Regulator Trade-Off Calculation Audit

Regulator Trade-Off Calculation Audit

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

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Ada ADA · CALCULATION AUDIT

Regulator Trade-Off Calculation Audit

A linear regulator (LDO) passes the full load current through and burns the leftover voltage as heat, while a switching buck trades that voltage for current instead. The chapter’s worked rail delivers 100 mA at 3.3 V from a full 4.2 V Li-ion cell, where the LDO runs at about 79% efficiency and, active five minutes an hour, wastes enough to cost roughly 29 mAh a day against a 90% buck. This audit carries both converters through the full duty cycle to weigh the regulator trade: which choice actually wastes less?

Companion to the chapter Hardware Optimization Strategies — every number here comes from that chapter.

Ada: A regulator choice is a conservation-of-energy check. Use the load power, then carry the converter and quiescent losses through the full duty cycle before deciding which part is efficient.

See the relationship before changing it

The figure reads from left to right. The blue card is active time each hour. The middle card applies this page's rule. The green card is daily charge saved by the buck. 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 active time each hour, so the numeric fixture does not switch without explanation.

Active time each hour changes daily charge saved by the buck An input card leads through the rule saving = 7.2517 mA x active minutes x 24 / 60 to the daily charge saved by the buck result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. Longer active time repeats the buck converter current saving more often.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 5 min.

  2. 2

    Name the relationship. saving = 7.2517 mA x active minutes x 24 / 60

  3. 3

    Substitute with units. 7.2517 x 5 x 24 / 60 = 14.50 mAh/day

  4. 4

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

Predict, then change active time each hour

Try Predict the direction of saving = 7.2517 mA x active minutes x 24 / 60. Test another active time each hour, then compare daily charge saved by the buck.

5 min
Chapter baseline
Daily charge saved by the buck

Observe Longer active time repeats the buck converter current saving more often. Reset active time each hour to 5 and compare daily charge saved by the buck.

Explain Longer active time repeats the buck converter current saving more often.

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 active time each hour moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only active time each hour. It is not a field prediction.

TryRecompute Heavy-load rail: compare battery current, not only efficiency. The chapter's worked rail is 3.3 V at 100 mA from a full 4.2 V Li-ion cell.
ObserveTrack Daily active loss: integrate the rail over time. The chapter's rail is active for 5 minutes/hour , which is 2 hours/day .
ExplainExplain Sleep rail: quiescent current can reverse the decision. The sleep example uses a 10 uA load, a buck controller drawing 15 uA , and a low-Iq LDO drawing 1 uA .

Ready: use the stated baseline inputs, then compare each displayed result.

1. Heavy-load rail: compare battery current, not only efficiency. The chapter's worked rail is 3.3 V at 100 mA from a full 4.2 V Li-ion cell.

P_load = 3.3 V x 100 mA = 330 mW
LDO input = 4.2 V x 100 mA = 420 mW; waste = 420 - 330 = 90 mW; efficiency = 330 / 420 = 0.785714 = 78.6%
90% buck input = 330 mW / 0.90 = 366.667 mW; battery current = 366.667 / 4.2 = 87.302 mA; loss = 36.667 mW

The buck therefore draws 100 - 87.302 = 12.698 mA less from the battery during this phase, or 12.7% less current after final rounding.

2. Daily active loss: integrate the rail over time. The chapter's rail is active for 5 minutes/hour, which is 2 hours/day.

LDO daily waste = 90 mW x 2 h = 180 mWh/day
Buck daily loss = 36.667 mW x 2 h = 73.333 mWh/day; saving = 180 - 73.333 = 106.667 mWh/day
Cell-equivalent saving = 106.667 mWh / 3.7 V = 28.829 mAh/day, which rounds to about 29 mAh/day

3. Sleep rail: quiescent current can reverse the decision. The sleep example uses a 10 uA load, a buck controller drawing 15 uA, and a low-Iq LDO drawing 1 uA.

Buck sleep draw = 10 + 15 = 25 uA; effective light-load efficiency = 10 / 25 = 40.0%
Low-Iq LDO sleep draw = 10 + 1 = 11 uA; effective light-load efficiency = 10 / 11 = 90.9%
For 23.76 h asleep: buck = 25 x 23.76 = 594.0 uAh; LDO = 11 x 23.76 = 261.36 uAh; saving = 332.64 uAh/day

The physics lesson is why the chapter separates heavy-load and light-load regimes: the best active regulator can be the wrong sleep regulator when its own controller current is larger than the load.

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

Technical boundaries. This regulator ledger deliberately does not simulate converter transients, efficiency curves, thermal limits, or battery-voltage sag. It uses the fixed 3.3 V load, 4.2 V source, 100 mA draw, 90% buck efficiency, quiescent currents, and duty cycle.