
Application Hints
(Continued)
A 40W/8
X
, 60W/4
X
AMPLIFIER
Given:
Power Output
40W/8
X
60W/4
X
1V Max
Input Sensitivity
Input Impedance
100k
Bandwidth
20 Hz–20 kHz
g
0.25 dB
Equations (1) and (2) give:
40W/8
X
60W/4
X
Therefore the supply required is:
g
30.3V
@
3.16A, reducing to . . .
g
26.9V
@
5.48A
With 15% regulation and high line we get
g
38.3V using
equation (3).
V
OPeak
e
25.3V
V
OPeak
e
21.9V
I
OPeak
e
3.16A
I
OPeak
e
5.48A
The minimum gain from equation (4) is:
A
V
t
18
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.
R
f1
e
5.1k
R
f2
e
100k
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
Opeak
). The MJE712,
MJE722 driver pair are 80V transistors with a minimum beta
of 50 at 250 mA. This output combination guarantees I
Opeak
with 5 mA from the LM391.
R
IN
e
100k
C
f
e
10
m
F
C
C
e
5 pF
Output transistor heat sink requirements are found using
equations (7), (9), and (10):
P
D
e
0.4 (60)
e
24W
i
JA
s
200
b
55
24
i
SA
s
6.0
b
1.1
b
1.0
e
3.9
§
C/W
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:
(7)
e
6.0
§
C/W for T
AMAX
e
55
§
C
(9)
(10)
P
DRIVER
e
24
20
e
1.2W
(8)
i
JA
s
150
b
55
1.2
e
79
§
C/W
(9)
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
E
that is good for
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.
SOA 2N5882, 2N5880
TL/H/7146–10
The desired data points are:
V
M
e
80V
Since the break voltage is not equal to the supply, we will
use two resistors to replace R
3
and move V
B
.
V
B
e
47V
I
L
e
3A
I
ê
L
e
11A
Circuit Used
TL/H/7146–11
Thevenin Equivalent
Where: R
TH
e
R
A
3
ll
R
B
3
V
TH
e
V
b
D
R
A
3
R
A
3
a
R
B
3
(
TL/H/7146–12
10