LM391N-100 National Semiconductor, LM391N-100 Datasheet - Page 8

• Manufacturer Part Number: LM391N-100
• Description: IC AMP AUDIO PWR MONO AB 16DIP
• Manufacturer: National Semiconductor
• Type: Class ABr

Specifications of LM391N-100

• Output Type: 1-Channel (Mono)
• Features: Shutdown, Thermal Protection
• Mounting Type: Through Hole
• Package / Case: 16-DIP (0.300", 7.62mm)
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Voltage - Supply: -
• Max Output Power X Channels @ Load: -
• Other names: *LM391N-100; LM391N100

Application Hints
A 20W 8
Givens
Equations (1) and (2) give
Therefore the supply required is
With 15% regulation and high line we get
tion (3)
Sensitivity and equation (4) set minimum gain
We will use a gain of 20 with resulting sensitivity of 632 mV
Letting R
For low DC offsets at the output we let R
for R
The bandwidth requirement must be stated as a pole i e
the 3 dB frequency Five times away from a pole gives 0 17
dB down which is better than the required 0 25 dB There-
fore
Power Output
Input Sensitivity
Input Impedance
Bandwidth
20W 8
30W 4
g
g
23V
21V
f 1
gives
IN
2 24A reducing to
3 87A
equal 100k gives the required input impedance
30W 4
R
f 1
V
V
OP
OP
f
e
h
A
e
V t
20
e
e
100k
AMPLIFIER
f
20k
L
17 9V
15 5V
b
R
e
20
f 2
1
c
20
1
5
e
e
c
5
(Continued)
e
100k
8
5 26k use 5 1k
e
e
4 Hz
I
I
100 kHz
20 Hz–20 kHz
OP
OP
12 65
e
e
2 24A
3 87A
f 2
g
e
29V from equa-
100k Solving
20W into 8
30W into 4
g
0 25 dB
1V Max
100k
8
Solving for C
The recommended value for C
larger This gives a gain-bandwidth product of 6 4 MHz and
a resulting bandwidth of 320 kHz better than required
The breakdown voltage requirement is set by the maximum
supply we need a minimum of 58V and will use 60V We
must now select a 60V power transistor with reasonable
beta at I
are 60V 60W transistors with a minimum beta of 30 at 4A
The driver transistor must supply the base drive given 5 mA
drive from the LM391 The MJE711 MJE721 complementa-
ry driver transistors are 60V devices with a minimum beta of
40 at 200 mA The driver transistors should be much faster
(higher f
on the output will prevent instability
To find the heat sink required for each output transistor we
use equations (7) (9) and (10)
If both transistors are mounted on one heat sink the thermal
resistance should be halved to 2 4 C W
The maximum average power dissipation in each driver is
found using equation (8)
Using equation (9)
JA s
Opeak
T
150 C
) than the output transistors to insure that the R-C
C
f
SA s
f t
P
3 87A The TIP42 TIP41 complementary pair
12
DRIVER(MAX)
b
JA s
2 R
55 C
P
7 9
D
1
155
f 1
e
b
e
f
L
0 4 (30)
0 4
2 1
b
e
7 9 C W for T
e
55
7 8 F use 10 F
b
C
12
30
1 0
e
e
is 5 pF for gains of 20 or
e
237 C W
e
12W
400 mW
4 8 C W
AMAX
e
55 C
(10)
(7)
(9)