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Erik Strand
pit
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1a1c4c73
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1a1c4c73
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6 years ago
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Erik Strand
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Answer 15.1
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@@ -9,22 +9,152 @@ title: Problem Set 12
{:.question}
Show that the circuits in Figures 15.1 and 15.2 differentiate, integrate, sum, and difference.
I'll take as given that op amps with negative feedback keep their inputs at the same voltage. I also
assume ideal op-amps that draw no current from their inputs. All the proofs follow from
[
Kirchhoff's
current law
](
https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws
)
(
aka
conservation of charge).
I write $$R_f$$ to refer to the resistor in the feedback path.
For the integrator,
$$
\b
egin{align
*
}
\f
rac{V_i}{R} &= - C
\f
rac{dV_o}{dt}
\\
\f
rac{dV_o}{dt} &=
\f
rac{-V_i}{R C}
\\
V_o &= -
\f
rac{1}{RC}
\i
nt V_i
\m
athrm{d}t
\e
nd{align
*
}
$$
For the differentiator,
$$
\b
egin{align
*
}
\f
rac{-V_o}{R} &= C
\f
rac{dV_i}{dt}
\\
V_o &= -RC
\f
rac{dV_i}{dt}
\e
nd{align
*
}
$$
For the summing amplifier,
$$
\b
egin{align
*
}
\f
rac{V_1 + V_2 + V_3}{R_i} &= -
\f
rac{V_o}{R_f}
\\
V_o &= -
\f
rac{R_f}{R_i} (V_1 + V_2 + V_3)
\e
nd{align
*
}
$$
For the difference amplifier, neither input terminal is grounded. Call their voltage $$V_i$$.
Applying Kirchhoff's current law to the non-inverting input,
$$
\b
egin{align
*
}
\f
rac{V_2 - V_i}{R_i} &=
\f
rac{V_i}{R_f}
\\
\f
rac{V_2}{R_i} &= V_i
\l
eft(
\f
rac{1}{R_i} +
\f
rac{1}{R_f}
\r
ight)
\e
nd{align
*
}
$$
Then looking at the inverting input,
$$
\b
egin{align
*
}
\f
rac{V_1 - V_i}{R_i} &=
\f
rac{V_i - V_o}{R_f}
\\
\f
rac{V_1}{R_i} -
\f
rac{V_o}{R_f} &= V_i
\l
eft(
\f
rac{1}{R_i} +
\f
rac{1}{R_f}
\r
ight)
\\
\f
rac{V_1}{R_i} -
\f
rac{V_o}{R_f} &=
\f
rac{V_2}{R_i}
\\
V_o &=
\f
rac{R_f}{R_i} (V_2 - V_1)
\e
nd{align
*
}
$$
### (b)
{:.question}
Design a non-inverting op-amp amplifier. Why are they used less commonly than inverting ones?
Connect the signal you want to amplify to the non-inverting input. Then create a voltage divider,
connected to ground on one end, the inverting input in the middle, and the output on the other end.
I'll refer to the resistor connected to ground as $$R_g$$, and the feedback resistor $$R_f$$ as
before.
$$
\b
egin{align
*
}
\f
rac{V_o - V_i}{R_f} &=
\f
rac{V_i}{R_g}
\\
\f
rac{V_o}{R_f} &= V_i
\f
rac{R_f + R_g}{R_f R_g}
\\
V_o &= V_i
\l
eft( 1 +
\f
rac{R_f}{R_g}
\r
ight)
\e
nd{align
*
}
$$
A potential downside is that the gain must be at least unity. A potential upside is that the input
impedance is very high. And, of course, the output isn't inverted.
### (c)
{:.question}
Design a transimpedance (voltage out proportional to current in) and a transconductance (current out
proportional to voltage in) op-amp circuit.
For a transimpedance amplifier, replace the input resistor in an inverting amplifier with a wire.
Then the current flowing through the feedback resistor is equal to the current at the input. So $$I
= V_o / R_f$$, and $$V_o = R_f I$$.
For a transconductance amplifier, replace the feedback resistor with a wire. Then the current out
is $$I = -V_i / R_i$$. If you want a non-inverting transconductance amplifier, connect the input
straight to the non-inverting terminal and connect the inverting terminal to ground via a resistor
$$R_g$$. Then $$I = V_i / R_g$$.
### (d)
{:.question}
Derive equation (15.16).
The currents flowing through $$R_1$$, $$R_o$$, and $$C$$ must be equal. Call this current $$I$$
(I'll take left to right to be positive for all currents). Because of the negative feedback, the
inverting input of the op-amp is a virtual ground. Recall that $$Q = CV$$, so $$dQ/dt = I = C
dV/dt$$. Thus we can express $$I$$ three ways:
$$
\b
egin{align
*
}
I &=
\f
rac{V_{PD}}{R_1}
\\
&= -
\f
rac{V_{RC}}{R_0}
\\
&= C
\l
eft(
\f
rac{dV_{RC}}{dt} -
\f
rac{dV_F}{dt}
\r
ight)
\e
nd{align
*
}
$$
I'm using $$V_{RC}$$ to denote the voltage at the junction between $$R_O$$ and $$C$$.
Differentiating the first two lines yields
$$
\b
egin{align
*
}
\f
rac{dI}{dt} &=
\f
rac{1}{R_1}
\f
rac{dV_{PD}}{dt}
\\
&= -
\f
rac{1}{R_0}
\f
rac{dV_{RC}}{dt}
\e
nd{align
*
}
$$
Thus
$$
\f
rac{dV_{RC}}{dt} = -
\f
rac{R_0}{R_1}
\f
rac{dV_{PD}}{dt}
$$
As such we can write the current through the capacitor as
$$
I = C
\l
eft( -
\f
rac{R_0}{R_1}
\f
rac{dV_{PD}}{dt} -
\f
rac{dV_F}{dt}
\r
ight)
$$
This can still be equated
$$
C
\l
eft( -
\f
rac{R_0}{R_1}
\f
rac{dV_{PD}}{dt} -
\f
rac{dV_F}{dt}
\r
ight) =
\f
rac{V_{PD}}{R_1}
$$
So all that remains is to solve for $$dV_F/dt$$.
$$
\f
rac{dV_F}{dt} =
\f
rac{V_{PD}}{R_1 C} +
\f
rac{R_0}{R_1}
\f
rac{dV_{PD}}{dt}
$$
This differs in sign from (15.16) because the book makes the positive current direction right to
left.
## (15.2)
{:.question}
...
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