Week 2: D.C. Intro
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Transcript Week 2: D.C. Intro
SMV ELECTRIC
TUTORIALS
Nicolo Maganzini, Geronimo Fiilippini, Aditya Kuroodi
2015
Relevant Course(s): EE10, EE11L
2
RESISTORS IN
NETWORKS
What are we learning?
Learn about the math behind networks of resistors.
Current and Voltage laws.
Predicting/designing circuits that have specific values of
Current, Voltage, Resistance
Learn about some very important structures of networks
Parallel and series
How are they used?
CAUTION: Math involved.
3
Resistors in Networks
In Circuit Schematics:
In Real Life:
4
Resistor Network Calculations
- Series Networks
You have this circuit: R1 = 1 Ohm, R2 = 2 Ohm, R3 = 3 Ohm, V = 6V
How can you apply Ohm’s law to find out how much current is
flowing?
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Series Resistors Equation.
This is called a series connection:
Equivalent Resistance = R1 + R2 + R3 + R4
Since there is only one path for electrons, there is only
one current value in the part of the circuit with the
series connection.
Try it yourselves! (next slide)
6
The circuit we’re building:
R1 = 100 Ohm
R2 = 220 Ohms
R3 = 300 Ohms
Battery = 9V
Measure current at nodes 1,2.
Write them down. Check that they are equal.
Measure voltages V1(across R1), V2 (across R2), V3 (across
R3), across the battery.
Calculate:
V1/R1, V2/R2, V3/R3 What should these be equal to?
V1+V2+V3 What should this be equal to?
(V1+V2+V3)/(R1+R2+R3) What should this be equal to?
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Parallel Networks
Current has multiple paths it can take.
It will split according to the resistance in each path.
Path with lower resistance gets most current.
Path with higher resistance gets less current.
If resistances are equal, all paths have the same current.
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Let’s combine the two!
Then add it to this one!
This is in Parallel:
Find it’s equivalent
Split circuit between parallel and series parts.
Simplify the parallel part and add it to the series part.
Parallel part simplification:
Overall equation for resistance:
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Sample Problem
Calculate the current flowing out of the battery in this
circuit:
R1 = 100 Ohms
R2 = 150 Ohms
R3 = 200 Ohms
Battery = 9V
10
Kirchoff current and voltage
laws
How do we analyze more complicated circuits?
There are some physics laws that we can apply to
circuits that allows us to find equations: Kirchoff laws.
Steps:
1) Apply Laws
2) Find Equations
3) Solve equations to find current, voltage and
resistance.
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Kirchoff Voltage Law (KVL)
What the law says:
The sum of all voltages in a loop must be equal to zero.
Example of how we use it:
Vbatt = 9V.
V1 = 2V
V2 = 3V
R3 = 4 Ohms
Find the current in the circuit.
12
Kirchoff Voltage Law (KVL)
Step 1) Apply law:
The voltage produced by the battery is equal to the voltage
dropped by each resistor.
Step 2) Find Equation:
Vbatt = V1+V2+V3 Know Vbatt, V1, V2; Find V3
I = V3/R3 Know V3 and R3, Find I.
Step 3) Solve:
V3 = 9-2-3 = 4V
I = 4/4 = 1A
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Kirchoff Current Law (KCL)
What the law says:
The sum of all currents entering and exiting a node must
be zero.
Example of how we use it:
R1 = 100 Ohms.
R2 = 200 Ohms
R3 = 200 Ohms.
Current through R1 = 1A
Find voltage of battery.
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Kirchoff Current Law (KCL)
Step 1) Apply Laws:
Current flowing into node 2 from R2 and R3 must be equal
to current flowing out towards R1.
Current flowing in R2 and R3 must be equal because
resistances are equal (200 ohm)
Sum of voltages must be equal to the battery voltage
Step 2) equations:
I1 = I2 + I3
I2 = I3
V1+V2 = V1 + V3 = Vbattery
Step 3) solve:
1 = ½ + ½ I2 = I3 = ½ A
V1 = I1 R1 = 100V
V2 = V3 = ½ x 200 = 100V
Vbatt = 100 + 100 = 200V
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Using series connections to
make a sensor
Potential divider equation:
𝑉𝑜𝑢𝑡
= 𝑉𝑏𝑎𝑡𝑡𝑒𝑟𝑦
𝑅2
𝑅1 +𝑅2
VERY IMPORTANT EQUATION.
Pseudo-Derivation
If Resistance values are constant, then Vout will be
constant.
What if the resistance of one resistor changes with
temperature or light? How does Vout change?
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CAPACITORS AND SIGNAL
FILTERING
Ohm’s Law for Capacitors
Voltage across resistor depends on value of current at that
instant in time:
𝑉 = 𝐼𝑅
Voltage across capacitor depends on how fast the current is
changing:
𝑉
= 𝐼 𝑋𝑐
where
𝑋𝑐 =
1
= Capacitive
2𝜋𝑓𝐶
Reactance
𝑉
=
𝐼
2𝜋𝑓𝐶
V is maximum voltage across capacitor, I is maximum current
through capacitor, C is capacitance, f is frequency of signal.
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Low Pass Filter
Remember potential divider equation?
Voltage across R2 is given by:
𝑅2
1 +𝑅2
𝑉𝑜𝑢𝑡 = 𝑉𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑅
1
Now substitute R2 with reactance 𝑋𝑐 = 2𝜋𝑓𝐶
Can you do it?
1
𝑉𝑜𝑢𝑡 = 𝑉𝑏𝑎𝑡𝑡𝑒𝑟𝑦 2𝜋𝑓𝐶𝑅+1
Increase resistance = lower 𝑉𝑜𝑢𝑡
Increase capacitance = lower 𝑉𝑜𝑢𝑡
Increase frequency = lower 𝑽𝒐𝒖𝒕
This is a LOW PASS FILTER.
Can tune 𝑅 and 𝐶 to cancel out the right
requencies.
𝑉𝑜𝑢𝑡
𝑉
High Pass Filter
What if we turn around the circuit, so that capacitor is
on the top?
𝐶
𝑅
𝑉
𝑅2
1 +𝑅2
𝑉𝑜𝑢𝑡 = 𝑉 𝑅
𝑉𝑜𝑢𝑡
1
substitute 𝑋𝑐 = 2𝜋𝑓𝐶 for 𝑅1 .
Can you simplify it?
This time low frequencies are
attenuated
This is a high pass filter.
Note: if signal has f = 0, it is
completely eliminated
DC is blocked. Only signals that
change in time make it through
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R = 1 kOhm
C = 0.22 microF
What have we learned?
If the signal has a certain frequency, we can make an R-C
circuit that cancels the signal out.
If a signal has more than one frequency, such as noise:
Can clean it up using an R-C filter designed to cancel out all
frequencies lower than a certain amount.
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