The braking problem – energy analysis

Posted on September 26, 2008


I’ve been just talking about regenerative braking for too long. It’s time to define the requirements.

When trying to dimension ultracapacitors or batteries for regen braking, we have to take into account the energy and power of the braking in it’s worst case: from top speed down to zero, and in the shortest time possible. Let’s start with the energy (I will post again later about the power).

So, I already set the top speed at 120 km/h, and that makes it easy to calculate the total energy generated by a full-stop braking: it is equal to the total accumulated kinetic energy of the car (well, not really, the aerodynamic drag helps the braking a little – but I’ll go for the worst case for now).

So the total energy of my car at 120 km/h is:

  • E = 1/2 . m . V^2 [J]
  • E @120 = 0,5 x 1230 kg x 33^2 m/s
  • E @120 = 669.735,00 J = 670 kJ = 670 kWs
  • E @120 = 669.735,00 Ws / 3600 s = 186,04 Wh

186 Wh is easily managed into a small battery pack, whether for acceleration or braking.

Now, for the first phase of my project, I’m only considering rear-wheel motors; this means that the full braking energy will be divided by the two axles, and unevenly so, as usual. A “typical” stock car distribution of braking forces between front and rear axles is 65% / 35%, so:

  • E br@120 (rear) = 0,35 x 186,04 = 65,11 Wh

However, we know that the electromechanical system is not perfect; in fact, there are people that say regenerative braking is almost useless because of all the losses. Assuming a generator + converter loss of 15%, we get:

  • E br@120(rear)(effective) = 0,85 x 65,11 = 55,35 Wh

Hmm… we spent 186 Wh to accelerate the car up to 120km/h and all we get back from braking is 55 Wh. So, the regenerating efficiency is:

  • Eff br@120 = 55,35 Wh / 186,04 Wh = 0,2975 = 30%

But we still have to factor in the losses of the battery itself, and consider that the energy spent on speed maintenance (air drag and rolling resistance at constant speed) is never recovered. So, I’m looking at a 30% maximum potential autonomy increase, but I know it will be less. It all depends on the driving circuit and habits.

Ok, 55 Wh is pretty ridiculous for an EV battery. But if we are talking about a “panic stop”, it is delivered at a very high rate. How high? I don’t know, so I’m taking a guess and estimating that my car could stop from 120 km/h in 10 seconds. Just for the sake of argument. This means the average braking power would be:

  • P br@120(rear)(avg) = 55 Wh x 3600 s / 10 s = 19.926,00 W = 20 kW

Although 20 kW is a perfectly acceptable power level for an EV battery discharge, it may be excessive as a charging rate. And a “panic stop” involves a much higher power at the beginning of braking than at the end, so the real maximum braking power may be several times larger than 20 kW. This means I will probably need a few ultracapacitors to take the power hit.

So, how many ultracaps are we talking about? Here I have to extrapolate even further. Luckily, my new friend Guy gave me a hand here and picked out a 70 F / 2,1 V capacitor from Digikey as an example.

I have no idea what voltage the capacitor bank finishes its charge after the braking. But that is irrelevant; let’s assume we put all the caps in parallel, forcing the voltage down to a maximum of 2.1 V – then the corresponding capacitance would have to be:

  • E = 55,35 Wh x 3600 s = 143.910,00 J
  • E = 1/2 . C . sqr(U) <=>  C = 2 . E / sqr(U)
  • C total@2,1V = 2 x 143.910,00 J  / sqr(2,1 V) = 65,2 kF

And the necessary number of capacitors would be:

  • N @2,1V = 65.265,31 F / 70 F = 933 capacitors

Ordering 1000 caps, the price would be 6,912 USD per unit, or:

  • Total Cap Cost @2,1V = 1000 x 6,912 USD = 6.912,00 USD

Yes, 7000 USD is expensive, but still realistically achievable.

Now, something tells me I shouldn’t be getting away with putting all those caps in parallel… 😉

The way I see it, I’ll always need some controller in between the motor and the capacitor bank to regulate the motor braking current and the battery charging current; why can’t it work at such a low voltage as 2,1V? The only disadvantage I see is having a bidirectional “buck/boost” converter there, which would be expensive to build and complicated to manage, and probably inefficient. Another disadvantage would be the very large currents that circulate between capacitor bank and motor, and even between capacitors – also not good for efficiency.

So let’s try another approach. Let’s put enough caps in series to make them withstand 100 V and call that a module.

  • N caps / module @100V = 100 V / 2,1 V = 48 capacitors / module
  • C / module @100V = 70 F / 48 = 1,458 F / module
  • C total@100V = 2 x 143.910,00 J  / sqr(100 V) = 28,782 F
  • N modules @100V = 28,782 / 1,458 = 20 modules
  • N @100V = 20 * 48 = 960 caps
  • Total Cap Cost @100V = 1000 x 6,912 USD = 6.912,00 USD

So, here we are at the same place again. 6.912 USD or 4.713 EUR at today’s rate.

When it comes to Energy, it doesn’t really matter how you wire the capacitors, they store the same. The small difference between the two options is due to rounding imprecision. Parallel or series is only a question of how much Voltage and Current you need for your circuit.

The fundamental question is: is 15% ~ 25% increase in autonomy worth $7.000 (€5.000) to you?

I bet most DIY builders will answer NO.

So, another new friend (John) suggested a composite solution: tie an Ultracap bank to a Lithium Phosphate bank and then to a Lead Acid bank… like this:

  1. High energy / Low Power store: The Lead-Acid bank will provide a large and cheap(er) energy store, but will not be able to accept very high (dis)charging power;
  2. High Power / Low Energy store: The Ultracap bank will provide an expensive but very high power energy store, but it’s limited capacity will not handle the larger energy bursts;
  3. Medium Energy / Medium Power store: The Lithium-Phosphate bank will provide a compromise between the two, allowing the energy from the Ultracaps to flow to the Lead-Acid bank only after crossing the Lithium-Phosphate.

It’s much like a large electronic circuit: you have your big power supply feeding a group of circuits, then each bus has it’s own big capacitor, then each consumer device has it’s own small capacitor. Distributed energy system. Divide and conquer. 🙂 This way we reduce the system cost by reducing the amount of High Power components, without lowering the requirements.

As with any composite system, the problem here is the correct dimensioning. Where is the sweet spot? How large must each of the banks be, in order to make the overall solution financially acceptable?

That will be the subject of another study….

——- Addendum ——-

At the request of John, here goes the math for the full 4-wheel-drive system with industrial-grade efficiency (95%).

  • E k@120 = 669.735,00 J
  • E br@120(4×4,effective) = 0,95 x 669.735,00 = 636,3 kJ

The example capacitor implementation would cost:

  • C total@2,1V = 2 x 636.248,25 J  / 2,1^2 V = 288,5 kF
  • N @2,1V = 288.547,96 F / 70 F = 4.123 capacitors!!!
  • Total Cap Cost @2,1V = 4.122 x 6,912 USD = 28.492,5 USD
  • …which constitutes a mere fantasy at any level.

Now, replace the caps with the LiFePo4 batteries that John suggested (3,2V/40Ah/87,99USD):

  • E k@120 = 669.735,00 Ws / 3600 s = 186,04 Wh
  • E br@120(4×4,effective) = 0,95 x 186,04 = 176,74 Wh
  • Q total@3,2V = 176,738 Wh / 3,2V = 55,23 Ah
  • N @3,2V =  55,23 Ah / 40 Ah = 2 batteries (with 27,5% spare charge!)
  • Total battery cost @ 3,2V = 2 * 87,99USD = 175,98 USD
  • …which is 2 orders of magnitude below the cost of the caps.

But these calculations do not account for the charging current limit of the batteries. If we take the average power for a 10 second braking and the continuous charging current limit of these batteries (3C=120A), we get:

  • P br@120(4×4,effective,avg) = 176,738 Wh x 3600 s / 10 s = 63.625,68 W = 64 kW
  • U br@120(4×4,effective,avg) = 63.625,68 W / 120 A = 530,21 V
  • N @530,21V = 530,21 V / 3,2 V = 166 batteries!!!
  • Or
  • I br@120(4×4,effective,avg) = 63.625,68 W / 3,2 V = 19,9 kA
  • N @3,2V = 19.883,025 A / 120 A = 166 batteries all the same.
  • Total LiFePo4 cost = 166 * 87,99 USD = 14.606,34 USD
  • …which just means that the chosen batteries are too big: they support too little power and too much energy, and the usual (H)EV manufacturer approach of many small cells is the only way to support higher power.

It remains to do a recalculation with a smaller LiFePo4 cell size (like A123’s cells that can charge at 4,5C)… but I bet a composite solution with ultra caps and batteries must be devised…

Meanwhile, I’ve come across a new type of Lithium chemistry battery that is (almost) commercially available and promises even better performance than LiFePO4: Lithium-Titanate (LiTO). Toshiba’s “SCiB” 24V/4.2Ah pack apparently delivers the same performance as a capacitor; they state charging currents in the order of 12C, which is far better than the available LiFePO4’s 3~5C.

So, redoing the power math with these SCiBs (24V/4,2Ah/50A):

  • P br@120(4×4,effective,avg) = 64kW
  • U br@120(4×4,effective,avg) = 63.625,68W / 50A = 1,3 kV
  • N @1272,51V = 1.272,51V / 24V= 53 batteries
  • Since the price has not been published yet, I’m going to extrapolate the total cost. I’ll imagine these new batteries will cost 25% more than an equivalent 25,9V/4Ah Lithium-Polymer pack: 1,25 x 200 USD = 250 USD.
  • Total LiTO batt cost = 53 * 250 USD = 13.250 USD
  • And they would pack a nice energy capacity of 53 x 4,2Ah x 24V = 5,3kWh, which would be sufficient for 5.342,4Wh / 176,74 Wh = 30 full-stops/accelerations 0~120 km/h!

Well…13.250 USD in batteries is better than 28.500 USD in caps, right? Especially if the batteries have a life expectancy of 3000 cycles…

The conclusion to this article is that it is trivial to store the Energy required for an acceleration from zero to top speed or resulting from a full-stop braking; however, even when contemplating very modest values of average braking power, it is quite challenging to harness the full regenerative braking Power.

I will post again exclusively on this subject. Please keep your power questions and suggestions in the box until then. 😉