Datasheet

LM4880
SNAS103C –NOVEMBER 1995–REVISED MAY 2013 www.ti.com
In addition to system cost and size, click and pop performance is effected by the size of the input coupling
capacitor, C
i
. A larger input coupling capacitor requires more charge to reach its quiescent DC voltage (normally
1/2 V
DD
.) This charge comes from the output via the feedback and is apt to create pops upon device enable.
Thus, by minimizing the capacitor size based on necessary low frequency response, turn-on pops can be
minimized.
Besides minimizing the input and output capacitor sizes, careful consideration should be paid to the bypass
capacitor size. The bypass capacitor, C
B
, is the most critical component to minimize turn-on pops since it
determines how fast the LM4880 turns on. The slower the LM4880's outputs ramp to their quiescent DC voltage
(nominally 1/2 V
DD
), the smaller the turn-on pop. Choosing C
B
equal to 1.0 μF along with a small value of C
i
(in
the range of 0.1 μF to 0.39 μF), should produce a virtually clickless and popless shutdown function. While the
device will function properly, (no oscillations or motorboating), with C
B
equal to 0.1 μF, the device will be much
more susceptible to turn-on clicks and pops. Thus, a value of C
B
equal to 1.0 μF or larger is recommended in all
but the most cost sensitive designs.
AUDIO POWER AMPLIFIER DESIGN
Design a Dual 200 mW/8Ω Audio Amplifier
Given:
Power Output: 200 mWrms
Load Impedance: 8Ω
Input Level: 1 Vrms (max)
Input Impedance: 20 kΩ
Bandwidth: 100 Hz–20 kHz ± 0.50 dB
A designer must first determine the needed supply rail to obtain the specified output power. Calculating the
required supply rail involves knowing two parameters, V
opeak
and also the dropout voltage. As shown in Typical
Performance Characteristics, the dropout voltage is typically 0.5V. V
opeak
can be determined from Equation 3.
(3)
For 200 mW of output power into an 8Ω load, the required V
opeak
is 1.79V. Since this is a single supply
application, the minimum supply voltage is twice the sum of V
opeak
and V
od
. Since 5V is a standard supply
voltage in most applications, it is chosen for the supply rail. Extra supply voltage creates headroom that allows
the LM4880 to reproduce peaks in excess of 200 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. Remember that the maximum power dissipation value from Equation 1
must be multiplied by two since there are two independent amplifiers inside the package.
Once the power dissipation equations have been addressed, the required gain can be determined from
Equation 4.
(4)
A
V
= −R
F
/R
i
(5)
From Equation 4, the minimum gain is: A
V
= −1.26
Since the desired input impedance was 20 kΩ, and with a gain of −1.26, a value of 27 kΩ is designated for R
f
,
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
External Components Description, both R
i
in conjunction with C
i
, and C
o
with R
L
, create first order 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 frequency of 20 Hz is used in the following equations to ensure that the response if better than 0.5
dB down at 100 Hz.
C
i
≥ 1/(2π*20kΩ*20Hz) = 0.397 μF; use 0.39 μF
12 Submit Documentation Feedback Copyright © 1995–2013, Texas Instruments Incorporated
Product Folder Links: LM4880