LM388 The schematic shows that both inputs are biased to ground Application Circuit

Part No :
LM388 circuit
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LM388N-1  
National Semiconductor Corporation

find LM388 datasheet
LM388 circuit
Application Hints
(Continued)
is open If pins 2 and 6 are bypassed then R as low as 2 kX
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 kX Resistor The base current of the input transis-
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 kX it will contribute very little
additional offset (about 2 5 mV at the input 50 mV at the
output) If the dc source resistance is less than 10 kX then
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 kX Resistor between pins 2 and 6) it is necessary to
bypass the unused input preventing degradation of gain
and possible instabilities This is done with a 0 1
mF
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
a
R2 set the amount of base current available to the
output Transistor The maximum output current divided by
beta is the value required for the current in R1 and R2
(R1
a
R2)
e
b
O
(V
S
2)
b
V
BE
I
O MAX
Good design values are V
BE
e
0 7V and
b
O
e
100
Example 1 watt into 8X load with V
S
e
12V
I
O MAX
e
(R1
a
R2)
e
0
100

2 P
O
e
500 mA
R
L
(12 2)
b
0 7
e
1060X
05
J
To keep the current in R2 constant during positive swing
B
is added As the output swings positive C
B
lifts
R1 and R2 above the supply maintaining a constant voltage
across R2 To minimize the value of C
B
R1
e
R2 The pole
due to C
B
and R1 and R2 is usually set equal to the pole
due to the output coupling Capacitor and the load This
gives
C
B
j
C
4C
c
j
c
b
O
25
Example for 100 Hz pole and R
L
e
8X C
c
e
200
mF
and
C
B
e
8
mF
if R1 is made a Diode and R2 increased to give
the same current C
B
can be decreased by about a factor of
4 as in
Figure 4
For reduced component count the load can replace R1 The
value of (R1
a
R2) is the same so R2 is increased Now C
B
is both the coupling and the bootstrapping Capacitor (see
Figure 2
)
Typical Applications
TL H 7846 – 3
TL H 7846 – 4
FIGURE 1 Load Returned to Ground
(Amplifier with Gain
e
20)
FIGURE 2 Load Returned to V
S
(Amplifier with Gain
e
20)
4