Switch Loss, Battery Cost, and Heat

Switch Loss, Battery Cost, and Heat

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

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Ada ADA · CALCULATION AUDIT

Switch Loss, Battery Cost, and Heat

The chapter’s transistor choice is physics plus accounting: a 2 A motor switched by a BJT with beta 25, a 12 V valve drawing 300 mA for ten 30-second actuations a day against a 730-day target, and a 62 deg-C/W package. The channel turns current into heat, the drive spends battery, and the package temperature says whether the switch is safe. This audit weighs switch loss, battery cost, and heat from the chapter’s own values.

Companion to the chapter Transistor Selection — every number here comes from that chapter.

See the relationship before changing it

The figure reads from left to right. The blue card is rds(on). The middle card applies the page rule. The green card is package rise. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

Rds(on) changes package rise An input card leads through the rule rise = (2 A)^2 x Rds(on) / 1000 x 62 deg C/W to the package rise result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. The package rise combines channel resistance, current, and thermal resistance.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 28 mOhm.

  2. 2

    Name the relationship. rise = (2 A)^2 x Rds(on) / 1000 x 62 deg C/W

  3. 3

    Substitute with units. 4 x 28 / 1000 x 62 = 6.94 deg C

  4. 4

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

Predict, then change rds(on)

Try Predict the direction of rise = (2 A)^2 x Rds(on) / 1000 x 62 deg C/W. Test another rds(on), then compare package rise.

28 mOhm
Chapter baseline
Package rise

Observe The package rise combines channel resistance, current, and thermal resistance. Reset rds(on) to 28 and compare package rise.

Explain The package rise combines channel resistance, current, and thermal resistance.

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 rds(on) moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only rds(on). It is not a field prediction.

TryThe chapter’s transistor choice is physics plus accounting: a 2 A motor switched by a BJT with beta 25 , a 12 V valve drawing 300 mA for ten 30-second actuations a day against a 730-day target, and a 62 deg-C/W package. Use Check derivation.
ObserveThe displayed ledger resolves 2 A, beta 25, 12 V, 300 mA, 30-second at full precision. This audit weighs switch loss, battery cost, and heat from the chapter’s own values. Check derivation shows this.
ExplainThe BJT pays both one watt of saturation loss and 0.344 W of base-drive loss, whereas the MOSFET pays 0.112 W of I-squared-R loss; the resulting 12x gap also changes package rise from 83.3 C to 6.94 C. Check derivation confirms it.

Ada: A transistor choice is physics plus accounting. The channel or junction turns current into heat, the gate or base drive spends battery current, and the package temperature tells you whether the switch is merely rated on paper or actually safe in the enclosure. The chapter runs two separate worked examples – a 2 A motor comparison and a 300 mA valve – so let me re-derive both to full precision.

  • BJT base current (2 A motor, beta 25). I_B = I_C / beta = 2.00 / 25 = 0.0800 A = 80.0 mA – already past what a typical GPIO pin can source, which is why the chapter assumes a driver stage.
  • BJT heat, conduction plus drive. Conduction: P_CE = V_CE(sat) x I_C = 0.5 x 2.00 = 1.00 W. Base drive from a 5 V driver with a 0.7 V V_BE drop: P_base = (5.00 - 0.700) x 0.0800 = 0.344 W. Total: 1.00 + 0.344 = 1.344 W.
  • MOSFET channel loss for the same load. IRLZ44N at Rds(on) = 28 mOhm @ Vgs = 5 V: P = I_D^2 x R_DS(on) = (2.00)^2 x 0.0280 = 0.112 W.
  • Efficiency check. 1.344 / 0.112 = 12.0 – the chapter’s “12x more efficient” holds to three significant figures.
  • Thermal meaning at 62 deg-C/W. MOSFET: 0.112 x 62 = 6.94 deg-C rise. BJT: 1.344 x 62 = 83.3 deg-C rise – before any outdoor ambient margin, which is the real argument for the MOSFET, not just the wattage.
  • The 300 mA valve’s daily MOSFET loss. At Rds(on) = 35 mOhm (the same IRLZ44N re-specified at the valve’s 3.3 V gate drive): P = (0.300)^2 x 0.0350 = 0.00315 W = 3.15 mW. Active time is 10 x 30 s = 300 s = 0.0833 h per day, so daily energy is 0.00315 x 0.0833 = 0.0002625 Wh = 0.2625 mWh/day.
  • Two-year battery cost. Over 730 days: 0.0002625 x 730 = 0.1916 Wh, rounding to the chapter’s 0.192 Wh. At the 6 V battery string: 0.1916 / 6 = 0.03194 Ah = 31.94 mAh. Against 2500 mAh of AA capacity: 31.94 / 2500 = 1.28% of the string – negligible, exactly as the chapter claims.
  • Package temperature rise for the valve MOSFET. 0.00315 x 62 = 0.1953 deg-C – effectively unmeasurable, consistent with the chapter’s own “essentially unmeasurable” validation note.

Every figure reproduces the chapter’s own two deep-dive tables exactly. The two worked examples also cross-check each other: the same IRLZ44N part carries a lower Rds(on) at the higher 5 V gate drive (28 mOhm) and a higher Rds(on) at the lower 3.3 V gate drive (35 mOhm) – the correct direction for a MOSFET’s gate-voltage dependence, not a contradiction between the two scenarios.

Technical boundaries
The switch ledger deliberately does not simulate switching loss, gate-charge waveforms, thermal coupling, transient SOA, motor inrush, or datasheet spread; it compares stated steady conduction and drive losses.

Work the audit first, then check the displayed derivation.

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