Voltage Build-Up and Critical Resistance

Chapter 27 computed the generated EMF from the flux. In a separately excited machine that is straightforward: the field current is set externally, so the flux is known. In a self-excited machine it is not, and the difficulty is circular.

The escape is that the machine does not start with zero of everything. The pole iron retains a little flux from its last use — residual magnetism — and that is enough to begin a process which then sustains itself.

This chapter answers three questions:

• Where does the build-up stop? At the operating point of Section 28-5.

• When does it fail to start at all? Above the critical field resistance of Section 28-6, or below the critical speed of Section 28-7.

• What do you do about it? Section 28-9.

Chapter 6 established that a ferromagnetic material follows a hysteresis loop, and that when the magnetising force is removed the flux density does not return to zero but to the remanent value \(B_r\). The poles of a machine that has run before therefore retain a small flux.

Two features of this loop deserve attention. It is positive feedback, so it is fast — build-up takes a second or two, limited by the field winding's time constant rather than by anything magnetic. And it is self-limiting, because the curvature of the magnetisation characteristic guarantees that the gain falls below unity at some point.

The open-circuit characteristic (OCC), also called the magnetisation characteristic or no-load saturation curve, is a plot of the generated EMF against field current, at a stated constant speed, with the armature on open circuit.

It is obtained by driving the machine at constant speed, supplying its field from a separate variable source, and recording the terminal voltage as the field current is raised from zero. Since the armature carries no current, there is no \(I_aR_a\) drop and the reading is \(E_g\) itself.

Three features are worth naming:

• The curve does not pass through the origin. At zero field current it reads 8 V — the residual EMF.

• The initial portion is straight, because the air gap dominates the reluctance while the iron is unsaturated. Here \(E_g \propto I_f\).

• It then bends over as the iron saturates, and further field current buys very little extra EMF.

The OCC says what the armature can generate for a given field current. The field circuit says what field current a given voltage will drive. Both must be satisfied at once.

On open circuit the whole terminal voltage appears across the shunt field, so by Ohm's law

Note that \(R_{sh}\) here means the whole field-circuit resistance — the shunt winding plus any rheostat in series with it. That is what the operator actually controls.

Build-up stops where the two curves cross.

Steepening the field line moves the operating point down the curve. Steepen it far enough and the line no longer crosses the OCC at all — except very near the origin, where the voltage is only the residual.

The reason is simple once seen on the graph. For build-up to start, the OCC must lie above the field line near the origin, so that the small residual EMF drives a field current larger than needed to sustain it. If the line is steeper than the initial slope of the curve, that is never true, and the process cannot get started.

The same failure can be produced without touching the rheostat, simply by running the machine too slowly.

From Section 28-3, lowering the speed scales every ordinate of the OCC by \(N'/N\). Its initial slope — and therefore the critical resistance — scales in the same proportion:

Build-up fails when the critical resistance falls to the actual field resistance, which defines the critical speed:

The two failures are the same failure seen from opposite sides. Build-up requires the field line to be flatter than the OCC's initial slope; raising \(R_{sh}\) steepens the line, while lowering the speed flattens the curve. Either can bring them into coincidence.

Condition 2 deserves comment because it involves two things that can each be wrong: the field connections and the direction of rotation. Reversing either one reverses the field current relative to the residual flux, and the machine will not excite. Reversing both restores correct operation — the same product-of-two-quantities logic as the torque reversal of Chapter 17.

A generator that runs but produces only a few volts is a common workshop problem. The diagnosis follows the three conditions.

One diagnostic distinguishes the causes quickly. Open the field circuit and measure the terminal voltage while running. If a few volts appear, the residual magnetism is present and the fault lies in the field circuit or its connections. If nothing appears, the residual has been lost and the field must be flashed.

• Self-excitation appears circular but is resolved by residual magnetism, which generates a few volts to start a positive-feedback process.

• Build-up is self-limiting because the magnetisation curve saturates. Without saturation there would be no operating point at all.

• The open-circuit characteristic plots \(E_g\) against \(I_f\) at constant speed. It starts above the origin at the residual voltage, rises linearly, then bends over.

• Every OCC ordinate scales with speed: \(E_g' = E_g(N'/N)\). The abscissa does not change.

• The field resistance line \(V = I_fR_{sh}\) is a straight line through the origin of slope \(R_{sh}\).

• The operating point is their intersection, and it is stable: the machine returns to it after any disturbance.

• The critical field resistance is the slope of the tangent from the origin to the initial straight portion of the OCC. Above it, no build-up.

• The critical speed is \(N_c = NR_{sh}/R_c\). Below it, no build-up.

• Three conditions are needed: residual magnetism; field connected to assist it; and resistance below critical (equivalently, speed above critical).

• A machine that has lost its residual magnetism is restored by flashing the field from a battery.

• Drawing the OCC through the origin. It starts at the residual voltage, and without that offset build-up could never begin.

• Drawing the field line not through the origin. Ohm's law gives \(V = I_fR_{sh}\) with no constant term.

• Taking the critical resistance as the winding's own resistance. It is a property of the machine's OCC and speed, not of the copper.

• Forgetting the rheostat. \(R_{sh}\) means the whole field-circuit resistance, winding plus rheostat.

• Scaling the abscissa of the OCC with speed. Only the ordinates scale; the same field current gives the same flux at any speed.

• Inverting the critical-speed formula. It is \(N_c = NR_{sh}/R_c\) — a higher field resistance needs a higher speed.

• Assuming reversing the rotation alone will fix a polarity problem. That merely swaps one failure for another; both rotation and field must be reversed together.

• Trying to build up on load. The armature drop and the load current prevent the voltage from rising; build up on open circuit first.

• Starting with the rheostat at maximum. That is the setting most likely to exceed the critical resistance. Start at minimum.

• Expecting a proportional voltage rise as the rheostat is cut. Saturation makes the return steeply diminishing.

Problems P28.1 to P28.6 use the OCC of Table 28.1, measured at 1000 rev/min.

1. P28.1 Find the no-load voltage for a field-circuit resistance of 175 \(\Omega\) at 1000 rev/min.

   Show answer

   Tabulating OCC minus \(175I_f\): at 0.6 A, \(114 - 105 = +9\); at 0.8 A, \(140 - 140 = 0\).

   \[I_f = 0.8~\mathrm{A}, \qquad V = (0.8)(175) = 140~\mathrm{V}\]

   The crossing falls exactly on a tabulated point.

2. P28.2 What is the critical speed for a field-circuit resistance of 100 \(\Omega\)?

   Show answer

   \[N_c = 1000\left(\frac{100}{200}\right) = 500~\mathrm{rev/min}\]

3. P28.3 The machine is run at 1200 rev/min with 200 \(\Omega\) in the field circuit. Will it build up, and to what voltage?

   Show answer

   At 1200 rev/min the OCC ordinates are multiplied by 1.2, so

   \[R_c' = (200)(1.2) = 240~\Omega \gt 200~\Omega\]

   so it will build up. Locating the crossing with ordinates \(1.2\times\) Table 28.1: at 0.8 A, \(168 - 160 = +8\); at 1.0 A, \(192 - 200 = -8\).

   \[I_f = 0.8 + (0.2)(0.5) = 0.9~\mathrm{A}, \qquad V = (0.9)(200) = 180~\mathrm{V}\]

   The same 200 \(\Omega\) that was exactly critical at 1000 rev/min gives a healthy 180 V at 1200.

4. P28.4 Find the critical resistance at 600 rev/min.

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   \[R_c' = (200)\left(\frac{600}{1000}\right) = 120~\Omega\]

   So at 600 rev/min the field circuit must be below 120 \(\Omega\) — a much tighter requirement than at rated speed.

5. P28.5 The field winding alone measures 90 \(\Omega\). What is the maximum rheostat resistance that still permits build-up at 1000 rev/min?

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   \[R_{\text{rheostat}} \lt R_c - R_{\text{winding}} = 200 - 90 = 110~\Omega\]

   In practice a margin is left, so a rheostat of about 80 to 90 \(\Omega\) would be specified.

6. P28.6 The field copper loss at the 150 \(\Omega\) operating point of Example 28.1 is compared with that at the 120 \(\Omega\) point of Example 28.2. Find both.

   Show answer

   \[P_f = VI_f: \qquad (168.75)(1.125) = 189.8~\mathrm{W}\]

   \[(189.33)(1.578) = 298.7~\mathrm{W}\]

   The voltage rose 12 % but the field loss rose 57 % — the cost of working further into saturation.

7. P28.7 A generator builds up correctly. Its drive is then reversed. What happens, and what single change restores operation?

   Show answer

   Reversing the rotation reverses the armature EMF, so the field current now opposes the residual flux. The machine will not build up; the residual voltage will in fact be driven down.

   The remedy: also reverse the field winding connections. With both reversed the field again assists the residual, and the machine builds up — to the opposite polarity from before.

8. P28.8 Why must a shunt generator be brought up to voltage on open circuit rather than on load?

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   On load the terminal voltage is reduced by the armature drop \(I_aR_a\) and by armature reaction. The shunt field sees that reduced voltage, so it draws less current than the field line predicts.

   In effect the load steepens the operative field line, and if the load is heavy enough the effective resistance exceeds critical and the machine never builds up at all. Building up on open circuit lets the flux establish itself first, after which the machine can hold its voltage under load.

9. P28.9 A generator that has stood idle for a year produces nothing at all when driven. Diagnose and prescribe.

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   Diagnosis: loss of residual magnetism. The distinguishing symptom is that the output is zero, not merely low — a machine with intact residual flux and a field fault would still show a few volts.

   Confirmation: open the field circuit and run the machine. If nothing appears at the terminals, the residual is gone.

   Remedy: flash the field. With the machine stationary, connect the field winding briefly to a battery of the correct polarity. A few seconds suffices; the pole iron then retains its remanent flux and the machine will excite normally.

10. P28.10 Explain why a generator's operating point should lie well up the saturated part of the OCC.

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    Low on the curve, the OCC and the field line are nearly parallel, so their crossing is ill-defined. A small change in speed, in field resistance from temperature rise, or in the curve itself moves the operating point a long way, and the voltage is unstable.

    High on the curve the OCC is nearly flat while the field line still rises, so they cross steeply. The same disturbance now moves the operating point very little. Saturation, which limits the voltage, is also what stabilises it.

    Example 28.5 illustrated the opposite case: working close to critical gave only 94 V, and that value would drift with any small change.