Electrical Basics

Thevenin's Theorem

Thevenin's Theorem Theorem, Thevenin's Nonlinear circuit Circuits, nonlinear Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The qualification of "linear" is identical to that found in the Superposition Theorem, where all the underlying equations must be linear (no exponents or roots). If we're dealing with passive components (such as resistors, and later, inductors and capacitors), this is true. However, there are some components (especially certain gas-discharge and semiconductor components) which are nonlinear: that is, their opposition to current changes with voltage and/or current. As such, we would call circuits containing these types of components, nonlinear circuits. Thevenin's Theorem is especially useful in analyzing power systems and other circuits where one particular resistor in the circuit (called the "load" resistor) is subject to change, and re-calculation of the circuit is necessary with each trial value of load resistance, to determine voltage across it and current through it. Let's take another look at our example circuit:

Let's suppose that we decide to designate R2 as the "load" resistor in this circuit. We already have four methods of analysis at our disposal (Branch Current, Mesh Current, Millman's Theorem, and Superposition Theorem) to use in determining voltage across R2 and current through R2, but each of these methods are time-consuming. Imagine repeating any of these methods over and over again to find what would happen if the load resistance changed (changing load resistance is very common in power systems, as multiple loads get switched on and off as needed. the total

resistance of their parallel connections changing depending on how many are connected at a time). This could potentially involve a lot of work! Equivalent circuit Circuit, equivalent Thevenin's Theorem makes this easy by temporarily removing the load resistance from the original circuit and reducing what's left to an equivalent circuit composed of a single voltage source and series resistance. The load resistance can then be re-connected to this "Thevenin equivalent circuit" and calculations carried out as if the whole network were nothing but a simple series circuit:

. . . after Thevenin conversion . . .

The "Thevenin Equivalent Circuit" is the electrical equivalent of B1, R1, R3, and B2 as seen from the two points where our load resistor (R2) connects. The Thevenin equivalent circuit, if correctly derived, will behave exactly the same as the original circuit formed by B1, R1, R3, and B2. In other words, the load resistor (R2) voltage and current should be exactly the same for the same value of load resistance in the two circuits. The load resistor R2 cannot "tell the difference" between the original network of B1, R1, R3, and

B2, and the Thevenin equivalent circuit of EThevenin, and RThevenin, provided that the values for EThevenin and RThevenin have been calculated correctly. The advantage in performing the "Thevenin conversion" to the simpler circuit, of course, is that it makes load voltage and load current so much easier to solve than in the original network. Calculating the equivalent Thevenin source voltage and series resistance is actually quite easy. First, the chosen load resistor is removed from the original circuit, replaced with a break (open circuit):

Next, the voltage between the two points where the load resistor used to be attached is determined. Use whatever analysis methods are at your disposal to do this. In this case, the original circuit with the load resistor removed is nothing more than a simple series circuit with opposing batteries, and so we can determine the voltage across the open load terminals by applying the rules of series circuits, Ohm's Law, and Kirchhoff's Voltage Law:

The voltage between the two load connection points can be figured from the one of the battery's voltage and one of the resistor's voltage drops, and comes out to 11.2 volts. This is our "Thevenin voltage" (EThevenin) in the equivalent circuit:

To find the Thevenin series resistance for our equivalent circuit, we need to take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure the resistance from one load terminal to the other:

With the removal of the two batteries, the total resistance measured at this location is equal to R1 and R3 in parallel: 0.8 Ω. This is our "Thevenin resistance" (RThevenin) for the equivalent circuit:

With the load resistor (2 Ω) attached between the connection points, we can determine voltage across it and current through it as though the whole network were nothing more than a simple series circuit:

Notice that the voltage and current figures for R2 (8 volts, 4 amps) are identical to those found using other methods of analysis. Also notice that the voltage and current figures for the Thevenin series resistance and the Thevenin source (total) do not apply to any component in the original, complex circuit. Thevenin's Theorem is only useful for determining what happens to a single resistor in a network: the load. The advantage, of course, is that you can quickly determine what would happen to that single resistor if it were of a value other than 2 Ω without having to go through a lot of analysis again. Just plug in that other value for the load resistor into the Thevenin equivalent circuit and a little bit of series circuit calculation will give you the result. • •

•

REVIEW: Thevenin's Theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source, series resistance, and series load. Steps to follow for Thevenin's Theorem:

•

•

•

•

(1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be. (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points. (3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor reattaches between the two open points of the equivalent circuit. (4) Analyze voltage and current for the load resistor following the rules for series circuits. Norton's Theorem

Norton's Theorem Theorem, Norton's Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Just as with Thevenin's Theorem, the qualification of "linear" is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots). Contrasting our original example circuit against the Norton equivalent: it looks something like this:

. . . after Norton conversion . . .

Current source Source, current Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current. As with Thevenin's Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin's Theorem are the steps used in Norton's Theorem to calculate the Norton source current (INorton) and Norton resistance (RNorton). As before, the first step is to identify the load resistance and remove it from the original circuit:

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a direct wire (short) connection between the load points and determine the resultant current. Note that this step is exactly opposite the respective step in Thevenin's Theorem, where we replaced the load resistor with a break (open circuit):

With zero voltage dropped between the load resistor connection points, the current through R1 is strictly a function of B1's voltage and R1's resistance: 7 amps (I=E/R). Likewise, the current through R3 is now strictly a function of B2's voltage and R3's resistance: 7 amps (I=E/R). The total current through the short between the load connection points is the sum of these two currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source current (INorton) in our equivalent circuit:

Remember, the arrow notation for a current source points in the direction opposite that of electron flow. Again, apologies for the confusion. For better or for worse, this is standard electronic symbol notation. Blame Mr. Franklin again! To calculate the Norton resistance (RNorton), we do the exact same thing as we did for calculating Thevenin resistance (RThevenin): take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure total resistance from one load connection point to the other:

Now our Norton equivalent circuit looks like this:

If we re-connect our original load resistance of 2 Ω, we can analyze the Norton circuit as a simple parallel arrangement:

As with the Thevenin equivalent circuit, the only useful information from this analysis is the voltage and current values for R2; the rest of the information is irrelevant to the original circuit. However, the same advantages seen with Thevenin's Theorem apply to Norton's as well: if we wish to analyze load resistor voltage and current over several different values of load resistance, we can use the Norton equivalent circuit again and again, applying nothing more complex than simple parallel circuit analysis to determine what's happening with each trial load. • •

REVIEW: Norton's Theorem is a way to reduce a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load.

• •

•

•

•

Steps to follow for Norton's Theorem: (1) Find the Norton source current by removing the load resistor from the original circuit and calculating current through a short (wire) jumping across the open connection points where the load resistor used to be. (2) Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points. (3) Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit. (4) Analyze voltage and current for the load resistor following the rules for parallel circuits Superposition Theorem

Superposition Theorem Theorem, Superposition Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman's certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious. The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all "superimposed" on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active. Let's look at our example circuit again and apply Superposition Theorem to it:

Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in effect. . .

. . . and one for the circuit with only the 7 volt battery in effect:

When re-drawing the circuit for series/parallel analysis with one source, all other voltage sources are replaced by wires (shorts), and all current sources with open circuits (breaks). Since we only have voltage sources (batteries) in our example circuit, we will replace every inactive source during analysis with a wire. Analyzing the circuit with only the 28 volt battery, we obtain the following values for voltage and current:

Analyzing the circuit with only the 7 volt battery, we obtain another set of values for voltage and current:

When superimposing these values of voltage and current, we have to be very careful to consider polarity (voltage drop) and direction (electron flow), as the values have to be added algebraically.

Applying these superimposed voltage figures to the circuit, the end result looks something like this:

Currents add up algebraically as well, and can either be superimposed as done with the resistor voltage drops, or simply calculated from the final voltage drops and respective resistances (I=E/R). Either way, the answers will be the same. Here I will show the superposition method applied to current:

Once again applying these superimposed figures to our circuit:

Quite simple and elegant, don't you think? It must be noted, though, that the Superposition Theorem works only for circuits that are reducible to series/parallel combinations for each of the power sources at a time (thus, this theorem is useless for analyzing an unbalanced bridge circuit), and it only works where the underlying equations are linear (no mathematical powers or roots). The requisite of linearity means that Superposition Theorem is only applicable for determining voltage and current, not power!!! Power dissipations, being nonlinear functions, do not algebraically add to an accurate total when only one source is considered at a time. The need for linearity also means this Theorem cannot be applied in circuits where the resistance of a component changes with voltage or current. Hence, networks containing components like lamps (incandescent or gasdischarge) or varistors could not be analyzed. Another prerequisite for Superposition Theorem is that all components must be "bilateral," meaning that they behave the same with electrons flowing

either direction through them. Resistors have no polarity-specific behavior, and so the circuits we've been studying so far all meet this criterion. The Superposition Theorem finds use in the study of alternating current (AC) circuits, and semiconductor (amplifier) circuits, where sometimes AC is often mixed (superimposed) with DC. Because AC voltage and current equations (Ohm's Law) are linear just like DC, we can use Superposition to analyze the circuit with just the DC power source, then just the AC power source, combining the results to tell what will happen with both AC and DC sources in effect. For now, though, Superposition will suffice as a break from having to do simultaneous equations to analyze a circuit. • •

•

REVIEW: The Superposition Theorem states that a circuit can be analyzed with only one source of power at a time, the corresponding component voltages and currents algebraically added to find out what they'll do with all power sources in effect. To negate all but one power source for analysis, replace any source of voltage (batteries) with a wire; replace any current source with an open (break). Maximum Power Transfer Theorem

Maximum Power Transfer Theorem Theorem, Maximum Power Transfer The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. If the load resistance is lower or higher than the Thevenin/Norton resistance of the source network, its dissipated power will be less than maximum. Impedance This is essentially what is aimed for in stereo system design, where speaker "impedance" is matched to amplifier "impedance" for maximum sound power output. Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly

overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance. Taking our Thevenin equivalent example circuit, the Maximum Power Transfer Theorem tells us that the load resistance resulting in greatest power dissipation is equal in value to the Thevenin resistance (in this case, 0.8 Ω):

With this value of load resistance, the dissipated power will be 39.2 watts:

If we were to try a lower value for the load resistance (0.5 Ω instead of 0.8 Ω, for example), our power dissipated by the load resistance would decrease:

Power dissipation increased for both the Thevenin resistance and the total circuit, but it decreased for the load resistor. Likewise, if we increase the load resistance (1.1 Ω instead of 0.8 Ω, for example), power dissipation will also be less than it was at 0.8 Ω exactly:

If you were designing a circuit for maximum power dissipation at the load resistance, this theorem would be very useful. Having reduced a network down to a Thevenin voltage and resistance (or Norton current and resistance), you simply set the load resistance equal to that Thevenin or Norton equivalent (or vice versa) to ensure maximum power dissipation at the load. Practical applications of this might include stereo amplifier design (seeking to maximize power delivered to speakers) or electric vehicle design (seeking to maximize power delivered to drive motor). • •

REVIEW: The Maximum Power Transfer Theorem states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power Mesh current method

Analysis, Mesh Current method Mesh Current analysis The Mesh Current Method is quite similar to the Branch Current method in that it uses simultaneous equations, Kirchhoff's Voltage Law, and Ohm's Law to determine unknown currents in a network. It differs from the Branch Current method in that it does not use Kirchhoff's Current Law, and it is usually able to solve a circuit with less unknown variables and less simultaneous equations, which is especially nice if you're forced to solve without a calculator. Let's see how this method works on the same example problem:

The first step in the Mesh Current method is to identify "loops" within the circuit encompassing all components. In our example circuit, the loop formed by B1, R1, and R2 will be the first while the loop formed by B2, R2, and R3 will be the second. The strangest part of the Mesh Current method is envisioning circulating currents in each of the loops. In fact, this method gets its name from the idea of these currents meshing together between loops like sets of spinning gears:

The choice of each current's direction is entirely arbitrary, just as in the Branch Current method, but the resulting equations are easier to solve if the currents are going the same direction through intersecting components (note how currents I1 and I2 are both going "up" through resistor R2, where they "mesh," or intersect). If the assumed direction of a mesh current is wrong, the answer for that current will have a negative value. Voltage polarity The next step is to label all voltage drop polarities across resistors according to the assumed directions of the mesh currents. Remember that the "upstream" end of a resistor will always be negative, and the "downstream" end of a resistor positive with respect to each other, since electrons are negatively charged. The battery polarities, of course, are dictated by their symbol orientations in the diagram, and may or may not "agree" with the resistor polarities (assumed current directions):

Using Kirchhoff's Voltage Law, we can now step around each of these loops, generating equations representative of the component voltage drops and polarities. As with the Branch Current method, we will denote a resistor's voltage drop as the product of the resistance (in ohms) and its respective mesh current (that quantity being unknown at this point). Where two currents mesh together, we will write that term in the equation with resistor current being the sum of the two meshing currents. Tracing the left loop of the circuit, starting from the upper-left corner and moving counter-clockwise (the choice of starting points and directions is ultimately irrelevant), counting polarity as if we had a voltmeter in hand, red lead on the point ahead and black lead on the point behind, we get this equation:

Notice that the middle term of the equation uses the sum of mesh currents I1 and I2 as the current through resistor R2. This is because mesh currents I1 and I2 are going the same direction through R2, and thus complement each other. Distributing the coefficient of 2 to the I1 and I2 terms, and then combining I1 terms in the equation, we can simplify as such:

At this time we have one equation with two unknowns. To be able to solve for two unknown mesh currents, we must have two equations. If we trace the other loop of the circuit, we can obtain another KVL equation and have enough data to solve for the two currents. Creature of habit that I am, I'll start at the upper-left hand corner of the right loop and trace counterclockwise:

Simplifying the equation as before, we end up with:

Now, with two equations, we can use one of several methods to mathematically solve for the unknown currents I1 and I2:

Knowing that these solutions are values for mesh currents, not branch currents, we must go back to our diagram to see how they fit together to give currents through all components:

The solution of -1 amp for I2 means that our initially assumed direction of current was incorrect. In actuality, I2 is flowing in a counter-clockwise direction at a value of (positive) 1 amp:

This change of current direction from what was first assumed will alter the polarity of the voltage drops across R2 and R3 due to current I2. From here, we can say that the current through R1 is 5 amps, with the voltage drop across R1 being the product of current and resistance (E=IR), 20 volts (positive on the left and negative on the right). Also, we can safely say that the current through R3 is 1 amp, with a voltage drop of 1 volt (E=IR), positive on the left and negative on the right. But what is happening at R2? Mesh current I1 is going "up" through R2, while mesh current I2 is going "down" through R2. To determine the actual current through R2, we must see how mesh currents I1 and I2 interact (in this case they're in opposition), and algebraically add them to arrive at a final value. Since I1 is going "up" at 5 amps, and I2 is going "down" at 1 amp, the real current through R2 must be a value of 4 amps, going "up:"

A current of 4 amps through R2's resistance of 2 Ω gives us a voltage drop of 8 volts (E=IR), positive on the top and negative on the bottom. The primary advantage of Mesh Current analysis is that it generally allows for the solution of a large network with fewer unknown values and fewer simultaneous equations. Our example problem took three equations to solve the Branch Current method and only two equations using the Mesh Current method. This advantage is much greater as networks increase in complexity:

To solve this network using Branch Currents, we'd have to establish five variables to account for each and every unique current in the circuit (I1 through I5). This would require five equations for solution, in the form of two KCL equations and three KVL equations (two equations for KCL at the nodes, and three equations for KVL in each loop):

I suppose if you have nothing better to do with your time than to solve for five unknown variables with five equations, you might not mind using the Branch Current method of analysis for this circuit. For those of us who have better things to do with our time, the Mesh Current method is a whole lot easier, requiring only three unknowns and three equations to solve:

Less equations to work with is a decided advantage, especially when performing simultaneous equation solution by hand (without a calculator). Wheatstone bridge, unbalanced Another type of circuit that lends itself well to Mesh Current is the unbalanced Wheatstone Bridge. Take this circuit, for example:

Since the ratios of R1/R4 and R2/R5 are unequal, we know that there will be voltage across resistor R3, and some amount of current through it. As discussed at the beginning of this chapter, this type of circuit is irreducible by normal series-parallel analysis, and may only be analyzed by some other method. We could apply the Branch Current method to this circuit, but it would require six currents (I1 through I6), leading to a very large set of simultaneous equations to solve. Using the Mesh Current method, though, we may solve for all currents and voltages with much fewer variables. The first step in the Mesh Current method is to draw just enough mesh currents to account for all components in the circuit. Looking at our bridge circuit, it should be obvious where to place two of these currents:

The directions of these mesh currents, of course, is arbitrary. However, two mesh currents is not enough in this circuit, because neither I1 nor I2 goes through the battery. So, we must add a third mesh current, I3:

Here, I have chosen I3 to loop from the bottom side of the battery, through R4, through R1, and back to the top side of the battery. This is not the only path I could have chosen for I3, but it seems the simplest. Now, we must label the resistor voltage drop polarities, following each of the assumed currents' directions:

Notice something very important here: at resistor R4, the polarities for the respective mesh currents do not agree. This is because those mesh currents (I2 and I3) are going through R4 in different directions. Normally, we try to avoid this when establishing our mesh current directions, but in a bridge circuit it is unavoidable: two of the mesh currents will inevitably clash through a component. This does not preclude the use of the Mesh Current method of analysis, but it does complicate it a bit. Generating a KVL equation for the top loop of the bridge, starting from the top node and tracing in a clockwise direction:

In this equation, we represent the common directions of currents by their sums through common resistors. For example, resistor R3, with a value of 100 Ω, has its voltage drop represented in the above KVL equation by the expression 100(I1 + I2), since both currents I1 and I2 go through R3 from right to left. The same may be said for resistor R1, with its voltage drop expression shown as 150(I1 + I3), since both I1 and I3 go from bottom to top through that resistor, and thus work together to generate its voltage drop. Generating a KVL equation for the bottom loop of the bridge will not be so easy, since we have two currents going against each other through resistor

R4. Here is how I do it (starting at the right-hand node, and tracing counterclockwise):

Note how the second term in the equation's original form has resistor R4's value of 300 Ω multiplied by the difference between I2 and I3 (I2 - I3). This is how we represent the combined effect of two mesh currents going in opposite directions through the same component. Choosing the appropriate mathematical signs is very important here: 300(I2 - I3) does not mean the same thing as 300(I3 - I2). I chose to write 300(I2 - I3) because I was thinking first of I2's effect (creating a positive voltage drop, measuring with an imaginary voltmeter across R4, red lead on the bottom and black lead on the top), and secondarily of I3's effect (creating a negative voltage drop, red lead on the bottom and black lead on the top). If I had thought in terms of I3's effect first and I2's effect secondarily, holding my imaginary voltmeter leads in the same positions (red on bottom and black on top), the expression would have been -300(I3 - I2). Note that this expression is mathematically equivalent to the first one: +300(I2 - I3). Well, that takes care of two equations, but I still need a third equation to complete my simultaneous equation set of three variables, three equations. This third equation must also include the battery's voltage, which up to this point does not appear in either two of the previous KVL equations. To generate this equation, I will trace a loop again with my imaginary voltmeter starting from the battery's bottom (negative) terminal, stepping clockwise (again, the direction in which I step is arbitrary, and does not need to be the same as the direction of the mesh current in that loop):

Solving for I1, I2, and I3 using whatever simultaneous equation method we prefer:

The negative value arrived at for I1 tells us that the assumed direction for that mesh current was incorrect. Thus, the actual current values through each resistor is as such:

Calculating voltage drops across each resistor:

A SPICE simulation will confirm the accuracy of our voltage calculations:

unbalanced wheatstone bridge v1 1 0 r1 1 2 150 r2 1 3 50 r3 2 3 100 r4 2 0 300 r5 3 0 250 .dc v1 24 24 1 .print dc v(1,2) v(1,3) v(3,2) v(2,0) v(3,0) .end v1 2.400E+01

v(1,2) 6.345E+00

v(1,3) 4.690E+00

v(3,2) 1.655E+00

v(2) 1.766E+01

v(3) 1.931E+01

• • • • •

• • • •

REVIEW: Steps to follow for the "Mesh Current" method of analysis: (1) Draw mesh currents in loops of circuit, enough to account for all components. (2) Label resistor voltage drop polarities based on assumed directions of mesh currents. (3) Write KVL equations for each loop of the circuit, substituting the product IR for E in each resistor term of the equation. Where two mesh currents intersect through a component, express the current as the algebraic sum of those two mesh currents (i.e. I1 + I2) if the currents go in the same direction through that component. If not, express the current as the difference (i.e. I1 - I2). (4) Solve for unknown mesh currents (simultaneous equations). (5) If any solution is negative, then the assumed current direction is wrong! (6) Algebraically add mesh currents to find current in components sharing multiple mesh currents. (7) Solve for voltage drops across all resistors (E=IR).

Aim: To design and simulate a RC Coupled Amplifier circuit. Components: Name

EDWin Components Used

Description

Number of components required

BC107

BC107A

Transistor

1

RES

RC05

Resistor

6

CAP

CASE-A600

Capacitor

3

VGEN

SMB_VGEN

Ac voltage source 1

VDC

SMB_VDC

Dc voltage source 1

GND

SMB_SPL0

Ground

2

Design & Theory DC Analysis A transistor amplifier is designed to work in the active region. A single stage amplifier is designed using voltage divider bias. In a voltage divider bias the base voltage, VB is given by

First we bias the circuit in the active region with dc conditions. i.e. Current through R1, IR1 is a large current of the order of 20 or 30 times that of IB.

Current through R2, IR2 is given by We select VCC as 10V.

Minimum Voltage Gain required = 100

Let ICQ = 2mA – ref. Manufacturer’s Datasheet

Applying Kirchoff’s Voltage Law for the output side of amplifier

Design of R1 & R2 For a fluctuation in IR1 and IR2 there will be small change in IB. For example as mentioned (ref. (5)), if IR1=21IB and a 5% change in IR1 occurs there will be only change in IB. Therefore the circuit will be stable against small changes in R1 and R2 due to temperature or tolerance.

Thus

And

=0.7+0.1VCC

From (8) From (9)

AC Analysis

Modifications required from the dc circuit are 1. Coupling capacitors (Cin, Cout) 2. Bypass Capacitor CE.

The function of the coupling capacitor is to isolate the amplifier input circuit from the source. Since capacitor blocks dc it does not allow the dc components from the input circuit to get to the base of transistor and to change the Q point.

Additional requirement if Cin is that it must pass the input ac signal unattenuated. Hence the lower cut-off frequency (LCF) of the amplifier is determined by the coupling capacitor. The coupling capacitor Cin along with resistance combinations shown in fig forms a high pass filter whose cut-off frequency;

The LCF due to Cin; At mid or high frequencies capacitor almost acts as a short circuit.

Then Vi is related to VS by;

.

Effect of Cout

The effect of Cout on LCF is given by;

where

.

Effect of CE CE is designed such that impedance of the capacitor must be less than of RE across which it is connected at LCF. The function of CE is to bypass the emitter resistor during ac operation other wise during amplification with increase or decrease in current, the drop across RE will increase or decrease and hence VBE decreases or increases opposing the change in current. So the function of RE is limited to dc where the capacitor will be open and RE will stabilize the circuit. In the case of ac the capacitor will take over current. Form (15)

hie = 1kΩ (datasheet)

Let fLin = 100Hz

where For pn junction current I is related to voltage V by equation

where η =1 for Ge and 2 for Si. where TC is Room temperature in ° C TK = 25+273=298K

VT – Voltage equivalent of temperature IS – Reverse saturation current

Now

Let fLE = 100 Hz

Note: - Some tolerances have been made in order to obtain standard values of components. All capacitors are designed to obtain a fair frequency response curve. Procedure: Creating the Library Elements To create a new part select EDWin 2000 Main -> Library -> Part Editor -> File -> New Part.

For e.g.: A transistor may be created in the following manner. Type in the information in the necessary fields (Prefix, Description, Type, Manufacturer etc.). The symbol name may be given as NPN. It may be created manually or using Wizard. The symbol may be drawn using ‘Create Graphic Item’ function tool and its option tools. A filled bar is drawn using option tools rectangle and filled item. The filled triangle drawn using the option tools triangle and filled item represents the emitter and the direction of electron flow. Place the entries using ‘Create Entry’ option tool of ‘Create Graphic Item’ function tool. Then entries may be named as Emitter, Base and Collector using function tool ‘Edit Entry Attributes’. Now place the Component Name and Description appropriately. In order to assign Simulation Names follow these steps. Select the ‘Properties’ function tool and its option tool ‘Symbol Properties’. Now select the desired simulation function from the list. Using ‘Edit Entry Attributes’ function tool click on each entry and assign proper Simulation Name to each pin. Save the Symbol/ Part either in user defined libraries and add them to the search sequence or in the Project Library itself. Component placement To place the components select Schematic Editor -> Components. The components may be loaded in any of the following ways. 1. Add Component: Type in the name of the component to be added to the project. On clicking accept an

image of the component gets tagged to the cursor. Click anywhere on the workspace to place it. On each click on the workspace the component gets repeated until the Esc key is pressed. 2. Library Explorer: Use Drag-drop or Send-to feature to load the component. 3. Library Browser: Use Drag-drop or Send-to feature to load the component. Place these loaded components in the appropriate position using the ‘Relocate’ function tool and its option tool(s). Wires / Nets connection To wire the components select Schematic Editor -> Wires & Buses. The ‘Route Wire’ function tool and its appropriate option tool(s) may be used for this purpose. Enable option tool ‘Pin-to-pin routing’ for starting and ending wiring from a component entry. Otherwise the routing may start very near to component entry which causes netlist problem. Prior to this ‘Instant Net Name’ and ‘Instant Wire Label’ may be enabled in ‘Preferences’ menu, so that the net name can be given as required and labels for the same can be placed without using any additional tool(s). Enable the option ‘Nodes’ in View menu, so that the proper assignment of net to the component entries can be ensured

Assign Values The instance parameter (Component value) may be assigned either in Schematic Editor or in Simulators. In Schematic editor this is performed using the function tool ‘Add/Change Component text’ and its option tool ‘Add / Change Value’. It is mandatory to use its option tool for this purpose and the simulators will ignore any values assigned without using this option tool. Preprocess The purpose of preprocessing is to establish the degree to which the circuit is ready for simulation. The results of preprocessing are displayed in the "EDWin 2000 Info " window. This should be done in both the simulators prior to performing any simulation. Simulation: Simulation is performed in both Mixed Mode and EDSpice Simulator. In Mixed Mode simulator we are performing Transient Analysis, DC Sweep and AC Sweep Analysis. Mixed Mode Analysis Transient Analysis In Transient Analysis, we perform the simulation of the circuit and analyze the output voltage with respect to time. For this maximum simulation time and the time limit are set in the corresponding parameters window. Then simulation is performed with ‘Display Waveform’ option enabled. The output of the amplifier is viewed in the Waveform viewer. The user also has the provision to simulate with print and plot outputs. DC Sweep Analysis

In this type of analysis, usually temperature is the sweeping parameter. The user has the provision of sweeping parameters of any two components at a time. AC Sweep Analysis AC Sweep Analysis is performed for analyzing the frequency response of the circuit. It is required to run AC Sweep Analysis after DC Sweep Analysis, so that the circuit operating parameters would have been stabilized and set. The process is similar to Transient analysis with difference in the parameter setup. Simulation Assign the parameters for each and every component using the function tool ‘Set parameters/ models’ and its option tool ‘Set parameter’. Place the test points appropriately so that the voltage or current of that particular node or net has to be displayed. Now place the waveform markers at the node or net where the waveform has to be observed. Preprocessing of the circuit must be carried out before simulating it. Waveform Viewer Set all the necessary values in the waveform viewer as set in a CRO. Result: The output waveform for various analyses is shown below. Mixed Mode Simulator

Transient Analysis

DC Sweep Analysis

AC Sweep Analysis

Emitter Follower An emitter follower circuit shown in the figure is widely used in AC amplification circuits. The input and output of the emitter follower are the base and the emitter, respectively, therefore this circuit is also called commoncollector circuit.

DC operating point

Solving these equations, we can get

,

and

AC small-signal equivalent circuit

We assume

and therefore can be ignored, and have

Voltage gain:

As

,

is smaller than but approximately equal to 1.

The input resistance:

The input resistance is can be found by

we have

in parallel with the resistance of the circuit to its right including the load . But as

, which

Comparing this with the input resistance of the common-emitter circuit that the emitter follower has a very large input resistance.

, we see

The output resistance:

The output resistance is can be found by

in parallel with the resistances of the circuit to its left including the source, which

, where

and we have

Alternatively, the output resistance of the emitter follower (treated as either a voltage or current source) can also be found as the ratio

of the open-circuit voltage

output voltage when the load is open-circuit circuit

and the short-circuit current , and

, where

is the

is the output current when the load is short-

.

As the voltage gain of the emitter follower is close to unity, the open-circuit output voltage is approximately the same as the source voltage

. The short-circuit current

Now the output resistance can be found as

The overall output resistance is

can be found as

Conclusion: The emitter follower is a circuit with deep negative feedback, i.e., all of its output become part of its input

is fed back to

. The fact that this is a negative feedback can be seen by:

Due to this deep negative feedback, the voltage gain of the emitter follower is smaller than unity. However, many other important properties of the circuit, such as the input and output resistances, are drastically improved. In fact the emitter follower acts as an impedance transformer with a ratio of resistance is

times greater than

and the output resistance is

, i.e., the input

times smaller than

.

Comparing this with the input resistance and output resistances of the common-emitter transistor circuit, we see that the emitter follower circuit has very favorable input/output resistances. Although the emitter follower does not amplify voltage, due to its high input resistance drawing little current from the source, and its low output resistance capable of driving heavy load, it is widely used as both the input and output stages for a multi-stage voltage amplification circuit.

Schematic Diagram

As shown in the schematic diagram here, the astable multivibrator simply extends the modification that converted the bistable multivibrator to a monostable version of the circuit. Now, both transistors are coupled to each other through capacitors. Whichever transistor is off at any moment cannot remain off indefinitely; its base will become forward biased as that capacitor charges towards +5 volts. Once that happens, that transistor will turn on, thereby turning the other one off. If we pick a moment when Q1 has just turned off and Q2 is on, then the left end of C2 is at -5 volts. This negative voltage decreases as C2 charges through R2 towards +5 volts. However, the moment C2 charges enough to provide forward bias to the base of Q1, Q1 turns on and the 5 volt drop in Q1's collector voltage is coupled through C1 to the base of Q2. This turns Q2 off at once. As we saw in the previous experiment, the time that Q1 remains on and Q2 remains off is 0.693RC, which for the component values shown here is about 1 second. Now Q2 is held off while C1 charges through R1, until Q2's base becomes forward biased. At that point the transistors switch states again and the whole thing starts over. There is no stable state where the circuit can come to rest, so this circuit is known as an astable multivibrator. The time Q2 remains off is set by R1 and C1, just as the time Q1 remains off is set by R2 and C2. For our circuit, the components are of the same values on each side, so the timing will be the same on each half of the cycle. This is not required; the two halves of the circuit can have totally different time intervals. They actually operate independently of each other, even though they work together. Since this particular circuit will spend about 1 second on each half cycle, the total cycle time, or period, is about 2 seconds. The operating frequency of the circuit is the reciprocal of the period, or 0.5 Hz.

Astable Multivibrator An astable multivibrator is also known as a FREE-RUNNING MULTIVIBRATOR. It is called free-running because it alternates between two different output voltage levels during the time it is on. The output remains at each voltage level for a definite period of time. If you looked at this output on an oscilloscope, you would see continuous square or rectangular waveforms. The astable multivibrator has two outputs, but NO inputs. Let's look at the multivibrator in figure 3-3 again. This is an astable multivibrator. The astable multivibrator is said to oscillate. To understand why the astable multivibrator oscillates, assume that transistor Q1 saturates and transistor Q2 cuts off when the circuit is energized. This situation is shown in figure 3-4. We assume Q1 saturates and Q2 is in cutoff because the circuit is symmetrical; that is, R1 = R4, R2 = R3, C1 = C2, and Q1 = Q2. It is impossible to tell which transistor will actually conduct when the circuit is energized. For this reason, either of the transistors may be assumed to conduct for circuit analysis purposes. Figure 3-4. - Astable multivibrator (Q1 saturated).

Essentially, all the current in the circuit flows through Q1; Q1 offers almost no resistance to current flow. Notice that capacitor C1 is charging. Since Q1 offers almost no resistance in its saturated state, the rate of charge of C1 depends only on the time constant of R2 and C1 (recall that TC = RC). Notice that the right-hand side of capacitor C1 is connected to the base of transistor Q2, which is now at cutoff. Let's analyze what is happening. The right-hand side of capacitor C1 is becoming increasingly negative. If the base of Q2 becomes sufficiently negative, Q2 will conduct. After a certain period of time, the base of Q2 will become sufficiently negative to cause Q2 to change states from cutoff to conduction. The time necessary for Q2 to become saturated is determined by the time constant R2C1. The next state is shown in figure 3-5. The negative voltage accumulated on the right side on capacitor C1 has caused Q2 to conduct. Now the following sequence of events takes place almost instantaneously. Q2 starts conducting and quickly saturates, and the voltage at output 2 changes from approximately -VCC to approximately 0 volts. This change in voltage is coupled through C2 to the base of Q1, forcing Q1 to cutoff. Now Q1 is in cutoff and Q2 is in saturation. This is the circuit situation shown in figure 3-6.

Figure 3-5. - Astable multivibrator.

Figure 3-6. - Astable multivibrator. (Q2 saturated).

Notice that figure 3-6 is the mirror image of figure 3-4. In figure 3-6 the left side of capacitor C2 becomes more negative at a rate determined by the time constant R3C2. As the left side of C2 becomes more negative, the base of Q1 also becomes more negative. When the base of Q1 becomes negative enough to allow Q1 to conduct, Q1 will again go into saturation. The resulting change in voltage at output 1 will cause Q2 to return to the cutoff state. Look at the output waveform from transistor Q2, as shown in figure 3-7. The output voltage (from either output of the multivibrator) alternates from approximately 0 volts to approximately -VCC, remaining in each state for a definite period of time. The time may range from a microsecond to as much as a second or two. In some applications, the time period of higher voltage (-VCC) and the time period of lower voltage (0 volts) will be equal. Other applications require differing higher- and lower-voltage times. For example, timing and gating circuits often have different pulse widths as shown in figure 3-8. Figure 3-7. - Square wave output from Q2.

Figure 3-8. - Rectangular waves.

FREQUENCY STABILITY. - Some astable multivibrators must have a high degree of frequency stability. One way to obtain a high degree of frequency stability is to apply triggers. Figure 3-9, view (A), shows the diagram of a triggered, astable multivibrator. At time T0, a negative input trigger to the base of Q1 (through C1) causes Q1 to go into saturation, which drives Q2 to cutoff. The circuit will remain in this condition as long as the base voltage of Q2 is positive. The length of time the base of Q2 will remain positive is determined by C3, R3, and R6. Observe the parallel paths for C3 to discharge. Figure 3-9A. - Triggered astable multivibrator and output.

View (B) of figure 3-9 shows the waveforms associated with the circuit. At time T1, Q2 comes out of cutoff and goes into saturation. Also, Q1 is caused to come out of

saturation and is cut off. The base voltage waveform of Q1 shows a positive potential that is holding Q1 at cutoff. This voltage would normally hold Q1 at cutoff until a point between T2 and T3. However, at time T2 another trigger is applied to the base of Q1, causing it to begin conducting. Q1 goes into saturation and Q2 is caused to cut off. This action repeats each time a trigger (T2, T4, T6) is applied. Figure 3-9B. - Triggered astable multivibrator and output.

The prt of the input triggers must be shorter than the natural free-running prt of the astable multivibrator, or the trigger prf must be slightly higher than the free-running prf of the circuit. This is to make certain the triggers control the prt of the output. Monostable Multivibrator The monostable multivibrator (sometimes called a ONE-SHOT MULTIVIBRATOR) is a square- or rectangular-wave generator with just one stable condition. With no input signal (quiescent condition) one amplifier conducts and the other is in cutoff. The monostable multivibrator is basically used for pulse stretching. It is used in computer logic systems and communication navigation equipment. The operation of the monostable multivibrator is relatively simple. The input is triggered with a pulse of voltage. The output changes from one voltage level to a different voltage level. The output remains at this new voltage level for a definite period of time. Then the circuit automatically reverts to its original condition and remains that way until another trigger pulse is applied to the input. The monostable multivibrator actually takes this series of input triggers and converts them to uniform square pulses, as shown in figure 3-10. All of the square output pulses are of the same amplitude and time duration.

Aim: To design and simulate an Astable Multivibrator circuit. Components: -

Name

EDWin Components Used

Description

Number of components required

BC107

BC107A

Transistor

2

RES

RC05

Resistor

4

CAP

CASE-A600

Capacitor

2

VDC

SMB_VDC

Dc voltage source 1

GND

SMB_SPL0

Ground

1

Theory: Astable Multivibrator is a two stage switching circuit in which the output of the first stage is fed to the input of the second stage and vice versa. The outputs of both the stages are complementary. This free running multivibrator generates square wave without any external triggering pulse. The circuit has two stable states and switches back and forth from one state to another, remaining in each state for a time depending upon the discharging of the capacitive circuit. The multivibrator is one form of relaxation oscillator, the frequency of which may be controlled by external synchronizing pulses. In our experiment we are using transistor, as the amplifying device and also it is a collector coupled multivibrator.

Figure shows the basic symmetrical astable multivibrator in which components in one half of a cycle of the circuit are identical to their counterpart in the other half. Square wave output can be obtained from the collector point of Q1 or Q2. Operation When supply voltage, VCC is applied, one transistor will conduct more than the other due to some circuit imbalance. Initially let us assume that Q1 is conducting and Q2 is cut-off. Then VC1, the output of Q1 is equal to VCESAT which is approximately zero and VC2 is equal to VCC. At this instant C1 charges exponentially with the time constant R1C1 towards the supply voltage through R1 and correspondingly VB2 also increases exponentially towards VCC. When VB2 crosses the coupling voltage Q2 starts conducting and VC2 falls to VCESAT. Also VB1 falls due to capacitive coupling between collector of Q2 and base of Q1, thereby driving Q1 into OFF state. The rise in voltage VC1 is coupled through C1 to the base of Q2 causing a small overshoot in voltage VB2. Thus Q1 is OFF and Q2 is ON. At this instant the voltage levels are: VB1 is negative, VC1=VCC, VB2=VBESAT and VC2=VCESAT. When Q1 is OFF and Q2 is ON the voltage VB1 increases exponentially with a time constant R2C2 towards VCC. Therefore Q1 is driven to saturation and Q2 to cut-off. Now the voltage levels are: VB1=VBESAT, VC1=VCESAT, VB2 is negative and VC2=VCC. From the above it is clear that when Q2 is ON the falling voltage VC2 permits the discharging of capacitor C2 which inturn drives Q1 into cut-off. The rising voltage of VC1 is fed back to the base of Q2 tending to turn it ON. This process is regenerative. Derivation of time period The charging equation for a capacitor is given by

Capacitor voltage,

Hence

where VC - the capacitor voltage, VINIT – the initial capacitor voltage, VFIN – the final capacitor voltage t – the time period of charging. R and C – the resistor and capacitor through which charging occurs. The capacitor discharges from –VCC to VCC. Therefore VIN=(-VCC), VFIN=VCC, VC≅ 0V. Substituting this in equation (2)

Taking natural logarithm

For a symmetrical astable multivibrator

Charging and discharging time periods are given by

From equation (5)

where T is the total time period. Since the multivibrator is symmetrical Design: Design Specifications

Manufacturer’s specifications

Applying KVL for the collector side of Q2.

since it is a symmetrical astable multivibrator Applying KVL for the base loop of the circuit

since it is a symmetrical astable multivibrator.

Design of C The total time period T is given by

since it is a symmetrical astable multivibrator.

From equations (1) and (2)

The free running frequency is given by

Assume the frequency as 100Hz.

Procedure:EDWin 2000 -> Schematic Editor: The circuit diagram is drawn by loading components from the library. Wiring and proper net assignment has been made. The values are assigned for relevant components.

EDWin 2000 -> Mixed Mode Simulator: The circuit is preprocessed. The desired test points and waveform markers are placed. The Transient Analysis parameters have been set. The Transient Analysis is executed and output observed in Waveform Viewer. Result:The output waveform may be observed in the waveform viewer.

Aim: To design and simulate a Monostable Multivibrator circuit. Components: -

Name

EDWin Components Used

Description

Number of components required

BC107

BC107A

Transistor

2

RES

RC05

Resistor

6

CAP

CASE-A600

Capacitor

2

DIODE

1N4007

Diode

1

VDC

SMB_VDC

Dc voltage source

2

VGEN

SMB_VGEN

Ac Voltage Source 1

GND

SMB_SPL0

Ground

3

Theory: A multivibrator in which one transistor is always conducting (i.e. in the ON state) and the other is nonconducting (i.e. in the OFF state) is called a monostable multivibrator. Monostable Multivibrator or one-shot multivibrator has one stable state and one quasi-stable state. i.e. When one transistor is conducting and the other is non-conducting, the circuit will remain in this stable state until the application of external trigger pulse. After a certain time the circuit will automatically switch back to the original stable state and remains there until another pulse is applied. The circuit of a transistor monostable multivibrator is shown in the figure. With the above circuit arrangement Q1 is at cut-off and Q2 is at saturation. This represents the stable state. The base of Q1 is kept at a negative potential to ensure that it is always OFF unless when trigger is applied. The triggering network consists of the voltage source, the input capacitor, R4, R5 and D1.

When a trigger pulse is applied, Q1 turns ON and the collector voltage of Q1 drops from VCC to the saturation voltage of 0.2V. This negative change is coupled to the base of Q2 by the capacitor which inturn causes Q2 to turn OFF. This represents the quasi-stable state. Now the capacitor starts charging towards VCC. When the capacitor voltage reaches 0.7V, transistor Q2 turns ON and Q1 switches back to the OFF state. Procedure:EDWin 2000 -> Schematic Editor: The circuit diagram is drawn by loading components from the library. Wiring and proper net assignment has been made. The values are assigned for relevant components.

EDWin 2000 -> Mixed Mode Simulator: The circuit is preprocessed. The desired test points and waveform markers are placed. The Transient Analysis parameters have been set. The Transient Analysis is executed and output observed in Waveform Viewer. Result:The output waveform may be observed in the waveform viewer.

Aim: To simulate a RC Phase Shift Oscillator circuit. Components: Name

EDWin Components Used

Description

Number of components required

TRANSISTOR

BC107A

Transistor

1

RES

RC05

Resistor

8

CAPACITOR

CAP

Capacitor

4

VDC

SMB_VDC

Dc voltage source 1

GND

SMB_SPL0

Ground

3

Theory: An oscillator is a circuit, which generates ac output signal without giving any input ac signal. This circuit is usually applied for audio frequencies only. The basic requirement for an oscillator is positive feedback. The operation of the RC Phase Shift Oscillator can be explained as follows. The starting voltage is provided by noise, which is produced due to random motion of electrons in resistors used in the circuit. The noise voltage contains almost all the sinusoidal frequencies. This low amplitude noise voltage gets amplified and appears at the output terminals. The amplified noise drives the feedback network which is the phase shift network. Because of this the feedback voltage is maximum at a particular frequency, which in turn represents the frequency of oscillation. Furthermore, the phase shift required for positive feedback is correct at this frequency only. The voltage gain of the amplifier with positive feedback is given by

From the above equation we can see that if

. The gain becomes infinity means that there is

output without any input. i.e. the amplifier becomes an oscillator. This condition is known as the Barkhausen criterion of oscillation. Thus the output contains only a single sinusoidal frequency. In the beginning, as the oscillator is switched on, the loop gain Aβ is greater than unity. The oscillations build up. Once a suitable level is reached the gain of the amplifier decreases, and the value of the loop gain decreases to unity. So the constant level oscillations are maintained. Satisfying the above conditions of oscillation the value of R and C for the phase shift network is selected such that each RC combination produces a phase shift of 60°. Thus the total phase shift produced by the three RC networks is 180°. Therefore at the specific frequency fo the total phase shift from the base of the transistor around the circuit and back to the base is 360° thereby satisfying Barkhausen criterion. We select R1=R2=R3∗ =R and C1=C2=C3=C The frequency of oscillation of RC Phase Shift Oscillator is given by

At this frequency, the feedback factor of the network is amplifier gain Procedure: -

for oscillator operation.

. In order that

it is required that the

EDWin 2000 -> Schematic Editor: The circuit diagram is drawn by loading components from the library. Wiring and proper net assignment has been made. The values are assigned for relevant components. EDWin 2000 -> Mixed Mode Simulator: The circuit is preprocessed. The waveform marker is placed at the output of the circuit. GND net is set as reference net. The Transient Analysis parameters have been set. The Transient Analysis is executed and output waveform is observed in Waveform Viewer.

EDWin 2000 -> EDSpice Simulator: The circuit is preprocessed. The waveform marker is placed at the output of the circuit. The Transient Analysis parameters are also set. The Transient Analysis is executed and output waveform is observed in Waveform Viewer.

Result: The output waveform may be observed in the waveform viewer.

CLAMPERS Certain applications in electronics require that the upper or lower extremity of a wave be fixed at a specific value. In such applications, a CLAMPING (or CLAMPER) circuit is used. A clamping circuit clamps or restrains either the upper or lower extremity of a waveform to a fixed dc potential. This circuit is also known as a DIRECT-CURRENT RESTORER or a BASE-LINE STABILIZER. Such circuits are used in test equipment, radar systems, electronic countermeasure systems, and sonar systems. Depending upon the equipment, you could find negative or positive clampers with or without bias. Figure 4-16, views (A), illustrates some examples of waveforms created by clampers. However, before we discuss clampers, we will review some relevant points about series RC circuits.

Clippers Power amplifiers, when driven out of their linear range of operation, sound particularly bad, and can produce damage to themselves or the transducers to which they are connected. The design of traditional protection circuits is complicated by the various performance, cost, and sonic tradeoffs involved. There is certainly no one right answer to the limiter puzzle. The circuits presented here, however, are designed to maintain a high level of sonic integrity, while remaining cost-effective.

These circuits combine active limiting with a diode- based clipper to provide excellent driver protection while avoiding the sonic degradation of simpler designs. An innovative nonlinear capacitor circuit further improves the sonic performance of the limiter. Elements of Clipping & Clamping Circuits

a) Limiters(clippers): –

Diodes can be used to clip off portions of signal voltages (above or below

certain levels).

–

Diode will become forward biased as soon as VA becomes larger than

VBIAS+0.7. –

When diode is forward biased, VA cannot become larger than VBIAS + 0.7 V!

–

Thus, the voltage across the load, RL, will also be equal to VBIAS + 0.7.

–

When diode is reverse biased, it appears as an open, so the output voltage is

the voltage of RL alone. - Desired voltage levels can be attained with a voltage divider.

–

We replace the voltage source with a resistive voltage divider.

VBIAS = R3/(R2 + R3) VSUPPLY

b) Diode Clampers –

A clamper adds a dc level to an ac voltage.

–

Also called dc restorers.

–

When input voltage goes initially negative, diode is forward biased.

–

Capacitor charges to near peak of inpt (Vp(in) – 0.7).

–

Right after the negative peak, diode is reverse biased (because cathode is held

near Vp(in) – 0.7 by charge on capacitor). –

Capacitor can only discharge through the RL.

–

Since RL has high resistance, the capacitor discharges very little each period.

–

Note that time constant should be large (at least 10 times the period of the

input voltage). –

Since capacitor retains charge, it acts like a battery in series with the input

voltage.

Hartely Oscillator

Aim: To design and simulate a Hartely Oscillator. Components: -

Name

EDWin Components Used

Description

Number of components required

VDC

VDC

DC Voltage Source

1

GND

SPL0

Ground

2

NPN

BF959

Transistor

1

L

IND2012C

Inductor

2

C

CASE-A600

Capacitor

4

R

RC05

Resistor

5

Circuit Diagram: -

Procedure for creating schematic diagram and simulating it 1. Open EDWinXP x.xx from Start ->Programs->EDWinXP Main. 2. Invoke the Schematic editor by double clicking the Project ->Main page.

3. Choose EDSpice Simulator from Preferences Menu Drawing the circuit diagram 1. The circuit can be drawn in the required page size like A4 (portrait, Landscape), A3 etc can be selected from Tools ->Page Format ->Define Page (first function tool). Loading/Placing of components Components can be loaded in the schematic page by selecting 1. View ->Toolbar ->Schematic View. 2. Select the option Browser from the Schematic View toolbar. Browser is displayed in the right hand side of the Schematic Editor. 3. In Browser, Select Resistors and choose Resistor, click on place button and place it on the screen. Similarly place Capacitor, transistor, Inductor from it group. 4. Select Ground from sources group and place it. 5. Select Voltage source from Sources group and place it. 6. The components can be placed in the grids with 100 mils (.1") and snap of 50 mils (.05")in the required position by relocation. Routing Connections 1. Tools ->Connections ->Connect components (First Function tool) {Connect the wires from pin to pin according to the circuit diagram.} while asked for entering Net name, give suitable names 2. Select Tools ->Connections ->Connection properties ÒClick on any connection and name it as required. Giving values: Preprocess 1. Select Simulation ->Preprocess 2. Select Tools ->Components ->Component properties ->Change simulation parameters ->Select any component and change it values. Resistor : R1:100K,R2:22K,RC:4.7K,RE:1K,RL:470 Ohms Capacitor : Cc1:100nF,CE:100nF,Cc2:100Nf. Inductor : L1:40µH,L2:0.4mH Transistor : BF959. Vcc : 10V Simulation using Instruments Select Tools ->Instruments ->set wave form Contents ->Voltage waveform ->Click on net OUT and place the waveform marker as shown above. Preprocess: 1. Select Simulation ->Preprocess 2. For setting up simulation time and analysis types, select Simulation ->Analysis. 3. In the window pos up, select Transient analysis from tree view. 4. Enter following values: Step : 1µ Final time : .5ms Results : Select waveform from drop down menu. 5. Click Accept 6. Expand the transient tree view and select waveform viewer, click on as marked. 7. Click Accept. 8. Click on Run button to run Analysis 9. Verify the results Page Notes (For creating Title Block) For the placing of Title Block in a Schematic page, we use the page note property Tools ->Page Notes ->Create Graphic item ->Create Text.

Aim: To design and simulate a Schmitt Trigger circuit. Components: -

Name

EDWin Components Used

Description

Number of components required

BC107

BC107A

Transistor

2

RES

RC05

Resistor

6

CAP

CASE-A600

Capacitor

1

VGEN

VGEN

Ac voltage source

1

VDC

VDC

Dc voltage source

1

GND

SPL0

Ground

1

Theory:Schmitt Trigger is an emitter coupled binary trigger circuit. It is termed a binary trigger circuit since two stable states occur- the transistor Q1 may be ON and Q2 OFF or vice versa. In the absence of an input to transistor Q1 the voltage divider network Rb2 and R1 along with Rc1 maintains the base of Q2 at a slightly positive potential relative to the emitter and thereby Q2 operates in the saturation region. Owing to the current flow in Q2, the voltage developed across the common emitter resistor, Re maintains Q1 at cut-off. Since the base of Q1 is at ground potential, it is negative relative to the emitter. Thus the stable state in the absence of a

signal is Q2 ON and Q1 OFF and the output voltage is in the low state. The switching action may be started by raising or lowering the bias on Q1. When an input sine wave is applied, as soon as the input voltage attains a value equal to the sum of the voltages across Rb1 and Re, Q1 turns ON since its base becomes more positive relative to the emitter. Q1 is driven to conduction in the saturation region. The collector voltage of Q1 drops, which inturn is coupled by the network Rb2-R1 to the base of Q2. This eliminates the forward bias on Q2 and hence it is driven to cut-off. This state persists as long as the input voltage is greater than the sum of the voltages across Rb1 and Re. When Q2 is driven to cut-off, output voltage switches to the difference between Vcc and the voltage across Rc2. When the input voltage drops below the sum of the voltages across Rb1 and Re Q1 turns OFF and by regenerative action Q2 again turns ON. The output voltage falls back to the sum of the voltages across Re and the saturation voltage of Q2. Thus a square wave is produced. The turn ON voltage is usually called the upper trigger point or UTP and the tun OFF voltage is called lower trigger point or LTP. UTP is always greater than LTP since the voltage required to turn ON a device is more than that required to turn it OFF. Design: Design Specifications

Manufacturers specifications

Initially Q1 is OFF and Q2 is ON

From equations (1) and (2)

Applying KVL to the output side of the circuit,

According to Ohm’s law

i.e.

When Q1 is ON and Q2 is OFF

Applying KVL

From equation (9)

The condition for stability is

We take

C1 should be properly selected in order to get perfect output. Select C1 as 100pF. Procedure: EDWinXP -> Schematic Editor: The circuit diagram is drawn by loading components from the library. Wiring and proper net assignment has been made. The values are assigned for relevant components.

EDWinXP -> Mixed Mode Simulator: The circuit is preprocessed. The desired test points and waveform markers are placed. The Transient Analysis parameters have been set. The Transient Analysis is executed and output observed in Waveform Viewer. Result: The output waveform may be observed in the waveform viewer.