Motor Load Calculation Audit

Motor Load Calculation Audit

Ada re-derives this chapter’s own numbers step by step, at full precision

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Ada ADA · CALCULATION AUDIT

Motor Load Calculation Audit

A 6 V gearmotor that draws 180 mA free-running pulls 600 mA under load and 1.8 A at stall, yet the example driver is rated only 700 mA — about 2.57 times below the stall case. Its loaded running power is already 3.6 W, and a 2.0 A spike is normal for 80 ms but signals a jam if it lasts 2 s, exposing the driver to 25 times the startup charge. This audit re-derives each figure and asks whether a driver sized to the tidy running point is a safety argument, or whether startup and blocked-load behavior must be sized on their own.

Companion to the chapter DC Motors — every number here comes from that chapter.

— current, power, and fault timing, ~4 minutes

The mathematics checks the electrical budget; the physics reminder is that current changes as the motor starts, loads, stalls, and generates back-EMF. A driver sized only for the usual running point is not a safety argument.

See the relationship before changing it

The figure reads from left to right. The blue card is loaded motor current. The middle card applies this page's rule. The green card is loaded electrical power. 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 loaded motor current, so the numeric fixture does not switch without explanation.

Loaded motor current changes loaded electrical power An input card leads through the rule motor power = 6 V x loaded current to the loaded electrical power result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. Loaded current raises electrical power while stall current remains a separate driver-sizing boundary.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 0.6 A.

  2. 2

    Name the relationship. motor power = 6 V x loaded current

  3. 3

    Substitute with units. 6 V x 0.60 A = 3.60 W

  4. 4

    Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change loaded motor current

Try Predict the direction of motor power = 6 V x loaded current. Test another loaded motor current, then compare loaded electrical power.

0.6 A
Chapter baseline
Loaded electrical power

Observe Loaded current raises electrical power while stall current remains a separate driver-sizing boundary. Reset loaded motor current to 0.6 and compare loaded electrical power.

Explain Loaded current raises electrical power while stall current remains a separate driver-sizing boundary.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only loaded motor current moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only loaded motor current. It is not a field prediction.

The chapter’s values

Use only the chapter values above: a 6 V gearmotor, 180 mA no-load current, 600 mA loaded running current, 1.8 A stall current, a 700 mA driver example, a 70% PWM command, and the 2.0 A current spike at 80 ms versus 2 s.

power = voltage x current; average_voltage = duty x supply; charge = current x time

The worked checks

Check Arithmetic Review meaning
Running power 6 V x 0.600 A = 3.6 W The loaded motor is already a multi-watt power path, not a GPIO load.
No-load versus loaded current 0.600 A / 0.180 A = 3.33 The real mechanism draws about 3.3 times the bench no-load current.
Stall versus driver rating 1.8 A / 0.700 A = 2.57 A 700 mA driver is about 2.6 times below the stall case, so the fault path must be explicit.
PWM average voltage 0.70 x 6 V = 4.2 V The command matches the chapter's average-voltage example, but it still does not prove shaft speed.
Startup spike charge 2.0 A x 0.080 s = 0.16 A*s A short spike can be normal if supply sag, thermal stress, and restart behavior are bounded.
Jam exposure 2.0 A x 2 s = 4.0 A*s; 4.0 / 0.16 = 25 The two-second jam exposes the driver and supply to 25 times the startup charge.

Audit conclusion: the chapter's own numbers support the review rule: size for startup and blocked-load behavior, not just the running current, and pair PWM commands with current, speed, limit, airflow, pressure, or timeout evidence.

Every number above is taken from the chapter’s own examples and re-derived step by step.

TryCompare loaded, stall, and startup-current cases against the 700 mA driver rating.
ObserveStall current is 1.8 A and a 2 A spike lasting 2 s carries 25x the startup charge at 80 ms.
ExplainA running-current rating cannot establish jam or startup safety.
Technical boundaries. The audit uses stated steady currents, voltage, and two rectangular spike durations. It does not model motor inductance, back-EMF dynamics, PWM ripple, driver current limiting, thermal impedance, mechanical inertia, supply droop, or jam detection tolerance.
Audit result

A 700 mA driver is below the 1.8 A stall case; 2 A for 2 s transfers 25 times the charge of 2 A for 80 ms.