Question

6. LED Alarm Circuit One day, you come back to your dorm to find that your favorite candy has been stolen. Determined to catch the perpetrator red-handed, you decide to put the candy inside a kitchen drawer. Using the following circuit design, you would like to turn on a light-emitting diode (LED) "alarm" if the kitchen drawer is opened. Rli LED 2.5V LED Note Rphoto is a photoresistor, which acts like a typical resistor but changes resistance based on the amount of light i is exposed to. This photoresistor is located inside the kitchen drawer, so we can tell when the drawer is opened or closed. VLED indicates the voltage across the LED; we will guide you through the IV behavior of this element later in the problem. The LED is located in your room (and connected to a long wire going to the kitchen), so that you can remotely tell when the kitchen drawer has been opened (a) We see that the op-amp is in a comparator configuration. What is V+, the voltage at the positive voltage input? Your answer should be written in terms of Rphoto and Rfixed (b) We now want to choose a value for Rfixed. From the photoresistor's datasheet, we see the resistance in "light" conditions (i.e. drawer open) is 12. In "dark" conditions (i.e. drawer closed), the resistance is 10k2. To ensure the comparator detects the light condition with more tolerance, we decide to design Rfixed so that V+ is 3 V under the "light" condition. Solve for the value of Rfixed to meet this specification. (c) Write down Vor with any conditions in terms of V. For simplicity, consider the case when V V and assume the op-amp is ideal. (d) Using your answers to the previous parts, write down Vou with the conditions on its output in terms of Rphoto You can substitute the value of R fixed you found in part (b). As before, you can assume that V+メV and the op-amp is ideal (e) From the design steps in the previous parts, we have designed a circuit that outputs non-zero voltage when the photoresistor is exposed to light (i.e. kitchen drawer open). We now want to design the LED portion of the circuit, so we get a visual alarm when the drawer is open. From the LED's datasheet, the forward voltage, VF is 3V. Essentially, if VLED is less than this voltage the LED won't light up and ILED will be 0 A Here is an idealized IV curve of this LED. The LED behaves in one of the following two modes: i. If the voltage across the LED is less than V 3V or if ILED<0A, then the LED acts like an open circuit. ii. If the voltage across the LED is VF = 3 V, then the LED acts like a voltage source, except that it only allows positive current flow (i.e. only in the direction of current marked on the circuit diagram). 40 20 0 VLED V) To avoid exceeding the power rating of the LED (and having it burn out), the LED is 20mA. Find the value of the current-limiting resistor, Rlim, such that when the photoresistor is in the "light" condition ILE 20mA value for

Answer 1

Given that (a) Op - map in comparator configuration voltage divides configuration so v_= (R_ fixed/R_ fixed + R_photo) times 5 (b) Light conditions Rightarrow R_ photo = 1 k Ohm Double conditions Rightarrow R_ photo = 10 k Ohm Light condition Rightarrow v_t = 3 v V_T = (R_ fixed/R_ fixed + R_ photo) times 5 \r\n under light condition , 3 = R_ fixed/R_ fixed + 1 k Ohm times 5 3(R_ fixed + 1 k Ohm) = 5 R_ fixed 3 k Ohm = 5 R_ fixed - 3 R_ fixed R_ fixed = 3/2 k Ohm (c) Op - amp as comparator v_0 = {v_cc: v_+ > v_- 0: v__ < v_- or v_0 = [5v: v_+ > 25 v 0: v_+ < 2.5 v (d) v_+ = [R_ fixed/R_ fixed + r_- photo] times 5v R_ fixed = 3/2 k Ohm from part (b) v_+ = (R_ fixed/3/2 k + R_ photo) times 5v \r\n condition v_+ > 25 v rightarrow v_0 = 5v 5 r^3/2 k/R_photo + 3/2 k > 2.5