Charging a supercap from a DC motor

[software]

I am trying to harvest some residual power from a gearbox DC motor that is already in place. I just need to drive a few LEDs off a supercap as efficiently as possible. I already have rectifier going into a supercap. The question is about getting max power from this existing gearbox.

Once the supercap is completely discharged the low resistance appears to be using max power from the motor. But once the supercap is partially charged the power consumption goes down and the gearbox is just spinning wastefully.

I want to use the max power that gearbox provides at when it's completely stalled, which is about 300ma. I understand the gearbox will not last long under such use and will break under wear.

I am looking for some sort of high efficiency supercapacitor charger circuit that keep input resistance low. Hopefully there is a circuit I can buy. Thank you.

[Zero999]

When the gearbox is completely stalled, the power will be close to zero, since no work is being done, all of the energy is being converted to heat. Maximum electrical power transfer will occur when the load resistance is equal to that of the motor. Maximum efficiency is a more tricky one, since efficiency goes down, when the current is very high due to increased I²R losses, yet it's also low under light loads, as the frictional losses in the gearbox, motor bearing, commutator and eddy losses in the armature dominate.

Use a boost converter to charge the capacitor to as higher voltage, as it's rated to.

When discharging, use a buck converter to provide a constant current to the LED, as the capacitor discharges.

The bridge rectifier should of course be made from Schottky diodes to minimise the forward voltage drop.

[software]

Thank you.
Yes, I got Schottky diodes and I have a buck converter off ebay to produce 5v or so.

"Use a boost converter to charge the capacitor to as higher voltage, as it's rated to."

Say, I have a 5v, 10F that I am trying to optimize for. What kind of boost converter should I be looking at? Approximately. Analog devices has a whole bunch of chips, can I just use of of them?
https://www.analog.com/en/parametricsearch/11413

[Zero999]

What open circuit voltage does the motor give?

What's the total LED forward voltage?

E = ¹/₂CV²

Suppose the LED only works down to 3.5V and the capacitor is charged to 5V

Energy stored in the capacitor when fully charged:
V= 5V
C = 10F
E = 0.5*10*5² = 5*25 = 125J

When the capacitor voltage is 3.5V
V = 3.5
E = 0.5*10*3.5² = 5*12.25 = 61.25J

That's a considerable amount of energy left in the capacitor  61.25/125 = 49% of the energy is still left over.

The percentage of energy left over is dependant on the charge voltage and cut-off points, so the above formula can be reduced to:
%E_LEFT = V_START²/V_END²

For maximum efficiency, the LED should work down to as lower voltage as possible. Another possibility is to use a higher voltage capacitor. A 2.5F 10V capacitor would store as much energy as a 10F 5V capacitor, but would only have 3.5²/10² = 12.25/100 = 12.25% of the energy left if discharged from 10V to 3.5V.

[software]

Open circuit voltage past rectifier is about 7 - 10 volts.
The supercaps as big as 10F do come in 10v rating. Do I just string a whole bunch of 1F in series? Will that keep internal resistance low as all of the capacitors charge partially? Am I looking at 10pcs roughly?
 

[Zero999]

Open circuit voltage past rectifier is about 7 - 10 volts.
The supercaps as big as 10F do come in 10v rating. Do I just string a whole bunch of 1F in series? Will that keep internal resistance low as all of the capacitors charge partially? Am I looking at 10pcs roughly?

It looks like you need to read up a bit on capacitors in series and parallel.
https://www.google.com/search?q=capacitors+in+series+and+paralle&ie=utf-8&oe=utf-8&client=firefox-b

In short, if the capacitors are all the same value, putting them in series divides the total capacitance by the number of units, so 10×10F capacitors in series is equivalent to one 1F capacitor. Due to the voltage divider effect, the voltage across each capacitor is also divided by the number of units, so if each 10F capacitor is rated to 5V, the large capacitor will be rated to 50V. Because E = ¹/₂CV² the energy capacity of the 50V 1F capacitor is ten times that of the 5V 10F capacitor.

Note that the above assumes that all of the capacitors are perfect. In reality each one will have a different equivalent parallel resistance, i.e. leakage current, so voltage sharing resistors are normally used to balance it out.

[software]

Thank you, @Hero999
I will read up on connecting capacitors, for sure.

Is something like this my best choice then?
https://www.ebay.com/itm/16V1F-2F-Farad-Capacitor-Module-2-7V-10F-Super-Capacitor-With-Protection-Board/222981780355
Thanks again.