*In which we work out some numbers on the amount to be gained
by recovering braking energy into the battery.*

The theory of regeneratve braking sounds like a good one: rather than slowing the car down by friction in the brakes, why not slow it down by turning the electric motor into a generator, and charging the batteries for free?

There are two scenarios where we use the brakes when driving a car: either because we want to slow down or stop; or when going down a hill to keep the speed below a certain limit. Each of these scenarios is examined in turn for an example car, based loosely on the Fiesta project.

The example car has the following characteristics for drag on a level surface:

Speed | Air Resistance | Total drag | Power | Miles per kWh |
---|---|---|---|---|

5mph | 2N | 127N | 282W | 18 |

15mph | 19N | 144N | 958W | 15 |

25mph | 52N | 177N | 1966W | 13 |

35mph | 102N | 227N | 3526W | 10 |

45mph | 168N | 293N | 5860W | 7.7 |

55mph | 251N | 376N | 9190W | 6.0 |

65mph | 350N | 476N | 13736W | 4.7 |

75mph | 467N | 592N | 19719W | 3.8 |

It has a maximum speed of 36m/s (just over 80mph) and weighs 800kg. It has a battery storing 20kWh, which gives a range of about 75miles at 75mph.

The amount of kinetic energy in the car is the maximum that can
be recovered by regenerative braking. The kinetic energy **E**
of a car (in joules), weighing **M**kg and travelling at
**v**m/s is well known to any student of physics; it is given
by:-

E = M v^{2}/2

For our example car at maximum speed, that gives a figure of 518,400 joules, which sounds like a lot. But the unit of energy we're using for the battery is the kWh, and one kWh is 3,600,000 joules. So the maximum energy recovered by braking from 80mph to a dead stop is about 0.15 of a kWh, or a little less than 2/3% of the total battery charge. At 75mph, it would give an extra half mile of range - assuming the recovery was perfect.

However, the emergency braking distance at 80mph is 400feet
(according to the figures from the UK police driving manual,
*Roadcraft*) so if we are slowing down at a more leisurely
pace, we'll cover a good fraction of that 2/3 mile during the
braking. Which in turn means that a lot of the braking will
actually be drag, not regeneration.

When descending, potential energy is converted into kinetic energy. If we take our car a mile up in an aeroplane and heave it over the back, the energy released on impact (neglecting air resistance) is given by the height multiplied by the weight multiplied by the acceleration of gravity. For our example car, this is 12,544,000 joules; or 3.5 kWh for a mile of straight descent. So to gain a kWh of PE we need to descend about 0.3 of a mile, or about 500 metres.

This power output for a mile of descent can be applied to the drag figures to give a speed of descent if we drop it - a terminal velocity. It's measured in hundreds of miles per hour.

Because of how potential energy works, it doesn't matter if the car is heaved out of an aeroplane or descends a really twisty road: the energy released for a mile of vertical descent is the same. However, the gradient of a road is defined as the amount of vertical descent for a horizontal distance: which means we can work out how much energy is released for a horizontal distance, and so estimate what speed the car would reach if we let it roll down the slope unchecked.

This may be used to create a table of the gradients that will maintain these speeds without additional power:-

Speed | Air Resistance | Total drag | Power | Miles per kWh | Coasting gradient |
---|---|---|---|---|---|

5mph | 2N | 127N | 282W | 18 | 1/60 |

15mph | 19N | 144N | 958W | 15 | 1/50 |

25mph | 52N | 177N | 1966W | 13 | 1/43 |

35mph | 102N | 227N | 3526W | 10 | 1/33 |

45mph | 168N | 293N | 5860W | 7.7 | 1/25 |

55mph | 251N | 376N | 9190W | 6.0 | 1/20 |

65mph | 350N | 476N | 13736W | 4.7 | 1/16 |

75mph | 467N | 592N | 19719W | 3.8 | 1/13 |

So braking is clearly necessary for quite a moderate slope. Which means that a regenerative braking system will recover power. But the question is, how much? Well, to recover 10% of the battery's energy, 2kWh, the journey has to climb about 900m, or about 3,000 feet. However, this climb can be interspersed with descents: so we recover a theoretical 10% of the range if we do one climb over a 3,000 foot hill, or over five 600 foot hills, or ten 300 foot hills, and so on.

The other question worth bearing in mind is how much energy the system is expected to recover. Tanking down a 1 in 8 slope at 75mph (33.3m/s), the energy released from PE is about 33kW. The energy used by drag is only 20kW - so the regenerative braking must absorb 13kW. For our 20kWh battery, (and assuming 100% efficiency) that's a current of about 0.65C - quite a fast charging rate.

Driving more cautiously at 45mph (20m/s) down a 1/4 slope, the energy released is about 31kW. Now the energy used by drag is about 6kW, so the extra 25kW needs to be transferred into the battery. This is a current at 100% efficiency of 1.2C, which is possibly too rapid for some battery technologies, and may impact battery cycle life.

Driving recklessly at 75mph down a 1/4 slope, the energy released is about 52kW. Taking off our 20kW for drag, we need to convert 32kW, which for our 20kWh battery is a charging current of 1.6C at 100% efficiency. That's a lot for a charging system to manage, and is likely to have an impact on cycle life of the batteries.

Certainly no fully charged battery can absorb charging currents of 1C without damage. The energy has to go somewhere. So if regenerative braking is to be used, there needs to be a way to shunt the energy away, to create the same retardation but without putting the energy into the battery. A large shunt resistor will be required.

In flat country, regenerative braking is not worthwhile. Stop-start traffic at motorway speeds would only give a few percent of improvement in range.

In moderately hilly country, regenerative braking could give up to 10% extra range. In very mountainous regions, the amount recovered could be considerably more.

Any regenerative braking system must include a resistor to dump the energy into, so that battery damage is prevented. The worst case is where the charging point is at the top of a hill - the result would certainly be battery damage, and possibly a fire.

Electrically, the purpose of a regenerative braking system is to draw a current from the motors, to create a retarding force, and to transfer this current to the batteries. The way this is generally done is with an inductor in series with the motor (this may be the field windings) which is shorted out to allow the current to build up. When the current has reached the appropriate level the short is replaced with the battery pack, which dumps the current into the battery and allows it to fall. Then when the current has fallen a bit the short is returned to build it up ... and so on.

The main limit to regenerative charging is the battery capacity.
Batteries have a *bulk charge* regime, at which a maximum
current can be applied. Damage may result if this current is
exceeded even for a short while, as it may be related to current
density on the electrodes, not heating - so this current must be the
maximum that the regenerative braking can use.

However, if the motor voltage is less than the battery voltage (quite likely) then the batteries will not be in circuit for a lot of the time, but the short will be applied instead. So the average current put into the batteries is normally going to be less than the maximum current for bulk charging. For this reason the maximum power recovery only occurs at high speed, when the motor voltage is at its highest.

The other problem with bulk charging is that the batteries are only allowed to be bulk-charged if certain conditions are met. These conditions will always include how much the battery is charged, and will often include other considerations, like temperature. For these reasons (imagine you live on the top of a hill) there needs to be a method of dumping the power recovered by regenerative braking. This method is a ballast resistor.

The ballast resistor is applied instead of the battery (or applied in parallel with the battery, with a diode to prevent the battery discharging into the ballast resistor) and so must drop less voltage than the maximum for bulk charging when the maximum current is applied. So for a battery system with a 180v maximum bulk-charge and a maximum regeneration current of 55A, the ballast resistor should be 180V/55A or 3.27Ω.

The power that the ballast resistor needs to absorb is the maximum current at the maximum voltage that the motor will deliver. So in the above example, if the motor delivers a maximum 120v, the ballast resistor should be rated for 120v x 55A, or 6.6kW. This has to be a continuous rating, as a long slope at full speed may dump heat into the ballast resistor for quite a while.

There should be a separate ballast resistor for each motor.

This page is part of an Open Source Electric
Car Project, and is written and maintained by Simon. At this stage
these pages are *constantly* under revision. Thoughts and
comments are welcome.