Starting of DC Motors and Starters

Chapter 34 established the relation that governs everything here:

In normal running \(E_b\) absorbs almost the whole supply voltage, leaving only a few volts to drive the current. At the instant of switching on, the armature is stationary, so \(E_b = 0\) and the full supply appears across the armature resistance alone.

Only very small machines — a fraction of a kilowatt, where \(R_a\) is relatively large — may be switched directly on. Everything else needs a starter.

The remedy is to insert resistance in series with the armature and remove it as the motor gains speed.

The limiting value \(I_{\max}\) is chosen at 1.5 to 2 times the full-load current — enough to give a useful accelerating torque, low enough that the winding and commutator survive the few seconds involved.

The starting resistance cannot simply be removed all at once, nor left in until the motor reaches full speed.

1. At the first stud the full resistance limits the current to \(I_{\max}\).
2. The motor accelerates, so \(E_b\) rises and the current falls.
3. When the current has fallen to a chosen lower value \(I_{\min}\), a section of resistance is cut out, and the current jumps back up to \(I_{\max}\).
4. The process repeats until all the resistance is out and the motor runs on \(R_a\) alone.

The current therefore oscillates between two limits, and the accelerating torque with it. More steps means a narrower band, smoother acceleration and a more expensive starter — the design trade-off of Section 36-4.

Let \(R_1, R_2, \ldots\) be the total circuit resistance at successive studs, so that \(R_1\) includes all the sections and \(R_{n+1} = R_a\).

\(n\) rarely comes out a whole number. It is rounded up, and then \(k\) is recomputed from \(k = (R_a/R_1)^{1/n}\) so that the progression lands exactly on \(R_a\). Rounding up raises \(k\), which narrows the current band — the design becomes gentler than specified, never harsher.

The commonest starter for shunt and compound motors. Its three terminals are marked L (line), F (field) and A (armature).

1. The handle is moved from the off position to the first stud, putting the whole starting resistance in series with the armature.
2. The shunt field is connected across the supply through the no-volt coil, so it receives full excitation from the outset — giving maximum torque per ampere.
3. The handle is moved slowly across the studs as the motor accelerates, cutting out one section at a time.
4. At the last stud the handle is held by the no-volt coil against the pull of a spring, and the starting resistance is entirely out of the armature circuit.

A series motor has no separate field circuit, so there is nothing to connect a third terminal to. Its starter has only two terminals and the holding coil carries the full line current.

The same coil still serves as a no-volt release on supply failure, since a lost supply also means lost current. One coil, two protections — but no overload protection from it, so a separate device is fitted.

• At rest \(E_b = 0\), so a motor switched directly on would draw \(V/R_a\) — 20 to 30 times rated, damaging the winding, the commutation, the coupling and the supply.

• A starter inserts \(R_{st} = V/I_{\max} - R_a\) and removes it as \(E_b\) builds up. \(I_{\max}\) is chosen at 1.5 to 2 times full load.

• The resistance is removed in steps, keeping the current between \(I_{\max}\) and \(I_{\min}\).

• The total resistances form a geometric progression of ratio \(k = I_{\min}/I_{\max}\), with \(R_1 = V/I_{\max}\) and \(R_{n+1} = R_a\).

• \(n = \ln(R_aI_{\max}/V)/\ln k\), always rounded up, after which \(k\) is recomputed as \((R_a/R_1)^{1/n}\).

• Sections are \(r_m = R_m(1 - k)\) and are graded, the first being the largest.

• The three-point starter has its no-volt coil in series with the shunt field, so it protects against supply failure and field failure. Its defect is that field-rheostat resistance also weakens the holding coil.

• The four-point starter gives the coil its own circuit, restoring the full speed range but losing field-failure protection.

• The overload release short-circuits the no-volt coil rather than breaking the main circuit directly.

• A two-point starter serves a series motor; its coil carries the line current and so doubles as a no-load release.

• Forgetting to subtract \(R_a\). The starter resistance is \(V/I_{\max} - R_a\), not \(V/I_{\max}\).

• Rounding \(n\) down. Round up; the last stud must land on \(R_a\), and rounding down widens the current swing.

• Forgetting to recompute \(k\) after rounding. The progression will otherwise miss \(R_a\).

• Making the sections equal. They form a geometric progression; the first is the largest by a wide margin.

• Confusing \(R_m\) with \(r_m\). \(R_m\) is the total circuit resistance at that stud, including \(R_a\); \(r_m\) is one section.

• Starting with the field rheostat at maximum. Full flux gives maximum torque per ampere and the lowest starting speed.

• Thinking the overload release breaks the main circuit. It short-circuits the no-volt coil, which then releases the handle.

• Believing the four-point starter is simply better. It gains the speed range and loses field-failure protection.

• Using a three-point starter on a series motor. There is no shunt field to connect to F.

• Leaving the starter resistance in circuit while running. It is a starting device, not a speed controller — that is Chapter 37.

For design problems find \(R_1\) and \(k\) first, then \(n\), then recompute \(k\) before working out the sections.

1. P36.1 A 250 V motor has \(R_a = 0.20~\Omega\) and full-load current 50 A. Find the direct starting current and the starter resistance to limit it to twice full load.

   Show answer

   \[I = \frac{250}{0.20} = 1250~\mathrm{A} = 25\ \text{times rated}\]

   \[R_1 = \frac{250}{100} = 2.50~\Omega, \qquad R_{st} = 2.50 - 0.20 = 2.30~\Omega\]

2. P36.2 For P36.1, the current is to be held between 100 A and 60 A. Find the number of steps.

   Show answer

   \[k = \frac{60}{100} = 0.60, \qquad \frac{R_a}{R_1} = \frac{0.20}{2.50} = 0.080\]

   \[n = \frac{\ln 0.080}{\ln 0.60} = \frac{-2.526}{-0.5108} = 4.945 \quad\Longrightarrow\quad 5\ \text{steps}\]

3. P36.3 For P36.2, find the corrected \(k\) and the new lower current limit.

   Show answer

   \[k = (0.080)^{1/5} = 0.6034\]

   \[I_{\min} = (0.6034)(100) = 60.3~\mathrm{A}\]

   Barely changed, because \(n\) was already close to a whole number.

4. P36.4 For P36.3, find the total resistance at each stud and the first two sections.

   Show answer

   \[R_1 = 2.500, \ R_2 = 1.509, \ R_3 = 0.9102, \ R_4 = 0.5492, \ R_5 = 0.3314, \ R_6 = 0.2000 = R_a \ \checkmark\]

   \[r_1 = 2.500 - 1.509 = 0.991~\Omega, \qquad r_2 = 1.509 - 0.910 = 0.599~\Omega\]

   Sections sum to \(2.500 - 0.200 = 2.300~\Omega\) \(\checkmark\)

5. P36.5 A three-point starter has a 40 \(\Omega\) holding coil and a 160 \(\Omega\) field on 200 V. The coil needs 0.80 A. What is the largest usable field rheostat?

   Show answer

   \[\frac{200}{200 + R} \ge 0.80 \quad\Longrightarrow\quad R \le \frac{200}{0.80} - 200 = 250 - 200 = 50~\Omega\]

   Only 50 \(\Omega\) may be inserted before the handle drops out.

6. P36.6 A motor's armature dissipates 500 W at full load. What does it dissipate at the first stud if \(I_{\max}\) is 1.8 times full load, and on direct starting at 25 times full load?

   Show answer

   \[\text{with starter: } (500)(1.8)^{2} = (500)(3.24) = 1620~\mathrm{W}\]

   \[\text{direct: } (500)(25)^{2} = (500)(625) = 312\,500~\mathrm{W} = 313~\mathrm{kW}\]

7. P36.7 Why must the field rheostat be at minimum resistance during starting?

   Show answer

   Minimum rheostat resistance means maximum field current and maximum flux. Since \(T_a = K_a\Phi I_a\) and the starting current is deliberately limited, maximum flux gives the greatest possible torque from that limited current.

   It also gives the lowest speed for a given \(E_b\), since \(N \propto E_b/\Phi\), so the motor accelerates under control instead of racing.

   On a three-point starter there is a third reason: the holding coil is in the same circuit, and minimum rheostat resistance gives it maximum current.

8. P36.8 Explain how the overload release works, and why it does not break the main circuit itself.

   Show answer

   The overload release is an electromagnet carrying the line current. On overload it lifts a pivoted armature which short-circuits the no-volt holding coil.

   The holding coil, deprived of current, releases the handle, and the spring returns it to the off position — disconnecting the motor and simultaneously reinserting the full starting resistance ready for the next start.

   Why not break the circuit directly? A contact interrupting the full armature current would arc badly and need to be substantial. Short-circuiting a small coil is a much lighter duty, and it reuses the disconnecting mechanism that already exists for the no-volt function.

9. P36.9 Compare the three-point and four-point starters, stating what each gains and loses.

   Show answer

   Three-point: the holding coil is in series with the shunt field. It therefore protects against both supply failure and field failure. But any field rheostat inserted for speed control also weakens the coil, so the speed range is limited — Example 36.4 found only half the rheostat usable.

   Four-point: the coil has its own circuit across the supply through terminal N. Its current is independent of the field rheostat, so the full speed range is available. But it no longer releases on field failure, and a separate field-failure relay is needed.

   Neither is simply better. The choice depends on whether wide field control is wanted.

10. P36.10 Why does a series motor's two-point starter give protection that a shunt motor's does not?

    Show answer

    Because its holding coil carries the armature current rather than the field current. If the mechanical load is removed, the current falls, the coil weakens, and the handle springs back — disconnecting the motor before it can run away.

    This is exactly the protection a series motor needs (Chapter 35), and it comes free from the series connection.

    A shunt motor's three-point starter has its coil in the field circuit, so it releases on field failure instead — which is the fault a shunt motor is vulnerable to. In each case the coil is placed in the circuit whose failure is most dangerous for that machine.