2102-487
Industrial Electronics
Op Amp Circuit
Amplifier Fundamentals
Amplifier is a two-port device that accepts an external applied signal, referred to as
input, an in turn produces a signal, referred to as output, that is proportional to the
input: output = A x input
+
Vi
-
Ri
Ro
+AVi
+
Vo
-
+
RS
+VS
Vi
-
Vi =
+AVi
Ri
Amplifier
Source
Voltage amplifier model
Ro
+
RL
VL
-
Load
Voltage amplifier with source and load
Ri
VS
Ri + Rs
VL =
RL
AVi
Ro + RL
Ri
VL
RL
=
A
VS Ro + RL Ri + Rs
This shows that
| VL / VS |<| A | due to the loading effect
Loading effect complicates life because each time we change the source or the
load , we need to recompute the overall gain, and we also have signal loss.
Amplifier Fundamentals
+
Vi
-
+AVi
+
Vo
-
Rs
+VS
Source
Ideal voltage amplifier model
+
Vi
-
+AVi
Amplifier
+
RL
VL
-
Load
Ideal Voltage amplifier with source and load
To eliminate loading effect, the voltage across Rs and Ro must be zero regardless of
Rs and RL. The only way to achieve this goal is b imposing Ri = ∞ and Ro =0
VL
=A
VS
regardless of Rs and RL.
In practice, the loading effect can be eliminated by the conditions of Ri >> Rs and
and Ro << RL
Operational Amplifier (Op Amp)
The operational amplifier (op amp) is a voltage amplifier having extremely high
gain. By combining with external components, op amp could be configured to
perform a variety of operations such as addition, subtraction, multiplication,
integration, and differentiation etc.
+VCC
+VCC
-
-
+
-
+
Vd Rd
Vn
-
Vn
-
+
+
+
Vo
-
Vp
-
+
+
+aVd
+
+
Vo
-
Vp
-
-VCC
-VCC
Op amp symbol
Ro
Simple practical Op amp model
Vp = Non-inverting input voltage, Vn = Inverting input voltage and Vo = output voltage
a = unloaded voltage gain.
V p − Vn = Vd
Vo = aVd = a (V p − Vn )
For the popular 741 op amp, Rd = 2 MΩ, a ~ 200,000, and Ro = 75 Ω
Ideal Op Amp
We define ideal op amp as being an ideal voltage amplifier with infinite gain.
For the ideal op amp,
a → ∞, imply Vp = Vn,
Rd = ∞, imply In = Ip = 0
Ro = 0
Ip = Non-inverting input current, In = Inverting input current
+VCC
The output voltages are constrained by the
following relationship
-
+
Vd
Vn
-
+
+
+
+aVd
− VCC ≤ Vo ≤ VCC
+
Vo
-
Vp
-
-VCC
Ideal practical Op amp model
Ideal Op Amp Rules:
1. No current flows in to either input terminal
2. There is no voltage difference between the two input terminals
Operational Amplifier (Op Amp)
Op Amp Characteristic Property Values
Property
Gain, a
Input Resistance, Ri
Output Resistance, Ro
Input voltage difference, Vp-Vn
Input current, i1 or i2
Output voltage limits
Typical Op Amp
Ideal Op Amp
>200,000
>2 MΩ
<75 Ω
<0.1 mV
<50 pA
|Vo| < VCC
∞
∞
0
0 (virtual short)
0 (virtual short)
|Vo| ≤ VCC
Analysis of Op Amp Circuit:
Vn
Vi
Vp
+
-
Vo
+
Find Vo/Vi
2 MΩ
Vi
+
-
Vp
75 Ω
+
-
a(Vp-Vn)
-
+
Ro
+aVd
+
-
Rout
Apply KVL:
But:
+
Vo
-
+
Vo
-
Ro
Ri
Vn
+
i
Vn
Vd Ri
Vp
-
Rin
-
+
− Vi + i ( Ri + Ro ) + a (V p − Vn ) = 0
i=
And we have: Vo = Vn
V p − Vn
Ri
and
Vi = V p
Voltage gain for the voltage follower
Vo
Ri
= 1−
Vi
Ro + (1 + a ) Ri
Analysis of Op Amp Circuit:
i
Find Input resistance:
Ro
Vn
Rin
Ri
2 MΩ
Vi
+
-
Vp
75 Ω
+
-
+
a(Vp-Vn)
A
Vn
Vp
+
-
iRi = V p − Vn
But:
Vo
-
And we have:
Rin =
Vi
= (1 + a ) Ri + Ro
i
Rout =
Find output resistance:
it
Apply KCL at A:
Ro
75 Ω
Ri
Vi
i
Apply KVL: − Vi + i ( Ri + Ro ) + a (V p − Vn ) = 0
Equivalent circuit for finding Rin
2 MΩ
Rin =
+ Vt
a(Vp-Vn)
-
Simple equivalent circuit for finding Rout
But:
− it +
Vn = 0
Ri
and
+
Vi = 0
Vt − A(V p − Vn )
Ro
V p = Vt
Ro + (a + 1) Ri
Vt
Ro Ri
V
Ro Ri
Rout = t =
it Ro + (1 + a ) Ri
it =
And we have:
Vn − V p
Vt
it
=0
Analysis of Op Amp Circuit: Ideal Op Amp
Vn
ip
Vi
Vp
+
+
Vo
-
+
Vd
Vn
-
+
-
Vp
-
+
+
+aVd
+
Vo
-
Rout
Rin
Using Rule 1: (no current flows into the op Amp inputs)
i p = in = 0
V p = Vi
Using Rule 2: (no voltage difference between inverting and non-inverting inputs)
V p = Vn = Vo
Input Impedance and Output Impedance
Rin = ∞ and Rout = 0
Vo = Vi
Analysis of Op Amp Circuit:
Using the parameters of 741 op amp, Ri = 2 MΩ, a ~ 200,000, and Ro = 75 Ω
Voltage gain:
Vo
= 1 − 5 × 10 −6 ≈ 1
Vi
Input resistance:
Rin = 400 × 109 Ω
Output resistance:
Rout = 375 µΩ
Comparison between Ideal and Practical Voltage Follower
Property
Voltage Gain
Input Resistance
Output Resistance
Typical Op Amp
1−
Ri
≈1
Ro + (1 + a ) Ri
Ideal Op Amp
1
(1 + a ) Ri + Ro = 400 GΩ
∞
Ro Ri
= 375 µΩ
Ro + (1 + a ) Ri
0
Inverting Amplifier
Use KCL at point A and apply Rule 1:
KCL
(no current flows into the inverting input)
Rf
R1
Vin
+
-
A
v A − vin v A − vout
+
=0
R1
Rf
+
+
Rearrange
Vout
 1
1   vin vout 
−
+
=0
vA  +
 R R   R R 
f   1
f 
 1
-
Apply Rule 2:(no voltage difference between inverting and non-inverting inputs)
Since V+ at zero volts, therefore V- is also at zero volts too.
vin vout
+
=0
R1 R f
Rf
vout
=−
vin
R1
Input Impedance and Output Impedance
Ri = R1 and Ro = 0
vA = 0
Non-inverting Amplifier
KCL
Use KCL at point A and apply Rule 1:
Rf
R1
A
+
Vin
+
-
v A v A − vout
+
=0
R1
Rf
+
Vout
-
Input Impedance and Output Impedance
Apply Rule 2:
vin = v A
Rf
vout
= 1+
vin
R1
Ri = R∞ and Ro = 0
Basic Application of the Op Amp
R2
R2
R1
Vi
R1
-
-
+
-
+
+
Vo
-
Inverting amplifier
Av = −
R2
R1
+
+
Vi
Vo
-
+
-
Non-Inverting amplifier
Av = 1 +
Ri = R1
Ri = ∞
Ro = 0
Ro = 0
R2
R1
Summing Amplifier: Mathematic Operation
i = i1 + i2 + i3
Use KCL and apply Rule 1:
i
i1
R
i2
R
i3
v1
v2
R
vA
v A − v1 v A − v2 v A − v3 v A − vout
+
+
+
=0
R
R
R
Rf
Rf
Since vA = 0 (Rule 2)
_
vB
+
+
vout
-
vout = −
Rf
R
( v1 + v2 + v3 )
Sum of v1, v2 and v3
v3
Difference Amplifier: Mathematic Operation
Use KCL and apply Rule 1:
R2
R1
R3
v1
v2
vA
v A − v1 v A − vout
+
=0
R1
R2
_
vB
+
R4
Since vA = vB (Rule 2) and
+
vout
-
 R4 
v2
v A = vB = 
 R3 + R4 
vout =
Substitute eq. (2) into eq. (1), we get
If R1 = R2 = R and R3 = R4 = Rf
(1)
vout =
Rf
R
R4  R2 
R
 + 1v2 − 2 v1
R3 + R4  R1 
R1
( v2 − v1 )
Difference of v1and v2
(2)
Differentiator and Integrator: Mathematic Operation
R
i
vout = −iR
C
i
_
+
vin
But
i=C
dvC
and
dt
vout
dvin
= − RC
dt
+
vout
-
Differentiator
i
C
R
vout = −vC
+ vc -
t
1
But vC (t ) = ∫ idt + vC (0) and
C0
_
i
+
vin
Integrator
vin = vC
+
vout
-
t
vout
vin = iR
1
=−
vin dt + vC (0)
∫
RC 0
Difference Amplifier: Superposition
R1
V1
R2
+
-
+
+
V2
+
-
R1
V1
R3
-
R2
R4  R2 
1 + V2
V1 +
R1
R3 + R4  R1 
-
R4
R1
R2
+
R3
Vo = −
Vo
R2
+
Vo = −
R2
V1
R1
R4
Ri1 = R1 and Ro = 0
+
+
Vo
-
V2
Vo =
+
-
R3
R4  R2 
1 + V2
R3 + R4  R1 
R4
+
Vo
-
Ri 2 = R3 + R4 and Ro = 0
Commode and Differential Mode
Vdm = V2 − V1
Differential mode input:
Vcm =
Common mode input:
Rearrange:
V1 = Vcm − Vdm / 2
R1
R2
R1
Vo
-
+
-
R3(=R1)
R4(=R2)
-
+
Vcm
+
-
-
-
+
+
+
V2
V2 = Vcm + Vdm / 2
+
V1
V2 + V1
2
Vdm/2
Vdm/2
R3(=R1)
R2
+
+
Vo
-
R4(=R2)
Difference amplifier, in terms of the common and differential-mode inputs
Commode and Differential Mode
Output for difference amplifier: Vo = AdmVdm + AcmVcm
Adm = amplification of differential input
Acm = amplification of common mode input
Commode mode rejection ratio: CMRR = 20 log10
R1
V1
+
-
Vo = A1V1 + A2V2
+
Vo
We have
-
+
-
R3(=R1)
Quality index
Using superposition, the output from
difference amp can be expressed as
R2
+
V2
Adm
Acm
R4(=R2)
Vo =
1
2
(V1 + V2 ) − 12 (V2 − V1 )
V2 = 12 (V1 + V2 ) + 12 (V2 − V1 )
V1 =
1
2
( A2 − A1 )(V2 − V1 ) + ( A2 + A1 ) 12 (V2 + V1 )
Therefore in this case,
Difference amplifier
Adm =
1
2
( A2 − A1 )
and
Acm = ( A1 + A2 )
Commode and Differential Mode
Ex The difference amplifier is constructed with an ideal Op Amp and 1% tolerance
resistors of nominal values 2.2 kΩ and 5.1 kΩ. The resistors were measured and
fond to have the following resistance values:
R1 = 2.195 kΩ,
R2 = 5.145 kΩ,
R3 = 2.215 kΩ,
R4 = 5.085 kΩ
Determine the gain of the differential amplifier and its CMRR
The design gain of this amplifier is
R1
V1
+
-
R2
A=
+
+
However, this value is based on the
assumption of equal resistor ratios:
R1 R3
=
R2 R4
Vo
-
V2
+
-
R3(=R1)
R4(=R2)
The more exaction expression for the
output voltage
Vo =
Difference amplifier
R2 5.1 kΩ
=
= 2.318
R1 2.2 kΩ
R4  R2 
R
 + 1V2 − 2 V1
R3 + R4  R1 
R1
Commode and Differential Mode
Vo = A1V1 + A2V2
Here
A2 =
R4  R2 
 + 1
R3 + R4  R1 
5.085
 5.145 
+ 1 = 2.329

2.215 + 5.085  2.195 
5.145
A1 = −
= −2.344
2.195
A2 =
The CMRR is
CMRR = 20 log10
and
Therefore
A1 = −
R2
R1
Adm = 2.337
Acm = −0.01464
Adm
= 20 log − 159.6 = 44.06 dB
Acm
This is only a moderately good differential amplifier. If physical resistors used for
R1 and R3 were exchanged, the resistor ratios in each gain path would be more
nearly exact:
5.085
 5.145 
+ 1 = 2.321

Adm = 2.322
2.195 + 5.085  2.215 
Therefore
Acm = −0.00186
5.145
A1 = −
= −2.323
2.215
A
CMRR = 20 log10 dm = 20 log − 1248.0 = 61.92 dB
Acm
A2 =
Non-Ideal Op Amp
Non-ideal characteristics
-+
•Finite input resistance
•Finite voltage gain
•Nonzero output resistance
•Output Saturation
•Maximum output current
Ibias
VOS
1/2IOS
•Input offset voltage, VOS
•Input bias current, Ibias
•Input offset current, IOS
Ro
rd
+aVd
+
Ibias
Vos - the difference in voltage between the Op Amp input terminals when the output
voltage is zero
Ibias - the average of the two input currents when the output voltage is zero
I bias =
1
2
(I
p
+ In )
IOS - the difference between the input currents
I OS = I n − I p
Non-Ideal Op Amp
R1
In
Rp
R2
Vn
From Ohm’s Law:
Vp = −I p Rp
+
Ip
+
Apply KCL at the inverting input:
Vn − 0 Vn − Eo
+
+ In = 0
R1
R2
Eo
-
Vp
Estimating the output error caused by the
input bias currents
By Op Amp action Vn = Vp ,
Eliminating Vn and Vp
 R 
Eo = 1 + 2  (R1 // R2 )I n − R p I p
 R1 
[
Rp can be specified to cancel the two terms in the brackets
R p = R1 // R 2
In this case, this reduces the output error to
 R 
Eo = 1 + 2 [(R1 // R2 )I OS ]
 R1 
]
Non-Ideal Op Amp
Upper saturation region:
Vo (V)
-
+
Vd
Linear region:
VSAT
+
VSAT
+
Vo
+
-
-
Vd
+aVd
+
+
+
Vo
-
VSATL/a
a
VSATH/a
Vd (µV)
Lower saturation region:
-
VSATL
Vd
+
+
+
VSAT
+
Vo
-
Non-Ideal Op Amp
Ex The 741 inverting amplifier is driven by a ±10 V peak to peak triangular wave.
Sketch and label Vi, Vo and Vn. If 741 is supplied with ±15 V and this maximum
output ±13 V.
10 V
R1
Vi
+
-
R2
10 kΩ
Vn
Vi
6.5V
20 kΩ
t
-
-6.5V
+
+
-10 V
Vo
-
13 V
Vo
-6.5V < Vi < 6.5V: the op amp is in the linear
region Vo =- 2Vi
Vi > 6.5V: the op amp is Saturation Vo = -13 V
Vn =
R2
R1
Vi +
VSATL
R1 + R2
R1 + R2
Vi < -6.5V: the op amp is Saturation Vo = 13 V
Vn =
R2
R1
Vi +
VSATH
R1 + R2
R1 + R2
-13 V
Vn
2.33 V
-2.33 V
t