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ECE 3144 Lecture 20

Dr. Rose Q. Hu

Electrical and Computer Engineering Department

Mississippi State University

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Thevenin and Norton theorems• Let us suppose that we need to only make a partial analysis of the a circuit. For

example, perhaps we need to determine the current, voltage and power delivered to a single “load” resistor by the remainder of the circuit., which may consist of a sizable number of sources and resistors.

• Thevenin theorem tells us that we can replace the entire linear network, exclusive of the load resistor, by an equivalent circuit that contains only an independent voltage source in series with a resistor in such a way that the current-voltage (i-v) relationship at the load resistor is unchanged.

• Norton theorem tells us that we can replace the entire network, exclusive of the load resistor, by an equivalent circuit that contains only an independent current source in parallel with a resistor in such a way that the current-voltage (i-v) relationship at the load resistor is unchanged.

(a) (b)

A complex network including a load resistor RL.

A Thevenin equivalent network

A Norton equivalent network

io

+

-vo

io

+

-

vo

(c)

io

+

-

vo

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Thevenin Theorem• This result for Thevenin theorem is not too surprising if we note that the Thevenin equivalent circuit can produce any linear (straight line) i-v characteristic for some choice of Thevenin voltage vTH(t) and Thevenin resistance RTH.

•A natural question is how to determine vTH(t) and RTH

for a particular circuit. Sometimes they are measured experimentally. Sometimes they are calculated. •In either case, it is helpful to note that if we connect no load and therefore io(t) = 0, then we can determine vTH(t)

from

where voc(t) is called the open circuit voltage

•If we short circuit the two terminals to force vo(t) = 0, then we get

•If vTH(t) ≠ 0, then ishortcircuit(t) ≠ 0 and we find

)()()( tvtvtv octopencircuiTH

)()()(

titiR

tvscitshortcircu

TH

TH

)(

)(

ti

tvR

sc

ocTH

io

+

-

vo

4

•During the measurement, one difficulty is that if we short the terminals of some circuits (a wall plug, for example), the circuit can be damaged by the resulting large current. In that case, we can simply place a resistor, R*, across the terminals that causes the terminal voltage to drop below the open-circuit voltage by a measurable amount voltage, v*(t). By the voltage divider rule, we see, in this case that

•Consider, now, the case in which vTH(t) = 0. Then, the Thevenin equivalent circuit is

simply a resistor. To measure the Thevenin resistance in this case, we can apply a voltage vS(t) (usually a constant) and measure the resulting current iS(t):

)()()(*

*

*

** tv

RR

Rtv

RR

Rtv oc

THTH

TH

)1)(

)((

**

tv

tvRR oc

TH

=>

)(

)(

ti

tvR

S

S

Thevenin Theorem: cont’d

5

(c)

io

+

-

vo

•Based on source transformation we have learned, we can determine Norton current iN(t) and Norton resistance RN

)(

)(

ti

tvRR

sc

ocTHN

)()()( titiR

vti scitshortcircu

TH

THN

Norton theorem

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Summary for Thevenin and Norton theorems

• Thevenin’s theorem requires that, for any linear circuit consisting of resistance elements and energy sources with an identified terminal pair, the circuit can be replaced by a series combination of an ideal voltage source vTH and a resistance RTH, where vTH is the open-circuit voltage at the two terminals and RTH is the ratio of the open-circuit voltage to the short-circuit current at the terminal pair.

• Norton’s theorem requires that, for any linear circuit consisting of resistance elements and energy sources with an identified terminal pair, the circuit can be replaced by a parallel combination of an ideal current source iN and a resistance RN, where iN is the short-circuit current at the two terminals and RN is the ratio of the open-circuit voltage to the short-circuit current at the terminal pair.

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Applications of Thevenin and Norton’s Theorems

• Note that this systematic transformation allows us to reduce the network to a simpler equivalent form with respect to some other circuit elements.

• Although this technique is applicable to networks containing dependent sources, it may not be as useful as other techniques and care must be taken not transform the part of the circuit which contains the control variables.

• How to apply these theorems depends on the structure of the circuits.

• Case 1: if only independent sources are present, we can calculate the open-circuit voltage and short circuit current and then the Thevenin equivalent resistance.

• Case 2: if both independent sources and dependent sources are present, we will calculate the open-circuit and short circuit current first. Then determine the Thevenin equivalent resistance.

• Case 3: For circuit only contains dependent sources, since both open-circuit and short-circuit current are zero (no independent sources here to provide the controlling variables), we cannot determine RTH in this case by using voc/isc. Remember during the discussion of Thevenin theorem, we can apply an external voltage source vS(t)

(usually a constant) and measure the resulting current iS(t) => RTH = vS(t)/iS(t)

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Case 1 example 1Find the Thevenin equivalent circuit at terminal pair a and b for the circuit shown.

This specific problem can be solved by using different approaches. We solve the problem by using source transformation technique.

Thus we have RTH = 12 and vTH = -8V

+

-

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Case 1 example 2Find Vo in the circuit given in figure (a).

Vo

Vo is the voltage across the 6k resistor. Thus 6k resistor can be considered as the load resistor and the rest network can be replaced by the equivalent Thevenin circuit.

The open circuit voltage Voc is found from figure (b).

(a)

(b)

Voc

Apply mesh analysis technique:

Mesh 1: -6 +4kI1+2k(I1-I2) = 0

Mesh 2: I2 = 2 mA => I1 = 5/3 mA

Applying KVL Voc = 4kI1+2kI2 =4*5/3 + 2*2 = 32/3 V

RTH can be derived by calculating the short circuit current Isc. Then RTH = Voc/Isc. For the circuits containing only independent sources, RTH can also be derived by zeroing out all sources as shown in (c).

I1

I2

RTH

RTH = (4k//2k)+2k = 10/3 k

(c)

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Case 1 example 2: cont’d

Thus the original circuit is equivalent to the network shown in figure (d).

Vo

Thus using voltage divider =>

Vkk

kVo

7

48)

3/106

6(

3

32

The equivalent Norton equivalent circuit is shown in figure (e).

(d)

(e)

Vo

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Case 2 example+ -Vo

4 K

2 K

4 K2 K

1 2 VVx /1 0 0 0

Vx

+

-

6 VFind Vo in the circuit shown using Thevenin’s Theorem.

-

Find Voc

+

-

Vx = 6*2k/(4k+2k) = 2VVy = 12-4k*Vx/1000 = 12-8= 4VVoc = Vx-Vy = -2V

Find Isc

(6-Vx)/4k=Vx/2k+IscIsc = Vx/1K+(Vx-12)/4k

Vx=> Isc = -0.19mA

Rth = Voc/Isc=10.67k

Vo = Voc*2k/(2k+10.67k) = -0.32V

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Case 3 example a

b

Find the Thevenin equivalent of the circuit given

Since the rightmost terminals are already open-circuit, i=0. Consequently, the dependent source is dead. For this network, since there is no independent source in the network, both voc and isc are zero. Apply a 1-A source iS externally, measure the voltage vS across the terminal pairs. Then we have RTH = vS/iS = vS.

+

-vS

We can see that i=-1A. Apply KVL nodal analysis at node a:

123

)1(5.1

SS vv

a

=>

vS = 0.6V =>

RTH = 0.6

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Problem solving strategy by using Thevenin theorem/Norton theorem

• Remove the load and the find the voltage across the open-circuit terminals, Voc. All the circuit analysis techniques presented (KVL/KCL, current/voltage divider, nodal/loop analysis, superposition, source transformation)

• Determine the Thevenin equivalent resistance of the network at the open terminals with the load removed. Three different types of circuits may encountered in determining the resistance RTH.

– If the circuits contains only independent sources, they are made zero by replacing voltage sources with short circuits and current sources with open circuits. RTH is then found by computing the resistance of the purely resistive network at the open terminals.

– If the circuit contains only dependent sources, an independent voltage or current source is applied at the open terminals and the corresponding current or voltage at these terminals is calculated. The voltage/current ratio at the terminals is the Thevenin equivalent resistance. Since there is no energy source, the open circuit voltage is zero here.

– If the circuits contains both independent and dependent sources, the open circuit terminals are shorted and the short-circuit current between these terminals are decided. The ratio of the open-circuit voltage to the short-circuit current is the Thevenin resistance RTH.

• The load is now connected to the Thevenin equivalent circuit, consisting of Voc in series with RTH, the desired output is obtained.

• The problem solving strategy for Norton theorem is essentially the same as that for the Thevenin theorem with the exception that we are dealing with the short-circuit current instead of open-circuit voltage.

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Homework for Lecture 20

• Problems 4.32, 4.33, 4.34, 4.35, 4.39, 4.47

• Exam 2 (covering Chapter 3 and 4) will be on March 20.