Negative Feedback with Capacitors
http://gaussmarkov.net
January 8, 2008
The web page Op-Amps 4: Negative Feedback, describes several characteristics of the gain of a circuit like this one. This document provides the mathematical derivations of these characteristics.
1
Subcircuits
First, each resistor-capacitor pair is described separately. This gives expressions
for V1 and V2 , the voltages across each pair as indicated in the schematic above.
Second, these voltages are combined to find the output and gain of the op-amp.
In the section that follows this one, the expression for gain is summarized in
terms of its maximum, symmetry, and corner frequencies.
1
1.1
Resistor and Capacitor in Parallel
A resistance R 1 and capacitance C 1 in parallel have the
voltage V1 across their leads so that
V1 = I R R 1
dV1
dt
I1 = I R + IC
I C = C1
by Ohm’s law for resistors,
by the capacitor law,
by Kirchoff’s current law,
where I R is the current through the resistor, I C is the current through the capacitor, and I 1 is the current through
the parallel combination. These combine to give the relationship between current I 1 and voltage V1 :
I 1 R 1 = V1 + τ1
(1)
dV1
dt
where τ1 = R 1 C 1 .
1.2
Resistor and Capacitor in Series
A resistance R 2 and a capacitance C 2 in series have the
same current I 2 so that
VR = I 2 R 2
(2)
dVC
I 2 = C2
dt
V2 = VC + VR
by Ohm’s law for resistors,
by the capacitor law,
by Kirchoff’s voltage law,
where VR is the voltage across the resistor, VC is the voltage across the capacitor, and V2 is the voltage across the
pair. These combine to give V2 in terms of the voltage VC
(3)
V2 = VC + τ2
dVC
dt
where τ2 = R 2 C 2 .
2
1.3
Op-Amp with Divided Negative Feedback
Consider an op-amp with split supply and divided negative feedback where R 1 and C 2 are in parallel between the
output and the inverting input and R 2 and C 2 are in series from the inverting input to ground. Then, in addition
to the relationships above,1
V2 = Vin− ≈ Vin+ = Vsource
op-amp with feedback
I1 = I2
Kirchoff’s current law
Vout = V1 + V2
Kirchoff’s voltage law
Now we can describe a solution strategy.
1. Vsource determines V2 .
2. V2 will determine I 2 through the combined behaviour of R2 and C2.
3. I 2 equals I 1 .
4. I 1 will determine V1 through the combined behaviour of R1 and C1.
5. Finally, V1 and V2 sum to give Vout .
Steps 2 and 4 will use the fact that the stationary solution to the differential
equation
d y( t)
a cos ω t + b sin ω t = y( t) + c
dt
is
y=
(4)
2
¡
¢
1
(a − bcω) cos tω + ( b + acω) sin tω .
2
2
1+ω c
Circuit Output
Step 1: Following the standard approach, let the input signal be a simple sine
wave so that
(5)
V2 = Vin+ = sin ω t
where ω is an abbreviation for 2π f where f is the frequency.
1 When V
out is not clipped by the power rails, then
¡
¢
Vout = 200, 000 × Vin+ − Vin−
implies that
Vin+ − Vin− =
Vout
≈ 0.
200, 000
3
Step 2: Then (3) gives the differential equation for VC
sin ω t = VC + τ2
dVC
dt
which has the stationary solution (using equation 4)
VC =
1
1 + ω2 τ22
(sin ω t − ωτ2 cos ω t) .
Therefore, using (2), the current through this capacitor is
dVC
dt
C2 ω
I 2 = C2
=
1 + ω2 τ22
(cos ω t + ωτ2 sin ω t) .
Steps 3 and 4: Kirchoff ’s current law also implies that I 2 = I 1 or, using (1),
ρτ2 ω
1 + ω2 τ22
(cos ω t + ωτ2 sin ω t) = V1 + τ1
dV1
dt
where ρ = R 1 /R 2 . Again using (4), the solution to this differential equation is
³¡
´
¢
ρτ2 ω
2
¢
¡
¢
(6)
V1 = ¡
1
−
ω
τ
τ
cos
t
ω
+
τ
+
τ
ω
sin
t
ω
.
(
)
1
2
1
2
1 + ω2 τ21 1 + ω2 τ22
Step 5: These solutions for V1 and V2 , equations (6) and (5), allow us to solve for
Vout :
Vout = V1 + V2
³¡
´
¢
ρτ2 ω
2
¢
¢
¡
1
−
ω
τ
τ
cos
t
ω
+
τ
+
τ
ω
sin
t
ω
+ sin ω t
=¡
(
)
1
2
1
2
1 + ω2 τ21 1 + ω2 τ22
Because we specified an input with an amplitude equal to one, the squared gain
of the circuit is the squared amplitude of Vout . The squared amplitude of Vout
equals the sum of the squared coefficients of the sin and cos terms.
!2
¡
¢ !2 Ã
ρτ2 ω 1 − ω2 τ1 τ2
(τ1 + τ2 ) ρτ2 ω2
¢¡
¢ + 1+ ¡
¢¡
¢
(gain) = ¡
1 + ω2 τ21 1 + ω2 τ22
1 + ω2 τ21 1 + ω2 τ22
¡
¢
2 (τ1 + τ2 ) + ρτ2 ρτ2 ω2
¡
¢
¡
¢ .
= 1+
1 + ω2 τ21 1 + ω2 τ22
Ã
(7)
(8)
2
Like the special case of a resistive voltage divider, this gain is greater than one.
As a function of frequency f = ω/(2π), the square root of (8) is the gain plotted
by LTSpice in this graphic, generated with the component values in the schematics above:
4
There are several things about this plot that hold for every gain-frequency
plot no matter what the component values are:
1. one maximum
2. unity gain at the lowest and highest frequencies
3. symmetry in log-frequency around the maximum
In addition, for component values typically found in stompbox circuits,
4. the maximum gains is approximately equal to the no-capacitor gain and
5. lower and upper corner frequencies summarize this shape.
The next section verifies these properties.
3
3.1
Properties of the Gain Function
Extreme Values
Consider the part of gain that depends on ω:
ω2
1
³
¡
¢¡
¢=
¡
¢−1 τ2 τ1 ´ .
2
2
2
2
1 + ω τ1 1 + ω τ 2
τ1 τ 2 ω 2 τ1 τ 2 + ω 2 τ 1 τ2
+ τ1 + τ2
Because the function
x + x−1 ,
x>0
has a unique global minimum equal to 2 at x = 1, this part of the gain function
has a unique global maximum over values of ω at
s
1
2
ω ∗ τ 1 τ2 = 1
⇐⇒
ω∗ =
τ1 τ2
5
so that the maximum squared gain is
¡
¢
µ
¶2
2 (τ1 + τ2 ) + ρτ2 ρτ2
τ2
1+
=
1
+
ρ
τ1 + τ 2
(τ1 + τ2 )2
at the frequency
1
f∗ =
2π
s
1
τ1 τ 2
.
This maximum
¡
¢2 is slightly modified from the pure resistive case where squared
gain is 1 + ρ for all frequencies.
The minima of the gain profile occur at zero and infinite frequencies where
the gain is unity because
lim
x→0+
3.2
1
1
= lim
= 0.
x + x−1 x→∞ x + x−1
Symmetry
Written in terms of log-frequencies, the gain function is symmetric around its
maximum because
ax + (ax)−1 = 10 z−b + 10−( z−b) = 10| z−b| + 10−| z−b|
where x = 10 z and a = 10−b .
3.3
Corner Frequencies
The corner frequencies are easier to study when you notice that they are very
nearly the corner frequencies for two special cases: when one or the other capacitor is not present. Here are two plots to show the idea. In the first one, the
green line shows the gain function when there is no C1 capacitor and the blue
line shows the gain functions when there is no C2.
6
Without C1, the output looks like something like a high pass filter. The difference
is that the highs are amplified rather than the lows being attenuated. Without
C2, we get more lows than highs. In both cases, there is range of frequencies
where the capacitor that is present ceases to have an appreciable effect. For
those frequencies the gain equals the gain for the resistive voltage divider.
The second plot adds a red line for the gain with both C1 and C2. The red line
lies almost exactly on the lower value of the two other lines. For the component
values that we are using, there are no frequencies where both capacitors are
effective. This is the reason that the red line agrees with parts of the green and
blue lines.
As a result, we can use corner frequencies of the blue and green lines to accurately approximate the corner frequencies of the red line. We will derive the two
special cases as limits of the general case.
When C 1 → 0, then it is as though there is no C1 cap. When C 2 → ∞, then it
is as though there is no C2 cap. These two cases correspond to τ1 → 0 and τ2 → 0,
respectively. The squared gain limits are
¡
¢
¡
¢
ρ ρ +2
2 (τ1 + τ2 ) + ρτ2 ρτ2 ω2
¢¡
¢ = 1+
lim 1 + ¡
τ 1 →0
1 + ω2 τ21 1 + ω2 τ22
1 + 1/ (ωτ2 )2
and
Ã
lim
τ2 →∞
!
¡
¢
¡
¢
2 (τ1 + τ2 ) + ρτ2 ρτ2
ρ ρ +2
¢¡
¢ = 1+
1+ ¡
.
1/ω + τ21 ω 1/ω + τ22 ω
1 + (ωτ1 )2
The maximum squared gain limits are
µ
lim 1 + ρ
τ 1 →0
and
µ
lim 1 + ρ
τ2 →∞
τ2
¶2
τ1 + τ2
τ2
τ1 + τ2
7
¶2
¡
¢2
= 1+ρ
¡
¢2
= 1+ρ .
In both cases, the maximum gain simplifies to the simpler, no-capacitor case.
The corner frequency for τ1 = 0 is given by the frequency that solves
ρ ρ +2
1 + ( ¡ 2 )2 ¢
1+1/ ω τ2
1
=
¡
¢2
2
1+ρ
Naturally, a solution exists only if the maximum gain is greater than 2 (or ρ > 1).
Using ω = 2π f ,
s
2
1
1− ¡
f2 =
¢2
2πτ2
1+ρ
Similarly, for the τ2 → ∞ case,
ρ (ρ +2)
1+(ωτ1 )2
1 + ρ2
1+
=
1
2
gives
f1 =
1
1
.
r
2πτ1
1− 2 2
(1+ρ )
The passband is the range of frequencies between these two corner frequencies. As ρ → 1+ , f 2 → 0 and f 1 → ∞ and the passband approaches all frequencies.
For ρ > 5, the factor depending on ρ can be ignored so that f 2 ≈ 1/ (2πτ2 ) and
f 1 ≈ 1/ (2πτ1 ). This corresponds to 14 dB gain.
These approximations to the corner frequencies work well as long as there is
an interval in which both capacitors have little effect. Such an interval contains
frequencies for which gain is approximately the resistive divider gain:
¡
¢
¡
¢
¡
¢2
ρ ρ +2
ρ ρ +2
1+ρ ≈ 1+
, 1+
.
1 + 1/ (ωτ2 )2
1 + (ωτ1 )2
We can solve these approximations as equalities to find values for the angular
frequency ω that attain a gain of β(1 + ρ )2 :
¡
¢−2
β− 1+ρ
2
¡
¢ ,
ω2 =
τ22 1 − β
¡
¢
1−β
1
2
ω1 =
¡
¢ .
τ21 β − 1 + ρ −2
We require ω2 ≤ ω1 which implies
τ1
τ2
≤
1−β
¡
¢−2 .
β− 1+ρ
Taking β ≈ 1, this requires
³
¡
¢−2 ´ τ1
≤ 1−β
1− 1+ρ
τ2
8
or
³
¡
¢−2 ´ f 2
1− 1+ρ
≤ 1 − β.
f1
9