Most Effective Way to Use Generator Output

ronv

Joined Nov 12, 2008
3,770
Thanks for your input, but I am not interested in just getting charging to work, I want to understand why what works works best from engineering analysis.
Yes, I think we understand that, but yor specs leave something to be desired.
I'm going to assume when you say pedal quite hard you mean 10mph and that the magnets are about 20 inches around the rim.
So I make the output 4.2 volts peak at 12 HZ. and around 30ma. peak.
Please correct me if I am wrong. :)

This chip is more complicated, more difficult to work with 10 pins v. 6, extra components are required to fit into the tiny available space and because it is from Linear, it is perhaps too expensive, costing > 12x as much and re-layout of my PCB.
You should go back and re-read the data sheet. While it has several features not all of them need to be used, so the components are the same.
I'm sorry you already laid out your PCB. :(
Even if this is the best chip for the job, I am looking to understand from engineering analysis why a cheaper, simpler solution like the LM27313 can be made to work as well and why the MCP1624 can cannot.
Probably the most important is the start/stop voltage, because below this you get nothing.
Start -- 27313 - 2.7v 1624 -.8v LT .4v
Stop -- 1.5v .35v .225v
Then maybe worry about how much power the chip uses:
27313 -2ma 1624 - 220 ua LT 200ua.

So you can google around and probably find lots of better ones by reading the data sheets. Try energy harvesting boost converter.
 

jamesd168

Joined Nov 8, 2014
21
I am not manufacturing energy to sell to the city.. I am riding a bicycle, and like most everyone else, the speed and length of my trips will vary to the occasion. But this problem must have a reasonable answer.

I don't know the charging effect at slow speeds, or even on average speed, if there was such a thing. Charging is dependent on how fast, but certainly this is dependent the way I bike.

It remains a question that can be answered by considering whether converting any available voltage >.9V at the capacitor is more efficient/effective than letting just the unboosted capacitor voltage rise to >battery voltage to force charging current into the battery, all this when a 150 ohm series copper loss resistor of the generator is always in the circuit.


At high speeds I can see that the voltage generated would be sufficient to charge the battery whenever the rectified voltage was greater than the battery voltage and so it is sufficient to force current into the battery. But I also know that even a higher speeds, that there is considerable energy in the capacitor left unused because the voltage is just below the threshold of charging, and this energy would be harvested by a boost converter.

But one must also consider what is going on with slow biking, maybe even consider the that slow speed biking is a considerable portion of the total time I spend biking and maybe boosting small voltages would be an efficient way to keep the battery charged.

I don't know the answer to your question. However I suggest you look into the way people are doing the energy conversion. if a manufacturer of similar product is doing it one way versus another, it's probably a good indication that one way is more efficient than the other?
 

Thread Starter

madsi

Joined Feb 13, 2015
107
Yes, I think we understand that, but yor specs leave something to be desired.
I'm going to assume when you say pedal quite hard you mean 10mph and that the magnets are about 20 inches around the rim.
So I make the output 4.2 volts peak at 12 HZ. and around 30ma. peak.
Please correct me if I am wrong. :)


You should go back and re-read the data sheet. While it has several features not all of them need to be used, so the components are the same.
I'm sorry you already laid out your PCB. :(

Probably the most important is the start/stop voltage, because below this you get nothing.
Start -- 27313 - 2.7v 1624 -.8v LT .4v
Stop -- 1.5v .35v .225v
Then maybe worry about how much power the chip uses:
27313 -2ma 1624 - 220 ua LT 200ua.

So you can google around and probably find lots of better ones by reading the data sheets. Try energy harvesting boost converter.

Thanks much, I appreciate your analysis, ronv.

The bike has 28-in wheels, 4 magnets 90-deg appart, located about 4-in from the axle.

No one has taken the question of peak power into question: The power stored in a capacitor is C*V*V/2. This means the power available increases with the square of Vin. Also consider that at a higher Vin, both the MCP1624 or TSD27313 MOSFET switch have a lower RDSon.

Is it to best advantage to have a higher voltage stored in the capacitor or allow the converter (MCP16214) to struggle with a pulsating voltage that barely gets a chance to rise above .9V before a few cycles of conversion cause the capacitor to exhaust all the energy.

I could add a tiny PNP transistor, emitter to the enable pin, collector to ground, base to a resistor to ground, base to cathode of a forward-biased LED to VIN through a high valued resistor. By selecting LEDs by color, red=1.6v, yellow1.8v, green=1.9v, white=3.3v (all are approx and depend on Vin), I could select the startup Vin to be any of these LED voltages+.55V base-emitter of the PNP.. or I could consider the 2.7V starting up LM27313.

In any of the scenarios, as soon as the converter starts up, (250uS after Vin >= .63V) is it significant in the name of all power harvested, that at high values of C*V*V/2, any available power would be harvested more efficiently?

This UV hold off scenario also clearly advantages itself of the largest C that can fit in the available space, so more pulses are delivered in more packets of charge energy before Vin falls below the startup Vin.
 
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Thread Starter

madsi

Joined Feb 13, 2015
107
Hmm.. Sounds kinda fast for me!

10 mile/hr * 5280 ft/mile * 12-in/ft * 1-hr/3600 Sec = 176 in/Sec = 14-ft 9-in/Sec.
Circumference of 28-in bike wheel = 3.14 * 28 -in = 88 in /revolution traveled.
176 in/88 in = 2 RPS, or 4 * 2 = 8 pulses per second.. 8 Hz.

But even at 8 PPS, rather than 8Hz, pulses have only about a 30% duty=cycle due to the separation of the 4 magnets, evenly spaced on the 28-in wheel spokes, the large enclosures waste lotsa space just for housing each magnet, so quite a bit of nothing is generated in the time between each magnet's passing the pickup.

Doesn't compare well to your estimate of 12 Hz?
 
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Thread Starter

madsi

Joined Feb 13, 2015
107
I don't know the answer to your question. However I suggest you look into the way people are doing the energy conversion. if a manufacturer of similar product is doing it one way versus another, it's probably a good indication that one way is more efficient than the other?
A capacitor question I am faced with pedaling my way through my small hamlet:

To charge or not to charge, that is the question!

Is it more normal to pulse little and pulse well, or pulse with pitifully poor power planted in the capacitor.


UV be damned, it is better to hold off between spurts?

What profits a converter to work so hard in desperate efforts to harvest so little?
 
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ronv

Joined Nov 12, 2008
3,770
Yes, all the necessary information is all here for engineering calcuation and analysis.
Maybe not the wheel diameter.:rolleyes:


If you are going below the start threshold all of that energy is lost. That is why I am in the always charge camp.
For the best performance you would want to use the largest capacitor you could charge above the threshold with the 150 ohms of the coil at some rpm you decide as I think the voltage will be lower the slower you go.
Then you would like to limit the current to the battery that would just keep it from turning off.
This of course varies with how fast you pedal. What you are looking for is the maximum power from the capacitor and thus the coil. This is mot as simple as setting a voltage or a current as it changes. But all is not lost. They make what is called maximum power point controllers. The LT part I posted has this. It will manage this for you.
Short of that. Just make the cap as big as you can at the speed you decide and set the voltage to 4.2 volts.
As far as Rds on goes. They all seem to be about the same and all have current limit.
Actually, the transformer is sounding better all the time as the diode drop is very high compared to the voltage.
 

Thread Starter

madsi

Joined Feb 13, 2015
107
What profits a - circuit designer - to work so hard in desperate efforts to harvest so little?
Do you think that not bothering to think much or bother to use math to solve a problem is engineering?
Do you think that is ok not to try to learn to understand, that all one needs to do is guess or blindly follow opinion, that any engineering effort fails to to solve problems because it just takes too much effort to deal with complicated things like fractions and multiplication?


Yes, all the necessary information is all here for engineering calcuation and analysis.
Maybe not the wheel diameter.:rolleyes:


The diameter of the wheel is not at all important to answer my questions about energy harvesting, which has to do with how well whatever small energy is available is by what calculated means is best harvested.

If you are going below the start threshold all of that energy is lost. That is why I am in the always charge camp.
For the best performance you would want to use the largest capacitor you could charge above the threshold with the 150 ohms of the coil at some rpm you decide as I think the voltage will be lower the slower you go.
Then you would like to limit the current to the battery that would just keep it from turning off.
This of course varies with how fast you pedal. What you are looking for is the maximum power from the capacitor and thus the coil. This is mot as simple as setting a voltage or a current as it changes. But all is not lost. They make what is called maximum power point controllers. The LT part I posted has this. It will manage this for you.
Short of that. Just make the cap as big as you can at the speed you decide and set the voltage to 4.2 volts.
As far as Rds on goes. They all seem to be about the same and all have current limit.
Actually, the transformer is sounding better all the time as the diode drop is very high compared to the voltage.
I don't want to just pay for devices that "manage this for you." I'd first like to consider the calculation of what works best and design my own, it might be just as simple as the suggestion I have stated above with using holdoff UV startup of the converter, math will tell, you do not.

You give no math to justify your opinion, for either the voltage you've chosen or why the economy of your choices of devices are chosen for the design of a converter best work and suggest the absurdity of possibly designing and constructing, or elsewise even finding some bulky, expensive phantom custom transformer to somehow help to do this job of energy harvesting better.
 
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ronv

Joined Nov 12, 2008
3,770
OK, Here's your math.
Set the start voltage to 2.5 volts. Pedal around all day at 2 MPH. How much energy did you harvest?
Now set it to .5 volts and do the same thing. Infinitely better.
If we can agree on this it is a start. :D
 

wayneh

Joined Sep 9, 2010
17,496
Do you think that not bothering to think much or bother to use math to solve a problem is engineering?
Sometimes, yes. It takes wisdom to know when to walk away, or to formulate a better plan.

Rejecting the advice of experts and eliminating options because they don't fit your preconceived notions is not engineering either.

You want math, so let's do some math. Suppose your pace is 10mph, giving you 8pps as you calculated in #64. I think 10mph is a reasonable value since even an out-of-shape person can maintain 10mph for minutes at a time, if not hours. It might be an overestimate for a casual rider in town though, if there are frequent tops and restarts.

We don't know what a single pulse looks like and thus we don't really have good information on how much energy is produced with each pulse, and it will vary in proportion to speed. (The emf produced in the coil is proportional to dM/dt, the rate of change of the magnetic field passing the coil, which will depend on the velocity of the magnet.) Let's make a simplifying assumption that the pulse resembles one cycle of a sine wave. Let's be generous and estimate the peak values using ronv's values in #61; 30mA peak. Let's further assume the 30mA is driven into a 150Ω load so that power is maximized. (It's more likely the 30mA is a short-circuit current, but let's be generous.) That means we have an emf of 9V driving 30mA current in 300Ω (the combined impedance of the coil and the load). Voltage across the load is 4.5V, and power to the load at the peak is thus 4.5V•30mA=135mW.

OK, so at the peak of the pulse, the power is 135mW. The rms voltage across the load during the cycle is 4.5/1.41 = 3.19V and the rms current is I=V/R = 3.19v/300Ω= 10.64mA. The rms power into the load is V^2/R = 3.19^2/150=67.9mW (Note that rms power is not Vrms • Irms.)

So how long is a cycle? Taking a wild guess based on your photo in #1, the complete cycle from magnet-approaching to magnet-receding from the coil takes place over a 2" span. Again, that's a wild guess. You said the magnets are at a 24" radius in #63 but I'm sure you meant 10" radius, giving a circumference of πD=3.1416•20=63". At 2 rps, the magnets are rotating at 126 inch/sec and a single pulse happens in 2/126 = 15.9ms. Combining all this, each pulse produces power at 67.9mW for 15.9ms, or 1.08mW•s. We get 8pps, or 8.6mW•s per second, average power 8.6mW.

Does this value make sense for flashing an LED? A normal LED requires about 60mW but a flash might require a duty cycle of only 10% or less. So this value is plausible.

The capacity of your 18650 cell is perhaps 3000mAh at 3.7V, or 11,100mWh. One full charge of your battery would require 11,100/8.6= 1,291 hours = 54 days [corrected units] of continuous riding at 10mph. And we've made no allowance for losses yet.

Most engineers I know would call this pissing into the wind.
 
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Alec_t

Joined Sep 17, 2013
14,280
In post #70 '54 hours' should read '54 days'.
It would help to inform the choice if we knew exactly what voltage and waveform the source/dynamo outputs at what rpm. Without that information you're shooting in the dark.
I suspect the waveform is something like this
DynoWave.gif
but it would be useful to know the amplitude at, say, 120 rpm (2 rps).
 
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cmartinez

Joined Jan 17, 2007
8,218
Why settle for a pulsed signal? I've always wondered if it would be practical to attach an homopolar generator (faraday disk generator) to a bicycle... Here's a an article describing how it works, math and all.
 

Thread Starter

madsi

Joined Feb 13, 2015
107
Sometimes, yes. It takes wisdom to know when to walk away, or to formulate a better plan.

Rejecting the advice of experts and eliminating options because they don't fit your preconceived notions is not engineering either.

You want math, so let's do some math. Suppose your pace is 10mph, giving you 8pps as you calculated in #64. I think 10mph is a reasonable value since even an out-of-shape person can maintain 10mph for minutes at a time, if not hours. It might be an overestimate for a casual rider in town though, if there are frequent tops and restarts.

We don't know what a single pulse looks like and thus we don't really have good information on how much energy is produced with each pulse, and it will vary in proportion to speed. (The emf produced in the coil is proportional to dM/dt, the rate of change of the magnetic field passing the coil, which will depend on the velocity of the magnet.) Let's make a simplifying assumption that the pulse resembles one cycle of a sine wave. Let's be generous and estimate the peak values using ronv's values in #61; 30mA peak. Let's further assume the 30mA is driven into a 150Ω load so that power is maximized. (It's more likely the 30mA is a short-circuit current, but let's be generous.) That means we have an emf of 9V driving 30mA current in 300Ω (the combined impedance of the coil and the load). Voltage across the load is 4.5V, and power to the load at the peak is thus 4.5V•30mA=135mW.

OK, so at the peak of the pulse, the power is 135mW. The rms voltage across the load during the cycle is 4.5/1.41 = 3.19V and the rms current is I=V/R = 3.19v/300Ω= 10.64mA. The rms power into the load is V^2/R = 3.19^2/150=67.9mW (Note that rms power is not Vrms • Irms.)

So how long is a cycle? Taking a wild guess based on your photo in #1, the complete cycle from magnet-approaching to magnet-receding from the coil takes place over a 2" span. Again, that's a wild guess. You said the magnets are at a 24" radius in #63 but I'm sure you meant 10" radius, giving a circumference of πD=3.1416•20=63". At 2 rps, the magnets are rotating at 126 inch/sec and a single pulse happens in 2/126 = 15.9ms. Combining all this, each pulse produces power at 67.9mW for 15.9ms, or 1.08mW•s. We get 8pps, or 8.6mW•s per second, average power 8.6mW.

Does this value make sense for flashing an LED? A normal LED requires about 60mW but a flash might require a duty cycle of only 10% or less. So this value is plausible.

The capacity of your 18650 cell is perhaps 3000mAh at 3.7V, or 11,100mWh. One full charge of your battery would require 11,100/8.6= 1,291 hours = 54 hours of continuous riding at 10mph. And we've made no allowance for losses yet.

Most engineers I know would call this pissing into the wind.
Not quite:

Firstly, thanks megawatts for your explanation supported by some math!

I was in error about radial magnet positioning. My bike has the extremely common 28-in rims, and the magnets are close to 3-in from axle center.
The battery I am using has a capacity that was carefully measured by monitoring charge current and time and integrating the result(mWH= I*dt , dt=1 sec) with a MCU , yielding a result of approx 1800mAH capacity with a 4.18V end of charge voltage. This lower than usual capacity is due to the battery being salvaged, it's original use was in a PC laptop battery pack, its rated capacity was clearly marked 2200mAH and its capacity has diminished over time.

There is, always during an average week, some daytime biking, just for fun if the sun doth shine for a few hours, or else just to do some daytime business and errands. I don't use the battery during the day so it can charge the whole time I am biking. I mostly reserve use of the flashlight function of my headlight to navigate the dark corridors of the backyard for a few seconds at night, just to get me to the well-lit street, then its flasher time.

When flashing, the MCU PWM' delivers a 12.5mSec flash at 20mA of LED current with a 1-sec off time. With a very efficient array of white LEDs, things are looking bright.

Finally, It is not at all necessary for the battery to be fully charged for me and my battery to function well.

A typical scenario per week is that I will ride my bike more more hours during the day than at night. This means flashing at 20mA for 12.5 mSec/Sec most of the time, but only at night, and use full power 100% duty cycle at 40mA for 30 to 60 seconds a few times at night during the week.

The ideal would be charging the battery to a full charge, but the idea is use the reserve provided by the battery for high-power only when safety or necessity demands it.

My goal in this project was to have a battery charging scheme that might even result, over all the dark days of winter, to not need to be removed for charging, but enlighten me as needed and otherwise flash itself in the most power conserving way that still surpasses the flash intensity of the ordinary and hard to see axle-mounted safety flashers originally delivered below eye level.

Now, I am going to use some math to see how the same calculations you've used would show if a fully charged battery would be able to do its intended job in winter.

That being said, the question remains, what is the optimal level to charge the 1500uf cap before conversion begins using the MCP1624 chip?
 
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wayneh

Joined Sep 9, 2010
17,496
Now, I am going to use some math to see how the same calculations you've used would show if a fully charged battery would be able to do its intended job without being taken and re-charged once all winter.
Now you're thinking!

I suspect that a mint battery would be up to the job. One problem with a degraded battery may be an elevated self-discharge rate. Maybe someone here has knowledge on whether this is a factor in old Li-ion cells. It's a huge problem for NiCd cells.
 

Thread Starter

madsi

Joined Feb 13, 2015
107
Now you're thinking!

I suspect that a mint battery would be up to the job. One problem with a degraded battery may be an elevated self-discharge rate. Maybe someone here has knowledge on whether this is a factor in old Li-ion cells. It's a huge problem for NiCd cells.
I have had many years of working with new and old Li-ion cells, and the main reason I fell in love with them are light weight, small size, they are rugged and fairly safe to work with if short-circuit conditions are never to happen.
I discovered that even older compromised capacity cells keep whatever charge they are capable of accepting seemingly ad-infinitum, unlike NiMH cells that self-discharge themselves silly. The exception are those Li-ion cells so severely abused by heat and over discharge that they don't charge at all. Many old cells also have an increase in internal resistance, but that is not a problem because of the low power requirements of most of my battery-powered projects.
I also have used old cells to super power a hacked handheld vacuum cleaner with good results and using a MCU and PWM power to the motor discharged the cells at C capacity or greater with reasonable life times of battery service.
 
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Thread Starter

madsi

Joined Feb 13, 2015
107
See post 69.
Thanks, I also thought of the cut and try ways to answer my question.

Sounds a good but not too easy to accomplish, certainly would provide a close estimation of what voltage works best by empirical methods, but I was hoping to do the calculation with circuit engineering analysis.
 
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ronv

Joined Nov 12, 2008
3,770
I think you are missing the basic problem. If the converter does not get enough voltage from the pulse it will not turn on. If this happens all of that energy is lost forever, so you want it to turn on at as low a voltage as possible. Once it turns on the capacitor makes no difference. It will stay on as long as it has enough voltage , then it will shut off. You cannot get more out of it than you put in. Please show me where I am wrong.
 

#12

Joined Nov 30, 2010
18,224
Thanks, I also thought of the cut and try ways to answer my question.

Sounds a good but not too easy to accomplish.
You've been here for 5 days. That looks like, "not too easy to accomplish" from my point of view.
As the worst resident pragmatist, I have stayed out of this because you aren't being practical.

Try this idea: If you did the physical tests, the answers would show you the constants required for the engineering analysis.
 

cmartinez

Joined Jan 17, 2007
8,218
If you did the physical tests, the answers would show you the constants required for the engineering analysis.
Sometimes it's best to work on a problem backwards... that is, empirical data is sometimes easier and cheaper to obtain than spending many engineer-hours studying and simulating all possibilities.
 
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