LM388 National Semiconductor, LM388 Datasheet - Page 4

• Manufacturer Part Number: LM388
• Description: LM388 1.5W Audio Power Amplifier
• Manufacturer: National Semiconductor

Application Hints
is open If pins 2 and 6 are bypassed then R as low as 2 k
can be used This restriction is because the amplifier is only
compensated for closed-loop gains greater than 9 V V
INPUT BIASING
The schematic shows that both inputs are biased to ground
with a 50 k
tors is about 250 nA so the inputs are at about 12 5 mV
when left open If the dc source resistance driving the
LM388 is higher than 250 k
additional offset (about 2 5 mV at the input 50 mV at the
output) If the dc source resistance is less than 10 k
shorting the unused input to ground will keep the offset low
(about 2 5 mV at the input 50 mV at the output) For dc
source resistances between these values we can eliminate
excess offset by putting a resistor from the unused input to
ground equal in value to the dc source resistance Of
course all offset problems are eliminated if the input is ca-
pacitively coupled
When using the LM388 with higher gains (bypassing the
1 35 k
bypass the unused input preventing degradation of gain
and possible instabilities This is done with a 0 1 F capaci-
tor or a short to ground depending on the dc source resist-
ance on the driven input
BOOTSTRAPPING
The base of the output transistor of the LM388 is brought
out to pin 9 for Bootstrapping The output stage of the am-
plifier during positive swing is shown in Figure 3 with its
external circuitry
R1
output transistor The maximum output current divided by
Typical Applications
a
R2 set the amount of base current available to the
resistor between pins 2 and 6) it is necessary to
FIGURE 1 Load Returned to Ground
resistor The base current of the input transis-
(Amplifier with Gain
(Continued)
it will contribute very little
e
20)
TL H 7846– 3
then
4
Figure 2 )
beta is the value required for the current in R1 and R2
Good design values are V
Example 1 watt into 8
To keep the current in R2 constant during positive swing
capacitor C
R1 and R2 above the supply maintaining a constant voltage
across R2 To minimize the value of C
due to C
due to the output coupling capacitor and the load This
gives
Example for 100 Hz pole and R
C
the same current C
4 as in Figure 4
For reduced component count the load can replace R1 The
value of (R1
is both the coupling and the bootstrapping capacitor (see
B
e
8 F if R1 is made a diode and R2 increased to give
(R1
B
and R1 and R2 is usually set equal to the pole
B
a
a
FIGURE 2 Load Returned to V
is added As the output swings positive C
(R1
R2)
I
R2) is the same so R2 is increased Now C
(Amplifier with Gain
O MAX
a
e
B
R2)
100
C
can be decreased by about a factor of
e
B
load with V
j
e
BE
(12 2)
2 P
4C
e
R
O
O
L
c
O
0 7V and
(V
L
0 5
j
e
S
e
b
S
I
C
25
O MAX
2)
500 mA
c
e
0 7
8
e
B
b
12V
R1
20)
C
V
O
e
BE
c
e
e
e
S
1060
R2 The pole
100
200 F and
TL H 7846 – 4
B
lifts
B