LM4902LD datasheet(15/18 Pages) NSC

15 / 18 page

Application Information (Continued)

F/Ri =AVD/2

(6)

From Equation 5, the minimum A

VD is 1.55; use AVD =2.

Since the desired input impedance was 20 k

Ω, and with a

VD of 2, a ratio of 1:1 of RF to Ri results in an allocation of

i =RF =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 pole gives

0.17 dB down from passband response which is better than

the required ±0.25 dB specified.

L = 100Hz/5 = 20Hz

H = 20kHz x 5 = 100kHz

As stated in the External Components section, R

i in con-

junction with C

i create a highpass filter.

i ≥ 1/(2π*20 kΩ*20 Hz) = 0.397µF;

use 0.39µF

The high frequency pole is determined by the product of the

desired high frequency pole, f

H, and the differential gain,

VD. With a AVD = 2 and fH = 100kHz, the resulting GBWP

= 100kHz which is much smaller than the LM4902 GBWP of

25MHz. This figure displays that if a designer has a need to

design an amplifier with a higher differential gain, the

LM4902 can still be used without running into bandwidth

problems.

LM4902LD Datasheet(PDF) 15 Page - National Semiconductor (TI)