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:

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:

Using this basic version of an Op-Amp, can you remember any of them?

V p=V n

i p=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.

I 1+ I 2+ I3=0

Vp

VnVo

Assuming this is an ideal Op-Amp; using our assumptions listed above and nodal analysis we can solve for V o:

Start analysis at V 1:

V 1−V s

R s+

V 1−V 0

R f=0

What can we do to solve for V 0?

We can assume that V 1=V p=V n whereV n=0

−R f V s−R sV 0=0

−R sV 0=R f V s

V 0=−R f V s

R s

Now it’s your turnExample 1:

Assuming the Op-Amp is ideal, find V 0 in terms of variables given:

Start by doing analysis at Vn and Vp:

V n−V 1

R1+

V n−V 0

R2+0=0

V p−V 2

R3+

V p

R4=0

Solve for Vp:

V p=R4 V 2

R3+R4

Assuming V p=V n:R4 V 2

R3+R4−V 1

R1+

R4 V 2

R3+R4−V 0

R2=0

V 0=[( R4

R3+R4 )( R1+R2

R1 )]V 2−( R2

R1 )V 1

Nodal Analysis

Plugging In

Solution

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.

G=V 0

V s

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:

V 0 ≤ A (V p−V n )

It is important to note that when finding V 0, regardless of ideal or not, if the value

of V 0 is above or below +¿

−¿V cc ¿¿ the output voltage becomes

+¿−¿V cc ¿

¿

Open Loop Gain A

Inverting - Input V n

Noninverting - Input V p

Output Voltage V o

Power Supply V cc

Example 2:

Perform analysis on the circuit below using the following steps:a) Assume the following values: V s=1.5 volts, R s=1.0 kΩ, R f=50.0 kΩ,

RL=500 Ω, and V cc=10volts.b) Assume the Op-Amp is ideal. c) Find V 0.d) Calculate the gain (G) of the Op-Amp.

Analysis at V 1:V 1−V s

R s+

V 1−V 0

R f=0

−1.5V1000 Ω

+−V 0

50000 Ω=0

V 0=75Volts

When comparing it to V cc we find that 75 volts is too large.

Therefore: V 0=10Volts

Finding Gain:

G=V 0

V s

G= 10 V1.5 V

G=6.67

Analysis

Solution

Gain

Step RL Circuits:

States within the circuit:

1. 0−¿¿ Immediately prior to switch being thrown2. 0+¿ ¿ Immediately after switch is thrown3. ∞ 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 isflipped 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) I 1(0)b) I 2(0)c) I 1(∞)d) I 2(∞)

1) Model each state accordingly:

2) Nodal analysis before switch has been thrown to find a & b:

V 1−308

+V 1

6+

V 1

3=0

V 1

8+

V 1

6+

V 1

3=3.75

15V 1

24=3.75 V 1=6 Volts

Nodal

3) Plugging in V 1into the equation V=IR:

I 1=VR

=66=1 Amp

I 2=VR

=63=2 Amp

4) If we properly drew the circuit for t = ∞, finding the currents here is simple:

V=IR V s = 0

I = 0

Step RC Circuits:

States within the circuit:

4. 0−¿¿ Immediately prior to switch being thrown5. 0+¿ ¿ Immediately after switch is thrown6. ∞ 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 isflipped 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.

A & B

C & D

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¿b) vc¿c) vc(∞)

Model each state accordingly:

A) We can do voltage division to find vc¿:

vc¿

vc¿

vc¿

B) If we use our assumptions from above:

vc¿ Not enough time to change the charge on the capacitor

vc¿

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:

Part A

Part B

Part C

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 (ϖ o¿.

Damping Coefficient for series RLC:α= R

2 L=s−1

Resonant Frequency:ϖ o=

1√LC

=s−1

Final Example:

For the series RLC circuit shown, determine values below if the switch opens at t = 0:A) I L¿B) vc¿C) I L (∞)D) vc(∞)E) αF) ϖ o

G) Determine Damping Response

Using the assumptions we have reviewed earlier, redraw the circuit for each state:

Nodal Analysis at t = 0−¿¿:

V 1−12.4

+V 1

1.2+0=0

V 1

.4+

V 1

1.2=30

3V 1+V 1=36

V 1=9Volts

With 0 resistance between V 1 and V c:V 1=V c

V c ¿

Knowing that the voltage through a capacitor can’t change immediately:V c ¿

V c ¿

No resistance between V 1 and V c:V=IR R = 0

I L¿

Knowing that the current through an inductor can’t change immediately:I L¿

I L¿

If the circuit was draw correctly, there should be an open circuit when t = ∞:

Open circuit ¿ I L (∞ )=vc (∞ )=0

Part A

Part B

Part C & D

Damping Coefficient for series RLC:

α= R2 L

= 1.2 R2(.1 H)

α=6 s−1

Resonant Frequency:

ϖ o=1

√LC= 1

√( .1 H )(.4 F)

ϖ o=5 s−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!

Part E

Part F & G