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 vD = 1.6 V. Its luminous intensity Iv varies directly with forward current and is described by Iv = 40iD ≈ millicandela (mcd) A series

Question

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 [latex]v_D = 1.6 ext{V}[/latex]. Its luminous intensity [latex]I_v[/latex] varies directly with forward current and is described by
[latex]I_v = 40i_D ≈ ext{millicandela (mcd)}[/latex]

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

Accepted Answer

By (1), we must have
[latex]i_D = \frac{I_v}{40} = \frac{1}{40} = 25 ext{mA}[/latex]

From KVL, we have
[latex]V_S = Ri_D + 1.6[/latex]

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

Chapter 2

Q. 2.SP.32

Q. 2.SP.32

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