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ECE 3144 Lecture 22

Dr. Rose Q. Hu

Electrical and Computer Engineering Department

Mississippi State University

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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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Operational Amplifier

•An operational amplifier is a linear circuit network. •Based on Thevenin theorem or Norton theorem, any linear circuit can be considered as equivalent to a Thevenin circuit or Norton circuit at a specified terminal pair . The following is the detailed modeling for an opamp. Ri is the input Thevenin resistance and Ro is the output Thevenin resistance.

->iout

i+ ->

i- ->

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Ideal Operational Amplifier• Now we want to derive the ideal conditions for the op-amp. • If an external network A is connected to the op-amp at the input terminals, then op-amp is considered as the load of the external network A.

• In order to gain maximum output voltage (v+-v-) for op-amp with any external network A => the input Thevenin resistance for the ideal op-amp is infinite. Ri-> => i+ = i- = 0• If an external network B is connected to the op-amp at the output terminals, the external network B is considered as the load of the op-amp.

• In order for the load to gain the maxim output vo for any external network B=> the output Thevenin resistance for the ideal op-amp is zero, Ro = 0.• Finally, the gain of an ideal operational amplifier is infinite => A-> and v+ = v- = 0.

+

-

vo

A

B

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Ideal Operational Amplifier

i+ = i- = 0 and v+ = v-

Ideal model for an operational amplifier

Modeling parameters

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Example 1: Unit gain operational amplifierRecall that, for ideal operational amplifiers,

vout = A(v+ - v-)

Because the ideal operational amplifier draws no current through its input terminals, (and hence develops no voltage drop across the input resistor, R), thus vin = v+

For this particular circuit v- = vout

So that vout = A(vin – vout) or (1+A)vout = Avin =>

As A-> , vout = vin

inout vA

Av

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The fact that the output voltage follows the input voltage shows that our circuit is an example of a voltage follower. The main desirable feature of such circuits is that they draw little current at their inputs, and exhibit high Thevenin resistances at their inputs, but exhibit low Thevenin resistances at their output terminals. High Thevenin resistance at input terminals can draw high input voltages for the op-amp while low Thevenin resistance at output terminals can relay the high voltage to the load network.

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Example 1: Unit gain operational amplifier

• An obvious question is: if vout = vin, why not just connect vin to vout directly via two parallel connection wires; why do we need to place an op-amp between them?

• Assume we connect voltage source (vin, R) with load network (RL) => vout = vin – iR. We can see that vout is always less than vin unless the load resistance RL is infinite. If we connect an ideal op-amp between the source and load network, no matter how much the load resistor is, we always have vout = vin

• Also unit gain opamp can be used to isolate one circuit from another. The energy supplied to the load network in the direct connection case must come from the source network, however with unit gain opamp the energy comes from the opamp power supply.

• As a general rule, when analyzing the opamp circuits, we write the nodal equations by using KCL at the opamp input terminals, using the ideal opamp conditions i+ = i- = 0 and v+ = v-

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Example 2 : Differential operational amplifier

Apply KCL in the “-” terminal(inverting terminal) of the opamp:

Apply KCL in the “+” terminal (noninverting terminal) of the opamp:

v+ = v-

(1)

(2)

(3)

Using equations (1), (2) and (3) to solve vout

The output of this amplifier clearly is proportional to the difference of the voltages of the driving circuits. For this reason, this amplifier is called a differential amplifier.

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DC analysis using Matlab

• See handouts

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Homework for Lecture 21

• Problems 3.69. Use Matlab to solve the problem. Please attach your Matlab code and results.