LM4880 datasheet(10/12 Pages) NSC

10 / 12 page

Application Information (Continued)

(4)

V = −RF/Ri

(5)

From

Equation (4), the minimum gain is::

V = −1.26

Since the desired input impedance was 20 k

Ω, and with a

gain of −1.26, a value of 27 k

Ω is designated for R

f, assum-

ing 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 section,

both R

i in conjunction with Ci, and Co with RL, create first or-

der high pass 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 fre-

quency of 20 Hz is used in the following equations to ensure

that the response if better than 0.5 dB down at 100 Hz.

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

o ≥ 1/(2π*8Ω*20Hz) = 995 µF; use 1000 µF

The high frequency pole is determined by the product of the

desired high frequency pole, f

H, and the closed-loop gain,

V. With a closed-loop gain magnitude of 1.35 and fH = 100

kHz, the resulting GBWP = 135 kHz which is much smaller

than the LM4880 GBWP of 12.5 MHz. This figure displays

that if a designer has a need top design an amplifier with a

higher gain, the LM4880 can still be used without running

into bandwidth limitations.

LM4880 bảng dữ liệu(PDF) 10 Page - National Semiconductor (TI)