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