24 VDC Solenoid Control Cable Voltage Drop
A skid-mounted package has a 24 VDC power supply with ±10% output tolerance feeding four solenoid valves through one 2-wire control cable, 250 ft from the panel to the junction box. Each solenoid draws 0.5 A at 24 V and the four coils are paralleled at the far end, so the cable carries 2.0 A. The valve datasheet requires the coil to see at least 85% of rated voltage to guarantee pickup and to hold reliably; below that the valves chatter and burn out. The warehouse stocks 14 AWG copper (3.07 Ω per 1000 ft) and 12 AWG copper (1.93 Ω per 1000 ft); 10 and 8 AWG must be ordered. Ignore termination resistance and supply ripple.
1. Determine the minimum allowable loop resistance and the maximum conductor resistance per 1000 ft, then state which stocked cable, if any, is acceptable.
Read the worked answer
Worst-case voltage at the load:
- Minimum supply: 24 × 0.90 = 21.6 V
- Minimum acceptable coil voltage: 0.85 × 24 = 20.4 V
- Permissible drop: 21.6 − 20.4 = 1.2 V
Loop resistance limit: R_loop ≤ 1.2 V / 2.0 A = 0.60 Ω
Per-1000-ft limit: the loop is 2 × 250 ft = 500 ft, so R ≤ 0.60 Ω × (1000/500) = 1.20 Ω/1000 ft
Check the stock:
- 14 AWG: 3.07 × 0.5 = 1.535 Ω → drop 3.07 V → 18.53 V at the coil. Fails badly.
- 12 AWG: 1.93 × 0.5 = 0.965 Ω → drop 1.93 V → 19.67 V. Fails (below 20.4 V).
- 10 AWG: 1.21 × 0.5 = 0.605 Ω → drop 1.21 V → 20.39 V. Marginal fail - only 0.01 V short; do not rely on it, since supply tolerance and termination resistance consume the margin.
- 8 AWG: 0.764 × 0.5 = 0.382 Ω → drop 0.76 V → 20.84 V. Passes with ~0.4 V margin.
Conclusion: neither stocked size works. Order 8 AWG, or use one of the alternatives below.
2. The project schedule cannot absorb a cable order. Identify two field-alterable fixes that work with the material on hand, and one that sounds attractive but does not actually solve the problem.
Read the worked answer
Fix 1 - parallel the 12 AWG. Two 12 AWG conductors per leg halves the loop resistance: 0.965/2 = 0.483 Ω → drop 0.97 V → 20.63 V at the coil. Passes (margin ~0.2 V). Requires both terminals to accept two conductors and a check that the paralleled ampacity still covers the 2 A (trivially yes).
Fix 2 - sequence the valves. If the control scheme energizes one valve at a time, the cable carries 0.5 A, not 2.0 A. With a single 12 AWG pair: drop = 0.5 × 0.965 = 0.48 V → 21.12 V at the coil. Passes comfortably. This is a controls-logic change, not a hardware change, and should be confirmed against the process sequence and against any condition where two valves must shift simultaneously.
Fix that does not work - raising the supply setpoint to 24.0 V ±1%. That requires a regulated supply, not the one specified, so it is not a field fix. Note that if such a supply were used: budget = 23.76 − 20.4 = 3.36 V → R ≤ 1.68 Ω → 3.36 Ω/1000 ft, and 14 AWG (3.07) would pass with drop 3.07 V → 20.69 V. This is worth flagging to the panel designer as the real root cause: the cable is fine for a tightly regulated supply and fails only because of the ±10% tolerance.
Also check: solenoid inrush is typically 3–5× holding current for a few tens of milliseconds - this affects supply sizing and any series fuse, not the steady-state drop that governs pickup.
3. The panel designer proposes to keep one 14 AWG cable run but change the supply to 26 V. Evaluate this against the coil's maximum voltage and state the governing constraint.
Read the worked answer
Worst case with a 26 V nominal supply at +10%: 28.6 V at the panel. Drop on 14 AWG at 2 A is 3.07 V → 25.5 V at the coil, which is 106% of rated - usually acceptable for a DC solenoid (most are rated ±10% continuous) but it must be checked against the specific coil datasheet; many are only rated to 110% and some are not rated for continuous overvoltage at high ambient (coil temperature rise scales with V²).
At the other extreme (26 V −10% = 23.4 V): 23.4 − 3.07 = 20.33 V, which is below the 20.4 V pickup requirement. So the 26 V trick fails on the low end anyway unless the supply tolerance is tightened.
Governing constraint: the design must satisfy both ends of the tolerance simultaneously: V_supply,min − I·R_loop ≥ 0.85 × 24 and V_supply,max − I·R_loop ≤ V_coil,max
With the 14 AWG loop (1.535 Ω → 3.07 V drop) this requires a supply window narrower than roughly 23.5 V to 26.4 V, which is impractical to guarantee in the field. The robust answer is to set the conductor so the drop is small relative to the tolerance band (8 AWG gives 0.76 V), and treat supply voltage as a controlled parameter rather than a tuning knob.