Max guidesA technician must decide whether predicted diode clamp is safe before changing current ratio on the real device. The result is unresolved until the rule and units are checked. Predict the direction first.
See the relationship before changing it
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 current ratio, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 8 times.
- 2
Name the relationship. clamp = 0.650 V + 0.0595 log10(current ratio)
- 3
Substitute with units. 0.650 + 0.0595 log10(8) = 0.704 V
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change current ratio
Try Predict the direction of clamp = 0.650 V + 0.0595 log10(current ratio). Test another current ratio, then compare predicted diode clamp.
Observe Diode voltage grows logarithmically even when current multiplies quickly. Reset current ratio to 8 and compare predicted diode clamp.
Explain A forward-biased diode carries exponentially more current for modest voltage increases, so the clamp voltage rises only logarithmically with current ratio.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. A diode is not a resistor
A resistor needs voltage proportional to current. A forward-biased silicon junction carries exponentially more current for each small voltage increase.
2. Compare two points
Start with the junction lawI≈IS e^(V/nVT) when forward current is well above IS.
Take a ratioI2/I1=e^((V2−V1)/nVT).
Undo the exponentialUse ln: V2−V1=nVT ln(I2/I1).
3. Read the room-temperature scale
A tenfold current change costs only about 59.5 mV in this ideal comparison.
4. Try the flyback current
TryMove the current from the 10 mA reference toward the chapter's 80 mA coil current.
ObserveAt 80 mA, the ratio is 8.00×, ΔV=0.0538 V, and the predicted clamp is 0.704 V. Current rose eightfold while voltage rose only about 8.3%.
ExplainThe logarithm compresses a large current ratio into a small voltage change. That is why “about 0.7 V” can be a useful first estimate over this range.
The diode comparison keeps temperature, the 10.0 mA reference at 0.650 V, and an ideality factor of one fixed while current ratio changes. It excludes series resistance, pulse ratings, recovery, heating, wiring inductance, part variation, coil dynamics, avalanche, and the transistor's safe operating limits.
- series resistance
- Needs separate evidence
- pulsed-current ratings
- Needs separate evidence
- recovery
- Needs separate evidence
- junction heating
- Needs separate evidence
- wiring inductance
- Needs separate evidence
- diode tolerance
- Needs separate evidence
- coil dynamics
- Needs separate evidence
- avalanche
- Needs separate evidence
- the transistor's safe operating area
- Needs separate evidence
Use field evidence or a deeper model before release.
5. Keep energy and clamp voltage separate
The chapter's 0.32 mJ coil event says how much magnetic energy must go somewhere. The diode equation estimates junction voltage at a stated current. Energy does not turn directly into a proportional clamp voltage.
6. Verify protection
Measure the switch-node peak and decay with a safe probe. Record coil current and inductance, diode part and temperature, repetition rate, transistor rating, wiring, and the final safe state.
7. Check yourself
What mathematical move turns an exponential into a voltage difference?
Why is 0.650 V a labelled reference, not a universal constant?
Does a 0.704 V estimate prove the transistor is safe?
The diode comparison keeps temperature, the 10.0 mA reference at 0.650 V, and an ideality factor of one fixed while current ratio changes. It excludes series resistance, pulse ratings, recovery, heating, wiring inductance, part variation, coil dynamics, avalanche, and the transistor's safe operating limits.
- 1N4148-class
- Example flyback diode class
- 10.0 mA
- Diode reference current
- 0.650 V
- Forward voltage at reference current
- 80 mA
- Coil current before switch-off
- 25.85 mV
- Thermal voltage at 300 K
- 2.079
- Natural logarithm of eightfold current ratio
- 0.0538 V
- Forward-voltage increase at 80 mA
- 0.704 V
- Predicted forward clamp at 80 mA
- 8×
- Coil-to-reference current ratio
- about 7.7% in the chapter's rounded comparison
- Voltage increase relative to final clamp
- 59.5 mV/decade
- Forward-voltage increase per tenfold current rise
- 8,000 V ideal spike
- Ideal unclamped coil switch-off voltage
- 0.32 mJ
- Stored magnetic energy in coil
Real clamps require datasheets and measurement.