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1 Lab 2: DC Power Supply and Current Measurements EE97 Spring 2024 Thursday 1.30 - 4.15 PM Lab 2: DC Power Supply and Current Measurements Samal Maleesha Wijendra Partner: John Wu Submission Date: 02/20/2024

2 Lab 2: DC Power Supply and Current Measurements Learning objectives After completing this lab, we must be able to, • Use a bench-top DC power supply. • Experimentally determine the I-V characteristic curve of a two-terminal device. • Measure current using a digital multimeter (DMM). This lab also reinforces students’ understanding of resistors in parallel or in series, and the voltage division formula. Experiment 1 1. How do you get a single 30V output from this supply? In other words, how do you set up the power supply so that one of the output terminals is 30v higher than another terminal? (Hint: you need to use two voltage outputs connected in series.) How about 55V? Test your setup with a voltmeter. Set the +25V terminals’ voltage to 25V and the +6v terminals’ voltage to +5v, connect – terminal of the 5V to +25v using a wire and when those two gets combined, it adds up to 30v. Measure the voltage from the common to + terminal of the +6v terminal. How about 55V ? Set +6V to 5V, +25V and -25V to its maximum voltage values. Connect -25V to +25 terminal using a cable and connect +25V terminal to negative terminal of +6V. Then measure from -25V terminal to +6V + terminal. It should give 55V voltage.

3 Lab 2: DC Power Supply and Current Measurements 2. How do you simultaneously get +12V, +5V and –5V (relative to a common terminal) from this supply? Test your setup with a voltmeter. We can set +25V to 12V and +6V to 5V and -25V to -5V. When those are set, we can simultaneously measure voltages relative to common terminal. 3. How do you simultaneously get +6V, -6V, and -2V (relative to a common terminal) from this supply? Again, test your setup. We can set +25V to 6V and +6V to 4V and -25V to -6V. We can simultaneously measure of 6V and -6V voltages relative to common terminal using +25V and -25V terminals. Use a cable and connect – terminal of +6V to -25V and from common to + terminal of 6V can measure -2V.

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4 Lab 2: DC Power Supply and Current Measurements Experiment 2 1. Set the +6V supply to 2V and set the current limit to 0.1A. Connect a 47Ω resistor across the +6V output terminals. Observe the voltage and current (as displayed on the power supply panel) before and after this 10Ω resistor is connected to the terminals. Explain these values using Ohm's law. Resistors true value = 46.75Ω Before After 2.02v 0.1A – circuit is open 2.01v 0.044A – circuit closed and current flowing through the resistor It acts as independent voltage source V = iR I = V/R I = 2.01V / 47Ω = 0.042A 2. Repeat the above step with a 10Ω resistor. Explain why the voltage dropped when the resistor is connected. Before After 2.02v 0.001A 0.978v 0.093A It acts as a independent current source. Its because the voltage is being divided among the resistors.

5 Lab 2: DC Power Supply and Current Measurements Experiment 3 (Smoke test) The purpose of this experiment is to show the meaning of the power rating of resistors. Before we start the experiment, calculate the power dissipation of a 10Ω resistor with 5V across it. The 1/4-watt resistor in the following experiment will get EXTREMELY HOT. P = iV = V 2 /R = 2.5W 1. Set the current limit for the +6V supply to 5A. Setup the limits. 2. Apply 1.58 V (from +6V supply) to a 1/4W, 10Ω resistor. Carefully observe its temperature. Set the voltage to 1.58V. 3. Now apply 5V (from +6V supply) to the same 1/4W, 10Ω resistor (4 or 5 seconds should be long enough). Record the current and observe the resistor. Do not touch the resistor! Set the voltage to 5V. • Resistor gets hot very fast. • 0.59A current flows through it.

6 Lab 2: DC Power Supply and Current Measurements This happens because power dissipation of this specific 10Ω resistor is low and due to that it gets very hot when this voltage is given to it. 4. Repeat the above step with a 10W, 10Ω resistor. Record the current and observe the resistor. • It gets hot but slowly. • Current is 0.51A This happens because power dissipation of this specific 10Ω resistor is high and due to that it gets hot slowly since it can dissipate heat faster when this voltage is given to it. Experiment 4 1. Measure the I-V relationship for a 12V light bulb. You may connect the light bulb to the output of the DC power supply (as shown in Figure 4) and read out the voltage and current directly from the display on the power supply. Sweep the voltage from 0 to 1V at 0.2V increments and from 1V to 12V at 1V increments. Record the readings in a table such as the one shown below. Since a light bulb is not sensitive to voltage polarity, you don't need to test the circuit with negative voltage. 2. Determine the resistance value (i.e., V÷I) and power (V×I) of the light bulb at each voltage step and enter the results into the table. Voltage (v) (v) Current (I) (mA) Resistance ( ? 𝑖 ⁄ ) (Ω) Power ( ? × 𝑖 ) (w) 0 0 0 0 0.2 24 8.33 0.0048 0.4 31 12.9 0.0124 0.6 36 16.67 0.0216 0.8 41 19.51 0.0328 1 46 21.74 0.046 2 67 29.85 0.0597 3 84 35.71 0.252 4 100 40 0.4 5 114 43.86 0.57 6 127 47.24 0.762 7 139 50.36 0.973 8 150 53.33 1.2 9 161 55.90 1.449 10 171 58.48 1.71

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7 Lab 2: DC Power Supply and Current Measurements 11 181 60.77 1.991 12 191 62.83 2.292 3. When you prepare your lab report, sketch a current vs. voltage graph and a resistance vs. voltage graph. From these graphs, you should see that the resistance of the light bulb is lower at a lower supply voltage. This is due to the change of temperature of the filament in the light bulb.

8 Lab 2: DC Power Supply and Current Measurements Experiment 5 In this experiment, we will use the circuit in Figure 6A to determine the current ILED versus voltage VLED curve. According to Ohm’s law and Kirchhoff's voltage law, VLED and ILED are related by the following equation. 𝑉𝑆𝑜?𝑟𝑐𝑒 = 𝑉𝐿?? + 470Ω × 𝐼𝐿??. Vsource is the voltage from the power supply and VLED will be measured by a DMM. Knowing Vsource and VLED, from the above equation, ILED can be calculated. In fact, ILED is the same current as the power supply current, ISource which is displayed on the power supply. However, the resolution of the current display on the power supply is not high enough for this experiment and, therefore, we should use a DMM to measure VLED and calculate ILED. 1. Set the +25-volt supply to zero volt and its current limit to 0.030 A. 2. Select a 470-ohm resistor from your kit for the current-limiting resistor. Measure and record the actual resistance of the selected resistor. Measured resistance = 463.80Ω

9 Lab 2: DC Power Supply and Current Measurements 3. Construct the circuit in Figure 6A or 6C using the +25-volt supply as the Vsource. Make sure that the Cathode (the shorter lead of LED) is connected to the negative terminal of the +25 supply, i.e., the COM terminal (see Figure 2). 4. Set up the DMM to the voltage measuring mode and connect the probes across the LED, for measuring VLED. 5. Sweep the supply voltage from 0V to 10V in steps. At each step, record Vsupply, Isupply, and VLED from the DMM. Comment on the brightness of the LED. Enter all data into a table similar to Table 2 shown below. You may start with 0.5V steps in the supply voltage. When the supply voltage reaches three or four volts, a 1V step should suffice to show the shape of the I-V curve. Power Supply 34405A DMM LED brightness Source voltage V source (v) Source current I (A) Diode voltage VLED (v) Calculated LED Current iLED (mA) 0 0 0 0 NA 0.5 0 .5 0 NA 1 0 1 0 NA 1.5 0 1.5 0 NA 2 0 1.62 0.00081 Start to glow 2.5 .001 1.68 0.00174 Glow 3 .002 1.7 0.00277 Glow

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10 Lab 2: DC Power Supply and Current Measurements 4 .004 1.74 0.00481 Glow 5 .006 1.77 0.00687 Glow 6 .008 1.79 0.00896 Glow 7 .011 1.82 0.01102 Glow 8 .013 1.84 0.01311 Glow 9 .015 1.85 0.01521 Glow 10 .017 1.87 0.01730 Glow 6. Reverse the polarity of the LED (i.e., turn the LED connections around), and repeat the steps above. Since the LED is reverse biased in this configuration, Isupply should remain zero and VLED should be the same as Vsource (why) throughout the 10-volt range of this experiment. Power Supply 34405A DMM LED brightness Source voltage V source (v) Source current I (A) Diode voltage VLED (v) Calculated LED Current iLED (mA) 0 0 0 0 NA 0.5 0 .5 0 NA 1 0 1 0 NA 1.5 0 1.5 0 NA 2 0 2 0 NA 2.5 0 2.5 0 NA 3 0 3 0 NA

11 Lab 2: DC Power Supply and Current Measurements 4 0 4 0 NA 5 0 5 0 NA 6 0 6 0 NA 7 0 7 0 NA 8 0 8 0 NA 9 0 9 0 NA 10 0 10 0 NA 7. Include a graph of LED current (ILED) versus LED voltage (VLED) when you prepare the lab report. The voltage axis (the horizontal axis) of your graph must span both the positive side (forward-biased) and the negative side (reverse-biased). The horizontal axis in your plot can be just from -3V to 3V since, on the negative VLED side (reverse-biased side), ILED is all zero and, on the positive side, the maximum VLED should be less than 3V.

12 Lab 2: DC Power Supply and Current Measurements Experiment 6 This experiment will demonstrate the reverse bias behavior of a Zener diode. 1. Repeat Experiment 5, using a Zener diode instead of an LED. Note that one side of the Zener diode is marked with a stripe or band. Is the striped end of the Zener diode the anode or the cathode? Hint: The voltage across the diode should not exceed 0.9V in the forward bias condition. Power Supply 34405A DMM LED brightness Source voltage V source (v) Source current I (A) Diode voltage VLED (v) Calculated LED Current iLED (mA) 0 0 0 O NA 0.5 0 0.5 0 NA 1 0 .74 0.0005 NA 1.5 .001 .77 0.0015 NA 2 .002 .79 0.0026 Start to glow 2.5 .003 .79 0.0036 Glow

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13 Lab 2: DC Power Supply and Current Measurements 3 .004 .803 0.0047 Glow 4 .006 .81 0.0068 Glow 5 .008 .82 0.0089 Glow 6 .011 .830 0.0110 Glow 7 .013 .834 0.0131 Glow 8 .015 .839 0.0152 Glow 9 .017 .843 0.0174 Glow 10 .019 .847 0.0195 Glow Reversed diode Power Supply 34405A DMM LED brightness Source voltage V source (v) Source current I (A) Diode voltage VLED (v) Calculated LED Current iLED (mA) 0 0 0 O NA 0.5 0 0.5 0 NA 1 0 1 0 NA 1.5 0 1.5 0 NA 2 0 2 0 NA 2.5 0 2.5 0 NA

14 Lab 2: DC Power Supply and Current Measurements 3 0 3 0 NA 4 0 4 0 NA 5 0 5 0 NA 6 0 5.89 0 Faded 7 .002 5.96 0.0022 Glow 8 .004 5.97 0.0043 Glow 9 .006 5.98 0.0064 Glow 10 .008 5.99 0.0085 Glow 2. From your plot of current vs. voltage, can you tell why this particular Zener is called a 6V Zener diode? This diode regulates current through it in reverse position only after its voltage across becomes more than 6V, That’s why its called 6V Zener diode.

15 Lab 2: DC Power Supply and Current Measurements Experiment 7 In this experiment, we will verify the voltages and currents you calculated in the pre-lab of Lab 1. 1. Build the circuit in Figure 13 on the breadboard using R1 = 1 kΩ, R2 = 2.2kΩ and R3 = 5.1 kΩ, and use the power supply to provide the 12V source. Measure voltage VA, I1, I2, and I3. Remember that you must ‘open’ the circuit before taking any current measurement. 2. Measure and REQ. Recall that REQ is the resistance of the overall circuit as seen by the voltage source. When measuring REQ, the 12V voltage source must be disconnected from the circuit. Compare the measured values to the calculated results from Lab 1. R total = 2.499kΩ Va I1 I2 I3

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16 Lab 2: DC Power Supply and Current Measurements Measured 7.28V 4.787mA 3.36mA 1.42mA Calculated 7.24V 4.72mA 3.3mA 1.42mA 3. Build the circuit in Figure 14 with, R1 = 2.2 kΩ, R2 = 2.7 kΩ, R3 = 1 kΩ and R4 = 5.1 kΩ and measure all values indicated and compare the measured values to the calculated results from Lab 1. R total = 2.01Ω I1 I2 I3 I4 I5 I6 VA(v ) VB(v ) VC(v ) VD(v ) Measure d 3.27m A 2.67m A 4.98m A .959m A - 1.7m A 5.94m A 7.08 7.08 4.91 4.91 Calculate d 3.2mA 2.6mA 5mA 1mA - 1.8m A 6mA 7 7 5 5