Labl datasheet Lab Part 2 KDG- R1=19-57152 22=0.474.89 R3-06-7539kr Ru-co.1.4920kn R1/R2= 41.29 Lab part 3 KDG toolov 123/14-4.53 V"1= 1.01&v Lab Party VA: 0.5200 V₁20.23√√ i' = 0-487 MA √u" 2:0·490v ji" = 0.004MA F:1 ileso F = 3/4 F= 1/2 iled: 0.487mA iled-1-36/nA Scanned with CamScanner/n ECE 220 Laboratory 1 Color Code, Wheatstone Bridge, the Loading Effect, and Dimmer Circuit Michael W. Marcellin The first portion of this document describes preparatory work to be completed in advance of the laboratory session. This portion of the work is referred to as the "Prelab." The second portion of this document contains a description of the experiments to be performed during the laboratory session. That portion of the document is referred to as the "Lab." When you do the lab: 1) Always wear safety glasses. 2) Bring a printed copy of the lab data sheet (appears later in this document). 3) On-campus students - bring your Cat Card and wear closed toed shoes. Prelab Report Requirements For your Prelab Report, turn in all calculations/answers requested, together with all PSpice printouts requested in each part. Within each part, the items to be turned in are labeled with the letters a), b), c), etc. This does not need to be a formal report. The calculations/answers can be hand written. The level of detail should be similar to homework. Please circle your answers. Give numeric answers using at least 3 significant digits. Do not leave your answers as fractions. Your entire report must be contained in a single PDF file, which you must upload to the appropriate folder under the Assignments tab on D2L. You may update your report prior to the due date by uploading a new version. It will overwrite the previous version. Prelab Part 1: Color Code Give the color code for each of the following resistor values: a) 36 kQ ±5% b) 47 ±10% c) 680 Q±10% d) 1 MQ ±5% Give the resistance and tolerance for resistors that have the following color codes: e) brown, black, red, gold f) yellow, black, red, silver g) green, blue, red, silver h) green, blue, green, gold Prelab Part 2: Wheatstone Bridge In class, we learned (or will learn) that the Wheatstone Bridge shown in Figure 1 below is "balanced" when R₁/R₂ = R3/R4. When the bridge is balanced, v5 = 0 regardless of the value of vs or R5. Vs R3 + R₁ R5 +15 R4 R2 Figure 1. Wheatstone bridge. Let R₁ = 20 k§, R₂ = 4.7 k, and R3 = 6.8 ΚΩ. a) Calculate the value of R required to balance the bridge. Prelab Part 3: The Loading Effect Consider Figure 2 as shown below. From class, we know that the 1.5 k and 4.7 K resistors form a voltage divider and that v₂ = V₁ 1.5 k = 4.133 = 1 V. 1.5 k+4.7 k = 0 4.7 ΚΩ + 4.133 V 1.5 kov₁ - +1 Figure 2. A voltage divider. Consider now Figure 3 below. The circuit in Figure 3 is the same circuit from Figure 2, but with two 470 resistors added on the right. Note that the 1.5 k and 4.7 k resistors no longer form a voltage divider due to the presence of the 470 2 resistors. In particular, the 1.5 kQ and the 4.7 K resistors are no longer in series. That is, they do not have the same current flowing through them. The 470 Q resistors "load" the original circuit which changes (reduces) the value of v₁ to some new value v₁. This reduction in v₁ is called “the loading effect." It is worth noting that the two 470 £ resistors do form a voltage divider, so that v₁₂ = 470 470+470 = 2 4.7 ΚΩ w 4700 w 4.133 V 1.5 ko vi 470 Ω ξυ +1 Figure 3. Illustration of the loading effect. a) Calculate v₁, v½, and i'. Note that v₁ is significantly different than v₁ = 1.0. Now consider Figure 4 below, which is the same as Figure 3, but with the 470 resistors replaced by 470 k resistors. b) Calculate v₁, v½, and i". In this case, you should find that v¼ ≈ v₁ = 1.0. The loading effect is negligible in this case due to the fact that i" ≈ 0, and so essentially the same current flows through the 1.5 kQ and 4.7 kQ resistors. 4.7 ΚΩ w i" 470 ΚΩ ww 4.133 V 1.5 κωχ ν ' + 470 kv Figure 4. Negligible load current. Prelab Part 4: Dimmer Circuit In this part, we make a dimmer circuit for a light emitting diode (LED). This circuit provides a practical example of unwanted consequences of the loading effect. In general, diodes are non-linear devices. The circuit symbol for a diode is as follows: K+ + v Figure 5. Circuit symbol for a diode. A diode can be thought of as a "one-way valve." That is, current can only flow in the forward direction, as shown in the figure above. Current only flows when voltage is positive according to the polarity in the figure above. No current flows if a negative voltage is applied. The terminal of the diode where current flows in (the left side in the figure above) is called the anode. The terminal where the current flows out (the right side in the figure) is called the cathode. Recall from class that resistors are linear, as described by Ohm's Law which is given by v = as shown in the following figure. iR, ט Figure 6. Resistors are linear. In contrast, over the range of currents of interest here, a diode has a non-linear voltage-versus- current characteristic as sketched in Figure 7 below. v .8 .6 88 1 mA +10 mA Figure 7. Voltage-versus-current for a D1N4002 diode. This is a standard diode (not an LED). You will learn more about diodes in future courses. In ECE 220 we make the following simplifying assumption: If current is flowing in the forward direction (as shown on the circuit symbol in Figure 5), the "forward voltage" is fixed at v = 0.7 V. The voltage-versus-current characteristic corresponding to the simplifying assumption is shown in Figure 8 below. Note the similarities and differences between Figures 7 and 8. 2 .8. 86 .6 -1 mA +10 mA Figure 8. Simplified voltage-versus-current for D1N4002. For our lab experiments, we will use an LH3330 LED (red), which has a similar voltage-versus- current characteristic but with a forward voltage that we will approximate by v = = 2.1 V (rather than 0.7 V). = 1.5 = = 100 Using what we have learned above, we can analyze the LED circuit given in Figure 9 below. Assuming that VLED = 2.1 V, we can use KVL to find that VR 1.5 V. Thus, i LED = İR 15 mA, which is a safe value for the LED. In contrast, if we change the 3.6 V supply to -3.6 V, then i LED = 0 (one-way valve). So, VR = 100i LED 0. Then, by KVL, we can find that VLED = −3.6 V. 100iR = = 100 Ω VR + 3.6 V +1 + iLED LED Figure 9. A simple LED circuit. From the discussion above, it should be clear that the resistor in the circuit of Figure 9 is used to set the value of the current that flows through the LED. Don't forget this. It is a favorite question used by potential employers during job interviews! Simple Dimmer Circuit We will now explore a circuit to control the intensity of light emitted by an LED. This intensity varies as a function of i LED We employ a simple design using a potentiometer (pot). A pot is a 3-terminal device. It is a