Question

You must first calculate and select appropriate values for R. R and Rc that will tun on the LEDs. Configuration A Configuration B RA Rc From the insights gained conducting these evaluations, you are then required to: Design, evaluate (with regard to the four criteria above) and recommend an LED array configuration that will provide a minimum of 250 lux at a distance of 20cm. You may assume that: The source voltage available to power the LEDs is Vs-4.8V LED characteristics are similar to "White LEDs found in the Datasheets folder of the Projects section on Moodle. You should use the Typical" value for the Forward Voltage (Vi) as the LED turn on voltage and the corresponding typical "Forward Current ()as the target operating current for the LEDs. .Each LED gives approximately 45 added lux at distance of 20cm when turned on. Permitted values of resistors in your designs must come from the "E12 range of values If current through an LED exceeds 25mA, the LED has a 50% chance of electrical failure. If power absorbed by a resistor exceeds 62.5 mW, the resistor has a 20% chance of failure. The "failure" of a component leads to it becoming an open circuit.

Answer 1

Datasheet of LED is not given, So I assumed it as a 5mm white LED with following electrical parameters to solve this question.

VLHW4100, VLHW4101-YLWU Vishay Semiconductors VISHAY. www.vishay.com = 25 °C, unless otherwise specified) OPTICAL AND ELECTRICAL CHARACTERIST VLHW4100, VLHW4101-YLWU, WHITE PARAMETER TEST CONDITION 누.20 mA 20 mA SYMBOL | MIN. | TYP. | MAX. | UNIT 4500 715011 250 mcd 5600 840011 250 mcd ly ly Luminous intensity Chromatically coordinate x acc. to CIE 1931 Chromatically coordinate y acc. to CIE 1931 VLHW4101 0.33 0.31 0.33 0.32 22.5 3.2 VLHW4100 VLHW4101-YLWU VLHW4100 VLHW4101-YLWU Angle of half intensity Forward voltage Reverse current Temperature coefficient of Vp Temperature coefficient of w deg 2.8 3.8 IF 20mA TCVF IF 20mA mV/K %/K -0.5

From datasheet we have Vf=3.2 V

For configuration A

For LEDs to get turn on in this condition, supply voltage should be equal to sum of voltage drop across resitor and voltage drop across LED.

Voltage drop across LED is 3*3.2=6.4V

Since our supply voltage is 4.8V, which is less than 6.4.So LEF will not get ON in this configuration for any value of resistor.

For configuration B

Supply voltage = voltage drop across LED + voltage drop across resistor.

4.8=3.2+ IR

IR=4.8-3.2=1.6

R=1.6/I

From datasheet, we have I= 25mA

ABSOLUTE MAXIMUM RATINGS VLHW4100, VLHW4101-YLWU PARAMETER Reverse voltage DC forward current Peak forward current Power dissipation Junction temperature Operating temperature range Storage temperature range Soldering temperature Thermal resistance junction/ambient 25 °C, unless otherwise specified) TES SYMBO VALUE UNIT 25 at 1 kHz, tp/T = 0.1 95 Tamb - 40 to+85 - 40 to85 ts5 s KW

So, R=1.6/25m=64Ohm

So, Ra=Rb=Rc=64Ohm

Power dissipation in resistor= IR=25m*64=1.6 Wa

For configuration C

Current through resistor=3*25mA= 75mA

By KVL: Supply voltage= IR+3.2

IR=4.8-3.2=1.6V

R=1.6V/75m=21.33Ohm

So R=21.33Ohm.

Power dissipation in resistor= IR=75m*21.33=1.6W.

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