Relay Test Lead Voltage Drop Calculator — Burden & Compliance
A relay test set can only drive a burden if it can supply the current at the voltage that burden demands. Lead resistance is part of that burden, and at high currents it is often the largest part. This calculator checks the loop before you connect, and sizes the lead.
Key takeaways
- Relay burden in VA at rated current converts to impedance: Z = VA / In2. A 5 VA relay at 5 A is 0.2 Ω.
- Lead loop resistance is R = 2 ρ L / A for an out-and-return pair. Both conductors carry the test current and both drop voltage.
- The set must supply V = I × (Zrelay + Rloop), or equivalently S = I2 × Ztotal.
- At 40 A a 2.5 mm² lead can drop several volts — often a large share of the relay voltage. That is how a healthy relay gets recorded as a failed pickup.
- The set’s VA rating is a power envelope, not a constant voltage. At its maximum current the available voltage is only P / Imax.
1. What this calculator does — and what it does not
It converts the relay burden to an impedance, adds the resistance of the test leads, and compares the resulting voltage and apparent power with the test set’s per-phase capability. It also recommends a lead cross-section for a target voltage drop and shows the maximum run length for the selected lead.
It is a steady-state loop calculation. It assumes the relay burden is constant, the leads are straight copper or aluminium conductors at one temperature, and the set’s output follows a constant-power envelope. It does not model relay burden variation with armature position, cable inductance, set derating, duty cycle or thermal limits. Where the set’s datasheet gives a V–I envelope, that envelope governs.
2. Relay test lead voltage drop calculator
Set data: VAE-430 3 × 420 VA at 40 A per phase; VAE-660 6 × 300 VA at 30 A per phase (VA-TEK data sheets). The constant-power envelope is an approximation — confirm the set’s V–I characteristic before relying on a marginal result.
3. How to read the result
Burden VA is only meaningful with its rated current
A relay’s burden is quoted in VA at a stated rated current, usually 1 A or 5 A. The impedance is Z = VA / In2, so the same 5 VA is 5 Ω at 1 A and 0.2 Ω at 5 A. Enter the rated current that belongs to the quoted burden, not the current you intend to inject.
The lead is a two-way circuit
Current leaves the set, passes through the relay and returns. Both conductors have resistance, so the loop resistance is twice the one-way lead resistance. The tool assumes a separate out-and-return pair; if the return is shared between phases, the loop is more complex and the result is optimistic.
Utilisation is a screening check, not a guarantee
A utilisation below 100% means the required apparent power is inside the set’s per-phase rating. It does not prove the set will hold that output indefinitely, stay within its thermal duty cycle or remain stable into an inductive load. A result close to 100% should be treated as marginal and confirmed against the datasheet envelope.
Recommended cross-section is a practical choice
The tool sizes the lead so the voltage dropped in the leads stays at or below your target percentage of the relay voltage. A 5% target keeps almost all of the set’s compliance voltage at the relay terminals. The recommendation is an engineering starting point; the lead must also be rated for the current, the duty cycle and the working voltage.
4. Formula and assumptions
| Quantity | Expression | Notes |
|---|---|---|
| Relay impedance | Z = VA / In2 | In is the rated current at which the burden is quoted. |
| Conductor resistivity | ρT = ρ20 × [1 + α(T − 20)] | Copper 0.0175 Ω·mm²/m, α = 0.00393; aluminium 0.0282, α = 0.00403. |
| One lead resistance | Rlead = ρT L / A | L one way in metres, A in mm². |
| Loop resistance | Rloop = 2 Rlead | Out and return. |
| Voltage required | V = I (Z + Rloop) | At the set terminals. |
| Apparent power | S = I V = I2 (Z + Rloop) | Compare with per-phase VA. |
| Lead drop share | I Rloop / V × 100% | How much of the set’s voltage is lost in the leads. |
5. Worked example — 5 VA relay, 10 m leads
Relay burden 5 VA at 5 A, so Z = 0.2 Ω. Copper leads, 10 m one way, 2.5 mm², 20 °C. At low current the loop is comfortable; at 40 A the same leads dominate the result.
| Test current | Lead CSA | Loop resistance | Lead drop | Total voltage | Apparent power | VAE-430 utilisation |
|---|---|---|---|---|---|---|
| 5 A | 2.5 mm² | 0.140 Ω | 0.70 V | 1.70 V | 8.5 VA | 2.0% |
| 40 A | 2.5 mm² | 0.140 Ω | 5.60 V | 13.6 V | 544 VA | 129.5% — exceeds |
| 40 A | 6 mm² | 0.058 Ω | 2.33 V | 10.3 V | 413 VA | 98.4% — marginal |
| 40 A | 10 mm² | 0.035 Ω | 1.40 V | 9.40 V | 376 VA | 89.5% |
The 2.5 mm² lead works at 5 A and fails at 40 A, even though the relay burden never changed. The lead drop and the relay voltage both scale with current, so their ratio stays the same — but the set’s available voltage falls as P / I. At 40 A there is only about 10.5 V to share, and the lead takes 5.6 V of it. That is the practical reason lead size belongs in the test plan, not in the toolbox.
6. Frequently asked questions
How do I convert relay burden VA to impedance?
Z = VA / In2, where In is the rated current at which the burden is quoted. A 5 VA burden at 5 A is 0.2 Ω; the same 5 VA at 1 A is 5 Ω. Using the wrong rated current changes the answer by a factor of 25.
Why is the lead length doubled?
Because the test current flows out to the relay and back to the set. Both conductors contribute resistance. The tool takes the one-way length and doubles it; if you already have the total loop length, halve it before entering it.
My set is rated 420 VA per phase — why does it fail at 40 A?
Because 420 VA at 40 A corresponds to only about 10.5 V. If the relay plus lead loop needs 13.6 V at 40 A, the required apparent power is about 544 VA, which is beyond that phase’s rating. The VA figure is a power envelope, not a voltage that is available at every current.
What lead cross-section should I use?
Size for the voltage drop you can tolerate at the highest test current, then check the current rating and the working voltage. The tool recommends a cross-section for a target percentage of the relay voltage; a 5% target is a reasonable starting point for timing tests.
Can long leads make a healthy relay appear to fail pickup?
Yes. The relay sees the voltage at its own terminals, which is lower than the set’s terminal voltage by the lead drop. If the set is already near its compliance limit, the relay may not pick up even though the set is operating correctly. Measure at the relay terminals where possible, and keep the lead run short or the cross-section large.
7. Where to go next
For the injection method and the classic mistakes, see secondary injection testing. For how channel count and per-phase power affect the choice of set, see how to choose a relay test set. The VAE-430 and VAE-660 data sheets give the per-phase power used here; for a configuration matched to your test plan, request a quote.
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