- 2082 Shrawan · 6 marks
Diode current and voltage, VDD 5 V, R 1 kΩ
Short answer
Iterative solution of diode law VD=0.7+0.05 ln(ID/1mA) and KVL ID=(5−VD)/1k yields VD≈0.772 V, ID≈4.23 mA.
Determine the current ID and diode voltage VD with VDD = 5 V and R = 1 kΩ. Assume that the current through the diode is 1 mA at a voltage of 0.7 V. Assume η = 2. [Figure: series circuit of source VDD, resistor R and a diode, with current ID and diode voltage VD]
Combine the diode's exponential law with KVL and solve by iteration.
Given
VDD = 5 V, R = 1 kΩ, ID = 1 mA at 0.7 V, η = 2, VT = 25 mV (η VT = 50 mV).
Equations
Diode: VD = 0.7 + η VT ln(ID/1 mA) = 0.7 + 0.05 ln(ID/1 mA)
KVL : ID = (VDD - VD)/R = (5 - VD)/1 kΩ
Iteration
| Step | VD (V) | ID (mA) = 5 - VD | New VD |
|---|---|---|---|
| 1 | 0.70 | 4.300 | 0.7729 |
| 2 | 0.7729 | 4.227 | 0.7721 |
| 3 | 0.7721 | 4.228 | 0.7721 |
Converged: VD = 0.772 V, ID = (5 - 0.772)/1 kΩ = 4.228 mA.
Answer: ID ≈ 4.23 mA, VD ≈ 0.772 V.
✓ Answer checked by Claude Opus 5.5 (high effort)
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