Transcript Week 2: D.C. Intro

SMV ELECTRIC
TUTORIALS
Nicolo Maganzini, Geronimo Fiilippini, Aditya Kuroodi
2015
Relevant Course(s): EE10, EE11L
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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?
5
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?
7
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
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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.
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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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