2/23/2011
section 2_8 Integrators and Differentiators
1/2
2.8 Integrators and Differentiators
Reading Assignment: 105-113
Op-amp circuits can also (and often do) implement reactive
elements such as inductors and capacitors.
HO: OP-AMP CIRCUITS WITH REACTIVE ELEMENTS
One important op-amp circuit is the inverting differentiator.
HO: THE INVERTING DIFFERENTIATOR
Likewise the inverting integrator.
HO: THE INVERTING INTEGRATOR
HO: AN APPLICATION OF THE INVERTING INTEGRATOR
Let’s do some examples of op-amp circuit analysis with reactive
elements.
EXAMPLE: A NON-INVERTING NETWORK
EXAMPLE: AN INVERTING NETWORK
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
section 2_8 Integrators and Differentiators
2/2
EXAMPLE: ANOTHER INVERTING NETWORK
EXAMPLE: A COMPLEX PROCESSING CIRCUIT
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
1/9
Op-Amp circuits with
reactive elements
Now let’s consider the case where the op-amp circuit includes reactive
elements:
i2 (t)
R2
C
vin(t)
v-
-
i1 (t)
ideal
v+
vout (t)
+
Q: Yikes! How do we analyze this?
A: Don’t panic! Remember, the relationship between vout and vin is linear, so we
can express the output as a convolution:
vout (t ) = L ⎣⎡vin (t ) ⎦⎤ =
Jim Stiles
t
∫ g (t − t ′) vin (t ′) dt ′
−∞
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
2/9
Just find the Eigen value
Q: I’m still panicking—how do we determine the impulse response g (t ) of this
circuit?
A: Say the input voltage vin (t ) is an Eigen function of linear, time-invariant
systems:
vin (t ) = e s t = e (
σ +j ω ) t
= eσ te j ω t
Then, the output voltage is just a scaled version of this input:
vout (t ) = L ⎡⎣e
− st
⎤=
⎦
t
∫ g (t − t ′) e
− st
dt ′ = G ( s ) e −st
−∞
where the “scaling factor” G ( s ) is the complex Eigen value of the linear
operator L .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
3/9
Express the input as a superposition of
eigen values (i.e., the Laplace transform)
Q: First of all, how could the input (and output) be this complex function e s t ?
Voltages are real-valued!
A: True, but the real-valued input and output functions can be expressed as a
weighted superposition of these complex Eigen functions!
+∞
vin ( s ) = ∫ vin (t ) e −s t dt
0
The Laplace transformÆ
+∞
vout ( s ) = ∫ vout (t ) e −s t dt
0
Such that:
vout ( s ) = G ( s )vin ( s )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
4/9
Find the eigen value from
circuit theory and impedance
Q: Still, I don’t know how to find the eigen value G ( s ) !
A: Remember, we can find G ( s ) by analyzing the circuit using the Eigen value of
each linear circuit element—a value we know as complex impedance!
v (s )
= Z (s )
i (s )
Jim Stiles
+
v (s )
−
Z (s )
i (s )
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
5/9
For example
For example, consider this amplifier in with the inverting configuration, where
the resistors have been replaced with complex impedances:
Z2(s)
Z1(s)
vin (s)
i2 (s)
0
v-
-
ideal
i1 (s)
0
v+
oc
vout
(s )
+
oc
vout
(s )
What is the open-circuit voltage gain Avo (s ) =
?
vin (s )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
6/9
The eigen value of this linear operator
From KCL:
i1 (s ) = i2 (s )
Z2(s)
Since v − (s ) = 0 , we find from Ohm’s Law :
i1 (s ) =
vin (s )
Z 1 (s )
Z1(s)
And also from Ohm’s Law:
vin (s)
oc
−vout
(s )
i2 (s ) =
Z 2 (s )
i1 (s)
i2 (s)
v-
-
ideal
v+
oc
vout
(s )
+
Equating the last two expressions:
oc
vin (s ) −vout
(s )
=
Z 1(s )
Z 2(s )
Rearranging, we find the open-circuit voltage gain:
oc
vout
(s )
Z (s )
Avo (s ) =
=− 2
vin (s )
Z 1(s )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
7/9
The result passes the sanity check
Note that this complex voltage gain Avo (s ) is the Eigen value G ( s ) of the linear
operator relating vin (t ) and vout (t ) :
vout (t ) = L ⎡⎣vin (t ) ⎤⎦
Note also that if the impedances Z 1(s ) and Z 2(s ) are real valued (i.e., they’re
resistors!):
Z 1(s ) = R1 + j 0
and
Z 2(s ) = R2 + j 0
Then the voltage gain simplifies to the familiar:
oc
vout
(s )
R
=− 2
Avo (s ) =
vin (s )
R1
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
8/9
Or, we can use the Fourier transform
Now, recall that the variable s is a complex frequency:
s = σ + jω .
If we set σ = 0 , then s = j ω , and the functions Z (s ) and Avo (s ) in the Laplace
domain can be written in the frequency (i.e., Fourier) domain!
Avo (ω ) = Avo (s ) σ = 0
And therefore, for the inverting configuration:
oc
vout
(ω )
Z (ω )
=− 2
Avo (ω ) =
vin (ω )
Z 1(ω )
Z 2(ω )
Z 1(ω )
vin (ω )
-
ideal
oc
vout
(ω )
+
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
Op amp circuits with reactive elements lecture
9/9
For the non-inverting
Likewise, for the non-inverting configuration, we find:
oc
vout
(ω )
Z (ω )
=1+ 2
Avo (ω ) =
vin (ω )
Z 1(ω )
oc
vout
(s )
Z (s )
=1+ 2
Avo (s ) =
vin (s )
Z 1(s )
Z 2(ω )
Z 1(ω )
-
ideal
vin (ω )
Jim Stiles
oc
vout
(ω )
+
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Differentiator lecture
1/8
The Inverting Differentiator
The circuit shown below is the inverting differentiator.
R
vin (s)
C
-
ideal
oc
vout
(s)
+
Since the circuit uses the inverting configuration, we can conclude that the
circuit transfer function is:
oc
vout
(s )
Z (s )
=− 2
G (s ) =
vin (s )
Z 1(s )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Differentiator lecture
2/8
Know the impedance; know the answer
For the capacitor, we know that its complex impedance is:
Z 1(s ) =
1
sC
And the complex impedance of the resistor is simply the real value:
Z 2(s ) = R
oc
Thus, the eigen value of the linear operator relating vin (t ) to vout
(t ) is:
G (s ) = −
Z 2(s )
R
=−
= −s RC
1
Z 1(s )
sC
In other words, the (Laplace transformed) output signal is related to the
(Laplace transformed) input signal as:
oc
vout
(s ) = − s (RC ) vin (s )
From our knowledge of Laplace Transforms, we know this means that the output
signal is proportional to the derivative of the input signal!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Differentiator lecture
3/8
Converting back to time domain
Taking the inverse Laplace Transform, we find:
oc
(t ) = −RC
vout
d vin (t )
dt
For example, if the input is:
vin (t ) = sinωt
then the output is:
d vin (t)
dt
d sinωt
= −RC
dt
= −ω RC cosωt
oc
(t ) = −RC
vout
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Differentiator lecture
4/8
Or, with Fourier analysis
We likewise could have determined this result using Fourier analysis (i.e.,
frequency domain):
oc
vout
(ω )
Z (ω )
R
=− 2
=−
= − j ω RC
G (ω ) =
vin (ω )
Z 1(ω )
1
jωC
(
)
Thus, the magnitude of the transfer function is:
G (ω ) = − jω RC
= ω RC
And since:
(
)
(
−j π
− j = e ( 2) = cos − π 2 + j sin − π 2
)
the phase of the transfer function is:
∠G (ω ) = − π
2
= −90D
Jim Stiles
radians
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Differentiator lecture
5/8
Look at the magnitude and phase
So given that:
oc
vout
(ω ) = G (ω ) vin (ω )
and:
oc
∠vout
(ω ) = ∠G (ω ) + ∠vin (ω )
we find for the input:
vin (t ) = sin ωt
where:
vin (ω ) = 1
and
∠vin (ω ) = 0
that the output of the inverting differentiator is:
oc
vout
(ω ) = G (ω ) vin (ω ) = ω RC
and:
oc
∠vout
(ω ) = ∠G (ω ) + ∠vin (ω ) = −90D + 0 = −90D
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Differentiator lecture
6/8
The result is the same!
Therefore, the output is:
(
oc
vout
(t ) = ω RC sin ωt − 90D
)
= −ω RC cosωt
Exactly the same result as before (using Laplace trasforms)!
If you are still unconvinced that this circuit is a differentiator, consider this
time-domain analysis.
i2 (t)
R
+ vc -
vin (t)
i1 (t)
C
v0
ideal
v+
Jim Stiles
oc
vout
(t )
+
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Differentiator lecture
7/8
Let’s do a time-domain analysis
From our elementary circuits
knowledge, we know that the
current through a capacitor
(i1(t)) is:
d vc (t )
i1(t ) = C
dt
R
+ vc -
vin (t)
i1 (t)
C
v0
-
ideal
v+
and from the circuit we see from
KVL that:
i 2 (t)
oc
vout
(t )
+
vc (t ) = vin (t ) − v − (t ) = vin (t )
therefore the input current is:
i1(t ) = C
Jim Stiles
d vin (t )
dt
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Differentiator lecture
8/8
Laplace, Fourier, time-domain:
the result it the same!
From KCL, we likewise know that:
i1 (t ) = i2 (t )
and from Ohm’s Law:
oc
oc
v1(t ) − vout
(t )
vout
(t )
i2(t ) =
=−
R
R
Combining the two previous equations:
oc
vout
(t ) = −i1(t ) R
R
+ vc -
vin (t)
i1 (t)
and thus:
C
v0
⎛
oc
(t ) = −i1(t ) R = − ⎜ C
vout
⎝
d vin (t ) ⎞
d vin (t )
R
RC
=
−
dt ⎟⎠
dt
-
ideal
v+
i2 (t)
oc
vout
(t )
+
The same result as before!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
1/8
The Inverting Integrator
The circuit shown below is the inverting integrator.
C
i2 (s)
vin (s)
R
i1 (s)
Jim Stiles
v-
-
ideal
v+
oc
vout
(s )
+
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
2/8
It’s the inverting configuration!
Since the circuit uses the inverting configuration, we can conclude that the
circuit transfer function is:
oc
(1 s C ) = −1
vout
(s )
Z (s )
=− 2
=−
G (s ) =
vin (s )
Z 1 (s )
R
s RC
In other words, the output signal is related to the input as:
oc
vout
(s ) =
−1 vin (s )
RC
s
From our knowledge of Laplace Transforms, we know this means that the output
signal is proportional to the integral of the input signal!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
3/8
The circuit integrates the input
Taking the inverse Laplace Transform, we find:
v (t ) =
oc
out
−1
RC
t
∫vin (t ′) dt ′
0
For example, if the input is:
vin (t ) = sinωt
then the output is:
v
Jim Stiles
oc
out
(t ) =
−1
RC
t
∫ sinωt dt ′ =
0
−1 −1
RC ω
cosωt =
The Univ. of Kansas
1
ω RC
cosωt
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
4/8
Or, in the Fourier domain
We likewise could have determined this result using Fourier Analysis (i.e.,
frequency domain):
oc
1 jω C )
(
vout
(ω )
Z 2(ω )
j
G (ω ) =
=−
=−
=
vin (ω )
Z 1(ω )
R
ω RC
Thus, the magnitude of the transfer function is:
G (ω ) =
And since:
j
1
=
ω RC
ω RC
( )
( )
j π
j = e ( 2) = cos π 2 + j sin π 2
the phase of the transfer function is:
∠G (ω ) = π
Jim Stiles
2
radians = 90D
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
5/8
Magnitude and phase
Given that:
oc
vout
(ω ) = G (ω ) vin (ω )
and:
oc
∠vout
(ω ) = ∠G (ω ) + ∠vin (ω )
we find for the input:
where:
vin (t ) = sinωt
vin (ω ) = 1
and
∠vin (ω ) = 0
that the output of the inverting integrator is:
oc
(ω ) = G (ω ) vin (ω ) =
vout
and:
Jim Stiles
1
ω RC
oc
∠vout
(ω ) = ∠G (ω ) + ∠vin (ω ) = 90D + 0 = 90D
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
6/8
See, it’s an integrator
Therefore:
oc
vout
(t ) =
=
1
ω RC
1
ω RC
(
sin ωt + 90D
)
cosωt
Exactly the same result as before!
If you are still unconvinced that this circuit is an integrator, consider this timedomain analysis.
i ( t)
2
C
+ vc -
vin(t)
R
i 1 ( t)
Jim Stiles
v-
-
i− = 0
v+
ideal
oc
vout
(t )
+
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
7/8
The time-domain solution
From our elementary circuits
knowledge, we know that the voltage
across a capacitor is:
vc (t ) =
1
t
∫ i (t ′) dt ′
C
2
0
and from the circuit we see that:
oc
oc
vc (t ) = v −(t ) − vout
(t ) = −vout
(t )
i 2 ( t)
therefore the output voltage is:
v (t ) = −
oc
out
1
C
t
∫ i2(t ′) dt ′
0
vin(t)
+ vc -
R
i 1 ( t)
Jim Stiles
C
The Univ. of Kansas
v-
-
i− = 0
v+
ideal
oc
vout
(t )
+
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
8/8
The same result no matter how we do it!
From KCL, we likewise know that:
i1(t ) = i2(t )
and from Ohm’s Law:
i1(t ) =
vin (t ) − v −(t ) vin (t )
=
R1
R1
i 2 ( t)
C
+ vc -
Therefore:
i2 (t ) =
vin (t )
R1
vin(t)
R
i 1 ( t)
and thus:
v (t ) =
oc
out
=
−1
C
−1
RC
t
∫ i (t ′) dt ′
2
v-
-
i− = 0
v+
ideal
oc
vout
(t )
+
0
t
∫vin (t ′) dt ′
0
The same result as before!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
An Application of the Inverting Integrator lecture
1/4
An Application of the Inverting
Integrator
Note the time average of a signal v (t) over some arbitrary time T
mathematically stated as:
average of v (t ) v (t ) =
1
T
is
T
∫v (t ) dt
0
Note that this is exactly the form of the output of an op-amp integrator!
We can use the inverting integrator to determine the time-averaged value of
some input signal v (t) over some arbitrary time T.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
An Application of the Inverting Integrator lecture
2/4
Make sure you see this!
For example, say we wish to determine the time-averaged value of the input
signal:
vin(t)
5
0
1
2
3
4
t
-5
I.E.,
0 <t < 2
2 <t < 3
t >3
⎧ 5
⎪
vin (t ) = ⎨−5
⎪ 0
⎩
The time average of this function over a period from 0 < t < T=3 is therefore:
3
1
5
vin (t ) = ∫ vin (t ) dt =
30
3
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
An Application of the Inverting Integrator lecture
3/4
This better make sense to you!
We could likewise determine this average using an inverting integrator. We
select a resistor R and a capacitor C such that the product RC = 3 seconds.
The output of this integrator would be:
⎧ 5t
⎪− 3
⎪
t
−1
⎪ 5t − 20
′
′
vout (t ) =
v
t
dt
=
(
)
⎨
in
3 ∫0
⎪ 3
⎪− 5
⎪ 3
⎩
vout (t )
0
−
−
Jim Stiles
1
2
3
0 <t < 2
2 <t < 3
t >3
4
t
5
3
10
3
The Univ. of Kansas
Dept. of EECS
2/23/2011
An Application of the Inverting Integrator lecture
4/4
We must sample a the correct time!
Note that the value of the output voltage at t =3 is:
3
−1
5
′
′
(
)
vout (t = 3) =
v
t
dt
=
−
in
3 ∫0
3
The time-averaged value (times –1)!
Thus, we can use the inverting integrator, along with a voltage sampler (e.g., A to
D converter) to determine the time-averaged value of a function over some
time period T.
vin (t)
vo (t)
t =T=RC
vout (t = T ) = −vout (t )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example An Inverting Network lecture
1/3
Example: An Inverting Network
Now let’s determine the complex transfer function of this circuit:
R2
C
vin
R1
v-
-
ideal
v+
Jim Stiles
oc
vout
+
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example An Inverting Network lecture
2/3
It’s the inverting configuration!
Note this circuit uses the inverting configuration, so that:
G (ω ) = −
Z 2 (ω )
Z 1 (ω )
where Z 1 = R1 , and:
Z2 = R2 1
jωC
=
R2
1 + jωR2C
Therefore, the transfer function of this circuit is:
oc
vout
(ω )
R
1
=− 2
G (ω ) =
vin (ω )
R1 1 + j ωR2C
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example An Inverting Network lecture
3/3
Another low-pass filter
Thus, the transfer function magnitude is:
2
⎛ R ⎞
G (ω ) = ⎜ − 2 ⎟
⎝ R1 ⎠
2
1
2
⎛
⎞
1 + ⎜ω
⎟
⎝ ω0 ⎠
where:
ω0 =
1
R2C
Thus, just as with the previous example, this circuit is a low-pass filter, with
cutoff frequency ω 0 and pass-band gain (R2 R1 ) .
2
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example A NonInverting Network lecture
1/4
Example: A NonInverting Network
oc
(ω ) vin (ω ) for the following
Let’s determine the transfer function G (ω ) = vout
circuit:
R2
R1
v-
vin
Jim Stiles
ideal
R3
i3
-
v+
i+=0
C
oc
vout
+
iC
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example A NonInverting Network lecture
2/4
Some enjoyable circuit analysis
From KCL, we know:
i3(ω ) = iC (ω ) + i+(ω ) = iC (ω ) + 0 = iC (ω )
where:
i3(ω ) =
vin (ω ) − v +(ω )
R3
and
iC (ω ) =
Equating, we find an expression
involving vin (ω ) and v2(ω ) only:
v +(ω ) − 0
⎛ 1
⎞
⎜ j ωC ⎟
⎝
⎠
= j ωC v +(ω )
R2
R1
vin (ω ) − v +(ω )
= j ωC v +(ω )
R3
and performing a little algebra, we
find:
v2(ω ) =
Jim Stiles
vin (ω )
1 + j ωR3C
v-
vin
ideal
R3
i3
The Univ. of Kansas
-
v+
i+=0
C
oc
vout
+
iC
Dept. of EECS
2/28/2011
Example A NonInverting Network lecture
3/4
No need to go further:
we have a template!
The remainder of the circuit is simply the non-inverting amplifier that we
studied earlier.
We know that:
⎛
oc
vout
(ω ) = ⎜ 1 +
⎝
R2 ⎞
⎟ v (ω )
R1 ⎠ +
Combining these two relationships, we
can determine the complex transfer
function for this circuit:
vin
vR3
v +( ω )
-
ideal
oc
vout
+
C
⎞
v (ω ) ⎛ R2 ⎞ ⎛
1
G (ω ) = out
= ⎜1 + ⎟ ⎜
⎟
vin (ω ) ⎝ R1 ⎠ ⎝ 1 + j ωR3C ⎠
Jim Stiles
R2
R1
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example A NonInverting Network lecture
4/4
It’s a low-pass filter!!!
The magnitude of this transfer function is therefore:
2
⎛
R ⎞
G (ω ) = ⎜ 1 + 2 ⎟
R1 ⎠
⎝
2
where:
ω0 =
1
2
⎛
⎞
1 + ⎜ω
⎟
⎝ ω0 ⎠
1
R3C
This is a low-pass filter—one with pass-band gain!
2
G (ω ) (dB)
⎛
⎜1 +
⎝
R2
R1
⎞
⎟
⎠
2
ω0
Jim Stiles
The Univ. of Kansas
logω
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
1/10
Example: Another
Inverting Network
Consider now the transfer function of this circuit:
R2
i2
vin
R3
i3
Jim Stiles
v3
iC
C
R1
v-
-
i1
v+
ideal
oc
vout
+
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
2/10
Some more enjoyable circuit analysis
To accomplish this analysis, we must first…
Wait! You don’t need to explain this to me.
It is obvious that we can divide this is circuit into
two pieces—the first being a complex voltage
divider and the second a non-inverting amplifier.
R2
i2
vin
R3
i3
Jim Stiles
v3
C
v3
R1
v-
i1
v+
iC
The Univ. of Kansas
-
ideal
oc
vout
+
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
3/10
Can we analyze the circuit this way?
The transfer function of the complex voltage divider is :
1
v3(ω )
jωC
=
vin (ω ) R + 1
3
jωC
=
1
1 + j ωR3C
and that of the inverting amplifier:
oc
vout
(ω )
R
=− 2
v3(ω )
R1
And so of course I have correctly
determined that the transfer
function of this circuit is:
oc
oc
vout
(ω ) vout
(ω ) v3(ω )
R
1
=
=− 2
vin (ω )
v3(ω ) vin (ω )
R1 1 + j ωR3C
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
4/10
No, we cannot
NO! This is not correct:
R
vo (ω )
1
≠− 2
vi (ω )
R1 1 + j ωR3C
The problem with the above “analysis” is that we cannot apply this complex
voltage divider equation to determine v3(ω ) :
1
v3(ω ) ≠
R3 +
jωC
1
vin (ω )
jωC
The reason of course is that the output of this voltage divider is not opencircuited, and thus current i3(ω ) ≠ iC (ω ) .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
5/10
My computer suspiciously crashed
while writing this (really, it did!)
We cannot divide this circuit into two independent pieces, we must analyze it as
one circuit.
R2
i2
vin
R3
i3
v3
C
R1
i1
iC
vv+
-
ideal
oc
vout
+
Of course what I meant to say was that we should
determine the impedance Z1 of input network, and
then use the inverting configuration equation
T (ω ) = − Z 2 Z 1 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
6/10
An even worse idea than Vista
NO! This idea is as bad as the last one!
We cannot specify an impedance for the input network:
vin
R3
i3
R1
v3
C
v-
i1
iC
After all, would we define this impedance as:
Z1 =
Jim Stiles
vin − v −
i3
or
Z1 =
vin − v −
i1
The Univ. of Kansas
???
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
7/10
Don’t look for templates:
trust what you know
R2
So, there is no easy or direct way to solve
this circuit, we must consult Mr. Kirchoff and
his laws!
vin
R3
i3
R1
v3
C
i2
v-
i1
v+
iC
-
ideal
vout
+
We know that i1 = i2 , where:
i1 =
v3 − v − v3
=
R1
R1
i2 =
and
v + − vout −vout
=
R2
R2
Combining these equations, we get the expected result:
vout = −
Jim Stiles
R2
v3
R1
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
8/10
Don’t forget virtual ground!
We must therefore determine v3 in terms of vi :
R3
vin
i3
R1
v3
i1
C
iC
virtual
ground
Note R1 and C are connected in parallel!
Thus, from voltage division, we find:
v3 =
R1
1
j ωC
R3 + ⎛⎜ R1 1 j ωC ⎞⎟
⎝
⎠
vin
where:
R1
Jim Stiles
1
R1
(
1
j ωC
)=
j ωC = R + 1
1
j ωC
R1
1 + j ω R1C
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
9/10
The Eigen value at last!
Performing some algebra, we find:
⎛
R1
v3 = ⎜⎜
⎝ (R1 + R3 ) + j ωR1R3C
and since:
vout =
−R2
R1
⎞
⎟⎟ vin
⎠
v3
we finally discover that:
G (ω ) =
Jim Stiles
vout (ω ) ⎛⎜
−R2
= ⎜⎜
⎜⎝ ( R1 + R3 ) + j ωR1R3C
vin (ω )
The Univ. of Kansas
⎞⎟
⎟⎟
⎠⎟
Dept. of EECS
2/28/2011
Example Another Inverting Network lecture
10/10
This again is a low-pass filter
We can rearrange this transfer function to find that this circuit is likewise a
low-pass filter with pass-band gain:
⎛
v (ω )
−R2 ⎜⎜
1
G (ω ) = out
=
⎜⎜
vin (ω )
R1 + R3 ⎜ 1 + j ( ω
⎝
ωo
)
⎞⎟
⎟⎟
⎟
⎟⎠⎟
where the cutoff frequency ω0 is:
ω0 =
I wish I had a
nickel for every
time my software
has crashed—oh
wait, I do!
Jim Stiles
1
⎛ R1R3 ⎞⎟
⎜⎜
⎟C
⎜⎝ R1 + R3 ⎠⎟
=
1
( R1 R3 )C
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example A Complex Processing Circuit lecture
1/2
Example: A Complex
Processing Circuit using the
Inverting Configuration
Note that we can combine inverting amplifiers to form a more complex
processing system.
For example, say we wish to take three input signals v1 (t ), v2 (t ), and v3 (t ) , and
process them such that the open-circuit output voltage is:
t
vout (t ) = 5v1(t ) + ∫ v2(t ′) dt ′ +
−∞
d v3(t )
dt
Assuming that we use ideal (or near ideal) op-amps, with an output resistance
equal to zero (or at least very small), we can realize the above signal processor
with the following circuit:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/28/2011
Example A Complex Processing Circuit lecture
2/2
This circuit performs this operation!
5K
v1 (t )
1K
-
10K
+
10K
10 μF
v2 (t )
100K
-
10K
vout (t )
-
+
+
100K
v3 (t )
100K
t
-
10K
vout (t ) = 5v1(t ) + ∫ v2(t ′) dt ′ +
−∞
d v3(t )
dt
+
Jim Stiles
The Univ. of Kansas
Dept. of EECS