Breadboards, Multimeters, and Resistors

EGR1301

Your Multimeter

leads

probes

pincer clips – good for workingwith Boe-Bot wiring

You will use the multimeter to understand and troubleshoot circuits, mostlymeasuring DC voltage, resistance and DC current.

turn knob to what youwould like to measure

(push these onto probes)

Measure VinVin will be the same as your power supply voltage. For the case below, 4 AAbatteries are used resulting in approximately 6 V (5.79 V to be more exact).

Vin = power supply voltage Vss = ground (negative side of battery)

Switch to DC Volts

Measure VddVdd 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 ~ 5V Vss = ground (negative side of battery)

Build the Series Circuit Below

470

220

the breadboard connectsthese resistors

5V

Select Resistors Find the 220 and the 470 resistors from your Boe-Bot kit.

Example: 470resistor:

4 = yellow7 = violetAdd 1 zero to 47 to make 470, so 1 = brown

Now, find the 220 resistor.

So, 470 = yellow, violet, brown

set multimeterto measure

R ~ 470

Check Resistance of Resistors

Compute the Voltage Drops Across the Two Resistors

Ohms Law: V = I R

R1 470 R2 220 Vdd 5 V

Req R1 R2 Req 690 Find the equivalent Resistance

IVdd

Req

I 0.0072A Find the current leaving Vdd

VR1 I R1 VR1 3.41V Find the voltage across the 470 resistor

VR2 I R2VR2 1.59V Find the voltage across the 220 resistor

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:

Your Multimeter

leads

probes

pincer clips – good for workingwith Boe-Bot wiring

You will use the multimeter to understand and troubleshoot circuits, mostlymeasuring DC voltage, resistance and DC current.

turn knob to what youwould like to measure

(push these onto probes)

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

V1 = _____

V2 = _____

V3 = _____

V1V2V3

Rises must balance drops!!!!

V = 3.41V

V = 1.59V

Compare Measurements to Theory

5V

+ -

470

220

+

+

-

Pretty close!

V = 3.41V

V = 1.59V

5V – 3.41V – 1.59V = 0

5V

+ -

470

220

+

+

-

Kirchoff’s Voltage Law (KVL)

Kirchoff’s Voltage Law says that the algebraicsum of voltages around any closed loop in a circuit is zero – we see that this is true for ourcircuit. 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.

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

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

Building an LED Circuit

Vdd = 5V

LED

5V

+ -

470

These circuit diagramsare equivalent

these holes are “connected”

Vdd =

LED

holes in this direction are not connected

Diagram from the Parallax Robotics book

Replace the 470 Resistor with the 10k Resistor

What happens and Why??

ANSWER: The smaller resistor (470) provides less resistance to current thanthe larger resistor (10k). Since more current passes through the smallerresistor, 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.