I'm new to electrical engineering, and probably asking the wrong question. Not sure how else to phrase it.

For the question, I'm going to use an example layout of simply a 9v battery attached to a toy DC motor.

From my understanding of my previous question, the voltage coming from the battery through the motor will be around 9v. The current will be from 0amps to however much the battery can supply without frying. What decides how much current goes through the motor?

  • 1
    \$\begingroup\$ The current will decrease with speed; the more force there is trying to slow the motor down, the slower it will go and the more current it will draw. \$\endgroup\$ Commented Nov 22, 2016 at 22:31

3 Answers 3


Some non-math beginner concepts:

All motors are generators. Suppose your permanent-magnet-type DC motor generates 9V when the shaft is spun at, say, 100Hz or 6000RPM.

This means that, when the motor is connected to a 9V battery, it will speed up to 6000RPM, then remain at that speed. When it's up to 6000RPM, it draws zero current (ideally.) That's assuming that bearing- and brush-friction is insignificant, and the rotor is in vacuum (so, no air is heated, with no wind blowing out of the holes in the case.) At max speed the current falls to zero because the motor(generator) is supplying 9V, and the battery is also supplying 9V of the same polarity. The two opposing voltages subtract to zero. The rotor keeps spinning at a constant RPM. And, other than friction effects, it doesn't need any energy from the power supply. For a real motor, the entire supply current is caused by unwanted friction (and also of course by any mechanical-load wattage being drawn from objects connected to the spinning shaft.)

In other words, it's not easy to calculate the current a real motor will draw from its power supply. The current depends on internal friction of bearings and brushes and wind/fan effects, as well as depending on the amount of work (wattage) the motor shaft is performing.

With DC motors, the RPM of the shaft is proportional to the drive voltage and the generated voltage. If a motor runs at 6000RPM when connected to 9V, it will run at 1/3 the speed if powered at 1/3 the volts (so, 2000RPM for a 3V power supply.) There really isn't any such thing as a 9V motor or a 24V motor, etc. Those are just the operating voltages for a minimum bearing-wear and maximum brushes-lifetime. Ideally, any DC motor can be run at much lower voltage and RPM than its spec-sheet voltage. And, any DC motor can be connected as a generator, and used as an RPM-sensor: the output voltage is proportional to rotor speed.

The rotor of any motor has mass, and can't speed up instantly. So, whenever you connect your DC motor to a 9V battery, it won't draw infinite current, and won't instantly jump to 6000RPM speed. It will however accelerate, and momentarily draw a large current while doing so.

All motors are flywheels. When first connected to a DC power supply, the motor speed increases, and kinetic energy is stored in its rotor. When hooked up and running with constant RPM, we can disconnect the supply, and the RPM will stay the same, and not instantly stop. (The rotor is an energy-storage device.) Then, depending on friction and work-wattage, the motor RPM will smoothly decrease, just like any flywheel which is performing work and losing its kinetic energy.

If we want to rapidly slow down a DC motor, we can disconnect it from its power supply, then connect it to a resistor. The motor acts as a generator (as it always does,) and drives a current in the resistor. The motor slows down, and the "braking resistor" gets hot.

In all, DC motors resemble capacitors, where the "charge" is the kinetic energy of the rotor-flywheel. DC motors draw a big current when first connected to a DC voltage. Then, the current decreases to nearly zero. Disconnect a spinning DC motor, and the power-supply voltage still remains on its terminals. What if we short those terminals? POW! A huge current appears briefly, and the rotor jerks to a sudden halt. We've discharged the stored mechanical energy, and the conductors end up at higher temperature.

  • \$\begingroup\$ Great explanation! Thanks for clearing up what the voltage thing means. \$\endgroup\$
    – pepperjack
    Commented Nov 23, 2016 at 22:03

Basically, the load on the motor determines the current. There are two main things to keep in mind:

1) The motor when turning generates a speed-proportional voltage, referred to as the back-emf, that opposes the applied battery voltage: E = Ke * speed, where Ke is the back-emf constant.

2) The torque produced by the motor is proportional to the current: T = Kt * I, where Kt is the torque constant. (Note: if SI units are used throughout, Ke and Kt are identical.)

The motor windings have resistance, so the current flowing causes a voltage drop. The motor will run at a speed where the back-emf plus the voltage drop due to the current in the motor windings equals the battery voltage, and where the motor torque equals the applied load.

Without any external load, the motor torque only has to counter rotational losses, such as bearing friction. So the current is small, and the back-emf is close to the battery voltage.

If a load is applied to the motor shaft, the motor will slow a little and the back-emf will decrease proportionally. This increased discrepancy between the battery volts and the back-emf forces more current in the windings, consequently the shaft torque increases. The speed drops and the current increases until the motor reaches equilibrium at a new operating point.

A plot of speed vs torque would give a sloping line; the slope of the line is indicative of the winding resistance

  • \$\begingroup\$ What units are on Kt and I? Also, what is speed-proportional voltage? Does that mean that the voltage-in can vary? \$\endgroup\$
    – pepperjack
    Commented Nov 23, 2016 at 0:27
  • \$\begingroup\$ Kt : Nm/A, I : A. If you vary the input voltage, the speed will change to match. (And if the torque load depends on speed, then the current will vary to the new torque). \$\endgroup\$
    – user16324
    Commented Nov 23, 2016 at 9:43
  • \$\begingroup\$ Current (I) is always measured in Amperes, Amps for short. Kt has units of torque per current, and torque has SI units Newton-meters, but other torque units may used. Here we use ounce-inches a lot, because, well, I'm not sure why. Tradition, I suppose. And yes, you can vary the input voltage, this is a way to adjust the speed of the motor; the only way if the motor has a permanent magnet field. For example, if the motor runs at 3000 rpm unloaded at 9V supply, it will run close to 2000 rpm at 6V supply. ('Close to' because a practical motor is not perfectly linear.) \$\endgroup\$
    – user28910
    Commented Nov 23, 2016 at 15:16

The major parameters are;

  • DC resistance of motor coil relative to the battery internal resistance
  • Battery voltage vs motor rated voltage
  • Motor load vs RPM
  • rate of acceleration
  • inertia determines how long it takes to accelerate or brake (short cct)

    • Stall or start current is "mucho" greater than max rated current
  • \$I_{start} = V_{IN}/DCR\$ for DC resistance of motor = 8x to 10x Imax

Rules of Thumb for efficient motors

  • Full Speed No Load \$I = 10 \text{% of Imax} \$ of rated load typ.
    • where kV/RPM depends on number of poles etc
  • Accelerating rate or rotational load , I drops with RPM due to back EMF from generator effect increasing voltage, rotation = Hz/V or kV/RPM (kV/(Hz*60)

    • thus current declines as well as Torque capability with rising RPM until equilibrium where Motor Torque capacity = load Torque
    • Imin = 1% of Start "surge" current at full voltage
    • thus current only rises slowly with steady {Volts and RPM}, due to losses
    • surge currents drop to 8x in large motors and <=5x in soft start motors
  • Imax only occurs at max load torque which depends on load profile vs speed

    • e.g. friction vs prop have different profiles
  • Thus current with no motor load goes from (8x to) 10x Imax to 10% Imax


  • usually 80%-85% of no load speed in a matched system
  • battery ESR should be <10% of DCR of motor coil to drive without motor load dropping battery voltage significantly where R ratio is inverse to Vbat ratio.
  • \$\begingroup\$ some users vote -1 need help in overcoming inability to communicate constructively \$\endgroup\$ Commented Nov 23, 2016 at 15:10

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