Pulsed Welding Heat Input Calculator
Enter peak and background current, voltage, duty cycle, pulse frequency, travel speed, and efficiency to estimate pulsed welding arc energy and heat input. The result also shows the travel speed that would meet a maximum heat-input target with the same waveform settings.
Efficiency-adjusted heat input (kJ/mm)0.37
- Duty-weighted average current (A)
- 95.5
- Duty-weighted average arc power (W)
- 1,239.5
- Pulse period (ms)
- 666.67
- Peak and background duration
- 233.33 ms peak, 433.33 ms background
- Electrical energy per pulse cycle (J)
- 826.33
- Arc energy before efficiency (kJ/mm)
- 0.62
- Target utilization
- 46.48%
- Minimum travel speed for target (mm/min)
- 55.78
- Target check
- Within target
Uses a two-level square pulse with measured or representative voltage for each phase.
Apply the 1.00 efficiency option when the governing document asks for unadjusted arc energy.
How to use this calculator
- Enter the peak and background current from measured values or qualified machine settings.
- Enter voltage for each phase, or repeat one representative arc voltage in both fields.
- Set peak time, pulse frequency, travel speed, and the efficiency convention required for the job.
- Compare the adjusted heat input with the target and use the minimum travel speed if the limit is exceeded.
How pulsed welding heat input is calculated
Pulsed welding alternates between a peak phase and a background phase, so one average current does not fully describe the waveform. This calculator duty-weights the electrical power in each phase first, then converts that average power into arc energy per length of weld.
Let Ip be peak current, Ib background current, Up peak voltage, Ub background voltage, d peak time in percent, f pulse frequency, v travel speed, eta thermal efficiency, and Qmax the target heat input. The peak duty fraction is D = d / 100.
The duty-weighted average current is Iavg = D x Ip + (1 - D) x Ib. The average arc power is Pavg = D x Up x Ip + (1 - D) x Ub x Ib. Arc energy before efficiency is Ea = 60 x Pavg / (1000 x v) in kJ/mm. Efficiency-adjusted heat input is Q = eta x Ea.
The pulse diagnostics use the same duty fraction. Pulse period is T = 1 / f seconds, peak duration is D x T, background duration is (1 - D) x T, and electrical energy per complete pulse cycle is Pavg / f joules. Frequency changes these durations and the energy in each cycle, but it does not change average heat input unless the phase power, duty cycle, or travel speed also changes.
What moves the result most
Average phase power and travel speed dominate the heat input. Raising peak current, peak voltage, or peak time increases the average power. Slower travel speed spreads the same power over less weld length, so heat input rises directly. If the same voltage is entered for both phases, the power equation reduces to that voltage multiplied by the duty-weighted average current.
The target-speed result reverses the heat-input equation: vtarget = 60 x eta x Pavg / (1000 x Qmax). When the target is a maximum limit, the actual travel speed must be at least that value if the other settings stay fixed.
What this calculator leaves out
The model is a two-level square pulse. It does not represent ramps, polarity changes, short-circuit transfer, synergic programs, proprietary waveforms, or direct accumulated-energy measurements. It also does not predict penetration, cooling time, distortion, microstructure, or mechanical properties. Use the efficiency factor and measurement convention required by the governing WPS, code, or engineering procedure.
Worked example
Suppose a pulsed GTAW setup uses 180 A and 14 V during peak, 50 A and 11 V during background, 35% peak time, 1.5 Hz pulse frequency, 120 mm/min travel speed, a 0.60 efficiency factor, and a 0.80 kJ/mm target.
The average current is 0.35 x 180 + 0.65 x 50 = 95.5 A. The average arc power is 0.35 x 14 x 180 + 0.65 x 11 x 50 = 1,239.5 W. At 1.5 Hz, one pulse period is 666.67 ms, with 233.33 ms at peak and 433.33 ms at background. The electrical energy per cycle is 826.33 J.
Arc energy is 60 x 1,239.5 / (1000 x 120) = 0.61975 kJ/mm. Applying 0.60 efficiency gives 0.37185 kJ/mm, which uses about 46.48% of the 0.80 kJ/mm target. The same settings would meet the target at about 55.78 mm/min or faster.
Common questions
How do I calculate average current for pulsed TIG welding?
Convert peak time to a fraction, multiply it by peak current, and add the background current multiplied by the remaining fraction. For example, 35% at 180 A and 65% at 50 A gives 95.5 A. Heat input should still average phase power when the peak and background voltages differ.
Why average peak and background power separately?
Electrical power is voltage times current. If voltage changes between the peak and background portions of the pulse, averaging current first and multiplying by one voltage can miss part of the waveform. Duty-weighting each phase power keeps the calculation traceable to the measured pulse settings.
Does pulse frequency change welding heat input?
Frequency changes the period length and the joules in each pulse cycle. It does not change average heat input by itself when duty cycle, phase current, phase voltage, and travel speed remain the same. In real welding, changing frequency can also change bead behavior or measured arc values, so use the values actually used for the weld.
What is the difference between arc energy and heat input?
Arc energy is the electrical energy per weld length before an efficiency factor. Heat input often means arc energy multiplied by a thermal-efficiency factor specified by a code, procedure, or engineering convention. Choose 1.00 when the requirement asks for unadjusted arc energy.
How do I find the travel speed needed to stay below a WPS heat-input limit?
Hold the waveform settings fixed and solve the heat-input equation for travel speed. This calculator reports that speed in mm/min. If the target is a maximum, welding slower than that value would exceed the target under the entered assumptions.