Welding cable voltage drop
A soft arc on a long lead is not the machine being difficult — it is volts lost on the way. Below: how many go at your current, length and cross-section, and how much of that turns into heat.
How it is worked out
One formula: ΔU = 2 · L · I · ρ / A, where L is
the length of one lead in metres, I the current in amps,
A the cross-section in mm², and ρ the resistivity of
the conductor: 0.0175 for copper, 0.0282 for aluminium.
The two in the formula is the return path. Current comes back through the work lead, so it travels twice the distance from machine to job. That is why the calculator asks for one lead, not for the sum of both.
The power lost in the cable is P = ΔU · I. At 200 A with
a 4 V drop that is 800 W — the cable heating instead of the metal
melting. That is where the warm lead after an hour's work comes from: not
a fault, just this arithmetic.
This calculator answers "what does the cable I already have cost me". The opposite question — "what cross-section do I need to stay within a given drop" — is answered by the cable cross-section calculator. The formula and the resistivities are the same, so the two always agree.
What counts as too much
| Drop | What it means in practice |
|---|---|
| up to 2 V | Not noticeable at the arc; the lead has margin |
| 2–4 V | Borderline. The arc still holds, but the cable runs warm |
| over 4 V | Soft arc, poor striking, electrode sticking on the start |
These bands are workshop practice, not a standard. They come from a simple comparison: a covered-electrode arc runs at roughly 25 V, so 4 V lost in the cable is a sixth of the voltage that is no longer where it is needed.
What to do when it comes out too high
A thicker lead works in direct proportion: double the cross-section and the drop halves. It is the one measure that always works.
A shorter run works just as strongly and is often cheaper: moving the machine to half the distance does what doubling the cable does. Surplus cable coiled in a ring does not shorten anything and, on alternating current, adds an inductive nuisance of its own.
The earth clamp is often more to blame than the cable. Contact resistance across rust, paint or a slack clamp fits into no formula and can eat more than ten metres of lead. The clamp goes on cleaned metal, as close to the joint as the job allows.
Aluminium instead of copper gives 60 % more drop at the same cross-section. Aluminium cable is cheaper, but to match copper it has to be two sizes thicker — and the price advantage disappears.
What next
Choosing a cross-section for a given drop — the cable cross-section calculator. How long the machine will run without a break — duty cycle. Running off a generator — generator for a welder. Why a wet lead and a bad earth are more than a soft arc — electrical safety.
How to use the calculator
- Welding current — the current you actually weld at, not the maximum on the plate. For a 3.2 mm electrode the current table gives 90–140 A, not 200.
- Length of one lead — the distance from the machine to the work along the cable route, not as the crow flies. The calculator doubles it for the return path.
- Cross-section — printed on the sheath: “H01N2-D 1×35” means 35 mm². Measure an unmarked cable rather than guess.
- Conductor material — copper or aluminium.
The result is three numbers: the drop in volts, its share of a typical arc voltage (the calculator takes 25 V) and the power heating the cable. The badge next to it rates the drop against the table above: up to 2 V fine, 2–4 V borderline, above 4 V too much.
Worked examples
| Situation | Inputs | Drop | Share | Power in cable |
|---|---|---|---|---|
| Stick on site | 200 A, 20 m, 25 mm² Cu | 5.6 V | 22 % | 1120 W |
| Same, 35 mm² cable | 200 A, 20 m, 35 mm² Cu | 4.0 V | 16 % | 800 W |
| Same, 50 mm² cable | 200 A, 20 m, 50 mm² Cu | 2.8 V | 11 % | 560 W |
| Machine moved closer | 200 A, 10 m, 25 mm² Cu | 2.8 V | 11 % | 560 W |
| TIG in the shop | 120 A, 10 m, 16 mm² Cu | ≈2.6 V | ≈11 % | ≈315 W |
| Aluminium cable | 200 A, 20 m, 35 mm² Al | ≈6.4 V | ≈26 % | ≈1290 W |
The first four rows show the rule from the section on fixes: going from 25 to 50 mm² and cutting the run from 20 to 10 m give exactly the same result. On site, moving the machine is often cheaper than buying cable. The last row shows that 35 mm² aluminium does worse than 25 mm² copper.
Unequal leads and extensions
The calculator assumes the welding lead and the earth lead are the same length. If they are not, enter their average: 25 m to the holder and 5 m of earth counts as 15 m. The formula adds both paths, so the average gives exactly the same answer.
When a lead is made of sections with different cross-sections, work out the drop of each section separately and add them. On top come the connectors, which the formula knows nothing about — so the calculator’s result is the lower bound of what is really lost.
A warm cable loses more
The resistivity of 0.0175 for copper applies at about 20 °C. Copper’s resistance rises by roughly 0.4 % per degree, so a cable warmed to 60 °C drops about 16 % more than a cold one. In practice a “borderline” lead turns into “too much” after an hour’s work in the sun. Spare cross-section is cheaper than it looks.
Calculation mistakes
- Adding both leads and entering the total as one lead. The result comes out twice too high.
- The plate current instead of the working current. A cable sized for the maximum is oversized for everyday work.
- Forgetting the interconnection to the wire feeder. On a MIG set with a separate feeder the current runs through the interconnecting cable, and its length belongs in the sum.
- Blaming the machine. Before sending it in for repair, work out the drop and check the earth clamp — see welder servicing.
Frequently asked questions
How many volts may a lead lose?
Workshop practice puts the ceiling at about 4 V at working current, and under 2 V is better. It is not a requirement of any standard, just the point beyond which the arc visibly changes: with roughly 25 V at the arc, every volt lost is four per cent gone.
Why is the length multiplied by two?
Because the current has to come back. The circuit closes through the work lead, so the metal carries current over twice the distance from machine to workpiece. The calculator asks for the length of one lead and doubles it itself — do not add both leads by hand.
Do joined extension leads change the result?
They do, and usually for the worse. Every connector adds contact resistance that this formula knows nothing about; a tarnished or loose plug can cost more than ten metres of cable. If the arc feels soft despite a correct cross-section, check the connectors and the earth clamp first.
What welding cable size for 20 m at 200 A?
For copper and 20 m per lead: 25 mm² drops 5.6 V (too much), 35 mm² 4.0 V (borderline), 50 mm² 2.8 V. For site work at 200 A, 50 mm² is sensible — or a shorter run: at 10 m, 25 mm² is enough.
What does a large voltage drop do to the arc?
Less voltage is left for the arc. With stick the arc gets shorter and softer, strikes badly and the rod sticks at the start; with MIG/MAG the arc runs colder, as if the voltage had been turned down. Meanwhile the cable heats up: 5.6 V at 200 A is more than a kilowatt lost.
How do I calculate if the earth lead is shorter than the welding lead?
Enter the average of the two: 25 m to the holder and 5 m of earth is 15 m. The formula adds both paths, so the average gives exactly the same result. Sections of different cross-section are calculated separately and the drops added.
Updated: