SME
Electronics · Problem type
Integrator output ramp
Use this for “An integrator has R = 10 kΩ and C = 1 μF. Find vₒᵤₜ 4 ms after 0.5 V is applied.”
Diagram
vₒᵤₜ
=
−
vᵢₙt
RC
vₒᵤₜ
=
−
0.5
V
×
4
ms
10 kΩ × 1 μF
vₒᵤₜ = −0.2 V
A
steady
input
pushes
a
steady
current
vᵢₙ
R
into
C,
so
the
output
ramps
down
in
a
straight
line.
Inside the rails: |vₒᵤₜ| ≤ 15 V.
This is an example. Type your own values to start a new problem.
vₒᵤₜ (V)
=
−
vᵢₙ (V)
×
t (ms)
R (kΩ)
×
C (μF)
Vₛₐₜ
Supply rail
example
Formulas
vₒᵤₜ
=
−vᵢₙ
×
t
R
×
C
−0.2
=
−0.5
×
4
10 × 1
Assumptions
•
Ideal op-amp: no current flows into either input, and v₊ = v₋ (the virtual short).
•
The capacitor starts uncharged, so the output starts at 0 V and the input stays constant.
•
R in kΩ times C in μF gives RC in ms.
•
The output stays inside the supply rails, ±Vₛₐₜ; past them it clips.
Step-by-step
WE KNOW
vᵢₙ = 0.5 V, R = 10 kΩ, C = 1 μF, t = 4 ms, Vₛₐₜ = 15 V
FIND
output voltage (vₒᵤₜ)
Step 1 · Find output voltage (vₒᵤₜ)
Use vₒᵤₜ = −vᵢₙ × t ÷ (R × C)
The
steady
current
vᵢₙ
R
charges
C
,
so
the
output
falls
in
a
straight
line.
vₒᵤₜ
=
−0.5
×
4
10 × 1
vₒᵤₜ
=
−
2
10
vₒᵤₜ = −0.2 V
Check
Put the numbers back in. Both sides should match.
−0.2
=
−0.5
×
4
10 × 1
✓
Topic check
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