Question 2 (continued) A light emitting diode (LED) is connected across the term

Question

Question 2 (continued) A light emitting diode (LED) is connected across the terminal of the one-port network as shown in Figure 2.2 The relationship between the current through the LED (ID) and the voltage across the LED (VD) can be modelled by the following piece-wise function: ID-0 amperes, for VD 1.5V lp = 0.01xVo . 0.015 amperes, for Vp>1.5V One-Port Network VD Figure 2.2: The one-port network with a LED connected across the terminal. (iii) Calculate Ip and Vo when the LED is connected across the terminal of the one-port network as shown in Figure 2.2 (2 marks) In a particular application, it is necessary to provide a specific light intensity using the LED; and therefore to set a precise lo of 5mA. It is proposed that this be achieved by including an additional series resistor (Rx) between the same one port network and the LED as shown in Figure 2.3. ID One-Port Network Figure 2.3: The one-port network with the LED and a series resistor R connected across the terminal. (iv) Is it possible to choose a value of R such that the current Io is equal to 5mA? If it is, calculate this value of R. If it is not possible, explain. (3 marks)

Explanation / Answer

(iii). Voltage across one port netwrok is nothing but voltage across the LED.

If Voltage across the one port network =VD < 1.5V

Then ID = 0A.

If Voltage across the one port network =VD >= 1.5V

Then ID = 100VD - 0.015.

(iv).

Voltage across one port netwrok V1.

So VD = V1 - Rx*ID

When ID = 5mA

VD = (0.005 + 0.015)/0.01 = 2V.

Rx = (V1 - 2) / 0.005 = 200(V1 - 2)ohm

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