Transcript v out = A(v + - v - s3.amazonaws.com

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
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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.
i+ ->
->iout
i- ->
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Ideal Operational Amplifier
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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.
A
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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.
+
B
vo
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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.
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Ideal Operational Amplifier
Ideal model for an operational amplifier
Modeling parameters
i+ = i- = 0 and v+ = v-
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Example 1: Unit gain operational amplifier
Recall 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
So that
vout = A(vin – vout) or (1+A)vout = Avin =>
As A-> , vout = vin
vout
v- = vout
A

vin
1 A
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:
(1)
Apply KCL in the “+” terminal
(noninverting terminal) of the opamp:
(2)
v+ = v -
(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.
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