Braking of DC Motors

A load can always be stopped by a friction brake. Electric braking is preferred wherever stopping is frequent, because of what friction braking costs.

The three methods differ only in where that electrical power goes:

1. The armature is disconnected from the supply and connected across a braking resistor \(R_B\).
2. The field remains excited from the supply, so the flux is unchanged.
3. The armature is still turning, so it still generates \(E = K_a\Phi\omega\) — but this EMF now drives current through \(R_B\) instead of opposing the supply.
4. The current has reversed relative to motoring, so the torque opposes rotation and the machine decelerates.

A series motor cannot simply be disconnected and thrown across a resistor. Its field carries the armature current, so reversing the armature connection would reverse the field too — and \(T = K_a\Phi I_a\) with both reversed is unchanged in sign.

In traction practice the difficulty is sidestepped by connecting the machines so that each armature excites the other's field, or by switching to separate excitation from an auxiliary supply.

Also called counter-current or reverse-current braking. The supply connections to the armature are reversed while the machine is still running.

One further point of practice: at standstill the torque is still substantial, so the supply must be disconnected at zero speed or the motor will simply accelerate in the reverse direction. A zero-speed switch or a centrifugal contact does this automatically.

The armature stays connected to the supply in its normal sense. Braking occurs when the back EMF exceeds the supply voltage, so that current flows out of the machine.

Two situations bring this about naturally:

• An overhauling load — a lift descending, a train on a down gradient — drives the machine above its no-load speed.

• The flux is increased or the supply voltage reduced, so that the existing speed becomes higher than the new no-load speed. In a Ward-Leonard set (Chapter 37) this is done simply by turning the generator field rheostat down.

For rheostatic braking the equation of motion can be solved exactly, and the result explains the exponential coast noted in Section 38-3.

Note what the time constant depends on. A smaller braking resistor gives a shorter \(\tau\) and a faster stop, but a larger initial current — the same trade as everywhere else in this chapter. Doubling \(K_a\Phi\) would quarter the time constant, which is why braking is always done at full field.

• Electric braking has no wearing parts and predictable effort, but cannot hold a load at rest — a mechanical brake is still needed.

• All methods work by making the machine a generator: the armature current is reversed while the flux is kept, so \(T = K_a\Phi I_a\) opposes the rotation.

• Rheostatic: armature across \(R_B\), field still excited. \(I_B = E/(R_a + R_B)\). The effort fades to zero at standstill.

• A series motor must have its field reversed relative to its armature, and will only brake if the total resistance is below the critical value for self-excitation.

• Plugging: supply reversed, so \(I_B = (V + E)/(R_a + R_B)\). Needs about twice the resistance, dissipates about three times the kinetic energy, and retains roughly half its braking current at standstill.

• Regenerative: requires \(E \gt V\), so the machine must be driven above its no-load speed. Returns \(VI_a\) to the supply; cannot bring a load to rest.

• Under rheostatic braking the speed decays exponentially with \(\tau = J(R_a + R_B)/(K_a\Phi)^{2}\).

• The kinetic energy divides between \(R_a\) and \(R_B\) in proportion to their resistances.

• Switching off the field with the armature. Without flux there is no EMF and no braking whatever.

• Reversing both field and armature. \(T = K_a\Phi I_a\) is then unchanged in sign — reverse one only.

• Using \(V\) instead of \(E\) for the rheostatic braking current. The supply is disconnected; only the back EMF drives current.

• Using \(E\) alone for plugging. The driving voltage is \(V + E\), nearly twice as much.

• Forgetting to disconnect at zero speed when plugging. The motor will accelerate in reverse.

• Expecting regeneration to stop a load. It fails at the no-load speed and below that the machine motors again.

• Assuming any supply will accept regenerated power. A simple rectifier will not.

• Thinking a larger braking resistor brakes harder. It gives less current and a longer time constant.

• Braking at reduced field. \(\tau \propto 1/(K_a\Phi)^{2}\), so weak field means a very slow stop.

• Expecting electric braking to hold a load stationary. All three methods need rotation to work.

Find \(E\) at the moment of braking first. Then ask which voltage drives the braking current — \(E\), \(V + E\) or \(E - V\).

1. P38.1 A 220 V shunt motor with \(R_a = 0.20~\Omega\) takes 50 A at 800 rev/min. Find \(R_B\) for rheostatic braking at 75 A initial current.

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   \[E = 220 - (50)(0.20) = 210~\mathrm{V}\]

   \[R_a + R_B = \frac{210}{75} = 2.800~\Omega \quad\Longrightarrow\quad R_B = 2.600~\Omega\]

2. P38.2 For P38.1, find \(R_B\) if the same 75 A is wanted by plugging instead.

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   \[V + E = 220 + 210 = 430~\mathrm{V}\]

   \[R_a + R_B = \frac{430}{75} = 5.733~\Omega \quad\Longrightarrow\quad R_B = 5.533~\Omega\]

   That is 2.13 times the rheostatic value.

3. P38.3 For P38.2, find the braking current at standstill and as a fraction of the initial value.

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   \[I = \frac{220}{5.733} = 38.4~\mathrm{A}, \qquad \frac{38.4}{75} = 51\,\%\]

   The same 51 % as Example 38.3, because the ratio is \(V/(V+E)\) in both cases.

4. P38.4 A 250 V motor with \(R_a = 0.30~\Omega\) has a no-load speed of 1000 rev/min. It is driven at 1150 rev/min by an overhauling load. Find the regenerated current and the power returned.

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   \[E = 250\left(\frac{1150}{1000}\right) = 287.5~\mathrm{V}\]

   \[I_a = \frac{287.5 - 250}{0.30} = \frac{37.5}{0.30} = 125~\mathrm{A}\]

   \[P = (250)(125) = 31.25~\mathrm{kW}\]

5. P38.5 A machine with \(K_a\Phi = 2.50\) and \(J = 20\) kg·m² is braked rheostatically with a total circuit resistance of 4.0 \(\Omega\). Find the time constant and the time to fall to 10 % of speed.

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   \[\tau = \frac{(20)(4.0)}{(2.50)^{2}} = \frac{80}{6.25} = 12.8~\mathrm{s}\]

   \[t = \tau\ln 10 = (12.8)(2.303) = 29.5~\mathrm{s}\]

6. P38.6 For P38.5, the initial speed is 1200 rev/min. Find the stored energy and the initial braking torque.

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   \[\omega_0 = \frac{2\pi(1200)}{60} = 125.66~\mathrm{rad/s}\]

   \[\tfrac{1}{2}J\omega_0^{2} = \tfrac{1}{2}(20)(125.66)^{2} = 157\,900~\mathrm{J} = 157.9~\mathrm{kJ}\]

   \[E = K_a\Phi\omega_0 = (2.50)(125.66) = 314.2~\mathrm{V}, \quad I_B = \frac{314.2}{4.0} = 78.5~\mathrm{A}\]

   \[T_B = (2.50)(78.5) = 196.3~\mathrm{N\,m}\]

7. P38.7 Why must the field remain excited during rheostatic braking?

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   Because the braking EMF is \(E = K_a\Phi\omega\) and the braking torque is \(T_B = K_a\Phi I_B\). Both are proportional to the flux, and the torque is proportional to \(\Phi^{2}\) once \(I_B = E/(R_a + R_B)\) is substituted.

   With the field switched off there is no EMF, no current and no braking torque at all — the machine simply coasts.

   The consequence is a safety matter, not merely a performance one. A hoist whose field is disconnected along with its armature has no restraint on a descending load.

8. P38.8 Why does plugging dissipate roughly three times the kinetic energy of the load?

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   Because the supply continues to deliver power throughout the braking, and that power goes into the braking resistor along with the kinetic energy.

   The kinetic energy \(\tfrac{1}{2}J\omega_0^{2}\) is returned by the decelerating load. On top of it the supply delivers \(VI\) for the whole braking period — and since \(I\) stays high right down to standstill, this contribution is roughly twice the kinetic energy.

   Rheostatic braking dissipates only the kinetic energy; regeneration dissipates almost none. Plugging is the most wasteful of the three, which is the price of being the fastest.

9. P38.9 A lift descends with a heavy load. Which braking method restrains it, and what happens as it approaches the floor?

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   Regenerative braking restrains the descent. The overhauling load drives the machine above its no-load speed, so \(E \gt V\) and current flows back into the supply — the potential energy of the descending load is recovered rather than wasted.

   As it approaches the floor the speed must be reduced below the no-load value, at which point regeneration ceases and the machine would motor again. Control passes to rheostatic braking for the final slowing, and then to a mechanical brake to hold the car level at the floor.

   All three are used in sequence, each in the region where it works.

10. P38.10 A machine tool must reverse rapidly many times a minute. Which method, and what is the penalty?

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    Plugging, because it is the only method that maintains substantial braking torque all the way to standstill — and because the supply reversal that brakes the motor is also what accelerates it the other way, so braking and reversing are a single operation.

    The penalty is energy. Each reversal dissipates about three times the kinetic energy of the drive, all of it in the braking resistor. At many reversals a minute the resistor's average dissipation becomes the limiting factor, and it must be generously rated and ventilated.

    A zero-speed switch is essential if the motor is to be stopped rather than reversed, since at standstill the plugging current still produces about half the initial braking torque.