Transcript Fabrication of a Centrifugal Pump

Breadboards, Multimeters, and
Resistors
EGR1301
Your Multimeter
pincer clips – good for working
with Boe-Bot wiring
(push these onto probes)
probes
leads
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
the breadboard connects
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
+ 220W
470W
DV = 3.41V
DV = 1.59V
SOLUTION:
Ohm s Law : V = I R
R1   4 70W

Req   R1 + R2
I 
R2   2 20W
Req  6 90W
Vd d   5 V
Find the equiv alent Resis tanc e
Vd d
Req
I  0 .00 72A
Find the c urrent leav ing Vdd
VR1   I R1
VR1  3 .41V
Find the v oltage across t he 470
W resis tor
VR2   I R2
VR2  1 .59V
Find the v oltage across t he 220
W resis tor
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
Your Multimeter
pincer clips – good for working
with Boe-Bot wiring
(push these onto probes)
probes
leads
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
5V
+
-
+
DV = 3.41V
470W
-
220W
+
-
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.
5V
DV = 3.41V
470W
-
220W
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 = 5V
+
-
LED
470W
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
up your LED.