I was pondering on the cut out issues at speed for EUC's, and i was wondering that the main issue is down to a sudden voltage drop from the power supply. now, what if EUC's had a small array...

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I was pondering on the cut out issues at speed for EUC's, and i was wondering that the main issue is down to a sudden voltage drop from the power supply. now, what if EUC's had a small array of super-capacitors that could provide a Life-Saver 1 shot boost to the voltage + current to force the rider back hard and warn the rider that their extra life is now depleted. When the rider is forced back, the motor starts to act as a generator so that should help the system stay stable until a sensible speed is reached.

You do not need an array of supercapacitors to fulfill the entire workload of the battery, just a boost to make up the voltage difference (Maybe 24V and 3A for 1 second on the larger systems (no i have not checked if these figures are correct, I am just mussing this idea)). If i understand correctly, 1F of capacitance = 1A per second, so it sounds like you need a lot of capacitors to pull this off... but then we realise that we still have the battery providing current which offsets the requirement.

I was also thinking that the capacitors would only be charged via the mains to prevent aditional strain on an already weak battery.

This is just my mussings and it may be the case that it is impracticle for many reasons that i do not understand from an electronics / cost point of view... but if someone out there has some Electrical knowledge far superiour to my 20 year old Secondary School curcuit training - then i will be interested in what they think...

11 comments

Liked by Carl Svanfelt and 2 others

Robin Ridge

2 years ago

Hi it takes around 5CR seconds to charge a capacitor where C is in Farads and R is in Ohms. Likewise a capacitor discharges at the same rate. The resistance in our case is that presented by the motor. You cannot discharge the capacitor but reduce its voltage to some percentage of its charge at which point you can discount it as being effective. Now the problems the resistance of your motor is quite low and the time you wish the capacitor to deliver is indeed several seconds. This leaves us with the capacitance in Farads. A capacitor of 1 Farad is huge, a 1 Farad capacitor rated 85 or 100V even more so. Full marks for lateral thinking though :-)

ReaverOfCheesecake

2 years ago

Robin Ridge Thank you for the feedback. Size / Weight may be the physical limitation here I do agree. I was thinking along the lines of using 5.5V 1F Capacitors (These can be picked up from RS for around £3.50, so these will be much cheaper in bulk)) in a liner array of 4-6 (since they are quite small) to make 22V - 34V capacitor banks. These can be placed inline with the battery to make up the short fall in voltage when it is needed (A control circuit will prevent these from being engauged during normal operation of the EUC).

Remember, we will never be able to build up a 84V - 100V array, and have it small / light enough to fit on an EUC as you have correctly pointed out. How many seconds does it take to initiate a tiltback on the larger EUC's, this is sub second IMHO so maybe 4F will be enough (I thinks this will be a boost of 4A peak that falls of linerly with the capacitors discharge). There is also control / protection circuitry to consider but this is more a cost issue rarther than weight.

Again, I am not an electrical engineer and i fully except that my wishful thinking may beak some key laws in thermal dynamics... but, on the other hand, something good can come from wacky ideas ;-)

Old Trapper

2 years ago

Will not work. Putting any Voltage of less than the battery voltage - in this case 24 in parallel to a battery to 100V - will shorten the battery and most likely result in an explosion, if a smart BMS dose not switch off the battery and isolate it from the remaining circuits. --> Faceplant and Fire at least.

Dualtron scooters provide for an external battery connector, allowing to backup the power at any choosen moment ... sadly very bulky electronics and very expensive too. Might not be reasonable on an EUC to ride with batterypack in rucksack tethered to the wheel :-)

Some more figures: 1 Farad = 1A in 1 sec by 1 V. Like @Reaver wrote, even a 1 F capacitor is huge. Just following your idea of 24 V by 3 A for 5 sec would result in 360 F and a huge flash ... not taken the resistor into account though :-)

My final thought for this thread: Supercapacitors work in a temperature range from -25 to 60 °C ... hmmm ... on steep inclines my RS gets warmer than that.

ReaverOfCheesecake

2 years ago

Old Trapper Oh crap... I forgot that it was 1F per Volt... So Q = CV... so my plan for 5.5V @ 1F actually = 0.1818...A. Sigh, I am clearly being a dumbass here... oh well... I guess that pesky Thermal Dynamics wins again ;-)

Old Trapper

2 years ago

ReaverOfCheesecake grin ... my younger son is currently studying physics and he often abuses me as a sparringspartner on upcoming exams ... we had this discussion thermal dynamics versus electrodynamics recently ... slight advantage on my side ... sorry :-)

ReaverOfCheesecake

2 years ago

I have found the solution to our problems. Presenting, the Micro Universe
https://static.wikia.nocookie.net/rickandmorty/images/2/25/S2e6_Microverse_Battery.png/revision/latest?cb=20160930055729

Old Trapper

2 years ago

I have found another solution ... well not really but on it's way :-)
https://youtu.be/V_f8Q2_Q_J0

Robin Ridge

2 years ago

ReaverOfCheesecake 5.5V 1 Farad Capacitor. (5.5V is only a small droop, if you had x2 1 F capacitors in series then you would have 11V available but only 0.5 F, i.e twice the voltage x4 the size of the capacitor) The problem with this approach is that the battery voltage droop is because the battery can no longer supply the current being demanded. The drop in voltage leads to a drop in current. If you topped up the voltage with a 5.5v capacitor the resulting increased current would simply lead to the remaining batteries drooping even more until the voltage was low enough to support the current the batteries are capable of delivering. That is you are pretty much back where you started.

ReaverOfCheesecake

2 years ago

Robin Ridge Yeah, your right. As i admitted to Old Trapper, I screwed up the math and forgot to apply the formula Q = CV. So in short, in series, 1F @ 34V would only give us 0.029A... which may as well be nothing, and the voltage would drop away almost instantly on a Wheel. We would need so many of these capacitor banks to run in parallel that the weight would be huge. It's a good job i'm not an electician :-)

Robin Ridge

2 years ago

ReaverOfCheesecake A few years from now we will see solid state lithium batteries, that will cause a revolution in the capabilities of PEV's generally.

ReaverOfCheesecake

2 years ago

Robin Ridge Yeah, I'm looking forward to that... Hopefully when the time comes to replace the batteries on my S-18, I can replace them with solid-states ;-)

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