Engineering

Voltage drop

Voltage lost in a copper wire run, as volts, a share of supply, and heat.

Inputs

FRC rules set minimum gauges per breaker size; check the current manual.

From the PDH to the device. Both wires are counted.

Result

Voltage drop

0.7623
Share of supply
6.353%
Voltage at the device
11.24
Heat in the wire
30.49
Round-trip resistance
19.06

Check this: More than 5% of the supply is lost in the wire. Consider a thicker gauge or a shorter run.

How it’s calculated

Formula

d=0.127 mm×92(36−n)/39,A=πd24d = 0.127\,\text{mm} \times 92^{(36 - n)/39}, \qquad A = \frac{\pi d^2}{4}
R=ρ 2LA,Vdrop=IR,Ploss=I2RR = \frac{\rho \, 2L}{A}, \qquad V_\text{drop} = I R, \qquad P_\text{loss} = I^2 R

With your numbers (converted to SI units)

d=0.127×92(36−12)/39=2.053 mmd = 0.127 \times 92^{(36 - 12)/39} = 2.053\,\text{mm}
R=1.724×10−8×2×1.829 m3.309×10−6 m2=0.01906 ΩR = \frac{1.724 \times 10^{-8} \times 2 \times 1.829\,\text{m}}{3.309 \times 10^{-6}\,\text{m}^2} = 0.01906\,\Omega
Vdrop=40 A×0.01906 Ω=0.7623 VV_\text{drop} = 40\,\text{A} \times 0.01906\,\Omega = 0.7623\,\text{V}

What it means

Every wire has resistance, so current through it costs voltage before it reaches the device. Motors are the main concern: losing 0.5 V on a run to a drive motor at high current costs top speed and pushing force. Shorter runs and thicker wire (smaller AWG number) cut the loss.

Assumptions

  • Solid copper at 20 °C. Stranded wire (what FRC robots use) is about 2–3% higher, and warm wire about 0.4% higher per °C.
  • Constant current: the load draws the same current whatever voltage reaches it.
  • Connectors and crimps add resistance on top of this.

Learn the concepts behind this on resources.simbotics.org · All formulas

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