is too slow for 8pps and requires a steady DC in order to function, then you need a design to do that.

 

The idea of making the source impedance match the load works well with transmission lines but maybe it is not always best for operation of a boost converter for energy harvesting.

It's easy to calculate. Just use a 300 ohm load and a 50 ohm load with the 150 ohm in the motor, then calculate the power in each.. I think this will make you a believer .

Adding 150 ohms in series to the boost converter (and removing 1500uf capacitor as well??) makes the whole circuit and boost converter worthless as a energy harvesting device.

But, but, you said the resistance of the coil was 150 ohms. So it is in series with the coil.

I would really like to see your simulation, especially interested in seeing the stored power versus startup voltage and capacitor size, that is the stored power in the capacitor(it is actually 4500uf) v. time and startup voltage with/without the 150 ohm in parallel.

It is attached. Here is the web site where you can download the program.
LTspice IV
http://www.linear.com/designtools/software/?gclid=CIik9q-478MCFVU2gQodxAMAcw#LTspice

Of course, you must consider that the boost converter is less than an ohm when active, otherwise essentially an open circuit.

You must use current limit to match the load for either case. See post above. So to the coil it will look like 150 ohms and to the battery a current source.

For Wayneh.
The things are pretty amazing. It starts in less than 200usec. and only draws ua. when not in use.

Madsi, I love it when people want to learn, but I think I will watch television for a while instead of talking to it.

 

Madsi said "simulate", not stimulate.

"

Oh, simulate.

I don't use simulators. I prefer the real thing.

 

It's easy to calculate. Just use a 300 ohm load and a 50 ohm load with the 150 ohm in the motor, then calculate the power in each.. I think this will make you a believer .



But, but, you said the resistance of the coil was 150 ohms. So it is in series with the coil.


It is attached. Here is the web site where you can download the program.
LTspice IV
http://www.linear.com/designtools/software/?gclid=CIik9q-478MCFVU2gQodxAMAcw#LTspice

You must use current limit to match the load for either case. See post above. So to the coil it will look like 150 ohms and to the battery a current source.

For Wayneh.
The things are pretty amazing. It starts in less than 200usec. and only draws ua. when not in use.

Madsi, I love it when people want to learn, but I think I will watch television for a while instead of talking to it.

Thank you again,

I downloaded LTSpice and the .asc file but all I get is a black screen when I run the simulation. I am a complete noob with this program.
What am I missing?

Enjoy your TV.

 

A very positive development. You got 10.16V open-circuit voltage and I guessed 9V. But your short circuit current was lower at 10mA than my guess of 30mA into a 150Ω load. Unfortunately that shortfall works against you in terms of total power available.

One thing not clear to me at all is the behavior of the converters, since I have not read the data sheets and I'm ignorant of these devices. I understand the cutoff voltage, but do they draw no significant current after they shut off? Once they turn on, is there a way to know how much current they will draw? You'll need this sort of information if you hope to simulate various scenarios.

If your converter was ideal, and very fast, then the capacitor would not serve any purpose and would actually introduce a loss. With no capacitor the converter would turn on "instantly" and efficiently process every portion of the pulse profile that it can, then shut off and draw no current until the next pulse arrives. On the other hand, if the converter is too slow for 8pps and requires a steady DC in order to function, then you need a design to do that.

The datasheet explains it all. The MC1624 draws only nanoamps when shut off. It also has internal current limiting depending on the supply voltage.
The capacitor is the main component in making the MC1624 converter work at all and does not need a steady DC to function, it startsup at .65V and will function until the voltage drops below.3V.

 

I'd experiment with using only a small capacitor, e.g.. 10µF on the input. It looks to me that the converter can start up plenty fast and you don't need to store the energy upstream of the converter - it'll handle the pulse in real time.

 

Click on the schematic, then mouse over the wire you want to scope and left click
If you want to se current place it over the component you want to se current in. You will see a little current probe. Click again.
Want to se power? Hold down Alr and mouse over the component. You should see a little thermometer. Click.
Want to see average power go up to the scope and do a control click on the power lable you want the average of.

 

I'd experiment with using only a small capacitor, e.g.. 10µF on the input. It looks to me that the converter can start up plenty fast and you don't need to store the energy upstream of the converter - it'll handle the pulse in real time.

The MCP1624 can startup in 250 uSec, no problem there. Once started, the converter attempts to store all available Vin current in the inductor 4.7uH and the 150 ohm source resistance of the generator limits the current to <10mA..result: Vin drops almost instantly (a few uSec) to less than .3V and the converter shuts down. This repeats and almost no power is delivered to the battery.
In any case the converter needs a some value bypass/storage capacitor to be stable and operate.

 

Click on the schematic, then mouse over the wire you want to scope and left click
If you want to se current place it over the component you want to se current in. You will see a little current probe. Click again.
Want to se power? Hold down Alr and mouse over the component. You should see a little thermometer. Click.
Want to see average power go up to the scope and do a control click on the power lable you want the average of.

Thanks again, that's a start.

Ok, but I wonder also how to set the converter voltage source to show the low duty cycle generator output 8PPS pulse train that is correctly shown in your simulation.
Power table..I don't see any stinking power table!

 

The MCP1624 can startup in about 500 nSec, no problem there. Once started, the converter attempts to store all available Vin current in the inductor 4.7uH in short pulses and without the 4500uf cap, the 150 ohm source resistance of the generator limits the current to <10mA..result: Vin drops almost instantly to zero and converter shuts down. This repeats and almost no power is delivered to the battery. Therefor the 4500uf cap is the essential energy storage component prior to getting the boost converter to harvest the power in the capacitor, to transfer the energy in the cap to the battery. I am actually using 4500uf(value corrected from 1500uF).

 

Once started, the converter attempts to store all available Vin current in the inductor 4.7uH and the 150 ohm source resistance of the generator limits the current to <10mA..result: Vin drops instantly to zero and converter shuts down.

How do we know this? I think you'd only need to limit the output current so that the device looks like a 150Ω load to your generator. This should prevent it from overloading the coil and smashing the voltage.

 

Here my scheme to harvest energy, this is what my post is about:

I want to harvest as much as the bubba generator output energy best as I can.

I think the best way to do this is use a bridge schottky rectifier and it directly connects through the generator's 150-ohm source resistance and then stores the rectified power in the largest (space limited) capacitor that will fit. Once the voltage of the capacitor (4500uf in this case) reaches a voltage between .9V and 4.2V, I want to harvest the energy stored in the capacitor using the boost converter.

The boost converter upon startup, presents in a few uSec <.5 ohm load to the generator/4500uF capacitor in any case of operation, until the inductor current has reached to the current limit of the MCP1624. The current limit of the MCP1624 is highest with the highest Vin.

I assert that adding any impedance matching 150 ohm resistor to this circuit to limit the 4500uf capacitor's load or the converter's mismatch on the generator by using a series/parallel passive resistor ohms is nonsense, a waste of energy.

Of course I could be proven wrong.

 

The MCP1624 can startup in 750 nSec, no problem there. Once started, the converter attempts to store all available Vin current in the inductor 4.7uH and the 150 ohm source resistance of the generator limits the current to <10mA..result: Vin drops instantly to zero and converter shuts down. This repeats and almost no power is delivered to the battery.
In any case the converter needs a bypass/storage capacitor to operate.

You have a misconception about how a boost regulator works. It does not just turn on and suck all the current out of the coil. It only turns on for a few 100 ns and the inductor keeps the current from rising very much in that time.

The start time is ~ 200 usec.. There is a nice curve in the data sheet.
I have to go out but I will send you a simulation so you can see the power transfer.
There is not a power table. Once you have the power waveform displayed on the screen go to the very top of the "scope" to the label. Then ctrl click.

 

The idea of making the source impedance match the load works well with transmission lines but maybe it is not always best for operation of a boost converter for energy harvesting.

Adding 150 ohms in series to the boost converter (and removing 1500uf capacitor as well??) makes the whole circuit and boost converter worthless as a energy harvesting device.

I would really like to see your simulation, especially interested in seeing the stored power versus startup voltage and capacitor size, that is the stored power in the capacitor(it is actually 4500uf) v. time and startup voltage with/without the 150 ohm in parallel. Of course, you must consider that the boost converter is less than an ohm when active, otherwise essentially an open circuit.

My experience with the very low power Sturmey-Archer hub dynamos, suggest that any attempt to use a boost converter is doomed to failure.

It looks as if using a step down converter to draw less current and allow the output voltage to "flourish" is the route to more total power output.

This might also work with your set up - but you need more magnets!

You might find a use somewhere for a supercapacitor, fast charge conversion like any capacitor - but starting at around half a Farad, they're virtually a battery for micropower applications. They do have very limited voltage range though.