Diode In A Dc Circuit

When a silicon diode conducts in forward bias, it can be modelled as a constant voltage drop of about 0.7 V across its terminals, regardless of the exact current, because its forward characteristic is very steep past the knee. This constant-drop model is an excellent approximation whenever the series resistor limits the current to a modest value, since the diode voltage then barely changes even if the current shifts somewhat. Applying Kirchhoff's voltage law around the loop, the battery voltage is shared between the diode drop and the resistor: (V_{battery} = V_{diode} + I R). The voltage available to drive current through the resistor is the battery voltage minus the diode drop, (5 - 0.7 = 4.3,\text{V}). Ohm's law then gives the current as (I = 4.3 / 430 = 0.01,\text{A} = 10,\text{mA}). The value 11.6 mA ignores the diode drop and divides the full 5 V by the resistor. The value 7 mA misreads 0.7 V as a current. The value 1.6 mA results from incorrectly adding the diode drop to the battery voltage. As a final check, the resistor voltage of (0.01 \times 430 = 4.3,\text{V}) plus the 0.7 V diode drop sums to exactly 5 V, satisfying the loop equation.