Datasheet

LM4881
www.ti.com SNAS001D –SEPTEMBER 1997–REVISED MAY 2013
Once the power dissipation equations have been addressed, the required gain can be determined from
Equation 4.
(4)
A
V
= R
f
/R
i
(5)
From Equation 4, the minimum gain is: A
V
= 1.26
Since the desired input impedance was 20 kΩ, and with a gain of 1.26, a value of 27 kΩ is designated for R
f
,
assuming 5% tolerance resistors. This combination results in a nominal gain of 1.35. 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 assuming a single pole roll-off. As stated in the
External Components Description section, both R
i
in conjunction with C
i
, and C
o
with R
L
, create first order
highpass filters. Thus to obtain the desired frequency low response of 100 Hz within ±0.5 dB, both poles must be
taken into consideration. The combination of two single order filters at the same frequency forms a second order
response. This results in a signal which is down 0.34 dB at five times away from the single order filter −3 dB
point. Thus, a frequency of 20 Hz is used in the following equations to ensure that the response is better than 0.5
dB down at 100 Hz.
C
i
≥ 1 / (2π * 20 kΩ * 20 Hz) = 0.397 µF; use 0.39 µF. (6)
C
o
≥ 1 / (2π * 8Ω * 20 Hz) = 995 µF; use 1000 µF. (7)
The high frequency pole is determined by the product of the desired high frequency pole, f
H
, and the closed-loop
gain, A
V
. With a closed-loop gain of 1.35 and f
H
= 100 kHz, the resulting GBWP = 135 kHz which is much
smaller than the LM4881 GBWP of 18 MHz. This figure displays that if a designer has a need to design an
amplifier with a higher gain, the LM4881 can still be used without running into bandwidth limitations.
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