Talk Morgan Forums Morgan Sports Cars Morgan Three Wheeler & Super 3 Battery Replacement?

The little boards are the Voltage regulators
You are in the US so using 110V
Look at the wattage V * I roughly [ It is not really true because there is a difference between AC and DC ] you use at home.
12V * 250A = 3kW
divide through 110V = 27A
Here in EU we use 230V which gives 13 Amps, these very short 3 cm 1,5mm2 cables are used everywhere for AC till 16A.
And we are talking about just seconds.
Only at charging the capacitors for the first time only the caps get warm the cables not.
For the connections between the caps and for the longer cables I used 6mm2

Thanks Herrie. It all makes sense. My only concern would be that I believe the maximum current density for copper is 95 A/mm2, so if you want to fully run the starter from the capacitors and not just augment the battery, then your 1.5 mm2 cables are under-sized. But if the battery is dead this won't work anyway!

Hi JohnMat
Thank you for your comment and your important question. I suppose more people will have the same question.
I will for sure put this question in my document, when people ask to publish it.
I can tell you that the wires stay cold at cranking and starting is fun and immediately.
I am not sure but I think you forgot the length of the cable.
Please have a look at the following calculation, hope I am correct and it clarifies the issue

Calculation
Rho CU = 1,68E-08 Ohm.m at 293K ; Length : lm = 3,00E-02 meter ( 3 cm ) ; Asurf = 0,0000015 m2 ; IA in Amps

R = Rho * lm/ Asurf - - - without Temperature validation
IA = V/R

IA = 12Volt * Asurf / ( Rho * lm )
Max Current at : 3 cm ; IA = 35714 Ampere = 35kA
Max Current at : 1 meter ; 1071 Ampere = 1kA

Hi Herrie - I think you are calculating current assuming the voltage drop across the wire is 12V. At 35 kA you would be dissipating 420 kW in a 3cm wire, so it would not last long, not that it matters here as we are talking about much lower currents.
I think in practice there are three considerations for wire size:
1. The maximum power you can dissipate in the wire before heat problems arise, commonly referred to as ampacity for steady state. Given the short duty cycles here, it may not be a concern.
2. The maximum voltage drop you can stand due to the resistance of the wire. If you use, say, a 0.25V drop in each wire as your target, you should use 0.25V instead of 12V in your math above, so the limit is 48x lower.
3. The electro migration limits of the copper itself. I thought that was the 95 A/mm2 I quoted, but on further googling the actual limit seems to be 10^4 to 10^5 A/mm^2, so not a concern.
Please correct me if I'm wrong.

Hi JohnMat
Thanks again for your reply.
Let us be practical.
It functions and the wires stay cold. And for me that’s not a surprise.
The nice thing is that I crank and the engine starts immediately without hesitation, which it did for sure not before I used a super-capacitor bank.

As answer on your thoughts, but difficult to find the right answer :

Take the capacitor bank as something stand alone.
Yes. The voltage drop is in a short time around 12V, see the calculation here-under.
So I do not understand your assumption of 0.25V drop.
Keep in mind : It is not a battery which do have small Voltage drops, but that's a total different device.
Our Battery has 40Ah and only max 500 cycles to max 50%.
This capacitor bank has max 945As = 0,26Ah is 0,007 times less than our Battery. So our battery can - without extra charge for itself - charge this Capacitor bank more often than his own cycles. And that's why your battery will live much longer too when parallel connected - a hybrid system - to a capacitor-bank..
The capacitor-bank charges faster, It de-charge faster. And will do that also faster than a Battery when not in use.
You can - and you will do - de-charge it 100.000 of times from 15V to zero and back.
And for cranking we only need power for a couple of seconds and after that a device which is charged very fast to 15V to do the same trick again and that's a capacitor

Some formula’s for a capacitor
C(Coulomb) = I ( current in Amp ) x delta Time (seconds )
Coulomb = Capacity ( Farad ) x Voltage
I x Delta T = Capacitance x Voltage = 63 F x 15 V = 945 As
63F is the total Capacity of the bank.
15V is the max Voltage for the balancing circuits.
So with a cranking amp of 350 you are finished in a couple of seconds. So yes - in my humble opinion - a drop of 0.25V will only give you a ‘quantity of charge transfer’ ( = Coulomb ) of 15As and that is a value where we are not talking about.

But the only interesting value for us and our small 40Ah CA 350 cranking battery is that it works - with my cable choice - for me and for mr. Tesla who bought Maxwell recently for his future cars. Sometimes he does - I know unbelievable - wise things too.
I searched for you on Youtube and you can find more examples.
Also it is impossible to solder thicker wires on the contact-points of these Capacitors of 450F.
You have to go to 3000F to get real terminals with bolts.

Hi Herrie - I think your solution is sound, and the physics check out. I disagree with some of your math, but it doesn't really matter if it works, and I agree it will!

I suspect it works because the energy stored in capacitors is transferred at a moderate rate to the battery before the starter is operated, effectively giving the latter a small top up surface charge. This assumes an adequate voltage difference.