Datasheet

LM1876
SNAS097C –MAY 1999–REVISED APRIL 2013 www.ti.com
To achieve a transient free power-up and power-down, the voltage seen at the input terminals should be ideally
the same. Such a signal will be common-mode in nature, and will be rejected by the LM1876. In Figure 4, the
resistor R
INP
serves to keep the inputs at the same potential by limiting the voltage difference possible between
the two nodes. This should significantly reduce any type of turn-on pop, due to an uneven charging of the
amplifier inputs. This charging is based on a specific application loading and thus, the system designer may need
to adjust these values for optimal performance.
As shown in Figure 4, the resistors labeled R
BI
help bias up the LM1876 off the half-supply node at the emitter of
the 2N3904. But due to the input and output coupling capacitors in the circuit, along with the negative feedback,
there are two different values of R
BI
, namely 10 kΩ and 200 kΩ. These resistors bring up the inputs at the same
rate resulting in a popless turn-on. Adjusting these resistors values slightly may reduce pops resulting from
power supplies that ramp extremely quick or exhibit overshoot during system turn-on.
AUDIO POWER AMPLlFIER DESIGN
Design a 15W/8Ω Audio Amplifier
Given:
Power Output 15 Wrms
Load Impedance 8Ω
Input Level 1 Vrms(max)
Input Impedance 47 kΩ
Bandwidth 20 Hz−20 kHz ±0.25 dB
A designer must first determine the power supply requirements in terms of both voltage and current needed to
obtain the specified output power. V
OPEAK
can be determined from Equation 4 and I
OPEAK
from Equation 5.
(4)
(5)
To determine the maximum supply voltage the following conditions must be considered. Add the dropout voltage
to the peak output swing V
OPEAK
, to get the supply rail at a current of I
OPEAK
. The regulation of the supply
determines the unloaded voltage which is usually about 15% higher. The supply voltage will also rise 10% during
high line conditions. Therefore the maximum supply voltage is obtained from the following equation.
Max supplies ≈ ± (V
OPEAK
+ V
OD
) (1 + regulation) (1.1) (6)
For 15W of output power into an 8Ω load, the required V
OPEAK
is 15.49V. A minimum supply rail of 20.5V results
from adding V
OPEAK
and V
OD
. With regulation, the maximum supplies are ±26V and the required I
OPEAK
is 1.94A
from Equation 5. It should be noted that for a dual 15W amplifier into an 8Ω load the I
OPEAK
drawn from the
supplies is twice 1.94 Apk or 3.88 Apk. At this point it is a good idea to check the Power Output vs Supply
Voltage to ensure that the required output power is obtainable from the device while maintaining low THD+N. In
addition, the designer should verify that with the required power supply voltage and load impedance, that the
required heatsink value θ
SA
is feasible given system cost and size constraints. Once the heatsink issues have
been addressed, the required gain can be determined from Equation 7.
(7)
From Equation 7, the minimum A
V
is: A
V
≥ 11.
By selecting a gain of 21, and with a feedback resistor, R
f
= 20 kΩ, the value of R
i
follows from Equation 8.
R
i
= R
f
(A
V
− 1) (8)
Thus with R
i
= 1 kΩ a non-inverting gain of 21 will result. Since the desired input impedance was 47 kΩ, a value
of 47 kΩ was selected for R
IN
. 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
i
in conjunction with C
i
create
a high-pass filter.
C
i
≥ 1/(2π * 1 kΩ * 4 Hz) = 39.8 μF; use 39 μF. (9)
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