## 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.

## Step-by-Step

## 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