## Q. 2.SP.33

The reverse breakdown voltage $V_R$ of the LED of Problem 2.32 is guaranteed by the manufacturer to be no lower than 3 V.    Knowing that the 5-V dc source may be inadvertently applied so as to reverse-bias the LED, we wish to add a Zener diode to ensure that reverse breakdown of the LED can never occur. A Zener diode is available with $V_Z = 4.2 \text{V}, I_Z = 30 \text{mA}$, and a forward drop of 0.6 V.    Describe the proper connection of the Zener in the circuit to protect the LED, and find the value of the luminous intensity that will result if $R$ is unchanged from Problem 2.32.

## Verified Solution

The Zener diode and LED should be connected in series to that the anode of one device connects to the cathode of the other.    Then, even if the 5-V source is connected in reverse, the reverse voltage across the LED will be less than $5 – 4.2 = 0.8 \text{V} < 3 \text{V}$.    When the dc source is connected to forward-bias the LED, we will have

$i_D = \frac{V_S – V_{F\text{LED}} – V_{FZ}}{R} = \frac{5 – 1.6 – 0.6}{136} = 20.6 \text{mA}$

so that            $I_v = 40i_D = (40)(20.6 \times 10^{-3}) = 0.824 \text{mcd}$