Battery Bruno guidesA field team faces an unresolved physical question: Why can a pull-up defeat a powered-down sensor rail? They must answer it before changing forward voltage on the real device. Predict the direction first.
See the relationship before changing it
The figure reads from left to right. The blue card is forward voltage. The middle card applies this page's relationship. The green card is naive current. 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 added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline for forward voltage is 0.6.
- 2
Name the relationship. Inaive=3.3/4700=0.702 mA Icorrected=(3.3-0.6)/4700=0.574 mA Qleak=0.574x2x60=68.9 mA·s Qprocessor=0.010 mAx60=0.600 mA·s Qleak/Qprocessor=114.9
- 3
Substitute the chapter fixture. Set forward voltage to 0.6. The page ledger gives naive current as 0.702 mA.
- 4
Read the result. Keep mA beside the value. Use it only inside the technical boundary on this page.
Predict, then change forward voltage
Try Predict the direction of naive current. Move one control, calculate, then check your prediction.
Observe Ohm's law still applies, but only to the resistor's share of the rail voltage. Reset the control to 0.6 and compare naive current.
Explain Only forward voltage moves here. The other chapter fixtures remain fixed.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. Start with the physical story
A live signal can feed a depowered device through its input protection diode. The input is not a zero-volt short: the forward-biased junction keeps part of the rail voltage.
2. Name every algebra move
Find the naive leakDivide the full rail by the pull-up.
Remove junction voltageSubtract Vf before dividing.
Price all linesMultiply current by two signal lines.
Integrate sleepMultiply by 60 seconds.
Compare floorsDivide leak charge by processor sleep charge.
3. Reproduce the chapter case
Icorrected=(3.3−0.6)/4700=0.574 mA
Qleak=0.574×2×60=68.9 mA·s
Qprocessor=0.010 mA×60=0.600 mA·s
Qleak/Qprocessor=114.9
The 18.2% correction matters to the quoted value but does not overturn the whole-island gating decision.
4. Try one real input
TryMove the diode drop and predict the corrected leakage.
ObserveA larger junction drop leaves less voltage across the resistor, so the leak falls.
ExplainOhm's law still applies, but only to the resistor's share of the rail voltage.
This is a constant-drop estimate, not a diode-characterisation model.
- Junction
- Real Vf changes with current and temperature.
- Return path
- The board must be inspected to confirm where back-power flows.
- Damage
- An energy estimate does not prove pin injection is safe.
Correct, not complete: use this to expose the leak, then verify the real circuit and ratings.
5. Use the result in the design
Gate rail, pull-ups, and signals together, or add isolation that keeps unpowered pins inside their injection-current ratings.
6. Record the evidence state
Keep rail state, pull-up values, line count, pin schematic, diode path, measured leakage, temperature, sleep interval, and injection limit.
7. Check yourself
Why not divide 3.3 V by 4.7 kΩ?
Does the correction make the leak harmless?
Does this model prove the pin is safe?
The arithmetic uses the chapter's pull-up and a catalog-typical constant junction drop.
- Junction
- Real Vf changes with current and temperature.
- Return path
- The board must be inspected to confirm where back-power flows.
- Damage
- An energy estimate does not prove pin injection is safe.
Correct, not complete: use this to expose the leak, then verify the real circuit and ratings.