1
\$\begingroup\$

I would like to get some advice regarding the following scenario: I want to build a cheap, small generator for an energy harvesting application. There are magnets on a rotating shaft and off-the-shelf inductor(s) are positioned (fixed) near this shaft. After the inductors the voltage is rectified and then a power management IC takes over (boost/buck, storage).

At 1000 rpm the system should provide enough power (~10 mW) for the circuit behind it. But the rpm can vary between 1000 and 20000. Once the rpm are higher than 1000 there is excess energy. Therefore I have a find a way to deal with this excess energy. Either get rid of the excess energy or maybe prevent the generation in the frist place.

Using multiple inductors and switch them on/off depending on the rpm was an idea. But this would take quite some effort with rpm detection and circuitry, depending on how many inductors I would use. So I am looking for easier solutions.

A mechanical solution to increase the gap between the inductors and magnets is not possible, bc both are fixed in place and there is not much room at all.

Zener diodes could be a thing, but dissipating the energy as heat may not be the most suitable way, bc the thermal conductivity of surrounding materials is not good.

Do you guys have any ideas how to deal with this issue?

\$\endgroup\$
6
  • \$\begingroup\$ If power is approx linear with rpm then the energy excess is modest . 10 mW x 20,000/1000 = 200 mW. || If you MUST limit energy, and it's not obvious why you need to, a voltage regulator will let the alternator produce voltage BUT only dissipate Valt x Iload. In most alternators Voc_RPM_max is several times V_rpm_useful_min BUT not eg 20 x as high. eg an alternator may make 5V at desired load but 20V at full RPM. So using a voltage regulator and only drawing what you need till only increase power by 4 x(in this example) Pl= Il x 5V. Palt = 20V x Il. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 5, 2019 at 13:37
  • \$\begingroup\$ The generator generates exactly as much electrical energy as you are consuming. If you don't need more energy just disconnect the DC-DC-convetrer for a while. If there is no electrical load connected to the generator it will not generate any electrical energy. I don't see the problem. \$\endgroup\$
    – Curd
    Commented Jul 5, 2019 at 13:55
  • \$\begingroup\$ Continuing refusal to not allow us to understand the real problems involved will result in a typical to and fro that wastes your time and ours and gives you some guidelines that may or may not be overly relevant. The choice is yours, but all too often we see a "wandering about" when some really useful input could have been given. . \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 5, 2019 at 14:45
  • \$\begingroup\$ I'll repeat something here I mentioned below as it is highly relevant. | An alternator can be designed to act somewhat like a constant current source at the voltage of interest - usually by saturating an iron or other core in the inductor(s) when a designed amp-turn limit is reached. This means that while the alternator may make several or even many times the desired voltage if RPM is high enough, the output current is limited to ~~= a designed value. SO If you want say 5V at 2 mA, if you place a voltage clamp that accepts excess energy when VAC reaches say 6V then ... \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 6, 2019 at 11:16
  • \$\begingroup\$ ... the total energy dissipated is not much more than the energy taken by the load worst case. VACpeak = VRMS x 1.414. | Vdc = VACpeak - 2 x diode drops = say -1 V. So eg to get 5VDC with 2 x Schottky diode drops VACpeak = 6V so VAC RMS = 6/1.414 =~ 4V. So a 4VAC at 1000 RPM alternator with a say 5 VAC_RMS clamp will have some "headroom" voltage above the bare minimum and dissipate a total of no more than 25% more than the maximum load power. At the power levels concerned this seems unlikely to be a problem. But .... . \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 6, 2019 at 11:17

2 Answers 2

Reset to default
0
\$\begingroup\$

This is an interesting question and probably has a good solution available once you supply all the information that is relevant that you know but that, so far, we don't.

SO - this is a brief interim answer based on what you have told us. If you add more detail and explanation the answer can be improved. Otherwise, this is about as good an answer as you can reasonably expect with the level of detail provided. (Others MAY give you an even better answer but you have no right at this stage to expect it :-) ).

______________________________

If power available is approx linear with rpm then the energy excess is modest.
10 mW x 20,000/1000 = 200 mW.
In most situations 200 mW wiould not be hard to dissipate with modest temperature rise.

Also, the alternator need not dissipate all the power that it CAN generate. Just because it CAN produce 200 mW does not mean that it MUST.

If you MUST limit energy, and it's not obvious why you need to, a voltage regulator will let the alternator produce voltage BUT only dissipate Valt x Iload. In most alternators Voc_RPM_max is several times V_rpm_useful_min
BUT not eg 20 x as high.
eg an alternator may make 5V at desired load but 20V at full RPM.
So using a voltage regulator and only drawing what you need till only increase power by 4 x(in this example)
P_load = I_load x 5V.
P_alt = 20 V x I_load

Tell us more!

\$\endgroup\$
5
  • \$\begingroup\$ For the induced voltage I used: V_ind = -NAΔB/Δt \$\endgroup\$
    – BendingBender
    Commented Jul 5, 2019 at 13:59
  • \$\begingroup\$ For the induced voltage I used: V_ind = -NxAxΔB/Δt N: turns of inductor A: cross-sectional area of inductor B: magnetic flux t: time of magnetic flux increase/decrease Therefore if the rpm (revolutions per minute of the shaft) increase the voltage increases linear. But Power is voltage*current. If the load resistance is constant, the current increases linear as well. Therefore 2xrpm = 2x voltage = 2x current = 4x power. With 20000 rpm it would be 20x20=400 times the power generated. Means 4 W and not 200 mW, or am I wrong? \$\endgroup\$
    – BendingBender
    Commented Jul 5, 2019 at 14:06
  • \$\begingroup\$ @BendingBender Assumption that can be changed is " ... if the load resistance is constant ..." -> As noted above, if you use a voltage regulator then the load will "see" a fixed voltage and load current is constant after Vdesired is reached. So IF voltage increases linearly with voltage (and it would be exceptionally unusual to have Vmax = 20 x Vwanted) then power will be 20 x (linear with V) and not square related. As we still no ~~= nowt about what you are doing, allowed complexity, cost, ... ANYTHING ... we know not if a small smps would be OK but, if so, then Pmax is a few times Pwanted. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 5, 2019 at 14:43
  • \$\begingroup\$ "P_load = I_load x 5V. P_alt = 20 V x I_load " Thank you Russell, I think this helped me. :) \$\endgroup\$
    – BendingBender
    Commented Jul 5, 2019 at 16:10
  • \$\begingroup\$ @BendingBender That 20V x I_load is for the example I gave where the alternator "tops out" at some voltage. This is usually the case. You can also clamp excess current so that the alternator saturates and this limiots voltage and power dissipated. You can .... MANY MANY things but as long as we have no clude as to what you are doing we can't offer much good advice. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 6, 2019 at 1:28
1
\$\begingroup\$

Don't get confused! RPM is the rotation speed measure. Rotation speed is proportional to the Electormotion Force, which is voltage (Volts), not power (mW).

Power is voltage times current (A) . What current will your generator provide? Well, in wide range, it will be the exact current you take, nothing more, nothing less. That's it.

If you come closer to the maximum power that your source can provide, the rotation will slow down reducing the EMF (voltage), and depending on your circuit the current may get lower, or try to keep the power constant and completely choke the rotation.

\$\endgroup\$
6
  • \$\begingroup\$ Gregory - based on what he says odds are that the enery hatrvesting is not the main power consumer. "Choking" the power source mechanically sounds both impractical and undesirable. I may be wrong :-). \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 6, 2019 at 1:30
  • \$\begingroup\$ I never voted for choking anything. Just said what would happen if be wanted more than is available. \$\endgroup\$
    – user76844
    Commented Jul 6, 2019 at 14:26
  • \$\begingroup\$ Gregory - try again :-) -> As he suggests that RPM is liable to be up to 20x the speed at which the required energy will be generated the odds of the load choking the source is small. If load is resistive then power increases and decreases as square of V . V is liable to be about proportional to speed. So at say 50% of speed load is 25% of required or less. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 7, 2019 at 7:53
  • \$\begingroup\$ I don't know a thing about the odds. But i can see you are confusing power, energy, voltage and current as well. \$\endgroup\$
    – user76844
    Commented Jul 7, 2019 at 7:55
  • \$\begingroup\$ Politely - No. I am not confusing any of those. (Not for the last 50 years or so :-)). With a simple rotating PM alternator voltage increases approximately linearly with RPM, tailing off as various secondary effects become dominant. For a constant current load - such as would be seen if a linear regulator fed a constant resistance load, then load power increases linearly with voltage. For a constant resistance load (no regulator) load power increases as load squared and current linearly. Energy is the time integral of power and is not overly relevant in this discussion. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jul 8, 2019 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.