1. Breadboards, Multimeters, and Resistors EGR1301

2. Your Multimeter leads probes pincer clips – good for working with Boe-Bot wiring You will use the multimeter to understand and troubleshoot circuits, mostly measuring DC voltage, resistance and DC current. turn knob to what you would like to measure (push these onto probes)

3. Measure V in Vin will be the same as your power supply voltage. For the case below, 4 AA batteries are used resulting in approximately 6 V (5.79 V to be more exact). Vin = power supply voltageVss = ground (negative side of battery) Switch to DC Volts

4. Measure Vdd Vdd will always be around 5V (it is 4.94 V here). A voltage regulator on the Board of Education reduces this voltage from Vin down to ~ 5V. The Boe-Bot operates on 5V DC. Vdd ~ 5VVss = ground (negative side of battery)

5. Build the Series Circuit Below 470  220  the breadboard connects these resistors 5V 

6. Select Resistors Find the 220  and the 470  resistors from your Boe-Bot kit. Example: 470  resistor: 4 = yellow 7 = violet Add 1 zero to 47 to make 470, so 1 = brown Now, find the 220  resistor. So, 470 = yellow, violet, brown

7. set multimeter to measure  R ~ 470  Check Resistance of Resistors

8. Compute the Voltage Drops Across the Two Resistors  V = 3.41V  V = 1.59V 470  220  5V  (Hint: compute the total current provided by the power source and then apply Ohm’s Law to each resistor) 5V – 3.41V – 1.59V = 0 Now, add the voltage rise of the power source (+5V) to the voltage drops across the resistors (negative numbers). SOLUTION:

9. Your Multimeter leads probes pincer clips – good for working with Boe-Bot wiring You will use the multimeter to understand and troubleshoot circuits, mostly measuring DC voltage, resistance and DC current. turn knob to what you would like to measure (push these onto probes)

10. Use Multimeter to Measure Voltages Around Loop (1) From Vdd to Vss (2) Across the 470  resistor (3) Across the 220  resistor Remember... a RESISTOR is a voltage DROP and a POWER SOURCE is a voltage RISE  V 1 = _____  V 2 = _____  V 3 = _____  V 1  V 2  V 3  Rises must balance drops!!!!

11.  V = 3.41V  V = 1.59V Compare Measurements to Theory Pretty close!

12.  V = 3.41V  V = 1.59V 5V – 3.41V – 1.59V = 0 Kirchoff’s Voltage Law (KVL) Kirchoff’s Voltage Law says that the algebraic sum of voltages around any closed loop in a circuit is zero – we see that this is true for our circuit. It is also true for very complex circuits. Notice that the 5V is DIVIDED between the two resistors, with the larger voltage drop occurring across the larger resistor.

13. Gustav Kirchoff (1824 – 1887) was a German physicist who made fundamental contributions to the understanding of electrical circuits and to the science of emission spectroscopy. He showed that when elements were heated to incandescence, they produce a characteristic signature allowing them to be identified. He wrote the laws for closed electric circuits in 1845 when he was a 21 year-old student. Photo: Library of Congress

14. Connecting an LED Electricity can only flow one way through an LED (or any diode). LED = Light Emitting Diode Diagram from the Parallax Robotics book

15. Building an LED Circuit Vdd = 5V These circuit diagrams are equivalent these holes are “connected” Vdd = LED holes in this direction are not connected Diagram from the Parallax Robotics book

16. Replace the 470  Resistor with the 10k  Resistor What happens and Why?? ANSWER: The smaller resistor (470  ) provides less resistance to current than the larger resistor (10k  ). Since more current passes through the smaller resistor, more current also passes through the LED making it brighter. What would happen if you forgot to put in a resistor? You would probably burn up your LED.