Datasheet

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v d O L IN o rm s in rm s

A 2 P R / V V / Vt 

LM4860

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SNAS096C –AUGUST 1994–REVISED MAY 2013

For 500 mW of output power into an 8Ω load, the required V

opeak

is 2.83V. A minimum supply rail of 3.53V results

from adding V

opeak

and V

. But 3.53V is not a standard voltage that exists in many applications and for this

reason, a supply rail of 5V is designated. Extra supply voltage creates dynamic headroom that allows the

LM4860 to reproduce peaks in excess of 500 mW without clipping the signal. At this time, the designer must

make sure that the power supply choice along with the output impedance does not violate the conditions

explained in POWER DISSIPATION.

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

from Equation 6.

(6)

From Equation 6, the minimum A

is: A

= 2

Since the desired input impedance was 20 kΩ, and with an A

of 2, a ratio of 1:1 of R

to R

results in an

allocation of R

= R

= 20 kΩ. Since the A

was less than 10, a feedback capacitor is not needed. 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 4 Hz and 100 kHz respectively. As

stated in External Components Description , R

in conjunction with C

create a highpass filter.

≥ 1/(2π * 20 kΩ * 4 Hz) = 1.98 μF; use 2.2 μF. (7)

The high frequency pole is determined by the product of the desired high frequency pole, f

, and the differential

gain, A

. With a A

= 2 and f

= 100 kHz, the resulting GBWP = 100 kHz which is much smaller than the

LM4860 GBWP of 7 MHz. This figure displays that if a designer has a need to design an amplifier with a higher

differential gain, the LM4860 can still be used without running into bandwidth problems.

Product Folder Links: LM4860