- 2075 Ashwin · 5 marks
Diode current and voltage, 0.1 V per decade
Short answer
Using VD=0.7+0.1 log10(ID/1mA) and KVL ID=(5−VD)/1k, iteration converges to VD≈0.763 V and ID≈4.24 mA.
Determine the current ID and the diode voltage VD with VDD=5 V and R= 1KΩ. Assume that the diode has a current of 1 mA at a voltage of 0.7 V and that its voltage drop changes by 0.1 V for every decade change in current. [Figure: series circuit of VDD, resistor R and a diode, with current ID and diode voltage VD]
The diode voltage changes by 0.1 V for every tenfold (decade) change in current, i.e.
VD = 0.7 + 0.1 log10(ID / 1 mA)
Together with KVL, ID = (VDD - VD)/R, solve by iteration.
Given
VDD = 5 V, R = 1 kΩ, VD = 0.7 V at 1 mA.
Iteration
| Step | VD guess (V) | ID = (5 - VD)/1 kΩ (mA) | New VD = 0.7 + 0.1 log10(ID) (V) |
|---|---|---|---|
| 1 | 0.700 | 4.300 | 0.7633 |
| 2 | 0.7633 | 4.237 | 0.7627 |
| 3 | 0.7627 | 4.237 | 0.7627 |
Converged.
ID = (5 - 0.763)/1 kΩ = 4.24 mA
Answer: ID ≈ 4.24 mA, VD ≈ 0.763 V.
✓ Answer checked by Claude Opus 5.5 (high effort)
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