Block Diagrams
ECE 2210
Western Michigan University
John Stahl
Scaler
The circuit is an inverting amplifier
which scales the input X.
We start by analyzing the circuit
and solve for the output-input
relationship.
The circuit multiplies the input by
the ratio of the resistors and
inverts the value.
The circuit will be represented as a
block we can use without having to
draw the circuit every time.
Y
0−𝑌 𝑋−0
=
𝑅𝑓
𝑅1
𝑌=−
X
𝑅𝑓
∙𝑋
𝑅1
𝑅𝑓
−
𝑅1
Y
Adder
The circuit is an
inverting adder. If
R1 = R2 equal then
the circuit adds two
voltages. The ratio
of the resistors
scales the final the
sum.
𝑋−0 𝑌−0 0−𝐸
+
=
𝑅1
𝑅2
𝑅𝑓
𝑅1 = 𝑅2 = 𝑅
𝑅𝑓
𝐸=−
(𝑋 + 𝑌)
𝑅
X
+
Y
−
𝑅𝑓
𝑅
E
Subtractor
The circuit is an inverting
subtractor. Let the resistors be:
𝑅2
𝑅𝐵
𝑅2
1+
∙
=
𝑅1 𝑅𝐵 + 𝑅𝐴 𝑅1
So the we can rewrite the
𝑋 − 𝑉𝑎 𝑉𝑎 − 𝐸
equation to subtract two
=
𝑅1
𝑅2
voltages. This works if all resistors
𝑅𝐵
are set to 1k.
𝑉𝑎 =
X
-
Y
𝑅2
−
𝑅1
E
𝑅𝐵 + 𝑅𝐴
∙𝑌
𝑅2
𝑅2
𝑅𝐵
−𝐸 =
∙ 𝑋 − (1 + ) ∙
∙𝑌
𝑅1
𝑅1 𝑅𝐵 + 𝑅𝐴
𝑅2
𝐸 = − ∙ (𝑋 − 𝑌)
𝑅1
Integrator
The circuit takes a voltage X and
integrates the value along with
scaling the output.
Practically, the speed of the
integration depends on the time
constant 𝜏=𝑅𝐶. The time needed
to finish integrating is
approximately 4t.
𝑋−0
𝑑𝑉𝑐
𝑖=
=C
𝑅
𝑑𝑡
𝑉𝑐 = 0 − 𝑌
𝑋
𝑑𝑌
−
=C
𝑅𝐶
𝑑𝑡
X
∫
−
1
𝑅𝐶
Y
𝑌=−
1
𝑅𝐶
𝑋 ∙ 𝑑𝑡
Differentiator
The circuit outputs the derivative
of X and scales the result by –RC.
𝑌−0
𝑑𝑉𝑐
𝑖=
=C
𝑅
𝑑𝑡
𝑉𝑐 = 𝑋 − 0
𝑑𝑋
𝑌 = −RC
𝑑𝑡
X
𝑑
𝑑𝑡
−𝑅𝐶
Y
𝑑𝑋
𝑌 = −RC
𝑑𝑡