Transcript Chapter 3 - Op Amps(PowerPoint Format)

The Operational Amplifier
The integrated circuit operational amplifier (IC op amp) is one of the most
widely used and most versatile electronic components. The device is found in
many modern electronic systems because of its versatility, ease of use, and cost
effectiveness. Various schematic symbols for an op amp are shown below.
op amp schematic symbols
•In reality, all three schematic symbols are the same. The symbol on the far
left explicitly shows connections to external power, while the two symbols on
the right do not show these connections but imply them, nevertheless.
•Op amps are active devices and, therefore, require an source of power (also
called bias voltages).
•The two symbols on the right are identical except for their orientation.
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Op amp connections
V++, positive power supply (bias) connection
inverting input
+ vout output
noninverting input
V--,negative power supply (bias) connection
Often, in order to make circuit diagrams simpler, the power supply
connections are not shown. However, as stated earlier, they must always
be made in practice.
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Op amp model
0 A
vid
+
+
+
vo
-
Avocvid
-
0 A
Ri is the input resistance of the op amp. Ri is very large, several 100M or
several thousand M . Ideally, Ri =  ( an open circuit). Because Ri is so
large, under normal operating conditions the currents into the inputs of the
op amp are negligible.
Ro is the output resistance of the op map. Ro is very small, less than 100.
Because Ro is so small, under normal operating conditions the output
voltage Vo is the same as the voltage at the dependent source (Vo  Avocvid) .
Avoc is the open-circuit voltage gain. Avoc is very large, typically 100,000 or
greater. Ideally, Avoc is infinite. When operating under normal conditions,
this large value of Avoc causes the differential input voltage, vid, to be very
small; ideally vid = 0.
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Ideal op amp assumptions;
0A
0v
+
0A
The input of an ideal op amp acts like both an open circuit (zero current)
and a short circuit (zero voltage) simultaneously.
Using these two assumptions along with circuit analysis techniques such as
Ohm’s law, KVL and KCL, we can derive equations which define the
operation of a number of basic and useful op amp circuits.
An op amp is often used to construct an amplifier. An amplifier is a circuit
which converts a small input signal into a larger signal at the output. While
amplification is not the only reason to use an amplifier, it is one of the
primary reasons.
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The inverting amplifier
The voltage at the noninverting
terminal is zero because the
terminal is tied to common.
I2
I1
+
vin
-
0A
0v
+
vo
-
By KVL, vin - I1R1 = 0  I1 = vin/R1
The voltage at the inverting
terminal is zero because there
is no voltage drop across the op
amp input. The inverting
terminal is at virtual ground
By KCL, I2 = I1= vin/R1
By KVL, vo + I2R2 = 0  vo = -I2R2 = -(vin/R1) R2
vo
R2
Av 

vin
R1
Note also that vin “sees” R1 ohms of resistance. The input resistance of the
inverting amplifier is R1.
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The noninverting amplifier
The input signal is applied at
the noninverting terminal
I2
I1
0A
vin
+
vin
-
+
vo
-
The voltage at the inverting
terminal is also vin since there
is no voltage drop across the op
amp input.
By KVL, vin - I1R1 = 0  I1 = vin/R1
By KCL, I2 = I1= vin/R1
By KVL, vo - I2R2 - I1R1 = 0  vo = I2(R1 + R2) = (vin/R1) (R1 + R2 )
vo  R 2 
Av 
 1 

vin 
R1 
Note also that vin “sees” an open circuit since the op amp input current is zero.
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Op amp circuits
The inverting amplifier
vo
R2
Av 

v in
R1
Iin
+
vin
-
+
vo
-
I in 
vo
 R in  R1
R1
The noninverting amplifier
v
 R2 
A v  o  1 

v in  R1 
Iin
+
vin
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+
vo
-
I in  0  R in  
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The buffer amplifier
+
vin
-
vin
+
vo
-
vo = vin
Q: Since there is no gain (vo = vin  gain = 1) is this circuit useful? If so, why?
A: Iin = 0 since the input is tied to the noninverting input of the op amp.
There is no power demanded from the input source. vin sees a very large
(ideally infinite) input resistance.
vo is the same value as vin, but any power demanded from vo comes from
the op amp (the op amp power supplies) and not the input signal.
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The mixer
If = I1 + I2
v1
v2
I1 
and I 2 
R1
R2
I1
I2
0v
+
vo
-
vo   R f  I f   R f  ( I1  I 2)
 v1 v 2 
vo   R f   

 R1 R 2 
Observations:
• The output is a weighted sum of the input signals, v1 and v2.
• If R1 = R2, vo is proportional to the sum of the inputs.
• The circuit is sometimes called an inverting adder, summer, or mixer.
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The difference (differential) amplifier. Vo = K(V2 - V1)
By VDR: V 
X
R2
V 2
R1  R 2
vX
Since the voltages at the noninverting and
inverting terminals are the same, the
voltage at the inverting terminal is also Vx
+
vX
-
Writing a nodal equation at the inverting
terminal yields:
VX  V 1 VX  Vo
1  V 1 Vo
 1

 0  VX  


R1
R2
 R1 R 2  R1 R 2
 R 2  R1  V 1 Vo
VX 


 R1R 2  R1 R 2
 R 2  V 2  R 2  R1  V 1 Vo




 R1  R 2  R1R 2  R1 R 2
R2
Vo 
V 2  V 1
R1
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Vo 
+
vo
-
R2
V 2  V 1
R1
The output is proportional
to the difference of the two
inputs.
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