SME
Electronics
Op-amp circuits
Use this for “An op-amp with GBW = 1 MHz has a gain of 20. Find its bandwidth.”
Diagram
The
closed-loop
gain
20
(26.0
dB)
meets
the
open-loop
line
at
1,000
20
=
50.0
kHz.
This is an example. Type your own values to start a new problem.
f3dB (kHz)
=
GBW (kHz)
G
fₘₐₓ (kHz)
=
SR (V/μs)
2π
×
Vₚ (V)
ffull
Full-swing bandwidth
calculated · kHz
Formulas
f3dB
=
GBW
G
50
=
1,000
20
fₘₐₓ
=
SR
×
1,000
2π × Vₚ
7.9577
=
0.5
×
1,000
2π × 10
ffull = min(f3dB, fₘₐₓ)
7.9577 = min(50, 7.9577)
Assumptions
•
A single-pole op-amp: above its low corner the open-loop gain is GBW ÷ f, falling 20 dB a decade.
•
G is the noise gain, 1 + Rf ÷ Rg; the closed-loop gain stays flat at G until it meets that line.
•
The slew rate (SR) is in volts per microsecond, so SR ÷ Vₚ is in MHz.
•
The smaller of f3dB and fₘₐₓ limits a full-swing sine.
Step-by-step
WE KNOW
GBW = 1,000 kHz, G = 20, SR = 0.5 V/μs, Vₚ = 10 V
FIND
bandwidth (f3dB), full-power frequency (fₘₐₓ), full-swing bandwidth (ffull)
Step 1 · Find bandwidth (f3dB)
Use f3dB = GBW ÷ G
The
flat
line
at
G
meets
the
open-loop
line
GBW
f
where
f
=
GBW
G
.
f3dB
=
1,000
20
f3dB = 50 kHz
Step 2 · Find full-power frequency (fₘₐₓ)
Use fₘₐₓ = SR × 1,000 ÷ (2π × Vₚ)
The
sine’s
steepest
slope
is
2πfVₚ
and
can’t
pass
SR;
V/μ
s
V
is
MHz,
×
1000
for
kHz.
fₘₐₓ
=
0.5
×
1,000
2π × 10
fₘₐₓ = 7.9577 kHz
Step 3 · Find full-swing bandwidth (ffull)
Use ffull = min(f3dB, fₘₐₓ)
A full swing needs both limits met, so the smaller one wins.
ffull = min(50, 7.9577)
ffull = 7.9577 kHz
Check
Put the numbers back in. Both sides should match.
50
=
1,000
20
✓
7.9577
=
0.5
×
1,000
2π × 10
✓
7.9577 = min(50, 7.9577) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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