LM391 National Semiconductor Corporation, LM391 Datasheet - Page 10

• Manufacturer Part Number: LM391
• Description: Audio Power Driver (discontinued)
• Manufacturer: National Semiconductor Corporation

Application Hints
A 40W 8
Given
Equations (1) and (2) give
Therefore the supply required is
With 15% regulation and high line we get
equation (3)
The minimum gain from equation (4) is
We select a gain of 20 resulting sensitivity is 900 mV
The input impedance and bandwidth are the same as the 20
watt amplifier so the components are the same
The maximum supplies dictate using 80V devices The
2N5882 2N5880 pair are 80V 160W transistors with a mini-
mum beta of 40 at 2A and 20 at 6A This corresponds to a
minimum beta of 22 5 at 5 5A (I
MJE722 driver pair are 80V transistors with a minimum beta
of 50 at 250 mA This output combination guarantees I
with 5 mA from the LM391
Output transistor heat sink requirements are found using
equations (7) (9) and (10)
For both output transistors on one heat sink the thermal
resistance should be 1 9 C W
Now using equation (8) we find the power dissipation in the
driver
Power Output
Input Sensitivity
Input Impedance
Bandwidth
40W 8
60W 4
g
g
30 3V
26 9V
JA s
R
R
f 1
f 2
e
e
60W 4
200
3 16A reducing to
5 48A
SA s
5 1k
100k
V
V
24
OPeak
OPeak
b
JA s
P
P
6 0
DRIVER
55
D
AMPLIFIER
e
150
b
e
R
C
e
e
IN
f
0 4 (60)
A
1 1
6 0 C W for T
1 2
e
25 3V
21 9V
b
V t
e
e
10 F
b
(Continued)
55
24
20
100k
18
1 0
e
e
e
20 Hz–20 kHz
79 C W
e
24W
1 2W
I
I
Opeak
OPeak
OPeak
3 9 C W
AMAX
C
C
) The MJE712
e
e
e
g
e
5 pF
3 16A
5 48A
38 3V using
55 C
g
40W 8
60W 4
0 25 dB
1V Max
Opeak
100k
(10)
(7)
(9)
(8)
(9)
10
Since a heat sink is required on the driver we should inves-
tigate the output stage thermal stability at the same time to
optimize the design If we find a value of R
the protection circuitry we can then use equation (5) to find
the heat sink required for the drivers
The SOA characteristics of the 2N5882 2N5880 transistors
are shown in the following curve along with a desired pro-
tection line
The desired data points are
Since the break voltage is not equal to the supply we will
use two resistors to replace R
V
M
e
80V
SOA 2N5882 2N5880
Thevenin Equivalent
V
B
e
Circuit Used
47V
TL H 7146 – 12
3
and move V
I
L
e
3A
Where R
V
TH
E
e
that is good for
B
TL H 7146 – 11
I
L
V
b
e
TH
TL H 7146 – 10
11A
e
R
A
3
R
R
a
A
3
A
3
ll
R
R
B
3
B
3