Julian, That Type of SuperCapacitors is Nonlinear, They have a much Higher Capacitance at Extremely Low Voltage, and much less Capacitance near their max rated Voltage, The Capacitor you tested has a capacitance of 55F from 2.5V to 1.5V and 67F from 2V to 1V, if you don't believe me watch your video, (2.5V 5:00 )(1.5V 5:55 )(2V 5:24 )(1V 6:31 ) To Prove Just How Extremely Nonlinear Those SuperCapacitors are drain one completely and charge one to 1V, place them in series and charge them in series to 2V, if they were linear they would each increase in Voltage by 0.5V, But They Won't ! Then drain the one that is charged and charge the the one that was drained and repeat to prove that the variance of capacitance between the two capacitors is not to blame, The reason for the nonlinearity is similar to the Miller effect in semiconductor materials the higher the reverse PN Junction voltage the greater the depletion region and the lower the capacitance for each additional volt applied, When you drain the capacitor and you think it is fully drained hook it to a multimeter and drain it the rest of the way through the current shunt that way you can monitor the current.
@AdamWelchUK
7 жыл бұрын
I can see a number of people aren't impressed with the method or measurement. Just wanted to say I really enjoyed this video. Sometimes we need to be reminded of what can be achieved with simple tools.
@JulianIlett
7 жыл бұрын
Thanks Adam - I knew this video would upset a few people - but the maths is all good :)
@ElmerFuddGun
7 жыл бұрын
3:07 - I am confused why for some videos you don't like to use your DMMs yet for others you will use many? This was a perfect example of why you should and it would also verify that the volt meter on the load was accurate. Your DMM should also show another digit so even better still.
@ElmerFuddGun
7 жыл бұрын
Also if the load doesn't display voltage unless it is on and thus a minimum .2 A load why not use a DMM and minimize your charging time by using the full current output from the PS to charge the capacitor? Seemed like a weird solution to a simple problem.
@JulianIlett
7 жыл бұрын
Yeah, it was a bit weird. Next time I'll use a voltmeter - and an ammeter to confirm the current draw.
@The.Doctor.Venkman
7 жыл бұрын
I don't think you'll 'see' any difference....
@ElmerFuddGun
7 жыл бұрын
I'm not so worried about the accuracy of the readings you obtained as much as using the right tool for the job. You wanted to monitor the voltage as the capacitor was charging and instead of putting a DMM on it you turned on the load and in turn that slowed down the charging. LOL... seems like a bit of a brain fart. ;-)
@paulgrimshaw6301
7 жыл бұрын
So many people commenting are confusing capacitance with capacity, and concluding that Julian's method was incorrect. They're different things, and Julian was correctly measuring capacitance (measured in Farads), NOT capacity. To better understand the concepts, a useful analogy for a capacitor is a cylindrical container holding water, with a tube attached to a hole in the base. In the analogy, the water is the equivalent of charge, the water level is the equivalent of voltage across the capacitor, the height of the cylinder the equivalent of the maximum rated voltage. The water flow rate through the tube is the equivalent of the current into/out of the capacitor. Exactly like a capacitor, if you have a constant flow of water through the tube into the cylinder (equivalent to a constant current charging the capacitor), then the water level (voltage) increases linearly. The equivalent of capacitance in this analogy is the cross-sectional area of the cylinder, NOT the total water capacity. As you can see, the cross-sectional area determines the rate at which the water height increases for a given input water flow rate, completely independent of the maximum water capacity. This is analogous with the capacitance of a capacitor determining the rate at which the voltage across the capacitor increases, for a given current, again unrelated to the maximum capacity. A battery on the other hand is not analogous to a capacitor, and there's nothing equivalent to capacitance in a battery.
@thinkdunson
6 жыл бұрын
the fact that 1 Farad can be equated to 1 Coulomb per Volt, would seem to indicate that Farads are in fact a capacity dependent on the maximum possible charge.
@RWBHere
7 жыл бұрын
Thanks, Julian. Yet another scam is revealed. I've suspected this one for a long time, but you've laid it bare here. There seems to be no end to the dishonesty of some manufacturers and sellers.
@sdchua1
7 жыл бұрын
Discharge from 2.5v to 1v you get 60F, there is 1v to go to totally discharge it, if you continue to discharge the rest of the 1v, you might get Farad calculation closer to 100F.
@danweecc
7 жыл бұрын
Have you checked how accurate the current draw is on the load? Might be worth testing it for accuracy and linearity. Another way would be to stick an ammeter in-line with the capacitor and see if it varies over the course of the experiment.
@cmuller1441
7 жыл бұрын
Daniel Wee exactly! I have the same device and there's a 20% error in the current that goes down only at high values (4A) Check my videos...
@raymondheath7668
7 жыл бұрын
There was a quick trick we used in the 80's for figuring out the approx vaule of series and parallel stacked caps for custom values, but fo the life of me I can't remember. I do know that back then a 20% capacitor was something one could only dream of....
@raymondheath7668
7 жыл бұрын
There ya go, thanks
@jessstuart7495
6 жыл бұрын
Here's a simple method that yields reasonably accurate results. 1. Discharge the cap completely. 2. Charge the cap with a fixed power supply through a known resistor. 3. Measure the time it takes for the cap to charge up to 0.5V. 4. Discharge the cap (for safety). If your supply voltage is much higher than 0.5V, the voltage across the resistor will not change very much as the cap charges to only a small voltage. The current through the resistor will almost be a strait line from it's initial value I_start = (Vs/R) to I_end = (Vs - 0.5)/R. The average current will be I_ave = (I_start + I_end)/2 = Vs/R - 0.5/(2R) You can use this average current to calculate the capacitance. C = (I_ave)*t/0.5 Just make sure your resistor has a high enough power rating to handle the current (P=I^2*R). Lets throw some numbers in there... Vs = 12V, R=47ohm. => I_ave = 12/47 - 0.5/(2*47) = 250mA If you're lazy, you could just estimate I_ave as 12V/47ohm = 255mA, but this introduces about a +2% error. If your capacitor takes 20 seconds to rise up from 0 to 0.5 V, then... C = (0.25)*20/0.5 = 1F
@JasonMasters
7 жыл бұрын
*scratches head* Isn't the capacity measured with reference to zero charge = zero volts? It seems to me that since you didn't discharge the capacitor to zero volts, you're only measuring part of its total capacity, which means that 60 farads is probably reasonable given the measurement method. Or am I missing something?
@volodumurkalunyak4651
7 жыл бұрын
In theory Farads does not depends on Volts. In practice ESR can make measured capacitor capasity less than real (that has not happened here)
@Conservator.
7 жыл бұрын
JasonMasters If Julian would have kept draining the capacitor to 0V that would have taken more seconds (so more Farads) but at the same time the voltage drop would have increased (so less Farads). These two differences would have balanced each other out resulting in the same capacitance.
@superdau
7 жыл бұрын
A 10 liter bucket is still a 10 liter bucket, even if you pour just 5 liters out of it (and measure how long it takes at a certain rate). *Capacitance* is how much "fits" into a capacitor, *charge* is how much is in it.
@Infinion
7 жыл бұрын
You may be mixing up Capacity and Capacitance. When you say capacity, you are referring to energy in Joules, which, to be accurate, you would need to discharge to 0V to measure the total stored energy, as you say. However, we are looking at Capacitance, which is charge per volt and, like Volodrumur Kalunyak says, does not depend on what voltage you start or end at. Julian took the formula for capacitance, C = q/V, where q is the charge expressed in Coulombs, and substituted q for the equation for coulombs expressed in amps and seconds so he could compare his measured value against the value claimed by the manufacturer. The equation for energy (E = 0.5*C*V^2) uses capacitance, so it is important to know this value if you want an accurate approximation of the energy in the cell.
@jort93z
7 жыл бұрын
well you can do the same calculation at 0.5V dropped and 1V dropped(since the stopwatch is on screen). you always get roughly the same result. i'd only think that it'd be slightly more precise if you discharge it more since that'd lower the amount of human error in pressing the button.
@chris1newbury
7 жыл бұрын
whats 1 coulomb in pipple and noist?
@shlomsi2000
7 жыл бұрын
1 Terra volts/year
@Djhg2000
7 жыл бұрын
About 1.3 with respect to ground, but only in the northern hemisphere.
@sootikins
7 жыл бұрын
Hmm... we have microfarads, nanofarads and so on. Perhaps we need to add a new unit: chinafarads. According to this video, 1 chinafarad = approx 0.6 farad.
@stuartofblyth
7 жыл бұрын
Richard, a coulomb is a measure of electric charge. It's the charge you find on 6.24 x 10^18 electrons (6,240,000,000,000,000,000). Get that many electors to flow through an ammeter every second and the ammeter will register a current of 1 amp.
@tudornedelcuplace
7 жыл бұрын
I agree with some of the comments. At 1v left into the supercapacitor, you still have a charge on it. So indeed if it's a linear function you don't have to discharge to 0v but you need to count for the charge left into the supercapacitor. So from other comments, it turns out you have a supercapacitor of approximate 100F. Thank you for going trough the effort to exhaling it to us.
@SimoWill75
7 жыл бұрын
If your car has a 50 litre fuel tank, and you drive for 10 kilometers and use 1 litre, do you need to drive the tank dry to know how many kilometers per tank? Same.
@Infinion
7 жыл бұрын
it is as SimoWill says, in this case we are looking for capacitance which is charge per volt, which is not much different from litres per kilometer. You might be mixing capacitance up with energy. Capacitance = q/V whereas Energy = 0.5*C*V^2. Julian took the formula for charge q, expressed in Coulombs C and plugged it into the formula for capacitance to calculate and compare his result with what was being claimed by the manufacturer. However, if we were looking for total energy in the cell, you would be correct to say that there would be unaccounted charge if you stopped at 1V.
@tudornedelcuplace
7 жыл бұрын
Indeed. I learned something. ... Thanks for clarifying.
@tengelgeer
7 жыл бұрын
Am I the only one who just looked at the time when it dropped 1V? :p Divide by 1 is so much easier :p
@Infinion
7 жыл бұрын
Would have made for a much shorter video :P
@chrispotts2010
7 жыл бұрын
I think you nailed it Julian. Can't wait to see you do this with your 700F supercapacitor, as I read an article today somewhere which says that no Chinese manufacturers make these with more than a 400F rating, so your measurements could prove or disprove this. Wouldn't you like to know?
@JulianIlett
7 жыл бұрын
+Chris Potts I would like to know. I'll test the 700F caps soon.
@TheGiselaSchumacher
7 жыл бұрын
I bought six of these exact supercaps from china a couple years ago. I tested each one individually, logging data with an Arduino and even plotting it in Excel. I went partial voltage ranges, full 2.7 to 0 volt ranges, and a whole slew of different currents. Each cap tested right around 60 farads....
@JulianIlett
7 жыл бұрын
+Ian McClellan-Johnson thanks Ian :)
@175griffin
7 жыл бұрын
like many of the comments here I was also confused and thought only a portion of the capacitance was measured, but if you extrapolate from the data you can see the voltage range doesn't matter (assuming the drop is linear). 90s to drop 1.5v = 150s to drop 2.5v. 150/2.5 still = 60 farads. Maybe this is wrong, but I always thought the charge and discharge curves of a capacitor was logorithmic, not linear.
@Infinion
7 жыл бұрын
If you just connected a capacitor directly to a constant voltage source, or connected a charged capacitor to a resistor the curve would look just like an exponential decay function similar to a logarithmic curve on a graph. However, Julian deliberately connected the charged capacitor to a constant current load, so we get a linear slope instead, which makes calculating capacitance much easier. as you say, the voltage range doesn't matter. The V in julians derived equation is a voltage range ΔV and can be evaluated as: Total Farads = (Constant Amps * Total Time)/(Start Voltage - End Voltage) when the slope is linear.
@175griffin
7 жыл бұрын
That makes sense. Still, it seemed to me that the voltage was changing slower as it got lower. Probably just issues with the calibration of the measuring devices. I think a scope or a data logging meter would give a more accurate and complete picture here.
@Infinion
7 жыл бұрын
agreed.
@DrGreenGiant
7 жыл бұрын
One of the things we've found at Loughborough University is that the capacity depends upon how long you charge for. I.e. the state of charge isn't purely based upon the measured voltage. Similar to old lead acid batteries. The reverse of this effect is if you charge the cap, short it for a second to take it to 0V. Then use a high impedance multimeter to measure the OCV, it will start increasing (recharging itself!)
@jonathancook4022
7 жыл бұрын
Thanks for helping me revise A-level physics. Its been 10 years since I studied such things
@bigjon1359
7 жыл бұрын
You are lucky to even get 60F. I got 12 of these cheap china caps and figured out right away that they must have been rebranded rejects. 10 of the 12 would not even hold their charge over night and only 2 were over 50% charge after 24 hours. I did get a refund from the seller which surprised me but was happy after refund but disappointed that I could not play around with them in my projects.
@jaaasgoed
7 жыл бұрын
You should get the datasheet of the cap. Usually the capacitance of a supercap is given at a specific discharge current. You might well get to 100F with say a 20mA discharge current.
@snnwstt
6 жыл бұрын
For supercap, the value C is not a constant versus V and versus Q. If they were, they would charge almost instantaneously under R = 0 in serial. Still, if C was constant, you could easily compute it from an oscilloscope since their charge/discharge in time will follow the exponential law with an RC circuit: 0.5 V = V (exp(-t/RC)), so RC = -t_1/2 * ln(0.5); knowing R will give you C by measuring t_1/2 which is the time required to get (to lose) half the constant voltage, Vcc, at the end of the group resistor-capacitor, time that you measure on the scope (quite easy on digital scope allowing to "track" the curve with a cursor: start anywhere on the curve and end when your end point is at 0.5 Vcc more than your starting point, the t_1/2 is displayed as the Delta t between these two points). I think that for supercap, the announced C value is for Q=0, or a discharged cap. You can try RC = - t_1/10 * ln( 1/10 ), with t_1/10 the time required to reach Vcc/ 10, if it takes too much time for t_1/2.
@jimsmindonline
7 жыл бұрын
It's probably been said already. The discharge slope on a supercapacitor probably isn't linear as it would be with a regular cap. Behaves more like a battery with a more curved slope. ... I think??? Would be interesting to see it plotted on a graph. You'd need to calculate the area under the voltage curve over time for the full discharge to get the exact value. By choosing a point in the middle of the curve you've got a roughly accurate figure. Plus the schottky is probably throwing things off a bit too.
@TheUnnamedGent
7 жыл бұрын
It's exponential if the resistance is constant. It's linear if the current is constant as in this case.
@electrodacus
7 жыл бұрын
Your test is good and around 60F is the real capacitance of that supercapacitor. The only thing missing from your equation is ESR but luckily you eliminated most of that by starting the DC load at 2.6 or 2.7V so the voltage drop on the ESR was mostly accounted for. (not sure if that was intentional or just luck :)) The other two important parameters of super capacitors are ESR (part of the stored energy will be lost internally during charge and discharge) and the self discharge leakage a sort of equivalent parallel resistance. From my experience those inexpensive China super-capacitors have a wide variation of both ESR and leakage current (that is for the exact same model of capacitor difference between samples) and because of that they are completely unusable in a series connection since they will never stay in balance. A good set of supercapacitors with all equal ESR and leakage once balanced they will stay very well in balance while with those bad ones you will continuously try to balance them but never be able to keep them in balance or at least close. I seen as much as 2 to 3x difference in ESR between samples of the same model and even larger up to 10 or 20x leakage difference with some extreme ones that where drooping over 1V in 24h trough self discharge. They will still be great value even at 60F if all of them (same model) will be 60F and have the same ESR and self discharge but because they are not equal they are useless to use in any serious application in series since a lot of energy will be spent trying to balance them all the time.
@JulianIlett
7 жыл бұрын
I agree Dacian. The 100F caps are super cheap - about $2 each. I'm not expecting miracles. But this is what I do. I stick cheap-ass crap together and see what results. My supercapacitor speaker won't rival the Hydrogen (Hydrogen Sulphide maybe) but if it can bang out a few tunes, I'll be happy :)
@electrodacus
7 жыл бұрын
The comment was mostly for your audience since I completely agree that for your entertainment and educational use they are good enough. I like supercapacitors even if they currently have no real benefit compared to lithium batteries in almost any application. I will not like people to get a bad impression about supercapacitors because of the low quality ones. Is good they are super cheap so people can play with them. I remember when I seem the first supercapacitor in a local electronic store around 1997 so about 20 years ago where the price label of the supercap was saying 3.3F and I was sure that was a mistake so I pointed out to the seller that wrote that label (was a high school friend that was just haired there) and he insisted that is a 3.3F and give me a closer look at the capacitor. Sure enough the marking on that rectangular capacitor was saying 3.3F at 5.5V and while the 5.5V was suspiciously low I still did not believed that can be a 3.3F capacitor so I asked him as a test to charge the capacitor from a new 4.5V battery (that was still a popular type of primary battery made of 3 cells) and then try to light a small 5V incandescent bulb used in those same flashlights that used those 4.5V batteries. I was of course proven wrong and that was a 3.3F capacitor made internally of 2x 2.7 6.6F capacitors in series and encased in a rectangular plastic case similar to large AC high voltage capacitors. They where cheap so I purchased bot units available in that store immediately and had fun for quite a few days. One of my first tests was to power a portable caste/radio player (a cheap walkman clone) and the supercap was able to power the radio for a few minutes and even able to power the cassette player for a few seconds. If you search for "Goldcap Matsushita 5.5V 3.3F" you will find some old eBay Germany adds for this capacitors and can see a photo. They are black rectangular with of course gold text :) PS: I just looked at this www.ebay.co.uk/itm/4-Piece-Capacitators-Goldcap-Matsushita-5-5V-3-3F-RM32-5-40x15x40-/322519706493 and it mentiones the date code as 1987 so I was probably purchasing very old (a decade old) stock in 1997 :) I had no idea.
@GogogoFolowMe
7 жыл бұрын
Thanks Julian, I bought a bunch of 500F Green-Caps from the Great Ebay and I'll test their real capacity with the same technique
@MusicalBox
7 жыл бұрын
I'm just an amateur here, but wouldn't it make more sens to mesure from completely full to completely empty ? I don't know if this analogy is correct, but if I want to know how much water a bottle can contain, I'll mesure all the water I can put in it or all the water I take out of it, if it was full.
@MusicalBox
7 жыл бұрын
Fester Blats : But if you fill a bottle that is not completely empty, then start empty it and then stop before it's completely empty, you can easily mesure that it takes for example one minute to take 300 ml from the bottle. But that won't tell you the total capacity of the bottle. Right ?
@ccc10156
7 жыл бұрын
1 Columb is 1 Amp sec at 1 Volt, or 1 Wattsecond. So when discharging 2,5 volts at 1 amp You are draining 2,5 Wattseconds. That is why you observe a discrepancy. You have to count wattseconds in order to count columbs.
@trptmbalmer
7 жыл бұрын
As others have pointed out, all that was done here was measuring a portion of the capacitor's actual capacity. 1.5V is about 55% of the capacitor's total voltage range, so a 60F result (60F is 60% of the capacitor's rating) is not only acceptable, it's probably within spec, too.
@Infinion
7 жыл бұрын
I think you are confusing capacity (energy expressed in Joules) with capacitance in Farads which most people do. You do not need to discharge from its total voltage range to find capacitance. You would do this if you wanted to find the total capacity in Joules. If you just plugged in the capacitance value that the manufacturer claimed, you'd get a completely misleading number for its capacity, so its important to verify these things yourself if you want to be accurate, rather than trusting the label from a foreign supplier who may have just inflated the numbers to increase the value of their product. Capacitance does not depend on what voltage you start or end at. As long as you discharge at a constant current, any start and end voltage will give you the total capacitance. This is because in our formula for capacitance, C = q/V, the V is actually a ΔV (voltage range Vstart - Vend) or in other words Farads = (constant Amps * total time)/(Start Voltage - End Voltage). So if we discharged the capacitor from 2.7V to 2.69V at 1mA we would eventually get 60F after 10 minutes.
@szbalogh
3 жыл бұрын
I am not good in electronics but I think it has some capacitance between 0 and 1V and 2.5 and 2.7V as well. This should be added to the calculated 60F.
@pauls0416
7 жыл бұрын
I don't know why I was thinking you have to integrate to get the number of farads (or even take the area of the triangle, or half the voltage range times time). Instead you used delta V (for the change in voltage) in place of V in the equation.
@Barrybados
7 жыл бұрын
Would be a good idea to check the charge capacity of that cap , Like how much do you put in to how much you get out .
@JulianIlett
7 жыл бұрын
A very similar amount I would hope.
@lubricatedgoat
7 жыл бұрын
Julian, do you think you could build an arduino arrangement to automate this process, such that you connect a cap and hit start? (maybe input some parameters if needed)
@davestech6357
7 жыл бұрын
The discharge is not linear across your voltage swing. Try it again to zero volts from a full charge. Also you would need voltage to run the load bank. A power supply in series could be used on the load.
@JulianIlett
7 жыл бұрын
The electronic DC load was separately powered (the white cable).
@davestech6357
7 жыл бұрын
Julian Ilett Hi what I meant is you won't be able to drop it down to zero volts. You would need some voltage in series with the capacitor. Probably wouldn't drop below .6 volts and still have a load.
@SuperBrainAK
7 жыл бұрын
The 150W load can read voltages all the time, plus it has alot better range, 10mA to 10A. I did a review on my channel also where to buy it reliably. for those interested, besides I need more views, the stuff I find is a bit more exciting IMO but that is me talking about the channel I created. I havent found a big car audio supercap bank yet, now THAT would be a great find!
@paulgrimshaw6301
7 жыл бұрын
Didn't you comment on a previous video that the supercapacitors are getting warm when being charged (or discharged?). This suggests that some of the energy supplied in a charge is being wasted, i.e.: you get less out than you put in. So it would be interesting to measure the capacitance by using a constant current charging the capacitor instead of discharging it, to see if there's a different, larger result.
@SuperBrainAK
7 жыл бұрын
thank you for the video on how to measure large capacitors with constant current loads! the ones that count Ah would be able to give you these figures.
@KX36
7 жыл бұрын
Did anyone else notice how the voltage dropped very unevenly - drop a bit, stop a bit. I don't think that constant current load was acting very constantly. Proper multimeters would probably have helped here.
@greggibson6519
7 жыл бұрын
Why is it that Most People Don't Understand that the Voltage Rating on a Capacitor is it's Maximum Safe Operating Voltage and Has Nothing to do with the Capacitance Rating, 1 Farid=1 Amp Second to 1 Volt, The only reason that Companies use low voltage caps is to save Money and make their products have a short lifespan, Engineered to Fail Just after the warranty expires or with SuperCapacitors the low voltage is the best there is at the moment.
@paulgrimshaw6301
7 жыл бұрын
Is it only me that's concerned that the voltage reading on that electronic load was not dropping linearly? It seemed to be alternately dropping rapidly for a few digits, then slowly. If the voltage measurement is accurate then that suggests that the load current wasn't really constant, putting the whole measurement process in question. Have you verified with an ammeter?
@uksuperrascal
7 жыл бұрын
I read most of the comments and no one seems to take into account the tolerance percentage of the capacitor, was it + or - 20% 10 % 5% of it true value in farads what ? - as this could be a loss or a gain to its true capacity when measured to be truly full or empty.
@steveng5503
7 жыл бұрын
I would like to see measurements again with different values of C, maybe branded caps for comparison?
@nadravface
7 жыл бұрын
You have mentioned that super-capacitors do recover after quick discharge. I wonder how much voltage it recovered after you turned load off ? if it is significant it can be the reason of measured low capacity
@nickpelov
7 жыл бұрын
you should always use DMM across the device (capacitor) terminals. Never trust DC loads, power supplies (especially chinese ones). Even if they are good (exception is when you are doing 4 terminal measurement) there will be voltage drop on the terminals, cables, traces ... In theory at 1 A the drop should be constant and you don't care for this experiment, but still ... Ow btw I noticed that while you were waiting you were calculating - very good way to kill time. I also like to do that sometimes. I made a checkpoint at 1.5V (1V drop) - about 55 seconds = 55F.
@davegeorge7094
7 жыл бұрын
Julian brother, I think when above one volt 4 times the power each time you double the cap voltage. 2v times 100 farads power used down to 1v gives you more power compared to1v to 0, see?
@petermoore9504
7 жыл бұрын
Doesnt it start with 3v at 1a per second not the difference between the starting and finish voltage?
@andreimanea6260
7 жыл бұрын
Dear Julian, it is verry nice from you showing us and explaining with such details and patience. I have some question which I preffer to email you if that is ok for you. Where I can find your email? Thanks and best regards! Andrei
@Lastaii
7 жыл бұрын
I wonder if the result would be different if you used multimeters to measure the voltage and current? That said, I've found those psu displays to be reasonably accurate, certainly within 5%, which isn't going to suddenly make that capacitor great ;)
@JulianIlett
7 жыл бұрын
Like you say, not different enough. I just know I'm gonna get told off for not connecting a multimeter across the capacitor. Ah well :)
@UpcycleElectronics
7 жыл бұрын
Telling off, telling off, telling off ;) there, it's done. no need for further bs. I could not care less. Thanks for the upload... and nice stopwatch :)
@andruloni
7 жыл бұрын
You don't need a time machine to mess with the timestamps, just a friendly person at Google. In fact, time machine is of no use for this task. Or you could release the video early for certain people via a direct link to an unlisted video.
@klauspetersen8593
7 жыл бұрын
I might be missing something, but if you get a capacitance of 60F for a voltage drop of 1.5V, wouldn't the capacitance be 60 / 1.5V * 2.7V = 108F for a voltage drop of 2.7V?
@danielfmeyer
7 жыл бұрын
Klaus Petersen Yeah, from what I understand, you're right. He only used a portion of the total capacitance, and that is what he calculated/measures - his use. So your math gives the total capacitance not just what was used.
@volodumurkalunyak4651
7 жыл бұрын
In theory Farads does not depends on Volts. In practice ESR can make measured capacitor capasity less than real (that has not happened here)
@kamil004
7 жыл бұрын
I also think that he should do a test from 0 to 2.7V
@CezaryAkakios
7 жыл бұрын
No, that's not how it works. The equation has time in the numerator and voltage swing in the denominator. If you change either one the other changes by a proportionate amount, and they cancel out such that there is no net change. Julian could have picked any starting and ending voltage to measure and would have gotten (roughly) the same result.
@klauspetersen8593
7 жыл бұрын
Think I finally got my head wrapped around this. Thanks!
7 жыл бұрын
whats the differens between a 2,7v 100f super cap and a regular car audio capasitor that is a million farad
@deangreenhough3479
7 жыл бұрын
Very interesting Julian Thank you for explaining the maths 😀👍
@JulianIlett
7 жыл бұрын
Thanks Dean. Nice simple maths this time :)
@deangreenhough3479
7 жыл бұрын
Julian Ilett I had those super caps laying around and loved your modification with the leds. So made 2 up and had great fun with them. Your take on things just seems to resonate and it's sinks in. Keep up the good work my friend 🙏
@MaxintRD
7 жыл бұрын
Hi Julian, I was wondering if you also measured some of your other supercaps that way. What was the outcome? Do all of your 100F supercaps turn out as 60F (for the 1.5V drop)? When I use my multimeter or transistor tester to measure regular capacitors, they very quite a bit.
@davegeorge7094
7 жыл бұрын
After constant charging of 2.7 and 2.5 volt type DL caps 5 or more in parallel I found leakage rate of under .001 amp.. the more in parallel the more the leak current. I don't think it's too bad compared to a super low impedance OSCAP alum.electrolytic @.001F
@MauroSedrani
7 жыл бұрын
Thanks Julian! Very interesting and simple method. I'm sure Galileo Galilei will approved! :)
@sylvainlazare1114
7 жыл бұрын
What is the distribution of values when measuring several supercaps, Julian?
@JulianIlett
7 жыл бұрын
I'll buy some more of those 100F capacitors and give that a try :)
@davelowe1977
7 жыл бұрын
Measure it during the charge cycle too, to calculate the efficiency.
@Psi105
7 жыл бұрын
100F capacitor discharged at 1A should drop at 1/100 = 0.01 volts per second. So 90seconds of discharge should be 0.9V lower than the start. So yeah, looks like that supercap is not 100F
@daque1960
7 жыл бұрын
You might want to set the discharge to .2 A and charge to 1.2 A for a net 1 amp charge rate and see how long that takes. I am wondering also if it has to "soak" a while at 2.7 v. If you charge it to 2.7 v quickly and disconnect the cap does the voltage drop as the chemicals stabilize ?
@JulianIlett
7 жыл бұрын
There's no chemical reaction going on inside a supercapacitor. By timing the voltage drop under a constant current load, the ESR is factored out.
@daque1960
7 жыл бұрын
I know there is not supposed to be any reaction, I just don't trust caps that big and that cheap to be everything they are supposed to be.😀
@DaffyDaffyDaffy33322
6 жыл бұрын
My "500F" caps were measured to be around 330F using a similar method. That's terrible even assuming a 20% tolerance. Do you recommend any sources for good quality caps?
@Firecul
7 жыл бұрын
I don't get why you did you measurment like this. If you don't use the full 2.7v-0v (or compensate in the maths) then all you are measuring is the capacity used to send 1A for 90s. You still have unmeasured capacity from the 2.7v-2.5v and 1.0v-0v. So only getting 60F would be about right. Also best to disconnect the power supply before testing as you'll be adding it's capacitors also. Small effect but still.
@jimthvac100
6 жыл бұрын
You don't need to go from Vmax to zero to measure capacitance. even a capacitor tester can check a capacitor quickly by looking at this same voltage and current change over time. it does not max out the capacitor and drain to zero to do a test. it would take forever and drain the on board battery to do that. Julian did it correctly.
@staevobr
7 жыл бұрын
Why didn't you use your oscilloscope?
@JulianIlett
7 жыл бұрын
I may do that in another video, just to see if the discharge slope is linear.
@staevobr
7 жыл бұрын
Julian Ilett The validity of your calculations is completely dependent on the slope being constant. Unless that is verified, the calculated capacitance is unreliable. You're measuring voltage over time. That's exactly what an oscilloscope is for. - Then again, I wouldn't be surprised if your calculations are correct and the capacitor marking is simply fraudulent.
@PanoWorks
7 жыл бұрын
But we have an oscilloscope - it's just in an awkward format; the measurement from the dummy load, and the stopwatch in Julian's hand. All the same, we can just plot the data points we get from these and see what happens: i.imgur.com/QvBDueE.png
@ElmerFuddGun
7 жыл бұрын
PanoWorks - ha! Nicely done. And fast!
@staevobr
7 жыл бұрын
PanoWorks Assuming the load readout is correct, of course.
@jimthvac100
6 жыл бұрын
Assuming this formula would work the same while charging the capacitor at 1 amp, then measure the time in seconds till voltage raised 1.5 volts.
@daab889
7 жыл бұрын
Who did you buy it from?
@suleymanhafizoglu6019
7 жыл бұрын
instead of using fixed current source why don't you use simple resistance. t = r *c and total time for discharge is 5t.
@JulianIlett
7 жыл бұрын
Discharging into a fixed resistance makes the maths more difficult.
@wouter481
7 жыл бұрын
please tell me if i am Wrong but 100f means the complete charge and you only measerd 1.5 of 2.7 v and that is 56% so it makes sence to measure only 60f or do i forget something?
@JulianIlett
7 жыл бұрын
True, I only measured a partial discharge, but that takes less time. The calculation for capacitance is a ratio (time / voltage change).
@Infinion
7 жыл бұрын
100F does not mean full charge. No matter what voltage a capacitor is charged to, it always has the same capacitance. Farads is equal to charge per volt just like fuel economy is litres per kilometer. You use capacitance to calculate total energy capacity in Joules. This formula is for energy = 0.5*C*V^2 . Total energy is 0.5*C*(Vstart - Vend)^2. Like Julian says, Capacitance does not depend on what voltage you start or end at, it is a ratio of time/change in voltage. As long as you discharge at a constant current, any start and end voltage will give you the total capacitance. This is because in our formula for capacitance, C = q/V, the V is actually a ΔV (voltage range Vstart - Vend) or in other words Farads = (constant Amps * total time)/(Start Voltage - End Voltage). So if we discharged the capacitor from 2.7V to 2.69V at 1mA we would eventually get 60F after 10 minutes.
@wbfranks
7 жыл бұрын
Verify the accuracy of you current load and recalculate. I have the exact same DC load and the current readout is off by more than 30%. Use your multimeter to measure the current and see what the results are.
@eliotmansfield
7 жыл бұрын
Interesting - I have the same device too, will check it.
@JulianIlett
7 жыл бұрын
Good call - I'll do that.
@eliotmansfield
7 жыл бұрын
I've tested mine and the display is dead on with my Brymen bm867s
@wbfranks
7 жыл бұрын
Took another look at mine and it's worse than I recalled. Actual current with 200ma indicated is 390ma and 800ma with 1 A indicated.. Current is also 130ma when off, even with control power at 0V, and load current limits at about 900ma regardless of increased setting. In addition to being uncalibrated, I suspect my unit may be defective, or a factory reject. It would appear the unit is intended to be software calibrated as there are no adjust pots one might normally expect. There also is no apparent way to access the maH and other displays. Anybody have a manual or know how it's supposed to work?
@maicod
7 жыл бұрын
what is the reason you're drawing 200 mA while charging the Capacitor ? I hoped you would test your new 700F caps this way :)
@iamdarkyoshi
7 жыл бұрын
Why not "calibrate" it with a supercap off of a reputable source?
@donaldfilbert4832
7 жыл бұрын
I'm curious why you didn't verify the value by just measuring the capacitance with a capacitance meter ??
@ElmerFuddGun
7 жыл бұрын
Capacitance meters just like resistance meters have a limited range over which they will work. If you check you will find this is many many magnitudes beyond what a commonly available meter can read.
@donaldfilbert4832
7 жыл бұрын
Right on !! Thanks for the clarification. I heard him say 100 Farads - but my mind thought - 100 uF !! {:>)
@ElmerFuddGun
7 жыл бұрын
LOL... ya your mind was only off by a few zeros... 6... ;-)
@peerappel2012
7 жыл бұрын
aaand? Will the next video be POSTBAG 100!!!?
@nickpelov
7 жыл бұрын
I wonder if the Nippon chemi con will show better results (if they are not fake)
@JulianIlett
7 жыл бұрын
I really hope they're genuine!
@electrodacus
7 жыл бұрын
You should just test them the same way. It will only take about 10x longer :). Usually those where genuine but used (dissembled from some prototype hybrid electric buses in the past not sure if it is the case now).
@madbstard1
7 жыл бұрын
What was the sound at about 10.30? Sounded like you were watching the Grand Prix :D
@JulianIlett
7 жыл бұрын
Pretty sure it was a single decker bus, speeded up 6 times :)
@jamieh9351
7 жыл бұрын
Did you try the same experiment starting from 2.7V and letting it run down to 0V?
@JulianIlett
7 жыл бұрын
I didn't do that, but it would have taken correspondingly longer and the end result would be the same capacitance.
@karlfoley
7 жыл бұрын
Would it? 2.7V to 0V would take longer, and give larger capacitance? Closer to the 100F stated.
@CezaryAkakios
7 жыл бұрын
No, the capacitance measured should not depend at all on the starting and ending points. The longer time taken is exactly cancelled out by the wider voltage swing. (Obviously any real capacitor is not an ideal device and will exhibit non-linearities, but they should be relatively small, certainly not on the order of half of the total capacitance.)
@ABaumstumpf
7 жыл бұрын
karl - as he said - the voltage drop should be pretty much linear, so dropping 2 V should take twice as long as dropping 1 V.
@kaoshavoc
7 жыл бұрын
He is extrapolating the capacitance from what he has drained. Given that it should drain in a liner fashion that is really easy to calculate as he did. So unlike a battery, you don't need to drain it all the way to see its full capacity, because batteries DO NOT drain linearly and thus have to be discharged or charged completely to find their capacitance. And above he said he might do a video using his scope to show that capacitors really do discharge in a linear fashion.
@renelefebvre53
7 жыл бұрын
I play with my supercapa ans I see a thing. When I charge it with a power supply , then the voltage rise 2.5V and rest constant, the currant must be decrease quikly, but no ! The currant take 2 minutes or more to décrease at 5% of the max that my alim give. And the voltage are constant à 2.5 V Perharps, in your test , you must waiting for the currant decrease before start the decharge ? sorry for my english, I'm an eat frogger :-)
@tonyavito208
5 жыл бұрын
i ADVISE YOU TO CHECK THE THEME NON LINEARITY OF ULTRA-CAPACITORS, YOUR EXPERIMENT WILL WORK, IF YOU ARE USING NORMAL CAPACITOR. HOWEVER WITH UC, THIS WONT WORK...
@WaltonPete
7 жыл бұрын
Should have used your scope and a basic frequency generator as per the video from w2aew.
@roberteliassen5020
7 жыл бұрын
Published on Jun 12, 2017, today, yet the comments are 2 or 3 days ago. Did you build a time machine? :)
@scwfan08
7 жыл бұрын
Robert Eliassen Video was up before but only Patreon supporters had access.
@roberteliassen5020
7 жыл бұрын
Ah! :)
@UpcycleElectronics
7 жыл бұрын
Robert Eliassen Patron supporters get access to videos early. Lots of KZitemrs do this. Basically they upload videos with a private setting. This allows them to give out the direct link on Patron. Anyone with the link can watch it. This lets the KZitemr respond to comments from people that are more genuinely interested and consistent, it issolates the content creator from the need to read and filter through all of the imbeciles and trolls that plague YT, and it allows them to see suggestions and requests from people that are actually making such requests possible. There are too many people on YT that don't understand the direct connection between comments sections and content creators. Really YT comments should be restricted to people that have uploaded content with a minimum number of views in my opinion.
@SpeccyMan
7 жыл бұрын
Upcycle Electronics So, by that argument you are suggesting non-patreon supporters are all less genuinely interested and consistent? Comments should be based on a viewer's opinions and should require no more than that since not every viewer is in a position to upload content! I have no problem with patreon subscribers getting an early view (fair is fair) but as for your other idea not only is it ludicrous it is totally unfair. My opinion is your opinion is flawed in that respect. I am a subscriber to Julian's channel and a regular viewer of his content so I ought to be free to comment as I see fit - in my opinion.
@UpcycleElectronics
7 жыл бұрын
Nick B I was saying, everything in KZitem comments gets attacked like an argument. I don't blame anyone for prioritizing dealing with those that pay to argue over those that don't. The content of any statement doesn't seem to matter in KZitem comments someone always seems to find a reason to argue over any stupid trivial aspect.
@twentycm
7 жыл бұрын
Thanks
@n2n8sda
7 жыл бұрын
I can't quite make out the brand name of the capacitor in the video but i'm going to go on a limb and say it's one of these generic Chinese capacitors? I'n which case i'm quite surprised it even has over half it's stated capacity, espeically going on the capacity claims of some of the lithium ion cells you can buy from there...
@JulianIlett
7 жыл бұрын
It's branded Nesscap, but I don't know if it's genuine.
@PhaQ2
7 жыл бұрын
The total capacity gets even worse when you run these caps in series. 6/2.7v 500F caps in series will give @ best 230F over all.
@Infinion
7 жыл бұрын
I think you're thinking of capacitance that decreases in series, not capacity. Total capacity (energy in joules) will stay the same regardless of whether they are connected in series or parallel. Six 2.7V 500F capacitors in series will have a total Capacitance of 83.333F and total energy (capacity) of 10,935 J at 16.2V. Six 2.7V 500F capacitors in parallel will have a total capacitance of 3000F and total energy of 10,935J.
@h07d0q
7 жыл бұрын
what is your components tester measuring?
@JulianIlett
7 жыл бұрын
It's an electronic DC load that displays voltage and current.
@h07d0q
7 жыл бұрын
Julian Ilett yes, but don't you have a DVM or the AVR component tester (also known as M-Tester) for comparison of the capacitance and testing the ESR?
@spudhead169
7 жыл бұрын
It dropped from 2.5V to 1.5V that's 1V @ 1A in 55 seconds, so 55F
@mal2ksc
7 жыл бұрын
It also dropped from 2.5V to 2V in 25 seconds in the first test, which would equal 50F. Clearly the voltage drop is _not_ linear, though it's not incredibly far from it. The difference between theory and practice is that in theory there is no difference while in practice, there is.
@spudhead169
7 жыл бұрын
We also don't know how responsive that CCL is, it could be a little laggy. Am I right in thinking the discharge across a fixed load resistance would produce an inverse square voltage drop?
@mal2ksc
7 жыл бұрын
A bit of lag in the readings shouldn't be a big problem because it would affect the timing at both ends of the stopwatch. If he was timing from full charge, it might matter, but since he's timing from a slightly lower voltage, the lag would delay both the 2.5V and 1V result, and we're dealing with tens to hundreds of seconds here, so it's probably not going to be sufficiently non-linear as to matter. Even the capacitor itself is only mildly non-linear into the constant-current load. I'm pretty sure you're right about the inverse square relation across a fixed resistance, but then poor Julian would have to sample many times and then use calculus to get Farads out of it.
@nightmareinaction629
7 жыл бұрын
Julian what did you study for college
@JulianIlett
7 жыл бұрын
Computing and control systems at Coventry Polytechnic (as it was back then)
@nightmareinaction629
7 жыл бұрын
Julian Ilett well I'm a senior next year in high school I love tech I work on the electronics in robotics class I do all the esc and motors and program the mixing in the controllers and do the binding j made my own ebike with a drone esc the was supposed to work on 6s lipo max just chaged the capacitors and made it work on 10s no problem but I still dont know what I want to do for college
@rontopsvoort7277
7 жыл бұрын
Julian, we all hate maths and so do you ;-).... why not just make it easier, let it drop 1V at a 1A discharge current. The time measured in seconds needed for that equals the capacity in F...
@JulianIlett
7 жыл бұрын
I'll do that next time - so elegant - seconds=Farads - I love it :)
@DoRC
7 жыл бұрын
Whereas your pocket watch definitely isn't garbage is also not particularly valuable.
@JulianIlett
7 жыл бұрын
That's a shame, I thought it might be Harrison ;)
@CornishMiner
7 жыл бұрын
The 100F is rated capacitance which is, for a DC voltage, 0.5 x C x V^2. In this case 0.5 x 90 x 1.5^2 = ~100. So not a disappointment at all.
@JulianIlett
7 жыл бұрын
The C in E=½CV ² is capacitance in Farads - the C in my equation is charge in Coulombs
@CornishMiner
7 жыл бұрын
Yep. My bad. Maybe you could hold the capacitor at 2.7V for a time with a constant voltage source to ensure it is fully charged - seems that is best practice. Otherwise your method looks to be standard procedure. Best wishes.
@theirisheditor
7 жыл бұрын
With the E=½CV ² formula, I end up with the 60F worked out from the video: At 2.5 volts, a 100F capacitor would hold 0.5 x 100 x 2.5² = 312.5J (E is Joules) At 1 volt, a 100F capacitor would hold 0.5 x 100 x 1² = 50J To discharge from 2.5V to 1V, a 100F capacitor will provide 312.5 - 50 = 262.5J of energy With a linear voltage decline of 2.5V to 1V, the average voltage is (2.5-1)/2 = 1.75V With a discharge period of 90 seconds from 2.5V to 1V at 1A, this capacitor provided 1.75 x 1 x 90 = 157.5J of energy 157.5/262.5 = 0.6 (60%) of what was expected. 60% of 100F is 60F. Expected discharge time for this capacitor from 2.5V to 1V at 1A would be 312.5 / (1.75 * 1) = 150 seconds.
@TheUnnamedGent
7 жыл бұрын
The comments on this video are my favourite. It is possible I'm a bad person.
@renelefebvre53
7 жыл бұрын
I buy some super capa of 700 F. after mesure like Julian, the real capacity is only 400 or 450F. if you have only 50% be happy 😠
@oreubens
7 жыл бұрын
Your result of 60F is off because 1) you didn't count the bit of charge when it was fully charged (2.7V down to 2.5 V). 2) You stopped at 1V, but it still had charge left at that point. So your test is only measuring part of the total capacity.
@realflow100
7 жыл бұрын
You didnt drain it from 2.7v to 0v. if you did you'd realize its Around 100 farads (You could probably squeeze out a bit more if you charged it to 2.75v i doubt it'd hurt the capacitor. Mine has been taken to 3.5v and back again and its still going strong. they seem to have a huge tolerance if your charging at a low rate. i was only charging at 400mA to 0.75A Also you need to keep it topped up at 2.7v for a while or charge it slightly over for it to hold at 2.7v for a while. then you will probably get 100 farads out of it
@Infinion
7 жыл бұрын
I think you're mixing capacitance in Farads up with energy in Joules. To calculate capacitance, you use the formula C = q/V, where q is the symbol for charge, expressed in units of Coulombs. He plugs the formula for Coulombs into this equation to solve for capacitance and to compare it with what was claimed by the manufacturer. the formula for energy is E = 0.5*C*V^2. If you wanted to know the total stored energy in Joules you would indeed discharge it from 2.7V to 0V. You could go higher but if the supercap's electrolyte is an ionic liquid then the electrolyte will break down and oxidize your electrodes (irreversible redox reactions), some supercaps, called pseudocapacitors, have the proper electrode metals that allow reversible faradaic reactions. A typical double layer capacitor can store additional energy at a higher voltage, but you will oxidize your electrode and will eventually have capacitance degradation.
@realflow100
7 жыл бұрын
I never said anything about joules! Where did you get that from?? And my super capacitor has not lost any capacitance. it just must be very tolerance of overcharging i guess. samwha green cap brand 100 farad 2.7v super capacitor. Normally i only go to 2.85v and hold it there for a minute then take off the charge so it will settle around 2.7v simply because i only have 1 super capacitor soon i'm going to be getting two super capacitors and i will be putting them in series each being 400 farads to use to make about 5v no need to overcharge them at all then
@Infinion
7 жыл бұрын
Sorry I should clarify. To calculate the capacitance in Farads, you do not need to drain from full to 0. You would do this if you wanted to find the total energy in Joules. Capacitance does not depend on what voltage you start or end at. As long as you discharge at a constant current, any start and end voltage will give you the total capacitance. If we discharged the capacitor from 2.7V to 2.69V at 1mA we would eventually get 60F after 10 minutes. This is because in our formula for capacitance, C = q/(Vstart -Vend) or in other words Farads = (Amps * total time)/(Start Voltage - End Voltage). I'm glad to hear you don't hold your supercap at an overcharged state for too long, that should limit the amount of faradaic reactions that oxidize your electrode metal and would explain why its capacitance hasn't degraded much!
@nightmareinaction629
7 жыл бұрын
speedy thingy lol it's. a timelapse
@Hasitier
7 жыл бұрын
As many comments say, you did not use alle capacity. If you correct the equation (divide by 1.5 and than multiply by 2.7) you will get your around 100F
@birnodin
7 жыл бұрын
"Hello, here is the police department! How can we help you?" "Well, I have a neighbor... He has many cylinders with wires on it... and.. some are beeping and counting down..." "Ok, I unterstand, we do what we can, do not go back in your house!" The next day prime minister May standing in front of Downing Street 10 informs the people: "Enough is enough now, dangerous subjects live in our towns and villiages. But yesterday we had a great success: We could wipe out one of these people! It is sure that he was planning something dangerous. Too bad he was not able to answer our question any more. Let us continue this path!"
@Infinion
7 жыл бұрын
that bad, eh?
@TheDutyPaid
7 жыл бұрын
Open one up in the name of science
@JulianIlett
7 жыл бұрын
I'm seriously considering hammering a nail through a fully charged supercap - maybe :)
@davidreynolds9649
7 жыл бұрын
Microwave it.
@flagpoleeip
7 жыл бұрын
Why would you not use a DVM?
@gaabinubatrafinulifilit122
7 жыл бұрын
I love these analog stopwatches. I once measured the length of a fart with one.
@stuartofblyth
7 жыл бұрын
I'm afraid the reasoning here is flawed. Imagine a bucket (the capacitor) filled with golf balls (electrons). Now you pull golf balls out of the bucket at the rate of 1 golf ball every second (1 amp) for 90 seconds. At the end of that time you have removed 90 golf balls (so many coulombs). But that doesn't tell you how many golf balls there were to begin with (the capacity), only that there was room for 90 golf balls and more! To measure the capacity of a capacitor you need to know how much electricity (measured in coulombs) you can put into it, and what it's voltage is when full. Only then can you apply the formula C=Q/V.
@doug7131
7 жыл бұрын
He is measuring capacitance not how much energy was stored in it. The discharge curve into constant current is linear so it doesn't matter what voltage range you use to test it. Running it right down to 0v would increase the time taken to discharge but the voltage range would be increased. These will cancel each other out giving you the same result that is in the video
@stuartofblyth
7 жыл бұрын
I never mentioned energy. The fact is that this experiment showed the capacitance of the capacitor to be 60F when it's nominally 100F. The only way to resolve the various arguments about why that should be so is to charge to the 2.7V the capacitor is rated at and see how much electricity you can draw until it is totally discharged. That's what capacitance is - how much electricity you can store at a given voltage. If I had the equipment I would obviously do the experiment myself. Experiment is king!
@doug7131
7 жыл бұрын
Fair enough. I doubt the result will change much with going from 2.7 - 0 instead of 2.7 - 1 but i agree that it is worth confirming. The lack of capacitance is most likely down to it being a no name eBay sourced capacitor. Also super caps tend to have very high capacity tolerances. Even branded ones from Nichicon and Panasonic have ±20% tolerances on their capacity.
@stuartofblyth
7 жыл бұрын
I wish I could think more quickly. We don't need to do the experiment again - we can use the video! Here are the readings, seconds paired with volts: 15/2.17; 30/1.88; 45/1.64; 60/1.42; 75/1.21; 90/1.01. Conclusion 1: the fall in voltage is NOT linear. Conclusion 2: using Mr Ilett's theory of C = Amps*secs/voltage fall gives varying values of capacitance. For example, at 15 seconds the voltage has fallen by 0.33V (2.50V - 2.17V). The capacitance is therefore 1*15/0.33 = 45F. Using Excel on the other readings gives 48F, 52F, 56F, 58F and 60F. Using corresponding figures from the 2nd run shows a capacitance between 42F and 57F. I rest my case.
@renxula
7 жыл бұрын
60F isn't capacity, it's capacitance. Capacity (like battery capacity in Watt-hours) is an amount of energy, but capacitance is the inverse of the rate at which its voltage changes when current goes in or out. But yes as others have pointed out, the supercap's capacitance changes depending on its voltage, so it's not a simple constant.
@vmailtk5
7 жыл бұрын
can you translate the Farads to Watt Hours?
@ABaumstumpf
7 жыл бұрын
No, cause they are not the same. Farads are Wattseconds divided by Volts squared. But if you want to know the storage capabilities of that Cap - you have 1/2*C*U² so 0.5*60*2.7² = 218 joule.
@vmailtk5
7 жыл бұрын
Sorry I meant Ah. I'm just trying for figure out what to replace a 12V 88Ah battery with using super capacitors.
@anullhandle
7 жыл бұрын
Assuming you need that full capacity not just the rate of power delivery for that battery size... 12volt*88amp hour=1056 watt hour 60sec*60min*1056 = 3,801,600 watt seconds aka joules J=1/2C*V^2 or C=2*J/V^2 C= 2*3801600/7.29 so little more than 1,000,000 farad assuming 2.7volt standard cap voltage. If you wanted a starting voltage close to a 12 volt battery 5 series caps would have a staring voltage of 13.5. Series capacitance of about 42,000 farad. In practice you couldn't really get 1056 watt hours out of a lead acid battery without killing it very quickly. Maybe 500 watt hours or so.
@malenottinicco
7 жыл бұрын
Why do you start the stopwatch at 2.5v?? The total capacity of the capacitor is measured from 2.7 to 0.0 volts, not from 2.5 to 1.0 volts, if the load doesn't work for less then 1v you need to do some math to count the remaining charge of the capacitor.
@Infinion
7 жыл бұрын
I think you're mixing capacitance in Farads up with energy in Joules. To calculate capacitance, you use the formula C = q/V, where q is the symbol for charge, expressed in units of Coulombs. He plugs the formula for coulombs into this equation to solve for capacitance and to compare it with what was claimed by the manufacturer. edit: just for completeness, the formula for energy is E = 0.5*C*V^2
@brucewoods9377
7 жыл бұрын
It is my understanding that both charging and discharging of a capacitor is EXPONENTIAL, not linear.
@Infinion
7 жыл бұрын
if connected to a voltage source or discharging across a resistor it is an exponential decay, but because Julian applied a constant current load, the discharging is linear.
@brucewoods9377
7 жыл бұрын
Infinion his "constant current load" is STILL RESISTIVE. However further to my comment he mentioned that the "Charge rate" slowed as it neared 2.5 volts, and slowed as it approached 1volt. Therefore validating my statement that it's charge/discharge was/is EXPONENTIAL. In addition, he only discharges it partially. So as an other commenter says, "a ten gallon bucket is still a ten gallon bucket even if you only discharge five gallons of its contents.
@Infinion
7 жыл бұрын
+Bruce Woods a constant current load functions exactly how it sounds. Like +KASA kasa says, it varies the resistance so the current remains fixed, and the voltage curve remains *linear*. Julian used a constant current load to discharge, so we can calculate exactly how long it would take to reach 1V from 2.5V. If it wasn't linear then this whole video would be moot and we wouldn't have found our 60F capacitance. Julian connected the supercapacitor directly to his supply while charging, so in that case it would be charging under a constant voltage and would show an exponential decay, but we weren't looking for capacitance while we were charging because of this reason. so to rephrase your statement, charging and discharging of a capacitor DEPENDS on the circuit it is connected to. It is considered Exponential decay if the source is a constant voltage, linear if constant current, or oscillatory decay if connected to an inductor.
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