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ECE 3144 Lecture 17

Dr. Rose Q. Hu

Electrical and Computer Engineering Department

Mississippi State University

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Review for Chapter 3• Nodal analysis for an N-node circuit

– Unknown parameters are nodal voltages– Apply KCL to each node– There are N-1 independent equations– 4 cases

• Loop analysis for an N-loop circuit– Unknown parameters are loop currents– Apply KVL to each loop– There are N independent equations– 4 cases

• Ideal op-amp– For an ideal op-amp, i+ = i- = 0 and v+ = v-

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Ideal Op-amp example

VO

+20

80K

10K

40

-

5V

40

For the first op-amp va = vb = 0Apply the KCL at terminal a

a

bcd

e

0804010

5

k

v

k

v

kec =>

402 ec vv

Apply the KCL at terminal d

(1)

=>02040

k

vv

k

v eoo

oe vv 32 (2)

For the second op-amp

vc = vd= vo(3)

There are three equations three unknowns =>

vo= -11.4 V

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Chapter 4: Additional analysis tools

• Circuit linearity

• Superpostition for linear circuits

• Thevenin’s rule and Norton rule

• Review for ideal op-amp

• Maximum power transfer

• Matlab and Pspice techniques

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Linear system• Linear element: a passive element that has a linear voltage-current

relationship- example: v(t) = Ri(t)

• Linear dependent source: a dependent current or voltage source whose output current or voltage is proportional only to the first power of a specified current or voltage variable in the circuit

• Linear circuit: a circuit composed entirely of independent sources, linear dependent sources and linear elements.

• All circuits we have analyzed are linear circuits

• For a linear system, it satisfies both additivity and homogeneity.

Linear systemx1(t) y1(t)

Linear systemx2(t) y2(t)

additivity

Linear systemx1(t)+ x2(t) y1(t)+ y2(t)

Linear systemx1(t) y1(t) Linear system

Ax1(t) Ay1(t)homogeneity

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Using Homogeneity to solve the circuit problem

For the circuit shown, we wish to determine the output current Io. Rather than approaching the problem in the straightforward way we have introduced and calculate I1 and I2, then Io. We will use Homogeneity to solve the problem. Remember that Homogeneity says if the single input is scaled by A, then the output is scaled by A too.

We just simply assume the output current Io=1 mA. This output current will yield a value for the input current source I. We will then find the scaling ratio of the  true current value I to the calculated one. The same scaling ratio times 1 mA will yield the actual value for Io.

If Io = 1mA, I3 = 0.5 mA (based on the current divider) =>I1 = 1.5mAR2 = 4k+8k = 12k. (The resistance to the left side of the input source)R1 = 2k+6k//3k = 4k (The resistance to the right side of the input source)Based on the current divider

34

12

1

2

2

1 k

k

R

R

I

I => I2 = 0.5 mA => I= I1+I2 = 2 mA

Thus the assumption of Io = 1mA produces a source current 2 mA. Since the actual source current is 6 mA, the scaling factor is 3. Thus the actual output current is 3 mA.

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Homework for Lecture 17

• Problems 4.3, 4.4

• Due Feb 25