Chapter 2

Q. 2.SP.32

A light-emitting diode (LED) has a greater forward voltage drop than does a common signal diode.    A typical LED can be modeled as a constant forward voltage drop v_D = 1.6  \text{V}.    Its luminous intensity I_v varies directly with forward current and is described by
I_v = 40i_D ≈ \text{millicandela (mcd)}

A series circuit consists of such an LED, a current-limiting resistor R, and a 5-V dc source V_S.    Find the value of R such that the luminous intensity is 1 mcd.


Verified Solution

By (1), we must have
i_D = \frac{I_v}{40} = \frac{1}{40} = 25  \text{mA}

From KVL, we have
V_S = Ri_D + 1.6

so that              R = \frac{V_S  –  1.6}{i_D} = \frac{5  –  1.6}{25  ×  10^{-3}} = 136  Ω