digilent thevenin

7/28/2019 Digilent Thevenin

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Overview

In previous chapters, we have seen that it is possible to characterize a circuit consisting of sourcesand resistors by the voltage-current (or i-v) characteristic seen at a pair of terminals of the circuit.When we do this, we have essentially simplified our description of the circuit from a detailed model ofthe internal circuit parameters to a simpler model which describes the overall behavior of the circuit asseen at the terminals of the circuit. This simpler model can then be used to simplify the analysisand/or design of the overall system.

In this chapter, we will formalize the above result as Thveninsand Nortons theorems. Using thesetheorems, we will be able to represent any linear circuit with an equivalent circuit consisting of a singleresistor and a source. Thvenins theorem replaces the linear circuit with a voltage source in serieswith a resistor, while Nortons theorem replaces the linear circuit with a current source in parallel witha resistor.

In this chapter, we will apply Thvenins and Nortons theorems to purely resistive networks.However, these theorems can be used to represent any circuit made up of linear elements.

Before beginning this chapter, you shouldbe able to:

After completing this chapter, you should beable to:

Represent a circuit in terms of its i-vcharacteristic (Chapter 1.7.3)

Represent a circuit as a two-terminalnetwork (Chapter 1.7.3)

Determine Thvenin and Norton equivalentcircuits for circuits containing power sourcesand resistors

Relate Thvenin and Norton equivalent

circuits to i-v characteristics of two-terminalnetworks

This chapter requires:

N/A

Consider the two interconnected circuits shown in Figure 1 below. The circuits are interconnected atthe two terminals a and b, as shown. Our goal is to replace circuit A in the system of Figure 1 with a

simpler circuit which has the same current-voltage characteristic as circuit A. That is, if we replacecircuit A with its simpler equivalent circuit, the operation of circuit B will be unaffected. We will makethe following assumptions about the overall system:

Circuit A is linear Circuit A has no dependent sources which are controlled by parameters within circuit B

Circuit B has no dependent sources which are controlled by parameters within circuit A


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1.7.4: Thvenins and Nortons Theorems

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Figure 1. Interconnected two-terminal circuits.

In chapter 1.7.3, we determined i-v characteristics for several example two-terminal circuits, using thesuperposition principle. We will follow the same basic approach here, except for a general linear two-terminal circuit, in order to develop Thvenins and Nortons theorems.

Thvenins Theorem:

First, we will kill all sources in circuit A and determine the voltage resulting from an applied current, asshown in Figure 2 below. With the sources killed, circuit A will look strictly like an equivalentresistance to any external circuitry. This equivalent resistance is designated as RTH in Figure 2. Thevoltage resulting from an applied current, with circuit A dead is:

iRvTH

=1 (1)

Figure 2. Circuit schematic with dead circuit.

Now we will determine the voltage resulting from re-activating circuit As sources and open-circuitingterminals a and b. We open-circuit the terminals a-b here since we presented equation (1) asresulting from a current source, rather than a voltage source. The circuit being examined is as shownin Figure 3. The voltage vOC is the open-circuit voltage.

Figure 3. Open-circuit response.


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Superimposing the two voltages above results in:

OCvvv += 1 (2)

or

OCTHviRv += (3)

Equation (3) is Thvenins theorem. It indicates that the voltage-current characteristic of any linearcircuit (with the exception noted below) can be duplicated by an independent voltage source in serieswith a resistance RTH, known as the Thvenin resistance. The voltage source has the magnitude vOCand the resistance is RTH, where vOC is the voltage seen across the circuits terminals if the terminalsare open-circuited and RTH is the equivalent resistance of the circuit seen from the two terminals, withall independent sources in the circuit killed. The equivalent Thvenin circuit is shown in Figure 4.

+

-

RTH

VOC v

+

-

i

Equivalent

Circuit

Figure 4. Thvenin equivalent circuit.

Procedure for determining Thvenin equivalent circuit:1. Identify the circuit and terminals for which the Thvenin equivalent circuit is desired.2. Kill the independent sources (do nothing to any dependent sources) in circuit and determine

the equivalent resistance RTH of the circuit. If there are no dependent sources, RTH is simplythe equivalent resistance of the resulting resistive network. Otherwise, one can apply anindependent current source at the terminals and determine the resulting voltage across theterminals; the voltage-to-current ratio is RTH.

3. Re-activate the sources and determine the open-circuit voltage VOC across the circuit

terminals. Use any analysis approach you choose to determine the open-circuit voltage.


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Example: Determine the Thvenin equivalent of the circuit below, as seen by the load, RL.

We want to create a Thvenin equivalent circuit of the circuit to the left of the terminals a-b. Theload resistor, RL, takes the place of circuit B in Figure 1.

The circuit has no dependent sources, so we kill the independent sources and determine theequivalent resistance seen by the load. The resulting circuit is shown below.

From the above figure, it can be seen that the Thvenin resistance RTH is a parallel combination of

a 3 resistor and a 6 resistor, in series with a 2 resistor. Thus, =++

= 42

36

)3)(6(TH

R .

The open-circuit voltage vOC is determined from the circuit below. We (arbitrarily) choose nodal

analysis to determine the open-circuit voltage. There is one independent voltage in the circuit; it islabeled as v0 in the circuit below. Since there is no current through the 2 resistor, vOC = v0.

Applying KCL at v0, we obtain: VvvvVv

AOC

6036

62 0

00===

+

+ . Thus, the Thvenin

equivalent circuit is on the left below. Re-introducing the load resistance, as shown on the rightbelow, allows us to easily analyze the overall circuit.


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Nortons Theorem:

The approach toward generating Nortons theorem is almost identical to the development of

Thvenins theorem, except that we apply superposition slightly differently. In Thvenins theorem,we looked at the voltage response to an input current; to develop Nortons theorem, we look at thecurrent response to an applied voltage. The procedure is provided below.

Once again, we kill all sources in circuit A, as shown in Figure 2 above but this time we determine thecurrent resulting from an applied voltage. With the sources killed, circuit A still looks like an equivalentresistance to any external circuitry. This equivalent resistance is designated as RTH in Figure 2. Thecurrent resulting from an applied voltage, with circuit A dead is:

THR

vi =1 (4)

Notice that equation (4) can be obtained by rearranging equation (1)

Now we will determine the current resulting from re-activating circuit As sources and short-circuitingterminals a and b. We short-circuit the terminals a-b here since we presented equation (4) asresulting from a voltage source. The circuit being examined is as shown in Figure 5. The current iSCis the short-circuit current. It is typical to assume that under short-circuit conditions the short-circuitcurrent enters the node at a; this is consistent with an assumption that circuit A is generating powerunder short-circuit conditions.

Figure 5. Short-circuit response.

Employing superposition, the current into the circuit is (notice the negative sign on the short-circuitcurrent, resulting from the definition of the direction of the short-circuit current opposite to the directionof the current i)

SCiii = 1 (5)

so

SC

TH

iR

vi = (6)

Equation (6) is Nortons theorem. It indicates that the voltage-current characteristic of any linearcircuit (with the exception noted below) can be duplicated by an independent current source in parallel


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with a resistance. The current source has the magnitude iSC and the resistance is RTH, where iSC isthe current seen at the circuits terminals if the terminals are short-circuited and RTH is the equivalentresistance of the circuit seen from the two terminals, with all independent sources in the circuit killed.The equivalent Norton circuit is shown in Figure 6.

Figure 6. Norton equivalent circuit.

Procedure for determining Norton equivalent circuit:

1. Identify the circuit and terminals for which the Norton equivalent circuit is desired.2. Determine the equivalent resistance RTH of the circuit. The approach for determining RTH is

the same for Norton circuits as Thvenin circuits.3. Re-activate the sources and determine the short-circuit current iSC across the circuit

terminals. Use any analysis approach you choose to determine the short-circuit current.


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Example: Determine the Norton equivalent of the circuit seen by the load, RL, in the circuit below.

This is the same circuit as our previous example. The Thvenin resistance, RTH, is thus the same

as calculated previously: RTH = 4. Removing the load resistance and placing a short-circuitbetween the nodes a and b, as shown below, allows us to calculate the short-circuit current, iSC.

Performing KCL at the node v0, results in:

AvVvv

236

6

2

000=

+

+

so

Vv 30 =

Ohms law can then be used to determine iSC:

AV

iSC

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3=

=

and the Norton equivalent circuit is shown on the left below. Replacing the load resistance resultsin the equivalent overall circuit shown to the right below.


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Source Transformations:

Circuit analysis can sometimes be simplified by the use of source transformations. Sourcetransformations are performed by noting that Thvenins and Nortons theorems provide two differentcircuits which provide essentially the same terminal characteristics. Thus, we can write a voltagesource which is in series with a resistance as a current source in parallel with the same resistance,and vice-versa. This is done as follows.

Equations (3) and (6) are both representations of the i-vcharacteristic of the same circuit.Rearranging equation (3) to solve for the current i results in:

TH

OC

THR

v

R

vi = (7)

Equating equations (6) and (7) leads to the conclusion that

TH

OC

SC

R

vi = (8)

Likewise, rearranging equation (6) to obtain an expression for v gives:

THSCTHRiRiv += (9)

Equating equations (9) and (3) results in:

THSCOCRiv = (10)

which is the same result as equation (8).

Equations (8) and (10) lead us to the conclusion that any circuit consisting of a voltage source inseries with a resistor can be transformed into a current source in parallel with the same resistance.Likewise, a current source in parallel with a resistance can be transformed into a voltage source in

Exceptions:

Not all circuits have Thvenin and Norton equivalent circuits. Exceptions are:

1. An ideal current source does not have a Thvenin equivalent circuit. (It cannot berepresented as a voltage source in series with a resistance.) It is, however, its ownNorton equivalent circuit.

2. An ideal voltage source does not have a Norton equivalent circuit. (It cannot berepresented as a current source in parallel with a resistance.) It is, however, its ownThvenin equivalent circuit.


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series with the same resistance. The values of the transformed sources must be scaled by theresistance value according to equations (8) and (10). The transformations are depicted in Figure 7.

R

VS

RIS

Figure 7. Source transformations.

Source transformations can simplify the analysis of some circuits significantly, especially circuitswhich consist of series and parallel combinations of resistors and independent sources. An exampleis provided below.


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Example: Determine the current iin the circuit shown below.

We can use a source transformation to replace the 9V source and 3 resistor series combination

with a 3A source in parallel with a 3 resistor. Likewise, the 2A source and 2 resistor parallel

combination can be replaced with a 4V source in series with a 2 resistor. After thesetransformations have been made, the parallel resistors can be combined as shown in the figurebelow.

The 3A source and 2 resistor parallel combination can be combined to a 6V source in series with

a 2 resistor, as shown below.

The current ican now be determined by direct application of Ohms law to the three series

resistors, so that AVV

i 25.0

242

46=

++

= .


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Voltage Current characteristics of Thvenin and Norton Circuits:

Previously, in Chapter 1.7.3, we noted that the i-v characteristics of linear two-terminal networkscontaining only sources and resistors are straight lines. We now look at the voltage-current

characteristics in terms of Thvenin and Norton equivalent circuits.

Equations (3) and (6) both provide a linear voltage-current characteristic as shown in Figure 8. Whenthe current into the circuit is zero (open-circuited conditions), the voltage across the terminals is theopen-circuit voltage, vOC. This is consistent with equation (3), evaluated at i= 0:

OCOCTHOCOCTHvvRviRv =+=+= 0 .

Likewise, under short-circuited conditions, the voltage differential across the terminals is zero andequation (6) readily provides:

scsc

TH

sc

TH

SC

iiRiR

v

i ===

0

which is consistent with Figure 8.

Figure 8. Voltage-current characteristic for Thvenin and Norton equivalent circuits.

Figure 8 is also consistent with equations (8) and (10) above, since graphically the slope of the line is

obviouslySC

OC

TH

i

vR = .

Figure 8 also indicates that there are three simple ways to create Thvenin and Norton equivalentcircuits:

1. Determine RTHand vOC. This provides the slope and y-intercept of the i-vcharacteristic. Thisapproach is outlined above as the method for creating a Thvenin equivalent circuit.


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2. Determine RTHand iSC. This provides the slope and x-intercept of the i-vcharacteristic. Thisapproach is outlined above as the method for creating a Norton equivalent circuit.

3. Determine vOCand iSC. The equivalent resistance RTHcan then be calculated fromSC

OC

TH

i

vR =

to determine the slope of the i-vcharacteristic. Either a Thvenin or Norton equivalent circuitcan then be created. This approach is not commonly used, since determining RTH theequivalent resistance of the circuit is usually easier than determining either vOCor iSC.

Note:

It should be emphasized that the Thvenin and Norton circuits are not independent entities.One can always be determined from the other via a source transformation. Thvenin andNorton circuits are simply two different ways of expressing the same voltage-currentcharacteristic.

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