ECE 3336
Introduction to Circuits & Electronics
Set #14
MORE on Operational Amplifiers
Fall 2012,
TUE&TH 5:30-7:00 pm
Dr. Wanda Wosik
1
Basics of Operational Amplifiers
Noninverting Case
We will focus on operational amplifiers, specifically on
•
Ideal Operational Amplifiers, definitions and requirements for
their ideal operation in noninverting configuration
•
Negative Feedback that allows for op-amp to be controlled by
external elements
2
Solving Op Amp Circuits
As for inverting configuration we will have two assumptions for the
analysis and design. We will again treat the op amps as ideal circuits.
We will again call these assumptions golden rules.
• The first assumption: i- = i+ = 0.
results from large resistances at
the inputs. Currents do not flow
into the op-amp.
• The second assumption v+≈vdeals with the output that makes
the input voltages equal v+≈v-.
This is realized by introducing
negative feedback loop, which
spans the output and the inverting
input.
iin=0A
negative feedback loop
3
A Note on the Second Assumption
The second golden rule v- = v+ results in the virtual short, or the summingpoint constraint. The constrain refers to the input voltages, which become the
same v- = v+ if there is the negative feedback and the open loop gain Av(OL) is
large.
Without negative feedback, even a small input voltage will cause
saturation of the output either at V+ or V-. That depends on the sign of vin.
Negative dc
power supply
NO NEGATIVE FEEDBACK yet
Inverting
Input
Output
Noninverting
Input
This is open loop
configuration
+ dc V supply
4
Op Amp Circuits with the Negative Feedback Loop
Negative feedback adds a portion of the output signal to the inverting
input. Since the signs of these voltages are opposite, the negative
feedback acts as if the signal applied to the input decreases.
The net result is that the output voltage can be controlled by the
external elements and does not saturate.
Negative feedback
For ideal op-amps we will apply two golden
rules to solve circuits
Golden Rules
1) i- = i+ = 0.
2) v- = v+. Virtual short
ideal
5
Op Amp in the Non-inverting Configuration
An op amp operates in the noninverting configuration when the input voltage is
applied to the noninverting terminal.
These comments are
identical as for the
inverting configuration
RF is the feedback resistor
Rs is the source resistor
ideal
•There is a negative
feedback thanks to RF
Av(OL)≈∞
•Negative feedback
gives the virtual short:
v-=v+. Since v+=Vs also
v-=Vs.
•The op-amp does not
draw currents iin=0A
6
Solving op-amp in the
Non-Inverting Configuration Closed Loop
As earlier, to find vout we have to find vRF.
To find vRF we have to know current iF which can be calculated from is.
The current is is given by the voltage v-=Vs and Rs.
Since we have golden
rules (iin=0, v+=v-)
is = iF
v+=vS
v- v S
is =
=
Rs Rs
i in = 0
-
0A
0A
v+=vS
ideal
v out - v
v out - v S
iF =
=
RF
RF
v S v out - v S
=
RS
RF
Av(OL)≈∞
R F v S = R S (v out - v S )
(R F + R S )v S = R S v out
Closed loop voltage GAIN:
vout
RF
= 1+
vs
RS
7
Significance of the Closed Loop Gain
The negative feedback loop, combined with ideal properties of the opamp (high open loop gain 105-107 and large input resistance)
ensures that
• the gain does not depend on the op amp
• the gain is the determined by a ratio of two resistors
connected to the op-amp.
v out
RF
= 1+
vs
Rs
No phase change
ideal
8
Voltage Follower
•
Important application of the noninverting configuration is obtained when
there is no resistance in the negative feedback loop.
RF=0Ω
Golden rules apply:
v+=v- and iin=0A
So, the voltage at the input is the same as the
voltage at the output vout=vS.
Do we gain anything here by doing that?
VS
ideal
VS
We do! We have a very large
input resistance of this circuit:
•Such op-amps do not show loading effects
(i.e. voltage drop due to low resistance
connected to an output of a circuit).
•They work as voltage follower but they also act
as impedance buffers.
9
The Differential Amplifier
• This is a combination of inverting and noninverting configuration.
As earlier we have negative feedback and the op-amp is ideal.
v1 - v +
i1 =
R1
i2=-i1
v-=v+
iin=0
v out - v - v out - v +
i2 =
=
R2
R2
vout
R2
R2
R1
= [-v1 +
v2 +
v2 ]
R1
R1 + R2
R1 + R2
Group and arrange:
v+ =
R2
v2
R1 + R2
v out
R2
= (v 2 - v1 )
R1
10
Instrumentation Amplifier (IA)
IA are made as integrated circuits
Now use the results from
differential op-amp
vout 2 - vout1 = -iR1 (2R2 + R1 )
vout1
iin=0
v -v
iR1 = - 2 1
R1
v out
iin=0
v1
iR1
v2
iin=0
v out 2 - v out1 = (v 2 - v1 )(
vout2
v out =
Advantages:
RF
=
(v out 2 - v out1 )
R
2R2 + R1
2R
) = (v 2 - v1 )(1+ 2 )
R1
R1
RF
2R
(v 2 - v1 )(1 + 2 )
R
R1
Very high input resistance
Very high common-mode-rejection-ratio CMMR (goal: CMMR  for perfectly matched
resistors. That results in vout≈0V for v1=v2)
11
Integrator
The integrating circuit was used earlier
v o (t) =
1
RC
ò v (t)dt
i
Now we add the op-amp and we get an integrator. It also constitutes a part of an
analog computer
The Golden Rules are used for the op-amp
iS (t) = -iF (t)
is (t ) =
iF (t) = CF
vs (t )
Rs
dv out (t)
dt
1
dv (t)
v s (t) = - out
RS CF
dt
Now we integrate both sides and we
have the integrator
Virtual
short
1
vout (t) = RSCF
t
ò v (t ')dt '
S
-¥
12
Differentiator
The differentiating circuit was used earlier
vo(t) = RC
dvi (t)
dt
Now we add the op-amp and we get a differentiator. It also constitutes a part of an
analog computer.
The Golden Rules are used for the op-amp
iS (t) = -iF (t)
v (t)
iF (t) = out
RF
iS (t) = CS
dv S (t)
dt
v out (t) = -RS CF
dv s (t)
dt
13
Active Filters
Vout ( jw )
Z
=- F
VS ( jw )
ZS
The concept of frequency dependence
of the signals seen in the filters
(remember that we had |H(j)|max=1 for
those filters) is here combined with the
signal amplification.
•We will use here the negative feedback
configuration
•We will also use impedances instead of
resistors
Vout ( jw )
Z
= 1+ F
VS ( jw )
ZS
We still have the same golden rules:
• no input currents (high Rin)
• virtual short
14
Active Low-Pass Filter
The voltage gain ALP is calculated using Golden Rules
ZF
Z
=- F
ZS
RS
1
RF
Z F = RF ||
=jwCF
1+ jwCF RF
A LP ( jw ) = -
ALP ( jw ) = 0V
Amplification
ZF
RF /RS
R /R
==- F S
w
ZS
1+ jwCF RF
1+ j
Cutoff frequency
1
w0 =
R FC F
w0
RF
w
- 20 log 1 + ( ) 2
RS
w0
Amplification
So the cutoff
| A LP ( jw ) |dB = 20 log
| A LP ( jw 0 ) |dB = 20 log
RF
- 20 log 2
RS
-3dB
frequency is
also the 3dB
frequency (as
before)
w
)
w0
ÐA LP ( jw ) = 0°- tan -1 (
Phase is just like for the simple filter
15
Negative feedback
Inverting configuration
Active High-Pass Filter
The voltage gain calculated using Golden Rules
A HP ( jw ) = -
ZF
R
= -- F
ZS
ZS
Z S = RS +
1
jwCS RS +1
=
jwCS
jwCS
w
Z
jwC S R F R S
jwC S R S
w0
A HP ( jw ) = - F = ×
=× RF / RS = ×R /R
w F S
ZS
1 + jwC S R S R S
1 + jwC S R S
1+ j
w0
j
cutoff
1
w0 =
RSC S
RF
w
w
+ 20 log - 20 log 1 + ( ) 2
RS
w0
w0
Amplification
R
w
| A HP ( jw 0 ) |dB = 20 log F + 20 log 0 - 20 log 2
RS
w0
| A HP ( jw ) |dB = 20 log
-3dB
3dB frequency
w
)
w0
Phase: ÐA HP ( jw ) = 90°- tan -1 (
Phase is just like for the simple high pass filter
16
Op-Amp as a Level Shifter
A useful circuit to adjust DC voltage level = to remove the DC offset from the signal
v s (t) = 1.8 + 0.1cos(wt)
Use the superposition principle (one source at a time)
220kΩ
v ou t = 10kΩ
Power supply
Potentiometer
RF
R
v sensor + (1 + F )V ref
Rs
Rs
inverting
noninverting
RF
R
[1.8 + 0.1cos(wt)] + (1 + F )V ref =
Rs
Rs
R
R
R
= - F [ 0.1cos(wt)] - F 1.8 + (1 + F )V ref
Rs
Rs
Rs
v ou t = -
We can design such
precision voltage sources
using Rp
We want this to be equal 0V
That gives Vref=1.714V
17
Negative feedback
Inverting configuration
Active Band-Pass Filter
The voltage gain ABP is again calculated using Golden Rules
jwC S R S + 1
1
ZF
Z
=
R
+
=
A BP ( jw ) = S
S
jwC S
jwC S
ZS
Z F = R F ||
RF / RS
1
=jwC F
1+ jwC F R F
1
Three characteristic frequencies
w1 =
w
R FC S
j
Z
jwC S R F
w1
A BP ( jw ) = - F = =1
w
w
ZS
(1 + jwC F R F )(1 + jwC S R S )
w HP =
(1 + j
)(1 + j
)
RSC S
w LP
w HP
1
w LP =
R FC F
Magnitude of ABP
| A BP ( jw ) |dB = 20 log
@1
Relations between
the frequencies
w
w 2
w 2
- 20 log 1 + (
) - 20 log 1 + (
)
w1
w LP
w HP
1 RS
R
= w LP S
RF CS RS
RF
C
1 CF
w1 =
= w HP F
R FC S C F
CS
w1 =
w
w
w
| A BP ( jw1 ) |dB = 20 log 1 - 20 log 1 + ( 1 ) 2 - 20 log 1 + ( 1 ) 2 = 0dB
w1
w LP
w HP
1 is the unity
gain frequency
18
Characteristic Frequencies in the Band-Pass Filters
The voltage gain has 3 characteristic frequencies: 1, LP and HP
w
w
× RF / RS
j
×C F /C S
jwC S R F × R S / R S
w HP
w LP
A BP ( jw ) = ==w
w
w
w
(1 + jwC F R F )(1 + jwC S R S )
(1 + j
)(1 + j
)
(1 + j
)(1 + j
)
w LP
w HP
w LP
w HP
j
=0
Gain around LP
| A BP ( jw LP ) |dB = 20 log
Gain around HP
=0
CF
w
w
w
+ 20 log LP - 20 log 1 + ( LP ) 2 - 20 log 1 + ( LP ) 2
CS
w LP
w LP
w HP
-3dB
| A BP ( jw HP ) |dB = 20 log
RF
w
w
w
+ 20 log HP - 20 log 1 + ( HP ) 2 - 20 log 1 + ( HP ) 2
RS
w LP
w LP
w HP
Cancel off 0
-3dB
So LP and HP are 3dB frequencies while 1 is the unity gain frequency
19
Bode Plots for the Active Band-Pass Filter
We can plot the magnitude of the voltage gain as a function of frequency
Linear scale
dB scale
LP
1
LP
HP
HP
1
Relations between the frequencies
1 RS
R
= w LP S
RF CS RS
RF
C
1 CF
w1 =
= w HP F
R FC S C F
CS
w1 =
45°
-45°
LP
HP
The phase is like for simple bandpass filters
w
w
) - tan -1 (
)
w LP
w HP
ÐA HP ( jw ) = 90°- tan -1 (
20
Limitations of the Op-Amps
Saturation of the voltage at the output occurs at about ±Vs.
Small signals at the input are required
21
Limitations of the Op-Amps
Frequency Response Limits refer to the voltage gain of the open loop
and closed loop configuration
Open loop gain decreases very
quickly with frequency
AV (OL ) ( jw ) =
A0
1+ j
w
w0
The voltage gain decreases in
the closed loop configuration
but the cutoff frequency
increases
The gain-bandwidth product is
constant K
A0w0 = A1w1 = A2w2 = K
22
Limitations of the Op-Amps
Slew rate limitation of op-amp means that the op-amp output voltage
does not respond with the same slope as the input signal
Increasing frequency means faster
changing or steeper slopes at the zero
crossing
Slew rate is limited by the
frequency and amplitude product
dvou t (t)
= wA = S0
dt max
As the result of limited slew there
is a distortion of the output signal.
23