A bipolar shunt generator has two magnetic poles and a field winding connected across its own armature terminals. Explore its rotating armature, stationary field and brushes, then use a simplified magnetisation curve to estimate the no-load operating point.
Bipolar Shunt Generator Interactive Calculator
Adjust the residual voltage, total field resistance and two curve parameters. All values refer to one fixed shaft speed. The drawing shows construction; the graph and results show equilibrium, not a timed start-up simulation.
Equation Used
- Fixed speed and field polarity that reinforces the residual magnetism.
- No external load; the armature still supplies field current. Armature resistance and brush drops are neglected.
- Exponential magnetisation is an illustrative fit, not measured data for a named machine.
- No time-domain field inductance, hysteresis, commutation, thermal or load-regulation model.
The intersection of the magnetization curve and field-resistance line gives the equilibrium voltage. With zero residual voltage and zero initial field current, this model remains at zero output. The initial-slope comparison does not establish the critical resistance of an actual machine.
Inside the two-pole shunt generator
A driven shaft carries the armature and commutator. Field coils surround stationary pole cores attached to the yoke. The brushes remain fixed while copper commutator segments pass beneath them. The cutaway separates those parts so rotation is visible.
The field circuit is connected across the armature terminals. Residual magnetism can initiate excitation when the shaft rotates and the field connection reinforces that magnetism. The calculator assumes that polarity and one fixed speed. The separate circuit drawing shows the shunt connection; it is not a wiring or restoration procedure.
What to explore
Move the field-resistance slider to change the straight line on the graph. Adjust the other inputs to change the curved magnetisation characteristic. Their intersection sets the modeled voltage and field current. This is a teaching model for understanding self-excitation, not a generator rating or a prediction of performance under load.
Calculation model
The entered curve is E = Eres + Esat[1 − exp(−If/Ik)]. Its limiting voltage is Eres + Esat. The resistance line is V = Rf If. The calculator solves their intersection, then finds field heating power Pf = V If.
The initial slope is Esat/Ik. The reported slope margin compares Rf with this number. An exponential curve with positive residual voltage has an intersection even above that slope; therefore a negative margin does not mean zero voltage. Real critical resistance should be assessed from a suitable measured magnetisation characteristic at the specified speed.
Exactly zero residual voltage with zero initial excitation remains at the zero solution. A different initial condition can lead to a different equilibrium when a nonzero solution exists. No start-up transient is calculated.
Worked example
With residual voltage 2 V, saturation increment 161.9 V, current scale 0.5 A and field resistance 200 Ω, the solution is approximately 110.0 V. Field current is about 0.55 A and field power about 60.5 W. These values come from the simplified no-load model.
The initial curve slope is 323.8 Ω, giving a slope comparison of 38.2%. This percentage does not certify excitation reliability, thermal capacity or permissible load.
What the model leaves out
Loading changes armature current, internal voltage drops and magnetic conditions. This calculator has no external-load input and does not predict voltage regulation, overload behaviour, efficiency or short-circuit protection. Dimensions and winding details in the illustration are schematic.
Use measured curves and the actual machine documentation for equipment assessment. Starting and servicing procedures depend on the machine’s winding arrangement, polarity and protection; this equilibrium calculation does not provide those procedures.
Frequently asked questions
Why does zero residual voltage give zero output?
The selected initial condition has no field current or residual flux to start excitation, so the model remains at zero output.
Does a negative slope margin mean failure?
It means the field-resistance line is steeper than the initial slope of this chosen curve. With positive residual voltage, this particular mathematical curve still has a low-voltage intersection. The result is not a universal pass/fail test.
Why do results stay constant while the shaft turns?
They describe an averaged steady operating point at one fixed speed. The slow rotation is a construction illustration, not instantaneous ripple or voltage build-up.
Does no-load mean no armature current?
No. With no external load, the armature still supplies the shunt field. The simplified calculation neglects its internal voltage drop.
References
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