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How does the discharge rate of the battery pack in an electric vehicle vary from 1C to 4C for example when the vehicle accelerates? It's the same electric motor and the same battery. How does the electric motor vary it's load on the battery?

guywithaquestion
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4 Answers4

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A motor can be modeled as a resistor, inductor, and voltage source in series. Your power source is probably a (probably variable) voltage source:

schematic

simulate this circuit – Schematic created using CircuitLab

The current through \$L_m\$ is what makes the motor go. The magnetic field in this inductance is what creates the torque in the motor. More current = stronger field = more torque.

\$V_m\$ is the voltage created by the rotor moving through the stator's magnetic field, by Faraday's law of induction. The faster the motor is spinning, the higher \$V_m\$ will be.

\$R_m\$ is the resistance of the windings. It is usually small, and desirably so, since it represents a loss that does nothing but make heat.

Say the vehicle is already moving fast. Since the current in the motor is constant, the voltage across \$L_m\$ is \$0V\$. We can consider it to be a short. Thus by Kirchoff's voltage law:

$$ V_{bat} - V_{Rm} - V_m = 0 $$

\$V_{Rm}\$ will be small, because there's not much current, because there's not much torque (just enough to overcome friction). So \$V_m\$ must be very close to \$V_{bat}\$, which makes sense, if the vehicle is already moving fast, and \$V_{bat}\$ is proportional to speed. This is why not much current is drawn from the battery when the vehicle is at a constant speed.

But, what if the vehicle isn't moving, but we apply the same voltage to the motor?

In this case, \$V_m\$ is zero, because the motor isn't turning. Initially, the current will be \$0A\$, but soon, it will rise to some maximum value limited only by \$R_m\$. At this point the current (and torque) are at the maximum. And let's say the rotor can't turn; it's locked in place. This is the stall current of the motor, when the full supply voltage is dropped over \$R_m\$, the current isn't changing (\$V_{Lm} = 0V\$) and the rotor isn't turning (\$V_m = 0V\$). Essentially, the battery is shorted by just a small resistor, \$R_m\$.

If the rotor isn't locked, that huge torque from the huge current will cause the motor to accelerate. (Torque is a rotating force, and force is the product of mass and acceleration: \$F=ma\$, but also \$a=F/m\$. If the torque does not exactly balance friction, there will be a force, and there must be acceleration). \$V_m\$ will become greater, thus reducing the voltage seen by \$L_m\$ and \$R_m\$, thus decreasing the current, until the vehicle has accelerated so much that the torque is just enough to balance friction (there is no net force), and the acceleration stops.

Here's an interesting consequence: If \$R_m = 0\Omega\$, and also \$V_{bat}\$ is an ideal voltage source, and also the wires are ideal, then the speed regulation of the motor is perfect. That is, if you put a brake on it, the motor's torque (and, the current drawn from \$V_{bat}\$) will increase exactly enough to maintain the same speed. Or, if you increase \$V_{bat}\$, this will result in an infinite torque, which will accelerate your vehicle instantly to the new speed at which \$V_m = V_{bat}\$. This is because with no resistance to otherwise limit the current, any difference between \$V_m\$ and \$V_{bat}\$ results in an infinite current (at least eventually; \$\frac{di}{dt}\$ is still limited by \$L_m\$).

Phil Frost
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  • thank you very much Phil. I don't want to seem unappreciative. this is the best answer i have seen to my "high school level" questions as some have described them. it will take me a few days to digest it and integrate with my current knowledge. – guywithaquestion Jul 02 '13 at 11:26
  • OK. Going through your great response, paragraph by paragraph: I understand that V(Lm) is 0 because the current is constant across Lm. (I have trouble distinguishing between Vm – based on Faraday’s law of induction and voltage across Lm, are they not the same?) I don’t fully understand why Vbat is proportional to speed. It seems to me like speed should be proportional to torque, which is proportional to current. – guywithaquestion Jul 02 '13 at 12:08
  • I understand that when motor isn’t turning (locked), V(Lm) = 0 because current still flows unchanged through the windings, but the motor is physically locked from turning so Vm = 0 and V(Rm) must be close to V(bat) (is this true?! Based on Kirchoff’s law). – guywithaquestion Jul 02 '13 at 12:09
  • Now we unlock the motor, Vm starts increasing, which based on Kirchoff’s law must mean that V(Lm) and V(Rm) decreases if the same battery powers the motor. I don’t understand the difference between Vm and V(Lm) like I said earlier, and I don’t understand how it is possible to reduce V(Rm) since you always have (the same?) current flowing through the same windings as long as the circuit is ON. And to top it off I don’t understand why unlocking the motor and its acceleration reduces the current.

    Also, how about the case when you accelerate from speed a to speed b.

     – guywithaquestion Jul 02 '13 at 12:09
  • @guywithaquestion torque is not proportional to speed. Put the vehicle against a brick wall and open the throttle all the way. You will push on the wall (high torque) but you won't go anywhere (low speed). If torque is in excess of drag forces, the vehicle will accelerate. If drag forces are in excess of torque, the vehicle will decelerate. Only when torque = drag is the speed constant, and only in this equilibrium condition is torque proportional to speed, to the extent that you have drag that increases with speed (like aerodynamic drag, but not many kinds of friction). – Phil Frost Jul 02 '13 at 12:28
  • @guywithaquestion as the speed of the motor increases, $V_m$ increases, opposing the voltage $V_m$, leaving less of that voltage to appear across $R_m$. By Ohm's law, less voltage across a resistor means less current, and since this is a series circuit, the current anywhere in this circuit is the same. So, the faster the motor turns, the more $V_m$ opposes $V_{bat}$, the less voltage across $R_m$, thus the less current in $R_m$, thus the less current drawn from the battery. – Phil Frost Jul 02 '13 at 12:39
  • wow, this is so counter intuitive to me, but i think i understand how Ohm's law and the fact that this is a series circuit reduces the current when Vm increases (which is inducted by motor). thank you very much – guywithaquestion Jul 02 '13 at 14:00
  • so basically what this means is that the rate of discharge of the battery (given a set motor) is governed by the magnetic field generated by the rotation of the motor it powers and more specifically by the Vm this magnetic field induces. Am i right? – guywithaquestion Jul 02 '13 at 14:01
  • @guywithaquestion The magnetic field is generated by the current through the inductor (it's an electromagnet); the motion is not relevant to the generation of the field (for common DC motors). The motion through this field is what makes $V_m$. – Phil Frost Jul 02 '13 at 14:14
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A vehicle moving along at a constant speed needs a certain amount of power to overcome losses such as friction. On an electric vehicle this would come from the battery in the form of current and voltage.

When you want to accelerate the vehicle you need more power to get it to move at greater speed and hence the consumption of current (coulombs per second) goes up as the voltage is fixed.

When you reach the higher speed you have to maintain it. This uses less power than acceleration but more than the lower speed consumption.

Its all down to Mr. Newton's third law - Force = mass x acceleration

JIm Dearden
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    Taking it a step further: mechanical power = force x velocity. Electrical power = electromotive force x current. By the law of conservation of energy, power at the motor equals power at the battery, plus (when accelerating) or minus (when braking regeneratively) losses. – Phil Frost Jul 01 '13 at 15:55
  • @PhilFrost very true +1 from me – JIm Dearden Jul 01 '13 at 15:59
  • Also, resistive losses in the electrical system are proportional to the square of current: $P=I^2 R$. A faster vehicle must not only overcome higher aerodynamic drag, but the electrical system for moving energy from the battery to the motor becomes less efficient. – Phil Frost Jul 01 '13 at 16:00
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Two words for you mate- MOTOR CONTROLLER

I will try to be as crude as possible.

Through the motor controller attached between your battery and motor you attach a throttle pedal. Now the throttle is nothing but a potentiometer. If you remember a potentiometer varies the resistance offered and thus the current changes. So if you press down your throttle pedal you increase current flow, and hence power supplied to the motor and hence speed. So the battery discharge rate depends on how hard you press your pedal.

Ayush
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What you are missing is that there is a power controller between the battery and the motor. This power controller decides how much current to allow to flow thru the motor. The battery is not directly connected to the motor. That would either overdrive the motor or drive it a maximum power.

To keep things efficient, the power controller actually applies lots of short full-power pulses to the motor. These are fast enough that the motor only feels the average. In effect, the power controller is a switch that is either full on (battery directly connected to the motor) or full off. This way it doesn't dissipate power on its own. By varying the fraction of time that the motor is on, it varies the effective power level.

Olin Lathrop
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  • ok so now i am a bit confused. based on the formulas coming from Phil Frost it seems the motor cannot overdrive because as Vm increases, V(Rm) decreases, which reduced the current in the circuit. What am I missing here? – guywithaquestion Jul 02 '13 at 14:08
  • @guywithaquestion for $V_m$ to increase, the motor must spin faster. It can only do this if the vehicle is light enough and there is little enough friction that the motor can accelerate it. If the motor can't accelerate, $V_m$ can't increase, and you draw a ton of current from the battery, possibly damaging (by overheating, probably) the battery or the motor. – Phil Frost Jul 02 '13 at 14:16