Operational amplifiers
Eugeniy E. Mikhailov
The College of William & Mary
Week 9
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Electronics 1
Week 9
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Operational amplifiers (Op-Amp)
V-
V+
−
+
Vout
A
Vout = A(V+ − V− ) thus sometimes called differential amplifiers
A is open loop gain
A is frequency dependent
A = 105 . . . 106 at DC
A → 0 at high frequency (roll off)
this limits operational bandwidth (typically in MHz . . . GHz range)
input impedances are high 106 . . . 1014 Ω
output impedances are low 0.1 . . . 10 Ω
however output current usually limited to 10 mA
it is super easy to design with them
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Electronics 1
Week 9
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If Op-Amps are so great why did we learn transistors?
some times one transistor is enough and op-amps are more
expensive
op-amps are made of transistors so to understand their limits we
need to know how transistors behave
op-amps require bipolar power supply
remember that op-amps cannot source a lot of current/power
while transistors can (recall our transistor voltage controlled switch
for a bulb)
So, combine op-amps and transistors for a power circuits. Otherwise
do your circuit with op-amps.
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Electronics 1
Week 9
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Very very bad amplifier !!!
Vin
+
−
Vout
A
Gain
Vout = AVin
But A depends on everything
temperature
power supply voltage
input voltage
frequency
. . . and so on
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Electronics 1
Week 9
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Non-inverting amplifier
R2
I1
R1
−
I2
Vin
Vout
+
Vout
A
R2
A
= 1+
Vin
R1
A + (1 +
R2
R1 )
Gain and impedances of ideal Op-Amp (A 1)
Gideal = 1 +
R2
R1
Zin = ∞, Zout = 0
notice that with negative feedback V+ ≈ V−
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Follower or Buffer
−
Vin
+
Vout =
Vout
A
A
Vin
A+1
Gain and impedances of ideal Op-Amp (A 1)
Gideal = 1
Zin = ∞, Zout = 0
notice that with negative feedback V+ ≈ V−
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Electronics 1
Week 9
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Op-amps golden rules
If negative feedback is applied and A(f ) 1 (open circuit gain at the
frequency of interest)
there is no current into the inputs
V− = V+
Gain of non ideal Op-Amp (A 6 1)
G = Gideal ||A =
Eugeniy Mikhailov (W&M)
Gideal A
Gideal + A
Electronics 1
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Inverting amplifier
R2
I2
R1
Vin
I1
−
+
Vout
A
Gain and impedances of ideal Op-Amp (A 1)
Gideal = −
R2
R1
Zin = R1 , Zout = 0
notice that with negative feedback V+ ≈ V−
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Electronics 1
Week 9
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Summing inverting amplifier
Vin1
Vin2
Vin3
VinN
R1
R2
Rf
R3
−
RN
+
Vout
A
for ideal Op-Amp (A 1)
Vout = −
Vin1 Vin2 Vin3
VinN
+
+
+ ··· +
R1
R2
R3
RN
Rf
ZinN = RN , Zout = 0
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Electronics 1
Week 9
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Differential amplifier
R2
Vin1
Vin2
R1
R3
−
+
Vout
A
R4
for ideal Op-Amp (A 1)
Vout =
R4 R1 + R2
R2
Vin2 −
Vin1
R1 R3 + R4
R1
Zout = 0
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Week 9
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