Insulation resistance calculator: temperature correction, PI and DAR
Correct a megohmmeter reading to 20 °C or 40 °C and calculate the polarization index (PI) and the dielectric absorption ratio (DAR) to IEEE 43.
Temperature correction, polarization index and DAR
Enter the megohmmeter readings and the insulation temperature. The tool corrects the resistance to the reference temperature and calculates the polarization index (PI) and the dielectric absorption ratio (DAR).
Result
Kt = 2^((T − Tref)/10): insulation resistance roughly halves for every 10 °C rise in temperature.
Indicative result. The calculation runs in your browser and is not sent to any server. This website may contain errors or omissions: before making any decision about an installation, the information must be checked and validated by a qualified engineer. Legal notice and terms of use
What this calculator does
It takes megohmmeter readings and returns four things: the resistance referred to the reference temperature, the polarization index, the dielectric absorption ratio and the comparison against the minimum recommended by IEEE Std 43. Everything runs in your browser; no reading is sent to any server.
The temperature correction
Conduction through an insulation is a thermally activated process: resistance falls as it warms up. The practical rule in IEEE 43 is that it roughly halves every 10 °C, which gives the factor
Kt = 2^((T − Tref) / 10)
and the corrected resistance R(Tref) = R(T) × Kt.
T is the temperature of the insulation — not of the air, nor of the frame surface — and Tref is the reference. IEEE 43 works at 40 °C. For transformers and cables it is common to refer readings to 20 °C; what matters is not which one you choose, but that you always use the same one so the history remains comparable.
The factor-of-two rule is an approximation. On very dry insulation and on large machines the real coefficient can depart from it appreciably; where the manufacturer publishes its own correction curve, that curve wins.
Polarization index and DAR
When the DC voltage is applied, three currents flow: the capacitive charging current, gone within seconds; the dielectric absorption current, decaying over minutes; and the leakage current, which is constant. Dry, clean insulation lets the first two die away, so the measured resistance climbs with time. Wet or contaminated insulation is dominated by leakage and the reading barely moves — or falls.
- DAR = R(60 s) / R(30 s). Quick check: below 1.25 the result is suspect, above 1.6 it is good.
- PI = R(10 min) / R(1 min). The reference figure: IEEE 43 asks for 2 or more on thermal class B and above.
Minimum values recommended by IEEE 43
The calculator compares the corrected one-minute resistance against the minimum for that winding:
| Winding type | Recommended minimum at 40 °C |
|---|---|
| Built before 1970 | kV + 1 (in MΩ) |
| Form-wound coils built after 1970 | 100 MΩ |
| Random-wound, or form-wound below 1 kV | 5 MΩ |
These are minima for returning to service, not quality targets: a healthy winding usually sits well above them.
Worked example
6.6 kV motor, form-wound coils from 1998. With the insulation at 18 °C the readings are 2.4 GΩ at one minute and 6.0 GΩ at ten minutes.
- Kt = 2^((18 − 40)/10) = 2^(−2.2) = 0.218
- R(40 °C) = 2,400 MΩ × 0.218 = 523 MΩ — well above the 100 MΩ required
- PI = 6,000 / 2,400 = 2.5 — dry insulation
The uncorrected reading, 2.4 GΩ, would have looked better still: almost five times the figure that is actually comparable with the machine’s history.
What this calculator does not do
It does not replace the judgement of whoever runs the test. It knows nothing about ambient humidity or surface contamination, which distort the reading when the guard terminal is not used; it cannot distinguish a localised defect from general ageing; and it does not say whether the applied test voltage was right for that winding, only which range IEEE 43 suggests. A high resistance does not rule out defects that only show up under AC either: that is what power factor, tan δ and partial discharge testing are for.
What measures it
The measurement is made with an insulation resistance tester able to hold the test voltage steady through the ten minutes of the PI and to record the curve. For AC diagnostics, partial discharge detectors cover what a DC test cannot see.
Frequently asked questions
- Why must insulation resistance be corrected for temperature?
- Because the resistance of an insulation roughly halves for every 10 °C rise in its temperature. Two readings on the same machine taken at 15 °C and at 45 °C can differ by a factor of eight without the insulation having changed at all. Uncorrected, a comparison against the machine’s history means nothing.
- What polarization index is acceptable?
- IEEE 43 considers a PI of 2 or more acceptable for thermal class B insulation and above. Below 1 the result points to moisture or contamination. On modern machines with very high one-minute readings (above 5,000 MΩ corrected) the standard itself warns that PI may stop being meaningful.
- What is the difference between PI and DAR?
- Both compare how the resistance evolves while the test voltage is held. DAR is the ratio of the 60-second reading to the 30-second one and works as a quick check. PI compares the 10-minute reading with the one-minute reading and is the reference figure, because by then the absorption current has died away.