Actuator Driver Power Path Calculation Audit
Actuator Driver Power Path Calculation Audit
Ada re-derives this chapter’s own numbers step by step, at full precision
ADA · CALCULATION AUDIT
Actuator Driver Power Path Calculation Audit
A greenhouse fan and an irrigation solenoid both run from 12 V, drawing 6 W and 4.8 W, but the number that decides reliability is the 2 A startup surge: pushed through 0.3 Ω of wire and connector resistance it drops 0.6 V, so the load sees only 11.4 V while the wiring itself burns 1.2 W. This audit re-derives each figure with the power law and Ohm’s law and asks whether that 0.6 V sag is trivial, or enough to brown out a marginal 3.3 V controller sharing the same rail.
Companion to the chapter How Actuators Work — every number here comes from that chapter.
Two equations govern every number on this page: the power law P = V × I for how much the load draws, and Ohm's law V = I × R for what the wiring steals on the way. Each value below is one the chapter already gave; I only carry the arithmetic through.
See the relationship before changing it
The figure reads from left to right. The blue card is startup current. The middle card applies this page's rule. The green card is wiring voltage drop. Walk the arrows once: set the input, apply the rule, then read the result with its unit.
The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only startup current, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 2 A.
- 2
Name the relationship. wiring drop = startup current x 0.30 ohm
- 3
Substitute with units. 2.0 A x 0.30 ohm = 0.60 V
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change startup current
Try Predict the direction of wiring drop = startup current x 0.30 ohm. Test another startup current, then compare wiring voltage drop.
Observe A larger startup surge loses more voltage in the same wire and connector resistance. Reset startup current to 2 and compare wiring voltage drop.
Explain A larger startup surge loses more voltage in the same wire and connector resistance.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. Running power is the power law.
The greenhouse fan and the irrigation solenoid both run from 12 V; only their currents differ:
2. The wiring is a resistor in series.
At the 2 A startup surge, 0.3 Ω of wire and connector resistance drops voltage before the driver even sees the load, so the load receives less than the nominal 12 V:
3. That lost voltage becomes heat, and a held coil spends energy.
The wiring burns I²R, and the solenoid's safe 5 s energize window sets its energy budget:
| Quantity | Arithmetic shown | Result |
| Fan running power | 12 × 0.5 | 6 W |
| Solenoid running power | 12 × 0.4 | 4.8 W |
| Startup wiring drop (Ohm's law) | 2 × 0.3 | 0.6 V |
| Voltage actually reaching the load | 12 − 0.6 | 11.4 V |
| Heat burned in the wiring (I²R) | 2² × 0.3 | 1.2 W |
| Solenoid energy over a 5 s hold | 4.8 × 5 | 24 J |
What this means for your design: the 0.6 V startup sag looks trivial, but it lands on the same rail that feeds a 3.3 V regulator, and 0.6 V of ripple during inrush is enough to brown out a marginal controller and reset it. The 1.2 W dissipated in a thin connector is exactly what makes it warm to the touch. Give the load its own supply sized to the startup current, not the running current, and read the 0.6 V as evidence that wiring resistance and inrush — not steady power — decide whether the board resets.
Every number above is taken from the chapter’s own examples and re-derived step by step.
Audit result
At startup the load sees 11.4 V and the wiring dissipates 1.2 W; whether the controller resets requires rail-transient measurement.