SME
Electronics · Problem type
Active low-pass filter
Use this for “An inverting low-pass has R₁ = 10 kΩ, Rf = 100 kΩ and C = 1.59 nF. Find its gain and cutoff.”
Diagram
A
=
−
Rf
R₁
=
−100
kΩ
10
kΩ
=
−10
fc
=
1
2π·Rf·C
=
1.001
kHz
Below
fc
the
capacitor
is
nearly
open
and
the
gain
is
−
Rf
R₁
;
above
it
C
shorts
Rf
and
the
gain
falls
20
dB
a
decade.
This is an example. Type your own values to start a new problem.
A
=
−
Rf (kΩ)
R₁ (kΩ)
fc (kHz)
=
1
2π
×
Rf (kΩ)
×
C (nF)
G
Passband gain in decibels
calculated · dB
Formulas
A
=
−
Rf
R₁
−10
=
−
100
10
G
=
20
×
log₁₀(−
A
)
20 = 20 × log₁₀(−(−10))
fc
=
1,000
2π × Rf ×
C
1.001
=
1,000
2π × 100 × 1.59
Assumptions
•
Ideal op-amp: no current flows into either input, and v₊ = v₋ (the virtual short).
•
C sits across Rf: well below fc it is nearly open, and above fc it shorts Rf.
•
Rf in kΩ times C in nF gives RfC in μs, so 1000 ÷ (2πRfC) is in kHz.
Step-by-step
WE KNOW
R₁ = 10 kΩ, Rf = 100 kΩ, C = 1.59 nF
FIND
passband gain (A), passband gain in decibels (G), cutoff frequency (fc)
Step 1 · Find passband gain (A)
Use A = −Rf ÷ R₁
Below
the
cutoff
C
is
open,
leaving
an
inverting
amplifier:
minus
the
resistor
ratio.
A
=
−
100
10
A
=
−10
Step 2 · Find passband gain in decibels (G)
Use G = 20 × log₁₀(−A)
The size of the gain in decibels; the minus sign is the 180° phase.
G
=
20
×
log₁₀(−(−10))
G
=
20
×
log₁₀(10)
G
=
20
dB
Step 3 · Find cutoff frequency (fc)
Use fc = 1,000 ÷ (2π × Rf × C)
Rf
and
C
set
the
corner;
kΩ
×
nF
is
μs,
and
1000
turns
it
into
kHz.
fc
=
1,000
2π × 100 × 1.59
fc = 1.001 kHz
Check
Put the numbers back in. Both sides should match.
−10
=
−
100
10
✓
20 = 20 × log₁₀(−(−10)) ✓
1.001
=
1,000
2π × 100 × 1.59
✓
Topic check
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