Datasheet

LM4862
www.ti.com SNAS102F –MAY 1997–REVISED MAY 2013
From Equation 5, the minimum A
VD
is 2; use A
VD
= 2.
Since the desired input impedance was 20 kΩ, and with a A
VD
of 2, a ratio of 1:1 of R
f
to R
i
results in an
allocation of R
i
= R
f
= 20 kΩ. The final design step is to address the bandwidth requirements which must be
stated as a pair of −3 dB frequency points. Five times away from a –3 dB point is 0.17 dB down from passband
response which is better than the required ±0.25 dB specified. This fact results in a low and high frequency pole
of 20 Hz and 100 kHz respectively. As stated in External Components Description , R
i
in conjunction with C
i
create a highpass filter.
C
i
≥ 1/(2π*20 kΩ*20 Hz) = 0.397 μF; use 0.39 μF. (7)
The high frequency pole is determined by the product of the desired high frequency pole, f
H
, and the differential
gain, A
VD
. With an A
VD
= 2 and f
H
= 100 kHz, the resulting GBWP = 100 kHz which is much smaller than the
LM4862 GBWP of 12.5 MHz. This figure displays that if a designer has a need to design an amplifier with a
higher differential gain, the LM4862 can still be used without running into bandwidth problems.
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