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# Breadboards, Multimeters, and Resistors - PowerPoint PPT Presentation

Breadboards, Multimeters, and Resistors. EGR1301. Your Multimeter. pincer clips – good for working with Boe-Bot wiring. probes. (push these onto probes). leads. turn knob to what you would like to measure. You will use the multimeter to understand and troubleshoot circuits, mostly

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Presentation Transcript

EGR1301

pincer clips – good for working

with Boe-Bot wiring

probes

(push these onto probes)

turn knob to what you

would like to measure

You will use the multimeter to understand and troubleshoot circuits, mostly

measuring DC voltage, resistance and DC current.

Measure Vin

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 voltage

Vss = ground (negative side of battery)

Switch to DC Volts

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 ~ 5V

Vss = ground (negative side of battery)

Build the Series Circuit Below

5V

+

-

220W

470W

these resistors

Select Resistors

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

Example: 470W resistor:

4 = yellow

7 = violet

Add 1 zero to 47 to make 470, so 1 = brown

So, 470 = yellow, violet, brown

Now, find the 220W resistor.

Check Resistance of Resistors

R ~ 470W

set multimeter

to measure W

Compute the Voltage Drops Across the Two Resistors

(Hint: compute the total current provided by the power source

and then apply Ohm’s Law to each resistor)

5V

+

-

DV = 3.41V

220W

470W

SOLUTION:

DV = 1.59V

Now, add the voltage rise of

the power source (+5V) to the

voltage drops across the resistors

(negative numbers).

5V – 3.41V – 1.59V = 0

pincer clips – good for working

with Boe-Bot wiring

probes

(push these onto probes)

turn knob to what you

would like to measure

You will use the multimeter to understand and troubleshoot circuits, mostly

measuring DC voltage, resistance and DC current.

Use Multimeter to Measure Voltages Around Loop

(1) From Vdd to Vss

DV1 = _____

(2) Across the 470W resistor

DV2 = _____

Remember . . .

a RESISTOR is a voltage DROP and

a POWER SOURCE is a voltage RISE

DV1 - DV2 - DV3 = ________

(3) Across the 220W resistor

DV3 = _____

Rises must balance drops!!!!

Compare Measurements to Theory

DV = 3.41V

DV = 1.59V

Pretty close!

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.

DV = 3.41V

5V – 3.41V – 1.59V = 0

DV = 1.59V

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

LED = Light Emitting Diode

Diagram from the Parallax Robotics book

Electricity can only flow one way through an LED (or any diode).

Building an LED Circuit

Vdd =

LED

These circuit diagrams

are equivalent

these holes are “connected”

Vdd = 5V

holes in this direction are not connected

Diagram from the Parallax Robotics book

Replace the 470W Resistor

with the 10kW Resistor

What happens and Why??

ANSWER: The smaller resistor (470W) provides less resistance to current than

the larger resistor (10kW). 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