
Application Information (Continued)
leads possess finite resistance, currents running through
them will cause finite voltage drops to exist. If two ground re-
turn lines tie into the same path at different points there will
be a voltage drop between them. The first figure below
shows a common ground example where the positive input
ground and the load ground are returned to the supply
ground point via the same wire. The addition of the finite wire
resistance, R
2, results in a voltage difference between the
two points as shown below.
The load current I
L will be much larger than input bias current
I
I, thus V1 will follow the output voltage directly, i.e. in phase.
Therefore the voltage appearing at the non-inverting input is
effectively positive feedback and the circuit may oscillate. If
there were only one device to worry about then the values of
R
1 and R2 would probably be small enough to be ignored;
however, several devices normally comprise a total system.
Any ground return of a separate device, whose output is in
phase, can feedback in a similar manner and cause instabili-
ties. Out of phase ground loops also are troublesome, caus-
ing unexpected gain and phase errors.
The solution to most ground loop problems is to always use
a single-point ground system, although this is sometimes im-
practical. The third figure below is an example of a
single-point ground system.
The single-point ground concept should be applied rigor-
ously to all components and all circuits when possible. Viola-
tions of single-point grounding are most common among
printed circuit board designs, since the circuit is surrounded
by large ground areas which invite the temptation to run a
device to the closest ground spot. As a final rule, make all
ground returns low resistance and low inductance by using
large wire and wide traces.
Occasionally, current in the output leads (which function as
antennas) can be coupled through the air to the amplifier in-
put, resulting in high-frequency oscillation. This normally
happens when the source impedance is high or the input
leads are long. The problem can be eliminated by placing a
small capacitor, C
C, (on the order of 50 pF to 500 pF) across
the LM2876 input terminals. Refer to the External Compo-
nents Description section relating to component interaction
with C
f.
REACTIVE LOADING
It is hard for most power amplifiers to drive highly capacitive
loads very effectively and normally results in oscillations or
ringing on the square wave response. If the output of the
LM2876 is connected directly to a capacitor with no series
resistance, the square wave response will exhibit ringing if
the capacitance is greater than about 0.2 F. If highly capaci-
tive loads are expected due to long speaker cables, a
method commonly employed to protect amplifiers from low
impedances at high frequencies is to couple to the load
through a 10
resistor in parallel with a 0.7 H inductor. The
inductor-resistor combination as shown in the Typical Appli-
cation Circuit isolates the feedback amplifier from the load
by providing high output impedance at high frequencies thus
allowing the 10
resistor to decouple the capacitive load and
reduce the Q of the series resonant circuit. The LR combina-
tion also provides low output impedance at low frequencies
thus shorting out the 10
resistor and allowing the amplifier
to drive the series RC load (large capacitive load due to long
speaker cables) directly.
GENERALIZED AUDIO POWER AMPLIFIER DESIGN
The system designer usually knows some of the following
parameters when starting an audio amplifier design:
Desired Power Output
Input Level
Input Impedance
Load Impedance
Maximum Supply Voltage
Bandwidth
The power output and load impedance determine the power
supply requirements, however, depending upon the applica-
tion some system designers may be limited to certain maxi-
mum supply voltages. If the designer does have a power
supply limitation, he should choose a practical load imped-
ance which would allow the amplifier to provide the desired
output power, keeping in mind the current limiting capabili-
ties of the device. In any case, the output signal swing and
current are found from (where P
O is the average output
power):
(5)
(6)
To determine the maximum supply voltage the following pa-
rameters must be considered. Add the dropout voltage (4V
for LM2876) to the peak output swing, V
opeak, to get the sup-
ply rail value (i.e. ± (V
opeak + Vod) at a current of Iopeak). The
regulation of the supply determines the unloaded voltage,
usually about 15% higher. 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 + Vod)(1 + regulation)(1.1)
(7)
The input sensitivity and the output power specs determine
the minimum required gain as depicted below:
(8)
DS011775-15
LM2876
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