LM4881MMSLASHNOPB datasheet(13/15 Pages) TI

13 / 15 page

LM4881

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SNAS001D – SEPTEMBER 1997 – REVISED MAY 2013

Once the power dissipation equations have been addressed, the required gain can be determined from

Equation 4.

(4)

AV = Rf/Ri

(5)

From Equation 4, the minimum gain is: AV = 1.26

Since the desired input impedance was 20 k

Ω, and with a gain of 1.26, a value of 27 kΩ is designated for Rf,

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 Ri in conjunction with C i, and Co with RL, 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.

Ci ≥ 1 / (2π * 20 kΩ * 20 Hz) = 0.397 µF; use 0.39 µF.

(6)

Co ≥ 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, fH, and the closed-loop

gain, A V. With a closed-loop gain of 1.35 and fH = 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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