What would happen if I short the terminals of a DC motor together while power is disconnected but it is still freewheeling?

According to several sources, it would brake the motor. This makes sense. But they also mention using an array of power resistors and not just shorting the terminals. What would happen if I just shorted the terminals?

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    \$\begingroup\$ That depends on the power of the motor. If it's powerful enough you could melt something. \$\endgroup\$ – Majenko Jul 11 '11 at 12:31

What they said ... plus / but:

When a short circuit is applies to a DC motor's terminals the rotor and any attached load will be braked rapidly. "Rapidly" is system dependant but as braking power may be somewhat above peak motor design power the braking will usually be significant.

In most cases this is a bearable thing to do if you find the result useful.

Braking power is about I^2R

  • where I = motor initial short circuit braking current (see below) and

  • R = resistance of circuit formed including motor-rotor resistance + wiring + brush resistance if relevant + any external resistance.

Applying a short circuit achieves the maximum motor braking that you can achieve without applying external reverse EMF (which some systems do). Many emergency stop systems use rotor shorting to achieve a "crash stop". The resultant current will probably be limited by core saturation (except in a few special cases where an aircore or very large air gaps are used.) As motors are generally designed to make reasonably efficient use of their magnetic material you will usually find that maximum shorted current due to core saturation is not vastly in excess of maximum rated design operating current. As others have noted, you can get situations where the energy that can be delivered is bad for the motors health but you are unlikely to be dealing with these unless you have a motor from a spare electric locomotive, forklift or generally seriously large piece of equipment.

You can "ease into this" by using the method below. I've specified 1 ohm for current measurement purposes but you can use whatever suits.

As a test try using a say 1 ohm resistor and observe the voltage across it when used as a motor brake. Current = I = V/R or here V/1 so I = V. Power dissipation will be I^R or for 1 ohm peak Wattage with be peak amps squared (or resistor Volts squared for a 1 ohm resistor. eg 10A peak motor current will temporarily produce 100 Watt into 1 ohm. You can often but power resistors of say 250 Watt rating in surplus stores for very modest sums. Even a ceramic bodied 10 Watt wire wound resistor should withstand many times its rated power for a few seconds. These are usually wire wound, but the inductance should be low enough as to not be relevant in this application.

Another excellent source of resistor element is Nichrome or Constantan (= Nickel Copper) or similar wire - either from an electrical distributor or the former from old electric heater elements. Electric heater element wire is typically rated for 10 Amps continuous (when it glows heater-bar-cherry-red). You can place multiple strands in parallel to reduce resistance. This is hard to solder by normal means. There are ways, but easy for "playing" is to clamp lengths in screw down terminal blocks.

A possibility is a light bulb of about correct ratings. Measure its cold resistance and establish its rated current by I = Watts_rated/Vrated. Note that the hot resistance will be several to many times the cold resistance. When a current step (or current die to a voltage step) is applied to a bulb it will initially present its cold resistance which will then increase as it warms up. Depending on the energy available and bulb rating the bulb may glow up to full brightness or may hardly glimmer. eg a 100 Watt 100 VAC incandescent bulb will be rated at 100 Watt/110 VAC ~= 1 Amp. It's hot resistance will be about R = V/I = 110 /1 =~100 Ohms. It's cold resistance will be able to be measured but may be in the say 5 to 30 Ohm range. If initial power into the bulb is say 100 Watt's it will "bright up" rapidly. If power in initially is say 10 Watts it will probably not get above a glimmer. Best analysis of what a bulb is doing would be by two channel data logger of Vbulb and I bulb and subsequent plotting of V & I and summing the VI product as the motor brakes. A carefully handles oscilloscope will give a fair idea and use of two meters and great care may be good enough.

Some SMALL wind turbines use rotor shorting as an overspeed brake when windspeeds get too fast for the rotor. When the motor is not saturated power out rises approximately as V x I or square of wind (or rotor) speed. When the machine magnetically saturates and becomes a near constant current source, power increases approximately linearly with rotor speed or wind speed. BUT as wind energy is proportional to rotor speed cubed, it is evident that there will be a maximum rotor speed beyond which input energy exceeds max available braking effort. If you are going to depend on rotor shorting for over-speed control then you really really really want to start rotor shorted braking well below the input/output crossover speed. Failure to do this may mean that a sudden gust pushes rotor speed above the critical limit and it will then happily run away. Runaway wind turbines in high velocity winds may be fun of sorts to watch if you don't own them and are standing somewhere very safe. If both of these do not apply use lots of safety margin.

Likely braking profile can be determined semi empirically as follows.

  1. This is the hard part :-). Calculate rotor and load stored energy. This is beyond the scope of this answer but is standard text book stuff. Factors include masses and the moment of inertia of rotating parts. The resultant stored energy will have terms in RPM^2 (probably) and some other factors.

  2. spin shorted rotor at various speeds and determine losses at given RPM. This could be done with a Dynamometer but some current measurements and circuit characteristics should suffice. Note that the rotor will heat under braking. This may or may not be significant. Also, a motor that has run for some while may have warm rotor windings prior to braking. These possibilities need to be included.

  3. Do either an analytical solution based on the above (easier) of write an interative program to determine speed/power loss curve. Something like an excel spread sheet will do this easily. Timestep can be altered to observe results.

For maximum safety of playing the motor can be connected to a 1 ohm (say) resistor and spun up using an external drive - eg drill press, battery hand drill (crude speed control) etc. Voltage across the load resistor gives current.


Your motor will be working as a generator - so-called "electrical braking". The circuit will be formed of the motor coil and whatever you connect to it. The current will depend on that circuit resistance.

Since the coil and the other components are connected sequentially the current will be equal in all parts of the circuit. If you short the motor the resistance will depend solely on the coil resistance. This can lead to rather high current which depending on exact motor design and its speed at the point when you start braking can heat the motor up which can lead to the coil burning or melting. Consider railway trains - they have to use massive resistors for electrical braking and those heat up considerably.


If you short the terminals, the kinetic energy will dissipate on motor parts.

  • windings will be heated
  • high current will flow through the brushes and cause arcing
  • in long terms, brushes will decay and create conductive dust on commutator ring
  • the commutator ring will eventually become permanent shorting point causing overcurrent
  • eventually power switches, controlling the motor will be overstressed and fail (for example: transistors)

Btw. Typical normal electronic regenerative break includes few parts like 68 Ohm resistor, power transistor and some voltage dividers and zener.

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    \$\begingroup\$ Your "BTW" lacks context. Can you expand that a bit? \$\endgroup\$ – Kevin Vermeer Jul 11 '11 at 21:28
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    \$\begingroup\$ Regenerative resistors typically used in servos with >100W output and more. DC power source is loaded with 3 phase bridge and regen circuit in parallel. When voltage exceeds nominal voltage of bus (say 55V>48V), or CPU decides to brake, the overvoltage sensor opens power transistor and high current flows through resisitor. For some reason, this area is full of useless patents, so its not easy to google for self-explaining schematics. \$\endgroup\$ – user924 Jul 12 '11 at 13:07

Consider what happens if you apply the full motor voltage when the motor is at rest. The full voltage will appear across the armature resistance which will dissipate maximum power. As the motor torque accelerates the mechanical load, the motor speed, hence the back-emf, rises and the current, hence the power in the armature falls. Eventually, the back-emf is almost equal to the input voltage and the power dissipated by the armature reaches an idle level.

Now consider removing the input voltage and shorting the armature. The full back-emf now appears across the armature which dissipates almost as much as it did when starting. Eventually, the motor torque slows the mechanical load and eventually the motor stops.

So the armature power dissipation follows approximately the same curve against time when starting or stopping. So if your motor can survive having the full motor voltage applied from rest, it can survive having it's armature shorted at full speed.

As sharptooth says, in trains, braking resistors can be used to dump the load power but the full motor voltage is not applied from rest. I'm not an expert on state-of-the-art train design but on old London tube trains, ballast resistors were connected in series with the armature and progressively switched-out as the train gathered speed.


A typical brush motor may be reasonably modeled as an ideal motor in series with a resistor and an inductor. An ideal motor will appear electrically as a zero-resistance voltage supply/clamp (capable of sourcing or sinking power) whose polarity and voltage is a constant multiple of rotational speed. It will produce convert torque to current and vice versa, with the torque being a constant multiple of the current. To figure out braking behavior, simply use the model with a resistor equal to the motor's DC resistance when stalled; the inductance can probably be ignored except when one is trying to rapidly switch the motor current on and off (e.g. with a PWM drive).

Shorting the leads of a motor will cause current to flow equal to the ratio of open-circuit voltage (at its present speed) to resistance. This will cause braking torque roughly equal in magnitude to the torque that would result if that voltage were externally applied to the motor while it was stalled; it will also dissipate the same amount of power in the motor windings as that stall scenario.


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