Electrical Part II
The purpose of these pages is to remind us of the immense amount of knowledge
we happily gained during the second half of our Electric Circuits class.
Discussion Topics:
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Assumptions of Ideal Op-Amps
Nodal analysis
Op-Amp solving
Finding output voltages
Calculating Op-Amp gain
Step RC circuits
Step RL circuits
Determining damping
Step RLC series circuits
Ideal Op-Amp Assumptions:
Vp
Vo
Vn
Using this basic version of an Op-Amp, can you remember any of them?



𝑉𝑝 = 𝑉𝑛
𝑖𝑝 = 0
𝑖𝑛 = 0
Another use for Nodal Analysis:
As Debbie stated during her terrific discussion, Nodal Analysis is defined by: At any junction
within a circuit, the sum of currents flowing into the node is equal to the sum of currents flowing
out of the node.
𝐼1 + 𝐼2 + 𝐼3 = 0
Assuming this is an ideal Op-Amp; using our assumptions listed above and nodal analysis we
can solve for 𝑉𝑜 :
Start analysis at 𝑉1 :
𝑉1 − 𝑉𝑠 𝑉1 − 𝑉0
+
=0
𝑅𝑠
𝑅𝑓
What can we do to solve for 𝑉0?

We can assume that 𝑉1 = 𝑉𝑝 = 𝑉𝑛 𝑤ℎ𝑒𝑟𝑒 𝑉𝑛 = 0
−𝑅𝑓 𝑉𝑠 − 𝑅𝑠 𝑉0 = 0
−𝑅𝑠 𝑉0 = 𝑅𝑓 𝑉𝑠
𝑉0 = −
𝑅𝑓 𝑉𝑠
𝑅𝑠
Now it’s your turn
Example 1:
Assuming the Op-Amp is ideal, find 𝑉0 in terms of variables given:
NodalbyAnalysis
Start
doing analysis at Vn and Vp:
𝑉𝑛 − 𝑉1 𝑉𝑛 − 𝑉0
+
+0=0
𝑅1
𝑅2
𝑉𝑝 − 𝑉2
𝑉𝑝
+
=0
𝑅3
𝑅4
Solve
Plugging
forInVp:
𝑉𝑝 =
𝑅4 𝑉2
𝑅3 + 𝑅4
Assuming 𝑉𝑝 = 𝑉𝑛 :
𝑅4 𝑉2
𝑅4 𝑉2
𝑅3 + 𝑅4 − 𝑉1 𝑅3 + 𝑅4 − 𝑉0
+
=0
𝑅1
𝑅2
Solution
𝑉0 = [(
𝑅4
𝑅1 + 𝑅2
𝑅2
)(
)] 𝑉2 − ( ) 𝑉1
𝑅3 + 𝑅4
𝑅1
𝑅1
Finding Voltage Gain & Output Voltage:
Gain (G): The ratio between the source voltage and the output voltage the Op-Amp produces.
The gain allows us to see the effect the Op-Amp had on the starting.
𝑉0
𝐺=
𝑉𝑠
Finding the Output Voltage can be done a lot easier when the Op-Amp is not assumed to be
ideal. By simply using this equation when you are given certain variables by using you can find
Output Voltage:
𝑉0 ≤ 𝐴(𝑉𝑝 − 𝑉𝑛 )

It is important to note that when finding 𝑉0, regardless of ideal or not, if the value
of 𝑉0 is above or below +−𝑉𝑐𝑐 the output voltage becomes +−𝑉𝑐𝑐
Open Loop Gain
A
Inverting - Input
𝑉𝑛
Noninverting - Input
𝑉𝑝
Output Voltage
𝑉𝑜
Power Supply
𝑉𝑐𝑐
Example 2:
Perform analysis on the circuit below using the following steps:
a) Assume the following values: 𝑉𝑠 = 1.5𝑣𝑜𝑙𝑡𝑠, 𝑅𝑠 = 1.0𝑘𝛺, 𝑅𝑓 = 50.0𝑘𝛺,
𝑅𝐿 = 500𝛺, and 𝑉𝑐𝑐 = 10𝑣𝑜𝑙𝑡𝑠.
b) Assume the Op-Amp is ideal.
c) Find 𝑉0.
d) Calculate the gain (G) of the Op-Amp.
Analysis at 𝑉1:
Analysis
𝑉1 − 𝑉𝑠 𝑉1 − 𝑉0
+
=0
𝑅𝑠
𝑅𝑓
−1.5𝑉
−𝑉0
+
=0
1000𝛺 50000𝛺
𝑉0 = 75𝑉𝑜𝑙𝑡𝑠
Solution
When comparing it to 𝑉𝑐𝑐 we find that 75 volts is too large.
Therefore: 𝑉0 = 10𝑉𝑜𝑙𝑡𝑠
Finding Gain:
Gain
𝐺=
𝐺=
𝑉0
𝑉𝑠
10𝑉
1.5𝑉
𝐺 = 6.67
Step RL Circuits:
States within the circuit:
1. 0−  Immediately prior to switch being thrown
2. 0+  Immediately after switch is thrown
3. ∞  After switch has been thrown for a long period of time
 These different states can be used for both currents and voltages within the
circuit.
How to model an inductor for each state:
Initial circuit: Imagine a voltage source is initially attached to the circuit, but when a switch is
flipped the voltage source is disconnected from the rest of the circuit.
0− when voltage source has been applied for a long time prior to switch being thrown:
0+  model the inductor like a current source:

Important: when finding current through the inductor at this time, the current at
this state will always equal the current at 0− . This is because current through the
inductor cannot change immediately when the switch is flipped.
∞  voltage source has been removed from circuit due to switch flipping so the resistance
inside the inductor goes to 0:
Once you are able to model the inductor correctly given the certain state you need to find, the
next step is using nodal analysis to solve for what needs to be found.
RL Example:
For the circuit determine:
a) 𝐼1 (0)
b) 𝐼2 (0)
c) 𝐼1 (∞)
d) 𝐼2 (∞)
1) Model each state accordingly:
Nodal
2)
Nodal analysis before switch has been thrown to find a & b:
𝑉1 − 30 𝑉1 𝑉1
+
+
=0
8
6
3
𝑉1 𝑉1 𝑉1
+
+
= 3.75
8
6
3
15 𝑉1
24
= 3.75  𝑉1 = 6 Volts
A&B
3) Plugging in 𝑉1 into the equation 𝑉 = 𝐼𝑅:
𝐼1 =
𝑉
6
= = 1 𝐴𝑚𝑝
𝑅
6
𝐼2 =
𝑉
6
= = 2 𝐴𝑚𝑝
𝑅
3
C4)& DIf we properly drew the circuit for t = ∞, finding the currents here is simple:
𝑉 = 𝐼𝑅  𝑉𝑠 = 0
I=0
Step RC Circuits:
States within the circuit:
4. 0−  Immediately prior to switch being thrown
5. 0+  Immediately after switch is thrown
6. ∞  After switch has been thrown for a long period of time
 These different states can be used for both currents and voltages within the
circuit.
How to model an Capacitor for each state:
Initial circuit: Imagine a voltage source is initially attached to the circuit, but when a switch is
flipped the voltage source is disconnected from the rest of the circuit.
0− when voltage source has been applied for a long time prior to switch being thrown:

This is depicting an open circuit. The capacitor reached its max charge and is not
allowing any current through this part of the circuit.
0+  model the capacitor like a DC voltage source:

Important: when finding voltage through the capacitor at this time, the voltage at
this state will always equal the voltage at 0− . This is because voltage through the
capacitor cannot change immediately when the switch is flipped.
∞  Voltage source has been removed from circuit due to switch flipping so the voltage within
the capacitor eventually falls to zero resulting in an open circuit:
RC Example:
The circuit in the figure was in steady state until the switch was moved from terminal 1 to
terminal 2 at t =0. Determine each of the following:
a) 𝑣𝑐 (0− )
b) 𝑣𝑐 (0+ )
c) 𝑣𝑐 (∞)
Model each state accordingly:
Part A
A) We can do voltage division to find 𝑣𝑐 (0− ):
𝑣𝑐 (0− ) =
𝑣𝑐 (0− ) =
𝑅2
𝑣
𝑅1 + 𝑅2 + 𝑅3 𝑠
3𝑘𝛺
(42 𝑉)
2𝑘𝛺 + 3𝑘𝛺 + 4𝑘𝛺
𝑣𝑐 (0− ) = 14 𝑉𝑜𝑙𝑡𝑠
PartB)B If we use our assumptions from above:
𝑣𝑐 (0+ ) = 𝑣𝑐 (0− )  Not enough time to change the charge on the capacitor
𝑣𝑐 (0+ ) = 14 𝑉𝑜𝑙𝑡𝑠
PartC)C If we were able to draw our circuit at each of the states correctly:
There is no power source connected to the circuit for a long period of time.
𝑣𝑐 (∞) = 0
Step RLC Circuits and Determining Damping:
RLC circuits include all the assumptions we have learned for both inductors and
capacitors. We are just putting the two components within the same circuit to make
it an RLC circuit.
 Although you can model RLC circuits in both parallel and in series,
we will only be working with series RLC circuits.
3 different kinds of damping responses:
1) Overdamped ( 𝛼 > 𝜛𝑜 )
2) Critically Damped (𝛼 = 𝜛𝑜 )
3) Underdamped (𝛼 < 𝜛𝑜 )
 You can determine which kind of damping the circuit is responding
with by finding the damping coefficient (α) and resonant frequency
(𝜛𝑜 ).
Damping Coefficient for series RLC:
𝑅
𝛼=
= 𝑠 −1
2𝐿
Resonant Frequency:
𝜛𝑜 =
1
√𝐿𝐶
= 𝑠 −1
Final Example:
For the series RLC circuit shown, determine values below if the switch opens at t = 0:
A) 𝐼𝐿 (0+ )
B) 𝑣𝑐 (0+ )
C) 𝐼𝐿 (∞)
D) 𝑣𝑐 (∞)
E) α
F) 𝜛𝑜
G) Determine Damping Response
Using the assumptions we have reviewed earlier, redraw the circuit for each state:
Part A
Nodal Analysis at t = 0− :
𝑉1 − 12
𝑉1
+
+0=0
.4
1.2
𝑉1
𝑉1
+
= 30
. 4 1.2
3𝑉1 + 𝑉1 = 36
𝑉1 = 9 𝑉𝑜𝑙𝑡𝑠
With 0 resistance between 𝑉1 and 𝑉𝑐 :
𝑉1 = 𝑉𝑐
𝑉𝑐 (0− ) = 9 𝑉𝑜𝑙𝑡𝑠
Knowing that the voltage through a capacitor can’t change immediately:
𝑉𝑐 (0− ) = 𝑉𝑐 (0+ )
𝑉𝑐 (0+ ) = 9 𝑉𝑜𝑙𝑡𝑠
Partresistance
B
No
between 𝑉1 and 𝑉𝑐 :
𝑉 = 𝐼𝑅  R = 0
𝐼𝐿 (0− ) = 0
Knowing that the current through an inductor can’t change immediately:
𝐼𝐿 (0− ) = 𝐼𝐿 (0+ )
𝐼𝐿 (0+ ) = 0 𝐴𝑚𝑝𝑠
Part
& D was draw correctly, there should be an open circuit when t = ∞:
If
theC circuit
Open circuit = 𝐼𝐿 (∞) = 𝑣𝑐 (∞) = 0
Part E
Damping Coefficient for series RLC:
𝑅
1.2𝑅
=
2𝐿
2(.1𝐻)
𝛼=
𝛼 = 6𝑠 −1
Part
F&G
Resonant
Frequency:
𝜛𝑜 =
1
√𝐿𝐶
=
1
√(. 1𝐻)(.4𝐹)
𝜛𝑜 = 5𝑠 −1
Given that 𝛼 > 𝜛𝑜 :
This circuit’s total response is Overdamped
Today we have had an immense amount of pleasure reviewing the ideas and assumptions behind
ideal op-amps, RC circuits, RL circuits, and RLC circuits. A long with these assumptions we
have brushed up on our ability to find output voltages, gain, resonant frequency, damping
coefficient, as well as determining which response the circuit is getting. I hope that you all had as
much fun reviewing this material as I did producing this spectacular presentation. You will no
longer need luck to pass the electrical section on the FE Exam seeing as Debbie and I have
prepared you to ace it. You’re most welcome!