First Order Circuits - Lecture notes 4 PDF

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EET 300 || Chapter 4 (C) Lesson Notes

1st Order Circuits

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9/24/2011

First order circuits Two types of 1st order circuits:

RC

and

RL

RC networks:

A

Perform KCL at node A. Solve for the voltage across the capacitor with respect to time.

0  0 

VC (t)  Vs R VC (t)  Vs R

t=0

+ -

vs

+

R

v C (t)

C

-

 iC (t)  C

VS 1 V (t )  C  R R C

d VC (t)

dt dVC (t)



VS RC



d VC (t) 1 VC (t)  dt RC

dt

A

RL networks:

t=0 Perform KVL and solve for the inductor current.

vs

0   VS  iL (t )R  L VS L



+ -

R i(t)

diL (t) dt

di (t) R iL (t)  L  dt L

L

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1st Order Circuits

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1st order Linear Ordinary Differential Equations (ODE) General form  Q ( x ) 

dy  P (x ) y dx

The solution to this general form can be written: t

y(t)  A  B e 

where :

A and B are constants determined by the circuit configuration and the source mag nitude.  is the time constant We can de termine A and B through two conditions: 1) The "initial" conditions 2) The "fin al" condition

Let y(0 )  the initial response

t  0 t  

and

y     the final response

We can obtain two equations with which we can use to determine A and B. The equations are .... 0

 y( 0  )  A  B e   AB 1 

y   A  B e   A  0

from which we can easily see that A  y  therefore

and

B  y(0 )  y    t

y(t )  y      y( 0 )  y     e   

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The time constant for the two different types of networks are:

RC network

  RThCTh 

RL network

LTh

RTh

It should be noted that it may be necessary to Thevenize a circuit to determine what these values are. CASE 1

RC Network - Charge Phase - Voltage

t=0

  RC

+ -

vs

y  VC ( t)

R C

t

Vc ( t)  Vc     Vc ( 0 )  Vc     e      y( 0 )  VC ( 0 )  0v  determined  y    VC ( )  VS  earlier t

Vc ( t)  VS   0v  VS  e RC t   VC ( t )  VS  1  e RC     

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  0 VC ( t  0 )  VS  1  e    0  1  





VC ( t   )  VS 1  e1  0.632VS

 V ( t  3 )  V 1  e V ( t  4 )  V  1  e V ( t  5 )  V  1  e

   0.950V   0.982V   0.99 3V

VC ( t  2 )  VS 1  e 2  0.865VS 3

C

S

C

S

S

4

5

C

S

S

S

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

1

2

3

4

5

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CASE 1 RC Network - Charge phase - Current (Recall that the capacitor acts as a short circuit at t = 0+) y(t)  ic ( t) (Also equals iR (t )) t

ic( t)  i c( )  i c( 0)  i c( )  e V y( 0 )  ic ( 0)  s R and y( )  0   RC



V  t ic( t)  0   s  0  e RC   R V t ic ( t )  s e RC current charge pha se R iC ( 0 )  iC ( ) 

VS

V 0  S e R 1 R

VS

iC (3  )  iC (5 ) 

e1  0.368

R VS R VS R

e 3  0.05

VS R VS

e 5  0 .00 7

R VS

iC (2 )  iC (4 ) 

VS R VS R

e2  0.135 e 4  0.018

VS R VS R

R 1 0.9 0.8 0.7 0.6 0.5 0.4 (.368) 0.3 0.2 0.1 0 0

RC network Current CHARGE phase (.135)

1

(.05)

2

3

(.018)

4

(.007)

5

(tau)

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1st Order Circuits

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CASE 1

RC Network -

Discharge phase - Voltage

Discharge phase - Voltage t

 Vc (t )  Vc     Vc (0 )  Vc   e  y( 0 )  VC ( 0 )  VS   y    VC ( )  0v  Note that it is assumed here that the c apacitor

had time to completely charge to Vs. CASE 1

RC Network -

Discharge phase - Current

Discharge phase - Current vc (t )  vc      vc (0  )  vc     e   V  y( 0  )  i( 0  )   S  R y     i( )  0A  

t



Note that it is assumed here that the capacitor had time to completely charge to Vs. Also note that the current d irection will reverse in the capacitor discharge , thus the change in sign of the current.  V  t i( t)  0A    S  0A e RC    R

i( t)  

VS  t e RC R

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1st Order Circuits

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Vc(t)

Let’s combine the charge and the discharge phase of our RC circuit. These curves are based on a time constant = 1. The magnitudes are a percentage of the max voltage or current.

1V

S

RC voltage waveform

0.8 V S 0.6 V

S

RC = 1 0.4 V S 0.2 V S 0 0

VS

ic(t)

R

2

1 0.8 0.6 0.4 0.2 0 -0.2 0 -0.4 -0.6 -0.8 -1

4

6

8

t, s

10

RC current waveform RC = 1 t, s 2

4

6

8

10

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1st Order Circuits

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CASE 2

RL Network - Charge Phase – Voltage t

y( t)  y( )   y ( 0 )  y( )  e Let y( t)  v( t) and   L v L ( 0 )  Vs

R v L ( )  0

and

v L ( 0 )  0  Vs  0  e v L ( t )  Vs e CASE 2

RL

 tR



t

L



RL charge phase

- Discharge Phase – Voltage

let y(t )  V(t )

a nd

vL (0 )   Vs

an d

  L

R vL ( )  0

Note the change in polarity on Vs. V(t )  V( )   V(0  )  V(  )  e  Rt

vL (0 )  0   Vs  0 e vL ( t )   Vs e

Rt

L

 Rt

L

L

RL discharge p hase

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CASE 2

RL Charge Phase - Current

y( t )  y( )   y( 0 )  y( )  e Let y( t )  i( t ) iL ( 0 )  0 i( t) 

Vs



R

iL ( ) 

a nd

Vs

R

 Vs   R t L e  0  R  R 

i(t) 

CASE 2

  L

a nd

t

Vs

 Rt   1  e L  RL charge phase R  

RL - Discharge Phase - Current

y( t )  y(  )   y(0  )  y(  )  e   L Let y(t )  i(t ) and iL (0  ) 

Vs



R

iL ( )  0

an d

R

t

 Vs  R t L  0 e i( t )  0   R   i(t) 

Vs

 Rt

R

e

L

RL discharge phase

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1st Order Circuits

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Charge

Discharge

 Rt  Vs  i( t )  1  e L  R  

Vs

IL(t)

VL (t )  Vse

1

 Rt

i( t ) 

R

VL (t )  -Vse

L

VS R

RL Current waveform

0.8 0.6 RL = 1 0.4 0.2 0

VL(t)

0

2

1V S 0.8 0.6 0.4 0.2

0 -0.2 0 -0.4 -0.6 -0.8 -1 V S

4

6

8

t, s

10

RL voltage waveform RL = 1 t, s 2

4

e

 Rt

6

8

10

L

 Rt

L

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1st Order Circuits

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Example: The following is a MultiSim example of an RL circuit. A current controlled voltage source has been used so that the current thru the inductor can be plotted in a voltage analysis.

Instead of using a switch, a TTL square-wave is used with a 50% duty cycle. It is set up for a time high of 5 time constants and a time low of 5 time constants so that an entire charge discharge cycle can be shown. (See figure to the right)

VS

R1

VL

2.2kohm L1 V1 66.667kHz 5V

3.3mH IL 3

V2 1 Ohm

0 Note that the Node IL is actually a voltage which will look exactly like it's CONTROLLING current IL.

L R 3.3mH  2.2k   1.5s

 

5  5 * 1.5s  7.5s 2 cycles needed to be able to show both charge and dischar g e 1 1   66.667 khz f  T  2 • 7.5s  15s  15 s T

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Let’s predict what the peaks and valleys of the waveforms will be.

Charge phase

Discharge phase

i  0   0

5v  2.27mA 2.2k v 0   5 i  

v   0

i  0    2.27mA i    0

v  0    -5 v    0

Before simulating the circuit you need to set up how long the simulation will run. The period (T) was calculated as 15μs . Now, we desire to have 2 cycles therefore we need to set the simulation time (TSTOP) to: 2T = 30μs (Entered as 30e-6).

Select “Simulate”, “Analyses”, “Transient Analysis” and make the required changes shown.

Since the inductor was initially “unfluxed” the initial conditions setting was set at “Set to zero”.

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A final step is to set the output nodes that need to be plotted. The desire is to create a graph showing both VL and IL, so in the Transient analysis window select the Output variables tab. Once there, add the $il and $vl variables to the “Selected variables for analysis” table.

And finally, select “Simulate” at the bottom of the Analysis screen.

When the Grapher screen appears and the simulation is complete, you should see a screen which looks like the one to the right. This figure has set the background to white vice the default of black to save on printer ink. In order to do this, select the “reverse colors” button as shown in the figure. Next you should note that the blue (circled) trace is a straight line. This is because currently the voltage and the current waveforms are sharing the same axis. This will be corrected on the following page.

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Single right click in the body of the graph and the window to the left will open. Select “Properties” and then the “Graph Properties” window (right) will open. This figure has the Left Axis selected. If yours doesn’t, then select the Left Axis tab. We will set the left axis for the VL waveform so change the Label to “Inductor Voltage (V)”. I will also modify the font but that is up to you except for the color of the label. An axis labels color should match the color of its trace. I have decided that the voltage trace will be blue which will be set later. Next the Right Axis should be prepared for the current waveform. Since the prediction was that the waveform would go from 0 to 2.27mA, set the waveform to go from 0 to 5ms, 5 ticks and 1 minor tick. Make sure that you enable the axis and name the axis.

5mA was selected to help keep it off of the voltage waveform.

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Finally, select the Traces tab. Select Trace 2 which in the example is IL (yours might be Trace 1). Assign the waveform to the Right axis, change its color to red and thicken the pen size to 3. Select Apply and then select the right axis by selecting trace 1.

Again change the traces color to Blue and the Pen size to 3. The waveform should already be assigned to the left axis. If you like you can change the label names to Voltage and Current. This will be used for the trace ID window.

Select “General” and set the “Grid on”. If you desire, you can darken the grid by setting the pen size to 1 and changing the color to black. Just make sure that the grid is lighter and thinner than the traces. Next, left click on the graph and select Page Properties. This will open the Page Properties window where you can change the title of the graph. While there, change the Tab Name from the default to Un-fluxed.

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1st Order Circuits

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Finally, select the “Bottom Axis” and change the Time axis as shown. You could use Minor Ticks 1 vice 2 but the time labels may be too close together (depending on how large to make the graph window in your document.

Your graph should now look something like this: Note that the Labels are identified by color.

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Copy the graph into the clipboard so that it can be included in your word processed document by clicking on the copy icon.

Now that you have the graph formatted the way you want, select the File/save on the GRAPHER menu and save the graph.

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In the next example, we will repeat the problem except that we will initially flux the inductor to a current of 7mA in the up direction.

In this case, the inductor is double left clicked and a “-.007" is entered in the initial condition block. Then place a check mark in the “Initial Condition” block.

In the transient analysis window, set the initial condition block to “User-defined”.

(One item which was not mentioned earlier was that the minimum number of points was changed from the default of 100 to 150 {you could go farther} in order to make the graph smooth out).

For this example, let’s also change the voltage source to 10 volts vice 5 volts.

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Before actually simulating the circuit, let’s predict what the waveforms will look like. A model including the initial condition (in the correct direction) is created and then the predictions are calculated.

2.2k 

+ 10v -

- V R+

i(t) by KVL 0  10v - VL  VR

+ 7mA VL

VL  10v  (7mA)(2.2k )  10v  15.4v

-

VL  25.4v y  t   y( )   y(0 )  y( )  e y  t  i  t for i  0    -7mA

and

i   

 Rt

L

10v  4.55mA 2.2k 2.2 kt

i  t  4.55mA  -7mA  4.55mA e  i  t   4.55m A - 11.55mAe t (666.67k) 

for

y  t  VL  t

VL  0    25.4v

a nd

VL     0v

VL  t   0 v  25.4 v  0v  e t ( 666.67k)  VL  t   2 5.4ve t(666.67k) 

3.3mH

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The resulting graph looks like that on the right. If the graph from the previous example is not present on the tab list, go to the File/open menu on the Grapher menu and open the previously saved graph file. Let’s go ahead and select the Page Properties for this graph and name the Tab “Fluxed”. While you there, change the graph name to RL Charge-Discharge (Fluxed) example.

Select the Un-fluxed tab to bring up the previous graph. You spent a lot of time formatting that graph to get it to look so good. Observe the toolbar for that graph. You will note the two icons that look like price tags (one with an arrow leaving it and one with an arrow entering it). (If these icons are not available then you need to click inside the graph first.) These icons are time savers. They enable you to apply MOST (not all) of the formatting options of this graph to the next graph.

Left click on the icon with the arrow leaving the tag. Select the Fluxed tab Left click on the icon with the arrow going into the tag

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Your graph now looks something like this. Now most of what is needed is to change the limits on the right and the left axis.

This graph demonstrates the fluxed result. One thing you have to watch when using the properties save and apply icons is that sometimes the traces have the wrong labels attached to them. Apply some common sense to your choice of which trace is which.

One thing which can be observed on this graph is how the graphs are not smooth curves. This will be even more observable when you enlarge the graphs. Let’s return to the simulation menu and change the max number of points to 500. The final graph will show that the curve is much smoother.

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Note that after the initial cycle, things start acting with the new values instead of values due to the initial flux. This also demonstrates that you should always observe more than one cycle in any simulator. While this first cycle was correct, the first cycles of waveforms in simulators are not always to be trusted. Also note the measurements tied to the two moveable measure lines. Click on the trace which you want the measurements to reflect. Or, you can go to the page properties menu and select “show all traces” and all traces will be represented in the window....