A transformer winding measures 16 GΩ to earth at 2.5 kV DC, and the question is whether a 5 kV test will produce a higher or lower resistance. The 5 kV value cannot be calculated from the 2.5 kV result alone. The tester calculates insulation resistance from applied voltage and measured current, while that current changes with insulation condition and elapsed test time.
Voltage, current, and the reported resistance
The number that matters is leakage current after the transient components have decayed for a defined interval. Insulation resistance follows R = V / I, but increasing voltage does not automatically increase resistance. It increases current; the way current increases determines the new resistance.
Using the corrected result of 16 GΩ, the implied current at 2.5 kV is:
I = 2,500 V / 16,000,000,000 Ω = 156.25 nA
If the insulation remains linear and its resistance stays at 16 GΩ, the current at 5 kV becomes:
I = 5,000 V / 16,000,000,000 Ω = 312.5 nA
Those figures describe the stabilized resistive component under the labeled constant-resistance assumption. They are not predictions of the total current immediately after voltage application.
| Quantity | Value or limit | Where to read or calculate it |
|---|---|---|
| Initial reported resistance |
16 GΩ, not 16 MΩ |
Insulation tester result at 2.5 kV |
| Applied test voltage | 2.5 kV DC |
Tester output setting |
| Derived current at 2.5 kV | 156.25 nA |
I = V / R, assuming 16 GΩ represents stabilized resistance |
| Derived current at 5 kV | 312.5 nA |
I = V / R, assuming resistance remains 16 GΩ |
| Comparison time | Generally one minute after each voltage application or step | Test timer |
| Screening concern | Resistance reduction greater than 25% | Compare equally timed readings |
Same-voltage trending versus step-voltage testing
Two approaches answer different maintenance questions. Routine trending repeats the established test voltage and timing to preserve a meaningful benchmark. A DC step-voltage test deliberately raises voltage and watches whether current remains proportional to voltage.
| Approach | Question answered | Strength | Primary limitation |
|---|---|---|---|
| Repeat at 2.5 kV | Has insulation condition changed since the earlier test? | Direct comparison when setup, temperature, timing, and connections match | Does not test voltage-dependent leakage behavior above 2.5 kV |
| Step from 2.5 kV to 5 kV | Does leakage current increase linearly as electrical stress rises? | Can expose nonlinear conduction associated with weak, contaminated, or wet insulation | Requires an approved test voltage, controlled timing, and interpretation appropriate to the transformer construction |
Use the same 2.5 kV test for routine benchmarking. Use a 5 kV step only when the transformer test procedure permits it and the objective is to investigate voltage dependence. IEEE 95 describes DC insulation testing of large AC rotating machinery; its step-voltage concepts explain the current-versus-voltage behavior, but transformer acceptance criteria must come from the applicable transformer documentation.
Current components behind the changing display
Immediately after DC voltage is applied, measured current contains capacitive charging current, absorption current, and resistive leakage current. Capacitive current is strongest during charging and falls rapidly. Absorption current also decays with time as dielectric polarization develops. Resistive leakage is the component used to judge the steady insulation path.
This is current and dielectric behavior, not a fixed arithmetic conversion between 2.5 kV and 5 kV. Reading one test after a few seconds and another after one minute can create an apparent resistance change even when the insulation has not changed. Holding each step for the same interval separates voltage response from timing response.
Healthy dry insulation commonly approaches proportional current rise over a normal test range: doubling voltage approximately doubles stabilized current, leaving calculated resistance about the same. Deteriorated, contaminated, or wet insulation can conduct disproportionately more current at higher stress, so calculated resistance falls. A higher displayed value may occur because of timing, temperature, surface condition, or instrument behavior; voltage alone does not require it.
Transformer condition and test selection
First identify whether the transformer is dry type or oil filled. The rotating-machine analogy is useful for dry winding insulation, but oil-filled transformer systems add liquid insulation, bushings, internal geometry, and moisture distribution. Select the test and acceptance limits from the transformer manufacturer's maintenance procedure rather than transferring rotating-machine limits directly.
Before raising the voltage, check the nameplate and service documentation for the approved DC insulation-test voltage. Also record winding under test, terminals tied together, terminals grounded, equipment disconnected from the winding, ambient and winding temperature, tester model or range, applied voltage, and reading time. Changes in any of these conditions weaken the comparison.
A result below 12 GΩ at 5 kV represents a reduction greater than 25% from 16 GΩ: 16 GΩ × 0.75 = 12 GΩ. Treat that as a cause for investigation under the stated screening rule, not as a universal pass/fail limit. The manufacturer's limits and the transformer's historical trend control the final disposition.
Controlled step-voltage procedure
- Isolate the transformer according to the site's electrical safety procedure. Disconnect paths that would place connected equipment, surge devices, instrumentation, or unintended parallel insulation in the measurement circuit.
- Confirm that 5 kV DC is an approved test level for the winding and transformer construction. If the approved voltage is unavailable, stop before applying 5 kV and obtain the manufacturer's test instruction.
- Define the measurement path between the selected winding and earth. Keep winding connections, guarding, grounding arrangement, lead routing, and surface cleanliness unchanged between steps.
- Apply 2.5 kV DC and record the resistance at one minute. Record current as well if the instrument provides it.
- Discharge the winding using the tester's prescribed method, then verify the winding is discharged before changing connections or touching conductors.
- Apply 5 kV DC with the same connection arrangement and record the resistance at one minute. Watch the current trend throughout the step; an accelerating current or falling resistance is more significant than a small display fluctuation.
- Discharge the winding again and verify the discharged state. Record both readings, test durations, temperatures, and test configuration together.
Result verification and recurring pitfalls
Calculate the resistance ratio as R5kV / R2.5kV. A ratio near 1.0 means current rose approximately in proportion to voltage. A ratio below 0.75 means the resistance fell by more than 25% from the 16 GΩ baseline and calls for investigation of moisture, contamination, surface leakage, connection errors, or insulation deterioration.
Verify a concerning result by repeating the test with identical timing after checking clean, dry terminal surfaces and lead separation. Confirm that the tester did not change range, reach its measurement ceiling, or report an unstable value. Temperature differences can shift insulation resistance substantially, so compare readings at matched conditions or apply only the correction method specified for the transformer.
Common errors include comparing 16 GΩ with the earlier mistaken unit of 16 MΩ, treating an early transient reading as a one-minute value, changing the grounded terminals between tests, and assuming that twice the voltage must produce twice the resistance. The linear healthy case predicts approximately the same resistance and twice the current.
FAQ
Can I calculate the 5 kV insulation resistance from a 2.5 kV reading?
No. You can calculate that 16 GΩ at 2.5 kV corresponds to 156.25 nA, but the 5 kV resistance requires the measured current at 5 kV.
Does doubling the test voltage double insulation resistance?
No. For linear insulation, current approximately doubles and resistance remains near 16 GΩ; disproportionate current growth makes the calculated resistance lower.
Can I compare readings taken at different times?
Use the same elapsed time for every reading, generally one minute for the comparison described here. Capacitive and absorption currents decay after voltage application and can distort differently timed results.
Does a drop below 12 GΩ require stopping the test?
It represents more than a 25% reduction from 16 GΩ and warrants investigation. Stop if current rises abnormally, resistance continues falling, or 5 kV is not explicitly approved for the transformer; keep the equipment out of service when the manufacturer's acceptance criteria are not met. Escalate to the transformer manufacturer's official support channel with the test configuration, timed readings, temperatures, and current trend.