- Hopkinson Test Definition: The Hopkinson test is defined as a method to test the efficiency of DC machines using two identical machines operating back-to-back.
- Back-to-Back Operation: This test uses one machine as a generator and the other as a motor to drive each other, needing an external power source to overcome internal losses.
- Efficiency Calculation: Efficiency is calculated by measuring the current from the generator and the external source, and accounting for various losses in the system.
- Advantages: It requires less power compared to full-load testing, allows for temperature monitoring, accounts for changes in iron losses, and helps determine efficiency at different loads.
- Disadvantages: Finding identical machines is difficult, equal loading is hard to maintain, separating iron losses is not possible, and operating at rated speed can be tricky.
The Hopkinson test is a regenerative, back-to-back test of two similar DC shunt machines coupled on the same shaft. One operates as a motor and supplies mechanical input to the generator. The generator returns most of that power electrically to the common circuit, so the mains supply only the combined losses.
The machines must have compatible ratings and the correct polarity. Before their armatures are connected in parallel, the generator voltage is adjusted to match the bus. The external voltage source then supplies loss power to the loaded motor-generator set. Because both machines can carry load current while little net power is drawn, the Hopkinson test is also used as a heat-run test.
Connection Diagram of Hopkinson’s Test
The Hopkinson test starts with the coupled set operating as a motor at no load. Adjust the motor shunt-field resistance to bring the set to the required speed. This high-power test must use rated protective equipment and a supervised laboratory procedure.
With the coupling switch open, adjust the generator field until its terminal voltage, polarity and connection sequence match the supply. The voltmeter across the open switch reads zero only when the shown voltage difference is zero. Close the switch under the authorised procedure, then adjust the two field currents to obtain the required circulating load while monitoring rated current, speed and temperature.
Calculation of Efficiency by Hopkinson’s Test
Let V be the common terminal voltage.
The current relationship is 
I1 is the generator output current shown in the diagram.
I2 is the current drawn from the external supply.
The generator electrical output is VI1………………(1)
For a quick estimate, assume that both machines have the same efficiency ‘η’.
The motor shaft output is 
Combining equations 1 and 2 gives:
For loss separation, the motor armature copper loss is 
.
Ra is the assumed common armature resistance at the test temperature.
I4 is the motor shunt-field current.
The motor shunt-field copper loss is VI4.
The generator armature copper loss is 
I3 is the generator shunt-field current.
The generator shunt-field copper loss is VI3.
Power drawn from the external supply is VI2.
Subtract the measured armature and field copper losses from that supply input to obtain the combined residual loss:
If the two similar machines are assigned equal residual losses, then:
Residual loss per machine = W/2
Efficiency of Generator
The generator’s assigned total loss is 
Generator output = VI1
Therefore, generator efficiency is:
Efficiency of Motor
The motor’s assigned total loss is 
The motor input-current relation is
Therefore, motor efficiency is:
Advantages of Hopkinson’s Test
The main advantages are:
- The mains supply provides only the combined losses, even though the individual machines can carry rated current. This makes long tests of large machines economical.
- Because the machines operate under load, temperature rise and commutation can be observed during a properly controlled heat run.
- Loaded operation includes effects such as armature-reaction flux distortion and stray-load loss that a no-load test cannot reproduce.
- Adjusting the field currents allows efficiency to be evaluated at several load points within the machines’ ratings.
Disadvantages of Hopkinson’s Test
The main limitations are:
- The method needs two compatible, closely matched machines for a valid Hopkinson’s test.
- The motor and generator do not necessarily carry equal load because their currents and field excitations differ.
- The calculation cannot directly separate the two machines’ iron and mechanical losses. Dividing residual loss equally is an approximation.
- Field adjustment changes speed and excitation, especially in smaller machines, so achieving rated current at rated speed can be difficult.
