LM4862 datasheet(11/18 Pages) TI

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LM4862

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SNAS102F – MAY 1997 – REVISED MAY 2013

From Equation 5, the minimum AVD is 2; use AVD = 2.

Since the desired input impedance was 20 k

Ω, and with a AVD of 2, a ratio of 1:1 of Rf to Ri results in an

allocation of Ri = 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 –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 , Ri in conjunction with Ci

create a highpass filter.

Ci ≥ 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, fH, and the differential

gain, AVD. With an AVD = 2 and fH = 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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