Help with Circuit for Power Generation

Discussion in 'The Projects Forum' started by hin, Apr 29, 2013.

  1. hin

    Thread Starter New Member

    Apr 27, 2013
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    Hi everyone!

    I am new to electronics and have been learning the basics. I am trying to make a power harvesting device that generates power from the kinetic energy of a person while they are moving. The kinetic energy of the person drives a small motor generator, the generated power gets stored in a capacitor and is used to power a lower power electric device (around 1.5v, 20ma).

    Is the circuit I have designed below workable?

    [​IMG]

    The DC motor will output between 0 and 4.5v depending on the speed the person moves. The DC motor can turn both clockwise and anti-clockwise so there is a rectifier bridge. And there is a blocking diode to prevent the current flowing back to the motor from the capacitor. The linear voltage regular ensures a 1.5v output.

    I expect that the small motor generator will produce very small amounts of power and my concern is that the rectifier bridge and blocking diodes will cause a significantly voltage drop even when using Schottky diodes and there will be further losses from the linear voltage regulator. Can anyone suggest if there might be a better way to do this to allow the best chance of working?

    Many thanks!
     
  2. LDC3

    Active Member

    Apr 27, 2013
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    What you have will not work well. Since you have about a voltage drop of 2.1 V across the diodes, and you need at least 1.8 V for the regulator, you will not see a controlled output until the generator develops about 4 V. You will probably see some output between 3 and 4 V, but it will not have a 1.5 V regulated output.
    Since the bridge diodes are configured the same way as your blocking diode, get rid of it. Now it is down to about 3.3 V.
     
  3. #12

    Expert

    Nov 30, 2010
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  4. #12

    Expert

    Nov 30, 2010
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    And then there are these:
     
  5. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    Surprised no one so far has mentioned that as the OP said "And there is a blocking diode to prevent the current flowing back to the motor from the capacitor." The diode is backward, it allows the flow back to the motor.
     
  6. #12

    Expert

    Nov 30, 2010
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    I think somebody did.

    The problem with the diode being backwards is eliminated by completely removing it.
     
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  7. hin

    Thread Starter New Member

    Apr 27, 2013
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    Thanks for the information guys, it's really helpful!

    I don't know what I was thinking when I added the blocking diode, as you all mentioned it's completely unnecessary.
     
  8. hin

    Thread Starter New Member

    Apr 27, 2013
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    #12, the Micrel chip looks like a good option to go for. I don't understand something, why they specify a drop rate of 0.04v and also specify a input voltage range of 2.3 V ~ 5.5 V, shouldn't the input range be 1.54 V ~ 5.5 V because that would also output 1.5v?

    Also thanks for the PDF's, although they are probably too technical for my knowledge right now. I will go and learn about DC-DC converters and op amps.
     
  9. wayneh

    Expert

    Sep 9, 2010
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    As noted, at low voltage you are battling with the relatively large power loss you will suffer in the rectifier. Losing 1.2V across a rectifier is bad enough in a 12V system - where rectifier losses are 10% of the useful power produced. If you're only getting ~1.5V of useful voltage, the rectifier is losing almost half of all the power you are capturing.

    Consider using mechanical means to: 1) keep the polarity the same, so that one diode - half the loss - could be eliminated, and 2) increase the motor speed to produce a higher voltage output. These changes don't increase the available power, but they do help send more of it into a useful load.

    An active rectifier (I'm guessing the Micrel chip is such a thing) can eliminate the diode drop but needs a voltage (but almost zero current or power) to control the gate of the internal MOSFET.
     
  10. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    I think the OP would be better off connecting the motor to a small step-up transformer since he is generating such small voltages to begin with. Then off-loading those higher voltages to a rectifier, regulator/capacitor and constant current control to a battery.
     
  11. LDC3

    Active Member

    Apr 27, 2013
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    There is some information that might be helpful. Are you using an AC alternator or a DC generator to capture the energy? If you are using the DC generator, then the step-up transformer would not be very useful since the change in current is not frequent enough to transfer the energy into the secondary.
     
  12. wayneh

    Expert

    Sep 9, 2010
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    A generator wound to produce higher voltage per rpm directly would be better than the "wrong" motor plus a transformer. Both work, but the former would be more efficient. Both would be FAR more efficient than the current arrangement if the OP is using a standard, non-active rectifier.
     
  13. hin

    Thread Starter New Member

    Apr 27, 2013
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    Thanks for all the info, here's some more data:

    I was going to use a DC motor. Specifically a Maxon 203893 (I choose this motor because its small and partly by guess work by looking at the rpm/V output). It has the following spec:

    Power rating: 1.2w
    Nominal voltage: 12v
    No load speed: 11400 rpm
    No load current: 5.43 mA
    Speed constant: 965 rpm/V
    Full datasheet: http://www.farnell.com/datasheets/483495.pdf

    With the currently designed gearing the motor will be rotated between 0 and 4500rpm depending on how fast the user moves. Which equates to voltage output range of around 0 to 4.7v. I may be able to increase the gearing but I'm not sure because it may require too much torque to get it moving.

    Also I could pick another motor with a lower speed constant (rpm/V) so that the voltage will increase but then the current output will decrease. One thing I don't understand is how to calculate the current output of a motor generate at a particular speed given that the load is a capacitor.

    Or do you think it would be better to go with an AC motor and step up transformer as suggested?
     
  14. LDC3

    Active Member

    Apr 27, 2013
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    I can see that you are just getting started.

    Not all motors can be used to generate power. If the motor uses just coils for generating magnetic fields, then it will not work as a generator since there are no magnetic fields unless a current is present; which only occurs with a voltage applied across the motor.

    You will either need to test the motor or ask the manufacturer if the motor can be used to generate power.
     
  15. wayneh

    Expert

    Sep 9, 2010
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    It's complicated. The ideal situation for transferring power from the generator to the load is when the impedance of the load matches that of the generator. The EMF of the generator goes up linearly with rpm, but impedance also increases with rpm. At "low" rpm, DC resistance of the windings is the dominant factor, but at "high" rpm the DC resistance is a minor factor compared to inductance. As I said, it's compacted.

    The capacitor is not the load, BTW. Once it's charged, it won't take any power. The load is the DC load applied downstream.

    IMHO, the best if not only way to sort it out is to actually measure voltage across a load versus load impedance (DC resistance for a purely resistive load) at a given rpm. Repeat the data at another rpm. These data will allow you to predict the result at any combination of rpm and load.
     
  16. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    Wouldn't a stepper motor be a better candidate for this? Its made for low speed, has permanent magnets and three phases. The bridge rectifier for three phase only uses two diodes vs four diodes for each phase, so the losses should be less. Using Schottky diodes to make the bridge rectifier should also help.
     
  17. hin

    Thread Starter New Member

    Apr 27, 2013
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    I just did some research into using a stepper motor as generator, however apparently they make very inefficient generators.

    My thinking was to use a core-less permanent magnet motor because they do not have a heavy stator so they will spin up fast and can also change direction quickly. I think I might have to do some experimentation to see what works.
     
  18. wayneh

    Expert

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    Sadly true, it's often the surest way to get the answers you need.

    When experiments are difficult or costly, be sure to exhaust what you can learn from the experience of others. You can be sure someone has been through all this before. The challenge is to find their results.

    I spent a previous life in R&D and I can tell you, ignoring the literature is one of the biggest wastes I've ever seen. It's slipshod but soooo common. Lots of reinventing the wheel out there.
     
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