Starter Circuit Voltage Drop: Design Check versus In-Service Measurement
A 28 V DC system on a piston twin. The engine-start circuit uses one 25 mm2 copper cable with a total run (positive feed plus ground return) of 11 m. Copper resistivity is 1.72 x 10^-8 ohm.m. The starter is rated 250 A at 28 V and the maintenance manual allows a maximum total voltage drop of 2.0 V in the start circuit. The aircraft cranks slowly, especially when the engine is hot, and the engineer measures 4.6 V between the battery positive post and the starter terminal while cranking, with the battery at 25.6 V open-circuit and the starter drawing 240 A. The cable feels warm after a few seconds and one terminal is discoloured. The battery has just passed a capacity check.
1. Calculate the design resistance of the cable and the expected voltage drop at rated starter current. Does the installation as designed comply with the manual limit?
Read the worked answer
Cross-section in SI units
A = 25 mm2 = 25 x 10^-6 m2 = 2.5 x 10^-5 m2
Resistance
R = rho x L / A = (1.72 x 10^-8 x 11) / (2.5 x 10^-5) R = 1.892 x 10^-7 / 2.5 x 10^-5 = 7.57 x 10^-3 ohm = 7.6 milliohm
Drop at rated current
V = I x R = 250 x 7.57 x 10^-3 = 1.89 V
Verdict: 1.89 V is below the 2.0 V limit, so the cable as designed complies, but with only about 0.11 V of margin. Any joint, corrosion, or undersized return path will push it over. The design is therefore intolerant of degradation, which matters when interpreting the in-service reading.
2. From the in-service measurement, calculate the total circuit resistance and the excess over design. Where should the fault be sought, and what measurements would you take next?
Read the worked answer
Total resistance in service
R = V / I = 4.6 / 240 = 19.2 milliohm
Excess over the cable
19.2 - 7.6 = 11.6 milliohm, about 2.5 times the cable resistance itself.
Where to look
A copper cable does not double its resistance in service unless it is badly corroded or has lost strands. An excess of this size concentrated in a few centimetres is characteristic of a termination, not of the cable run. Candidates, in order of likelihood given the discoloured terminal:
- Battery positive post to lug (corrosion, loose nut, paint or anodising under the lug).
- Lug to cable crimp (green corrosion, wicking, insufficient crimp).
- Starter solenoid stud and the solenoid contact itself.
- Engine earth strap to airframe, and battery negative to airframe.
Next measurements
Crank (or use a high-current load) and measure volt drop across each joint individually, probe to probe on the same conductor, not to airframe earth. A sound heavy-current joint should show a fraction of a volt. The joint showing the largest drop is the fault.
Remedy: disconnect, clean to bright metal, replace damaged lugs, re-crimp and torque, apply protective compound, then repeat the whole-circuit crank test.
3. Quantify the energy dissipated at 240 A and explain the airworthiness consequences. State your release decision.
Read the worked answer
Power dissipated
P = I^2 R = 240^2 x 0.0192 = 1,106 W in the whole circuit, of which the cable accounts for 240^2 x 7.57 x 10^-3 = 436 W and the remaining ~670 W is concentrated in the faulty joint.
Why that matters
- 670 W in a joint of a few square centimetres during a 5 to 10 second crank will char insulation and can ignite fuel, oil or hydraulic fluid in the engine bay. This is a fire risk, not just a performance issue.
- Terminal voltage at the starter = 25.6 - 4.6 = 21.0 V, an 18 percent loss. A series-wound starter motor draws current roughly in proportion to the difference between supply voltage and back EMF, so torque and cranking speed fall sharply. Hot-start problems are expected because resistance rises with temperature.
- Note that the drop is measured while current is only 240 A; at 250 A the drop would be about 4.8 V, leaving the starter below 21 V.
Decision
The aircraft is not releasable. Defect entry in the technical log, rectify the termination, then demonstrate by cranking volt-drop test that the total circuit drop is at or below 2.0 V with a serviceable battery at 24 V or more, and record the result. Replacing the battery alone would not have fixed this and would have hidden a fire risk.