17 / 32 · Concept
A diode model must match the question
Use a constant-drop approximation while identifying its limits.
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Direction alone is insufficient
A diode is a nonlinear two-terminal device. Forward current rises strongly with forward voltage; reverse current is usually small before breakdown. An ideal on/off diode, a constant forward-drop diode, and an exponential diode equation answer different levels of detail. Choose the simplest model that supports the requirement and state it explicitly.
For this lesson, use a constant forward drop of 0.7 V when the diode is on. This is an assumed model value, not a universal silicon voltage. Actual forward voltage depends on current, temperature, and device construction. LEDs and Schottky diodes have different behavior, and reverse breakdown must not be treated as ordinary blocking.
Current limiting example
A 5 V source drives a forward-biased diode through 1 kΩ. Assuming the diode is on at 0.7 V, the resistor drops 4.3 V and current is 4.3 mA. Resistor dissipation is mW; diode dissipation is approximately 3.01 mW. The source supplies 21.5 mW, matching their sum.
Reverse the diode in the same assumed model. Current becomes zero and the resistor drop becomes zero; the diode takes the source voltage in reverse. A physical device would still have leakage and a reverse-voltage rating. Our model does not prove that an arbitrary diode tolerates that condition.
Check the assumption after solving
With a 0.5 V source, calculating (0.5 − 0.7)/1 kΩ gives negative forward current. That contradicts the assumed on state. Under the constant-drop model the diode is off instead. Real exponential behavior is smoother, so a small physical forward current can exist below 0.7 V.
For a clamp or flyback circuit, document current path, maximum current, dissipated energy, and reverse voltage. A generic diode symbol cannot certify those limits. In later SPICE work, use a named model and compare its bias point with the hand model. A disagreement may reveal different model assumptions rather than a simulator error. The numerical example is deliberately low voltage and resistively limited so that the calculation is transparent.
Try it yourself
Using the 0.7 V model, find current and resistor power for a 3.3 V source and 680 Ω series resistor.
Read the explanation
I = (3.3 − 0.7)/680 = 3.823529 mA. PR = 2.6²/680 = 9.941176 mW. A real diode requires its actual current-voltage data.