A 24 V loop-powered transmitter at the far end of a 1000-foot run — checking it still gets enough terminal voltage at the worst-case 20 mA signal, and how much farther the same wire gauge could actually reach.
| Loop supply voltage | 24 V |
| Transmitter minimum terminal voltage (at 20 mA) | 12 V |
| Receiver / PLC input impedance | 250 Ω |
| Wire gauge | 22 AWG (16.14 Ω per 1000 ft) |
| One-way cable run | 1000 ft |
| Check | Requirement | Actual | Status |
|---|---|---|---|
| Transmitter terminal voltage at 20 mA | ≥ 12 V | 18.35 V | ✓ PASS |
| Headroom | ≥ 0 V | 6.35 V | ✓ PASS |
Key insight: The voltage budget check always has to use 20 mA (the worst case), not the loop's typical operating current — wire voltage drop is highest exactly when the loop is signaling its maximum value, which is precisely the condition that must never cause the transmitter to starve for voltage and clip its own output.
Every input in this example is editable in the live calculator — free, no signup.
Open 4-20mA Current Loop calculator →Any additional series resistance — an indicator, a second barrier, an isolator — adds directly to Rtotal in the same way the receiver resistance does, reducing both the transmitter's available terminal voltage and the maximum supportable cable length. This calculator's 'other Ω' input exists specifically to let additional loop components be included in the same budget rather than being overlooked.
Yes — resistance per unit length drops roughly by half with each two-gauge step up (e.g. 22 AWG to 18 AWG cuts resistance to about 40% of the 22 AWG value), which directly multiplies the maximum supportable length for the same voltage budget. For genuinely long runs approaching a wire gauge's practical limit, stepping up gauge is often more effective than trying to increase supply voltage, which has its own upper limits from device and barrier ratings.