Charging a supercap from a DC motor - Page 1
Forum section: electronics / beginners

author: software
author_rank: Newbie
timestamp: November 27, 2018, 08:52:36 pm
content: 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.

author: Zero999
author_rank: Super Contributor
timestamp: November 27, 2018, 10:17:35 pm
content: 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.

author: software
author_rank: Newbie
timestamp: November 27, 2018, 10:35:07 pm
content: 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

author: Zero999
author_rank: Super Contributor
timestamp: November 28, 2018, 09:41:04 am
content: What open circuit voltage does the motor give?What's the total LED forward voltage?E =1/2CV2Suppose the LED only works down to 3.5V and the capacitor is charged to 5VEnergy stored in the capacitor when fully charged:V= 5VC = 10FE = 0.5*10*52= 5*25 = 125JWhen the capacitor voltage is 3.5VV = 3.5E = 0.5*10*3.52= 5*12.25 = 61.25JThat'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:%ELEFT= VSTART2/VEND2For 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.

author: software
author_rank: Newbie
timestamp: November 28, 2018, 04:17:51 pm
content: 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?

author: Zero999
author_rank: Super Contributor
timestamp: November 28, 2018, 07:10:32 pm
content: Quote from: software on November 28, 2018, 04:17:51 pmIt 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-bIn 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 =1/2CV2the 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.

author: software
author_rank: Newbie
timestamp: November 28, 2018, 08:11:13 pm
content: Thank you, @Hero999I 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/222981780355Thanks again.

author: Capernicus
author_rank: Frequent Contributor
timestamp: April 12, 2021, 01:31:28 am
content: This is an old post, but its interesting me, does anyone want to continue this conversation?I was wondering with a capacitor in series with a dc motor, does the capacitor get charged, or is the electricity going into the motor to make the magnetic field going to suck it all up, and the capacitor stays empty? or vice versa, the capacitor steals the electricity from the motor, so when the cap is charging the motor gets no power.

author: Jwillis
author_rank: Super Contributor
timestamp: April 12, 2021, 05:18:06 am
content: Quote from: Capernicus on April 12, 2021, 01:31:28 amYou should have started a new thread but to late now.In answer to your query , a permanent magnet brushed motor will generate a unfiltered DC ripple at a frequency depending on the speed of the motor. Like the output of a full bridge rectifier. As the speed of the motor rises the voltage will rise as will the ripple frequency. Add capacitor in parallel and it smooth out the ripple . Stop the motor and the capacitor will discharge back into the armature.

author: Zero999
author_rank: Super Contributor
timestamp: April 12, 2021, 08:02:44 am
content: Nah, I don't mind people bumping old threads, if it's on topic and they have similar goals as the orignial poster.Yes, there is ripple on the voltage, when no capacitor is connected, but it's not as bad as a single phase bridge rectifier. The simplist DC motor has three poles on the armature, giving the same waveform as a three phase full bridge rectifier. Most larger motors have more poles on the armature, giving an even smoother waveform. A lot of the noise is from arcing in the commutator.A motor with a capacitor connected to it will mechanically behave like a flywheel. It will be difficult to start the shaft turning, as the discharged capacitor will act like a short circuit. As the capacitor charges, the shaft will become easier to turn, as the voltage on the capacitor approaches the average the back-EMF generated by the motor. If the shaft it left to freewheel, it will spin for longer, than it otherwise would, without the capacitor connected, as the capacitor will dischage back into the motor and power it.