Linear Circuit Analysis:

Charge and Current

Work, Power, Voltage, and
Resistance

Kirchoff's Laws

Y Delta, Node, and Loop

Circuits with Operational
Amplifiers

Network Theorems

Analysis of Diode Circuits

Capacitance and Inductance

First-Order Transient Circuits

AC Steady State Analysis

Steady State Power

The Power Factor

Introduction

adv.92g
circuits.92g
delta2yt.92g

Linear Circuit Analysis:
Analysis of Diode Circuits

For all of the examples on this page define the input voltage V_S.

V_S will be a triangular wave with an amplitude ranging from 10 to -10. Originating at t = 0. The input will have a period of 2 seconds. We will graph this wave with the TI-92, but first lets set the window variables. Time is represented on the x axis on the TI-92.

Press [Diamond] [WINDOW] to enter the graph range variable editing screen.
Enter 0 for xmin
Enter 2 for xmax
Enter .5 for xscl
Enter -10 for ymin
Enter 10 for ymax
Enter 1 for yscl
press [2nd] [QUIT] to return to the home screen.

Assuming you have the Cust() and PceWise programs from circuits.92g, follow these steps to define the input voltage.

Press [2nd] [Custom] to activate the custom menu.
Press [F2] [ENTER] [ENTER] to start the PceWise program.
Since our function does not start until time equals zero, graph the line 0 on the interval from negative infinity to 0.
Press [Enter] twice to go to the next range.

¼ of the period equals ½. To reach the maximum amplitude in ¼ of the period the slope equals +20 with a y intercept of zero.

Graph this part of the function from 0 to: 1/2.
Graph the line: 20x
Press [Enter] twice to go to the next range.

To reach the minimum amplitude in ¾ of the period the slope equals -20 with a y intercept of 20.

Graph this part of the function from 1/2 to: 3/2.
Graph the line: -20x+20
Press [Enter] twice to go to the next range.

To reach the axis in the period, the slope equals 20 with a y intercept of -40.

Graph this part of the function from 3/2 to: 2.
Graph the line: 20x-40
Press [Enter] twice to go to the final range.

To finish out this piecewise function we need to define the line from 2 to positive infinity.

Graph this part of the function from 2 to: infinity [2nd] [j].
Graph the line: 0
Press [Enter] twice to finish the program.

The program has written the piecewise function and stored it in y1(x).

To view the graph press [Diamond] [GRAPH].
Press [F1] [9] to bring up the graph formats screen.
Select SIMILE for the Graph Order.
Select ON for the Grid.
Press [ENTER] to redraw the graph.

The PceWise program, by default, always writes its functions into y1(x) with the variable name fun. Since we will be referencing this function several times and don't want it over written, we will need to move it to another spot for graphing.

From the Home Screen:

At the Entry Line, enter Define y2(x)=.
Press [2nd] [RCL]
In the Recall request box enter fun.
Press [ENTER] three times.

One look at the syntax used to enter piecewise curves may be enough to convince you to use the PceWise program.

Plot the waveform for the output of the circuit given the input V_s.
DeCarlo. page 179 prob 6.2

For the branch containing the diode, no current flows until V_s provides more voltage than the battery underneath the diode. Therefore this branch is an open circuit for V_s < 5V. The equation for the output voltage is:

When V_s is greater than the battery, then the diode closes the circuit and current is allowed to flow down this branch. By writing a KCL equation at the node above the diode as the sum of all currents leaving the node we get for V_s > 5V:

This gives only one equation and one unknown. Using the TI-92's built in zeros function we can solve for V_o.
We have already defined V_s for this circuit to be the graph stored in y2(x). So let's take the last answer from the Home Screen and make some changes.

Press the up arrow one time to highlight the last answer and press [ENTER] to move it to the Entry Line.
Remove the list braces.
To the end of the equation press [2nd] [|] vs [=] y2(x) [->] g5.
press [ENTER].

We stored this expression in the variable g5 to help remember that this is the output for V_s > 5 volts.
Now we need to find the value of V_s (or in our case, y2(x) ) where time (or in our case, x ) equals 5 volts. This is easily found using the solve function.

Using the PceWise program we can define the output for the entire range.

Press [2nd] [CUSTOM] [F2] [ENTER] [ENTER] to start the program.
From negative infinity to 1/4, Graph the line y2(x).
From 1/4 to 3/4, Graph the line g5.
This is the range where V_s > 5 volts.
From 3/4 to infinity, Graph the line y2(x).

After the program is finished press [Diamond] [GRAPH] to see the input signal and the output signal graphed. This graph shows how the diode limits the output of V_s in the positive region.

If we Zoom in on the positive part of the signal we can use the maximum function y1(x) to find that the highest allowed voltage by this circuit is 5.25 volts. This helps to show how diodes can be used to protect devices that could be damaged by receiving to high of an input voltage.

Plot the waveform for V_o of the circuit.
DeCarlo. page 179 prob 6.3

For the branches containing the diodes, no current flows unless V_s provides more voltage than the batteries underneath the diodes. Therefore these branches are open circuits for
-5V < V_s < 5V.
The equation for the output voltage is:

When V_s is greater than battery1, then diode1 closes the circuit and current is allowed to flow down this branch. By inspection at this node we get for
V_s > 5V:

When V_s is less than battery2, then diode2 closes the circuit and current is allowed to flow down this branch. By By inspection at this node we get for
V_s < -5V:

Since the node equations turned out to be so trivial all we need to do before using the PceWise program is to find the value of Vs (or in our case, y2(x) ) where time (or in our case, x ) equals +5 and -5 volts. This is easily found using the solve function.

Press [2nd] [CUSTOM] [F2] [ENTER] [ENTER] to start the program.

From negative infinity to 1/4, Graph the line y2(x).
This is a range where -5 volts < V_s < 5 volts.
From 1/4 to 3/4, Graph the line 5.
This is the range where V_s > 5 volts.
From 3/4 to 5/4, Graph the line y2(x).
This is a range where -5 volts < V_s < 5 volts.
From 5/4 to 7/4, Graph the line -5.
This is the range where V_s < -5 volts.
From 7/4 to positive infinity, Graph the line y2(x).
This is a range where -5 volts < V_s < 5 volts.

Press [Diamond] [Graph] to graph the voltages. You may have to press [F2] [B] [2] to zoom to the previous range that we set up earlier. This graph shows how with out the resistors the voltage output is limited to the finite value of the batteries. The same concept can be applied for an input voltage of a sin wave. This circuit will clip the voltages.

Plot the waveform for V_o of the circuit.
DeCarlo. page 179 prob 6.7

For the branches containing the diodes, no current flows unless V_s provides more voltage than the batteries underneath the diodes. Therefore these branches are open circuits for
-.5V < V_s < .5V.
The equation for the output voltage is:

When V_s is greater than battery1, then diode1 closes the branch and current is allowed to flow down this branch. Diode2 holds its branch open and no current is allowed across its branch. By writing a KCL equation at the node above diode1 as the sum of all currents leaving the node we get for V_s > 0.5V:

When V_s is less than battery2, then diode2 closes the circuit and current is allowed to flow down this branch. Diode1 holds its branch open and no current is allowed across its branch. By writing a KCL equation at the node above the diode as the sum of all currents leaving the node we get for V_s < -0.5V:

These expressions give only one equation and one unknown. Using the TI-92's built in zeros function we can solve for Vo.
We have already defined Vs for this circuit to be the graph stored in y2(x). So let's take the last answers from the Home Screen and make some changes.

Press the up arrow three times to highlight the last answer and press [ENTER] to move it to the Entry Line.
Remove the list braces.
To the end of the equation press [2nd] [|] vs [=] y2(x) [->] g5.
press [ENTER].

We stored this expression in the variable g5 to help remember that this is the output for V_s > 0.5 volts.

Press the up arrow three times to highlight the last answer and press [ENTER] to move it to the Entry Line.
Remove the list braces.
To the end of the equation press [2nd] [|] vs [=] y2(x) [->] L5.
press [ENTER].

We stored this expression in the variable L5 to help remember that this is the output for V_s < -0.5 volts.

All we need to do before using the PceWise program is to find the value of Vs (or in our case, y2(x) ) where time (or in our case, x ) equals +0.5 and -0.5 volts. This is easily found using the solve function.

Press [2nd] [CUSTOM] [F2] [ENTER] [ENTER] to start the PceWise program.

From negative infinity to .025, Graph the line y2(x).
This is a range where -0.5 volts < V_s < 0.5 volts.
From .025 to .975, Graph the line g5.
This is the range where V_s > 0.5 volts.
From .975 to 1.025, Graph the line y2(x).
This is a range where -0.5 volts < V_s < 0.5 volts.
From 1.025 to 1.975, Graph the line L5.
This is the range where V_s < -0.5 volts.
From 1.975 to positive infinity, Graph the line y2(x).
This is a range where -0.5 volts < V_s < 0.5 volts.

Press [Diamond] [Graph] to graph the voltages. This graph shows how the voltage output is limited to the finite value of the batteries. As well as, how the resistor values can cause a different slope from the input voltage. The same concept can be applied for an input voltage of a sin wave. This circuit will lessen the voltages.

Plot the waveform for V_o of the circuit.
DeCarlo. page 180 prob 6.8

For the branches containing the diodes, no current flows unless V_s provides more voltage than the batteries underneath the diodes. Therefore these branches are open circuits for
4.5V < V_s < 5.5V.
The equation for the output voltage is:

When V_s is greater than battery2, then diode2 closes the circuit and current is allowed to flow down this branch. Diode1 holds its branch open and no current is allowed across its branch. By writing a KCL equation at the node above the diode as the sum of all currents leaving the node we get for V_s > 5.5V:

Press the up arrow three times to highlight the last answer and press [ENTER] to move it to the Entry Line.
Remove the list braces.
To the end of the equation press [2nd] [|] vs [=] y2(x) [->] L45.
press [ENTER].

We stored this expression in the variable L45 to help remember that this is the output for V_s < 4.5 volts.

Press the up arrow three times to highlight the last answer and press [ENTER] to move it to the Entry Line.
Remove the list braces.
To the end of the equation press [2nd] [|] vs [=] y2(x) [->] g55.
press [ENTER].

We stored this expression in the variable g55 to help remember that this is the output for V_s > 5.5 volts.

All we need to do before using the PceWise program is to find the value of Vs (or in our case, y2(x) ) where time (or in our case, x ) equals +4.5 and +5.5 volts. This is easily found using the solve function.

Press [2nd] [CUSTOM] [F2] [ENTER] [ENTER] to start the PceWise program.

From negative infinity to .225, Graph the line L45(x).
This is a range where V_s < 4.5 volts.
From .225 to .275, Graph the line y2(x).
This is the range where 4.5 volts < V_s < 5.5 volts.
From .275 to .725, Graph the line g55.
This is a range where V_s < 5.5 volts.
From .725 to .775, Graph the line y2(x).
This is the range where 4.5 volts < V_s < 5.5 volts.
From .775 to positive infinity, Graph the line L45.
This is a range where V_s < 4.5 volts.

Linear Circuit Analysis: Analysis of Diode Circuits

Linear Circuit Analysis:
Analysis of Diode Circuits