Generator Characteristics and Voltage Build-Up

Build the ratio column. A field-resistance line is \(E_a = I_fR_{sh}\), a straight line through the origin of slope \(R_{sh}\). It passes through a given point of the OCC if and only if \(R_{sh}\) equals \(E_a/I_f\) there. Tabulating that ratio replaces the graphical construction entirely:

The critical resistance is the largest value in that column taken over the straight portion of the curve — the steepest line through the origin that still touches the OCC:

The first interval, 0 to 0.2 A, is the residual toe and is excluded from the construction by convention: any straight line through the origin crosses it somewhere, but the intersection is worth only a few volts and is not a build-up. Beyond the toe the ratio falls monotonically, so 150 Ω is the ceiling.

Find the operating point for \(R_{sh} = 100\ \Omega\). Look for the row whose ratio is 100 Ω:

Check it directly: \(1.8\times100 = 180\) V, and the OCC gives 180 V at 1.8 A. The two relations are satisfied simultaneously, which is what an operating point means.

Why that point is stable. Below 1.8 A the OCC lies above the field line, so the armature generates more than the field circuit needs and the current climbs; above 1.8 A the line lies above the curve, the field is over-driven and the current falls back. The machine is pushed towards the intersection from both sides, which is why build-up is self-terminating and why the final voltage is always found on the saturated part of the curve.

For a no-load voltage of 148 V, read the OCC backwards. That emf occurs at \(I_f = 1.2\) A, so the field line must pass through \((1.2,\ 148)\):

The field rheostat is therefore opened by \(123.3-100 = 23.3\) Ω. Note how little of the range is usable: everything between 123.3 Ω and the critical 150 Ω sits on the knee, where the voltage is very sensitive to the setting.

Speed stretches the OCC vertically and leaves the field line alone. At speed \(N\) every ordinate of the 1000 rpm curve is multiplied by \(N/1000\), so the critical resistance is multiplied by the same factor:

The critical speed is where the given resistance becomes critical. Set \(R_c(N_c) = R_{sh}\):

Below 1200 rpm the 180 Ω line is steeper than anything the stretched OCC can offer and the machine sits at a few volts. Note the symmetry with Problem 1: too much resistance at a given speed and too little speed at a given resistance are the same failure.

At 1500 rpm, work with the 1000 rpm table directly. The build-up condition is \(1.5\,E_{1000}(I_f) = 180\,I_f\), which rearranges to a condition on the tabulated curve alone:

In other words, running 1.5 times faster with 180 Ω is equivalent to running at 1000 rpm with 120 Ω. Only one curve is ever needed.

Locate the ratio 120 Ω in Problem 1's third column. It lies between 1.2 A (123.3 Ω) and 1.4 A (115.0 Ω). Interpolate linearly on the OCC over that interval, where \(E_{1000} = 148 + 65(I_f-1.2)\):

The terminal voltage is that current times the field resistance:

Compare the field resistance with the critical value. At 1000 rpm the critical resistance is 150 Ω and

The field line is steeper than the steepest part of the OCC, so beyond the residual toe it lies entirely above the curve. There is no operating point at any useful voltage.

What the voltmeter actually reads. The line still crosses the residual toe. On the first interval the OCC is \(E = 6 + 120I_f\), so

Twenty-four volts on a machine that should give 180 — the classic symptom. The residual flux drives a trickle of field current, that current adds a little flux, and the process stalls almost at once because the field circuit swallows the extra emf faster than the iron can supply it.

The speed needed with 160 Ω:

A 7% increase in speed converts a dead machine into a live one. Nothing else about it has changed.

At 1100 rpm, apply the equivalence of Problem 2: the condition becomes \(E_{1000}/I_f = 160/1.1 = 145.5\ \Omega\). That ratio lies between 0.6 A (146.7 Ω) and 0.8 A (140.0 Ω), where \(E_{1000} = 88 + 120(I_f-0.6)\):

Only 100 V, and sitting on the knee of the curve where a 1% change in speed or resistance moves the voltage by several volts. Just above the critical speed a generator excites, but it excites badly.

The other ways a shunt generator refuses to build up, all of which give the same near-zero voltmeter reading:

In a series generator there is only one current. The field winding is in the load line, so

The magnetisation curve, plotted against that single current, therefore is the internal characteristic: with armature reaction neglected, the emf generated on load at current \(I\) is the same as the emf generated on open circuit with \(I\) in the field. There is no output at all at zero current beyond the residual 12 V — a series generator cannot excite itself until a load is connected.

The external characteristic is the internal one less the resistance drop. Both windings carry \(I\):

At 40 A the table gives directly

The turning point. The external characteristic rises while the curve gains more than the drop costs, and turns over when

The slope passes through 0.5 Ω between 60 and 70 A, and the tabulated external voltage confirms it: 257 V at 60 A, falling to 256 V at 70 A. The maximum is 257 V at about 60 A, and the machine's rating — roughly 250 V, 60 A, 15 kW — is set by that peak.

Why it is no use as a constant-voltage source. Between no load and 20 A the terminal voltage swings from 12 V to 190 V — the output voltage is essentially a function of the load, which is the opposite of what a supply is asked to do. Worse, beyond the turnover the characteristic has negative slope, so the operating point can run away. What the machine is good for is exactly this rising characteristic: connected in series with a feeder, it adds volts in proportion to the current drawn and cancels the feeder's own \(IR\) drop. That is the series booster, and it is the only common application.

Compounding is measured at the terminals. Compare the two measured voltages:

Positive, so the machine is over-compounded: the series field more than makes up for the armature drop and the terminal voltage rises with load. A level-compounded machine would read 250 V at both ends, and an under-compounded one less than 250 V at full load.

No-load conditions. With no load current, the armature carries only the shunt field current, and in a long-shunt connection that current flows through the series field as well:

The drop is 0.07 V — on open circuit the generated emf and the terminal voltage are, for all practical purposes, the same number.

Full-load conditions. Now the shunt field is across the higher terminal voltage, and the armature carries the load as well:

The rise in emf is larger than the rise in terminal voltage, because the series field has to pay for the armature drop before it can raise the terminals at all:

Six per cent at the terminals costs 11.6% inside the machine. Of the 29 V of extra emf the series field must produce, 14.07 V is consumed by \(I_a(R_a+R_{se})\) and only the remaining 14.93 V — plus the 0.07 V no-load drop — reaches the load.

The divertor is a current divider. It sits in parallel with the series field, so

Seventy per cent of the armature current now excites the series field, so its ampere-turns drop by 30% and the compounding falls with them. This is how an over-compounded machine is trimmed on site: not by unwinding turns, but by shunting them.

Naming the three cases, for a machine of the same 250 V no-load voltage:

Three relations must hold together, and none of them can be solved first because each needs the others:

A higher terminal voltage gives more field current, which gives more emf, which gives a higher terminal voltage. The loop closes only at one value of \(V_t\), and the practical route is to guess and iterate.

Case (a): armature reaction ignored. Start from the no-load voltage of 180 V and work down. Take a first guess \(V_t = 168\) V:

Converged: \(V_t = 166.7\) V. Between 1.6 and 1.8 A the OCC is \(E_a = 172 + 40(I_f - 1.6)\), which is what the third column uses.

Confirm it algebraically. On that interval the whole system is linear. Writing \(I_f = V_t/100\),

Note how small a part the armature drop plays: \(I_aR_a = 26.67\times0.3 = 8.0\) V of the 13.3 V total fall. The other 5.3 V come from the field current itself dropping from 1.80 A to 1.67 A.

Case (b): include armature reaction. The emf is now read off the OCC at the effective field current \(I_f - 0.05\). Iterating again lands on the interval 1.4 to 1.6 A, where \(E_a = 161 + 55(x-1.4)\) with \(x = V_t/100 - 0.05\):

Check: \(170.8 - 8.0 = 162.8\) V &checkmark. A demagnetising effect worth only 0.05 A of field current costs 3.9 V at the terminals — three times its face value, because the lost volts reduce the field current, which loses more volts.

Voltage regulation, from the no-load 180 V of Problem 1 to the loaded 162.8 V:

Ten per cent droop for 25 A on a machine whose armature drop alone is 4.4% — the shunt generator's characteristic sags far more than its resistance suggests. That is the price of self-excitation.