Voltage drop and heat loss on the short cable between an isolating transformer and its fixture.
The secondary side runs at only a few volts, so even a short tail with a small conductor can take a large share of it. Enter the fixture, the secondary current and the tail, and the tool reports the drop, the extra voltage and power the transformer must supply, the transformer loading with and without the cable, and how other conductor sizes compare.
Real power drawn by the fixture on the secondary.
1.00 for plain lamps. Enter the driver's value for LED fixtures.
Follows the series current through the transformer ratio. 6.6 A for a 6.6 A / 6.6 A transformer.
Treated as VA. A 30/45 W transformer is a 45 VA unit. Check against your transformer data sheet.
Used when Custom is selected. Take it from your cable data sheet.
The transformer's own secondary lead is part of the circuit, so count it in the tail. Take its conductor size and length from the transformer data sheet (AC 150/5345-47C).
Copper resistance rises about 0.393 % per °C above 20 °C.
Your project's own design limit, not a standard value.
The same tail length re-run in each conductor size at your temperature. The last column is the longest one-way tail that stays within your drop limit.
| Conductor | Loop resistance (Ω) | Drop (V) | Drop (%) | Loss (W) | Transformer loading (%) | Longest tail (m) |
|---|
Every step is shown so the result can be checked by hand.
The fixture's apparent power divided by the secondary current gives the voltage across it.
The loop is out and back, so a one-way tail is doubled.
On a series isolating transformer the secondary current follows the primary current, so the drop appears as extra voltage the transformer must deliver, not as a lower lamp current, until the transformer runs out of capacity.
For PF = 1 this reduces to V_t = V_f + I × R. Loading is shown for the fixture alone and with the cable.
| Item | Value |
|---|---|
| 45 W lamp, PF 1.00, 6.6 A | V_f = 6.818 V |
| 4 mm², 5 m one-way (10 m loop), 20 °C | R = 0.0461 Ω |
| Voltage drop | 0.304 V |
| Drop of fixture voltage | 4.46 % |
| Cable loss | 2.01 W |
| Terminal voltage needed | 7.122 V |
| Loading on a 65 VA transformer | 69.2 % → 72.3 % |
It does not cover the primary cable (see Calculator #15), transformer regulation or magnetising losses (see #17), connectors, or cable reactance. Conductor resistances are nominal DC figures for copper, so confirm them against your cable data sheet.
FAA AC 150/5345-47 (series isolation transformers), AC 150/5345-26 (L-823 plug and receptacle connectors) and AC 150/5340-30 (design and installation of airport visual aids). IEC 61823 (AGL series transformers). ICAO Aerodrome Design Manual, Doc 9157, Part 5. Use the current edition of each.
Quick answers on voltage drop and heat loss on the short cable between an isolating transformer and its fixture.
It calculates voltage drop and heat loss on the short cable between an isolating transformer and its fixture. It reports the drop, the extra voltage and power the transformer must supply, the transformer loading with and without the cable, and how other conductor sizes compare.
The secondary side runs at only a few volts, so even a short tail with a small conductor can take a large share of it. The fixture voltage is its apparent power divided by the secondary current, so a 45 W lamp at 6.6 A sees only about 6.8 V.
You can enter the one-way tail length, which is doubled for the loop, or the total loop length, out and back, in metres or feet. The transformer's own secondary lead is part of the circuit, so count it in the tail, taking its conductor size and length from the transformer data sheet (AC 150/5345-47C).
The loop resistance is the resistance per km at 20 degrees C, corrected for temperature with 1 plus 0.00393 times the temperature minus 20, times the loop length divided by 1,000. The drop is the current times the resistance, the loss is the current squared times the resistance, and the drop percentage is the drop divided by the fixture voltage.
On a series isolating transformer the secondary current follows the primary current, so the drop appears as extra voltage the transformer must deliver, and not as a lower lamp current, until the transformer runs out of capacity. The terminal voltage needed is the square root of (fixture voltage times power factor plus current times resistance) squared plus (fixture voltage times sin phi) squared. For a power factor of 1 this reduces to the fixture voltage plus the current times the resistance.
It re-runs the same tail length in each conductor size at your temperature, and shows the loop resistance, drop, drop percentage, loss and transformer loading. The last column is the longest one-way tail that stays within your drop limit, which is your project's own design limit and not a standard value.
A 45 W lamp at power factor 1.00 and 6.6 A sees 6.818 V. A 4 mm squared tail of 5 m one-way, which is a 10 m loop at 20 degrees C, has a resistance of 0.0461 ohms. The drop is 0.304 V, which is 4.46 percent of the fixture voltage, the cable loss is 2.01 W and the terminal voltage needed is 7.122 V. A 65 VA transformer goes from 69.2 percent to 72.3 percent loading.
It does not cover the primary cable (see Calculator #15), transformer regulation or magnetising losses (see Calculator #17), connectors, or cable reactance. Conductor resistances are nominal DC figures for copper, so confirm them against your cable data sheet. Final design shall be verified against the cable and transformer manufacturers' data and the applicable project specification.
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