LM391 [NSC]
LM391 Audio Power Driver; LM391音频功率驱动器
| 型号： | LM391 |
| 厂家： | 
National Semiconductor |
| 描述： | LM391 Audio Power Driver |
| 文件： | 总12页 (文件大小：238K) |
| 中文： | 中文翻译 | 下载： | 下载PDF数据表文档文件 |
December 1994
LM391 Audio Power Driver
General Description
The LM391 audio power driver is designed to drive external
power transistors in 10 to 100 watt power amplifier designs.
High power supply voltage operation and true high fidelity
performance distinguish this IC. The LM391 is internally pro-
tected for output faults and thermal overloads; circuitry pro-
viding output transistor protection is user programmable.
Features
Y
g
High Supply Voltage
50V max
0.01%
3 mV
Y
Low Distortion
Y
Low Input Noise
Y
High Supply Rejection
90 dB
Y
Gain and Bandwidth Selectable
Y
Dual Slope SOA Protection
Y
Shutdown Pin
Equivalent Schematic and Connection Diagram
TL/H/7146–1
Dual-In-Line Package
TL/H/7146–2
Top View
Order Number LM391N-100
See NS Package Number N16A
C
1995 National Semiconductor Corporation
TL/H/7146
RRD-B30M115/Printed in U. S. A.
Absolute Maximum Ratings
If Military/Aerospace specified devices are required,
please contact the National Semiconductor Sales
Office/Distributors for availability and specifications.
Package Dissipation (Note 1)
Storage Temperature
1.39W
b
a
65 C to 150 C
§
0 C to 70 C
§
§
a
Operating Temperature
Lead Temp. (Soldering, 10 sec.)
Thermal Resistance
§
Supply Voltage
LM391N-100
260 C
§
a
50V or 100V
g
Input Voltage
Supply Voltage less 5V
1 mA
i
20 C/W
§
JC
Shutdown Current (Pin 14)
i
63 C/W
§
JA
a
a
b
and Vb
90% V
e
.)
MAX
e
e
Electrical Characteristics T
25 C (The following are for V
§
90% V
A
MAX
Parameter
Conditions
Min
Typ
Max
Units
Quiescent Current
LM391N-100
Current in Pin 15
e
mA
V
0
5
6
IN
Va
Vb
7
7
Va
Vb
5
V
V
b
a
b
a
Output Swing
Positive
Negative
5
Drive Current
Source (Pin 8)
Sink (Pin 5)
5
mA
mA
5
Noise (20 Hz–20 kHz)
Supply Rejection
Input Referred
Input Referred
3
mV
70
90
dB
e
e
Total Harmonic Distortion
f
f
1 kHz
20 kHz
0.01
0.10
%
%
0.25
Intermodulation Distortion
Open Loop Gain
60 Hz, 7 kHz, 4:1
0.01
5500
0.1
5
%
V/V
mA
mV
mV
mV
mA
mA
e
f
1 kHz
1000
Input Bias Current
1.0
20
Input Offset Voltage
Positive Current Limit V
BE
Pin 10–9
Pin 9–13
Pin 10
650
650
10
Negative Current Limit V
BE
Positive Current Limit Bias Current
Negative Current Limit Bias Current
Pin 14 Current Comments
100
100
Pin 13
10
Minimum pin 14 current required for shutdown is 0.5 mA, and must not exceed 1 mA.
Maximum pin 14 current for amplifier not shut down is 0.05 mA.
The typical shutdown switch point current is 0.2 mA.
Note 1: For operation in ambient temperatures above 25 C, the device must be derated based on a 150 C maximum junction temperature and a thermal resistance
§
§
of 90 C/W junction to ambient.
§
Typical Applications
TL/H/7146–3
FIGURE 1. LM391 with External ComponentsÐProtection Circuitry Not Shown
2
Typical Performance Characteristics
Total Harmonic Distortion vs
e
Total Harmonic Distortion vs
e
4X)
Output Power vs Supply Voltage
Frequency (R
8X)
Frequency (R
L
L
s
TL/H/7146–4
Pin Descriptions
Pin No.
Pin Name
Comments
a
b
1
2
Input
Input
Audio input
Feedback input
Sets the dominant pole
3
Compensation
Ripple Filter
Sink Output
BIAS
4
Improves negative supply rejection
Drives output devices and is emitter of AB bias V multiplier
5
BE
6
Base of V multiplier
BE
7
BIAS
Collector of V multiplier
BE
8
Source Output
Output Sense
Drives output devices
9
Biases the IC and is used in protection circuits
Base of positive side protection circuit transistor
Diode used for dual slope SOA protection
Diode used for dual slope SOA protection
Base of negative side protection circuit transistor
Shuts off amplifier when current is pulled out of pin
Positive supply
a
a
b
b
10
11
12
13
14
15
16
Current Limit
SOA Diode
SOA Diode
Current Limit
Shutdown
Va
Vb
Negative supply
3
External Components (Figure 1)
Component
Typical Value
Comments
C
IN
1 mF
Input coupling capacitor sets a low frequency pole with R
.
IN
1
e
f
L
2qR
C
IN IN
Sets input impedance and DC bias to input.
Feedback resistor; for minimum offset voltage at the output this should be equal to R
R
R
R
100k
100k
5.1k
IN
.
IN
f
2
Feedback resistor that works with R to set the voltage gain.
f
f
2
1
R
R
f
f
2
1
e
a
1
A
V
C
f
10 mF
Feedback capacitor. This reduces the gain to unity at DC for minimum offset voltage at the
.
output. Also sets a low frequency pole with R
f
1
1
e
f
L
2qR
C
f
f
1
C
C
5 pF
Compensation capacitor. Sets gain bandwidth product and a high frequency pole.
1
GBW
e
e
&
GBW
, f
h
2q5000C
Max f for stable design
h
A
V
500 kHz.
C
R
R
C
C
3.9k
AB bias resistor.
A
10k
AB bias potentiometer. Adjust to set bias current in the output stage.
B
0.1 mF
5 pF
Bypass capacitor for bias. This improves high frequency distortion and transient response.
AB
R
Ripple capacitor. This improves negative supply rejection at midband and high frequencies.
, if used, must equal C .
C
C
R
R
R
100X
2.7X
Bleed resistor. This removes stored charge in output transistors.
eb
Output compensation resistor. This resistor and C compensate the output stage. This value
O
will vary slightly for different output devices.
O
C
R
R
0.1 mF
0.3X
Output compensation capacitor. This works with R to form a zero that cancels f of the
O b
output power transistors.
O
Emitter degeneration resistor. This resistor gives thermal stability to the output stage
quiescent current. IRC PW5 type.
E
39k
Shutdown resistor. Sets the amount of current pulled out of pin 14 during shutdown.
Compensation capacitors for protection circuitry.
TH
C , C
2
1000 pF
Ê2
X
L
10X 5 mH
Used to isolate capacitive loads, usually 20 turns of wire wrapped around a 10X, 2W resistor.
ll
4
Application Hints
To prevent thermal runaway of the AB bias current the fol-
lowing equation must be valid:
GENERALIZED AUDIO POWER AMP DESIGN
Givens: Power Output
Load Impedance
a
R
E
(b
1)
MIN
s
i
(5)
JA
V
(K)
CEQMAX
Input Sensitivity
where:
is the thermal resistance of the driver transistor, junc-
Input Impedance
i
JA
tion to ambient, in C/W.
Bandwidth
§
is the emitter degeneration resistance in ohms.
The power output and load impedance determine the power
supply requirements. Output signal swing and current are
found from:
R
E
b
is that of the output transistor.
min
V
CEQMAX
equation (3).
is the highest possible value of one supply from
e
V
2 R P
L O
0
Opeak
(1)
(2)
K is the temperature coefficient of the driver base-emitter
voltage, typically 2 mV/ C.
2 P
R
O
e
§
Often the value of R is to be determined and equation (5)
is rearranged to be:
I
Opeak
0
L
E
Add 5 volts to the peak output swing (V ) for transistor
OP
a
5V) at a current
. The regulation of the supply determines the unload-
g
voltage to get the supplies, i.e.,
of I
(V
OP
i
(V ) K
JA CEQMAX
t
R
(6)
peak
E
a
b
1
MIN
ed voltage, usually about 15% higher. Supply voltage will
also rise 10% during high line conditions.
The maximum average power dissipation in each output
transistor is:
&
a
a
5) (1 regulation) (1.1) (3)
g
max supplies
(V
Opeak
e
P
0.4 P
OMAX
(7)
DMAX
The input sensitivity and output power specs determine the
required gain.
The power dissipation in the driver transistor is:
P
P
R
V
V
0
DMAX
O
L
ORMS
e
t
e
P
(8)
A
(4)
DRIVER(MAX)
V
b
V
MIN
IN
INRMS
Heat sink requirements are found using the following formu-
las:
Normally the gain is set between 20 and 200; for a 25 watt,
8 ohm amplifier this results in a sensitivity of 710 mV and 71
mV, respectively. The higher the gain, the higher the THD,
as can be seen from the characteristics curves. Higher gain
also results in more hum and noise at the output.
b
T
T
AMAX
JMAX
s
s
i
(9)
JA
P
D
b
b
i
CS
i
i
i
(10)
SA
JA
JC
The desired input impedance is set by R . Very high values
IN
can cause board layout problems and DC offsets at the out-
where:
T
is the maximum transistor junction temperature.
is the maximum ambient temperature.
put. The bandwidth requirements determine the size of C
f
and C as indicated in the external component listing.
C
jMAX
T
i
i
i
i
AMAX
The output transistors and drivers must have a breakdown
voltage greater than the voltage determined by equation (3).
The current gain of the drive and output device must be high
is thermal resistance junction to ambient.
is thermal resistance sink to ambient.
is thermal resistance junction to case.
JA
SA
JC
CS
enough to supply I
with 5 mA of drive from the LM391.
Opeak
is thermal resistance case to sink, typically 1 C/W for
§
The power transistors must be able to dissipate approxi-
mately 40% of the maximum output power; the drivers must
dissipate this amount divided by the current gain of the out-
puts. See the output transistor selection guide, Table A.
most mountings.
5
Application Hints (Continued)
PROTECTION CIRCUITRY
resistor is set to limit the current to less than 1 mA (the
absolute maximum). This resistor with the capacitor gives a
time constant of RC. The turn-ON delay is approximately 2
time constants.
The protection circuits of the LM391 are very flexible and
should be tailored to the output transistor’s safe operating
area. The protection V-I characteristics, circuitry, and resis-
tor formulas are described below. The diodes from the out-
put to each supply prevent the output voltage from exceed-
ing the supplies and harming the output transistors. The out-
put will do this if the protection circuitry is activated while
driving an inductive load.
Example:
Amplifier with maximum supply of 30V, like the 20W, 8X
example in the data sheet, requiring a delay of 1 second.
e
Time delay
2 RC
Max Va
1 mA
e
R
TURN-ON DELAY
It is often desirable to delay the turn-ON of the power ampli-
fier. This is easily implemented by putting a resistor in series
with a capacitor from pin 14 to ground. The value of the
So:
e
R
a 30V rating.
e
30k. Solving for C gives 16.7 mF. Use C
20 mF with
Protection Circuitry with External Components
Protection Characteristics
TL/H/7146–6
TL/H/7146–5
a
e
Protection Circuit Resistor Formulas (V
V
)
B
Type of Protection
R , R
E
R , R
Ê1
1
R , R
Ê2
2
R , R
3
Ê
Ê3
w
e
Current Limit
R
Not Required
Short
Not Required
Not Required
Va
E
E
I
L
b
w
V
w
w
Single Slope SOA
Protection
M
M
e
e
e
e
R
R
R
R
1 kX
1
2
2
I
w
#
#
J
J
L
Dual Slope SOA
b
w
V
e
b
1
Protection
a
R
E
R
1 kX
R
3
R
2
1
Ê
b
w
I
w
I
L R
Ð
(
L
E
e
(V
V
)
B
Note: w is the current limit V voltage, 650 mV. Assumptions: Va
transistors.
w, V
w. V is the load supply voltage. V is the maximum rated V of the output
a
ll
ll
BE
M
M
CE
6
OSCILLATIONS & GROUNDING
Application Hints (Continued)
Most power amplifiers work the first time they are turned on.
They also tend to oscillate and have excess THD. Most os-
cillation problems are due to inadequate supply bypassing
and/or ground loops. A 10 mF, 50V electrolytic on each
power supply will stop supply-related oscillations. However,
if the signal ground is used for these bypass caps the THD
is usually excessive. The signal ground must return to the
power supply alone, as must the output load ground. All
other groundsÐbypass, output R-C, protection, etc., can tie
together and then return to supply. This ground is called
high frequency ground. On the 40W amplifier schematic all
the grounds are labeled.
TRANSIENT INTERMODULATION DISTORTION
There has been a lot of interest in recent years about tran-
sient intermodulation distortion. Matti Otala of University of
Oulu, Oulu, Finland has published several papers on the
subject. The results of these investigations show that the
open loop pole of the power amplifier should be above 20
kHz.
To do this with the LM391 is easy. Put a 1 MX resistor from
pin 3 to the output and the open loop gain is reduced to
about 46 dB. Now the open loop pole is at 30 kHz. The
current in this resistor causes an offset in the input stage
that can be cancelled with a resistor from pin 4 to ground.
The resistor from pin 4 to ground should be 910 kX rather
than 1 MX to insure that the shutdown circuitry will operate
correctly. The slight difference in resistors results in about
15 mV of offset. The 40W, 8X amplifier schematic shows
the hookup of these two resistors.
Capacitive loads can cause instabilities, so they are isolated
from the amplifier with an inductor and resistor in the output
lead.
AB BIAS CURRENT
To reduce distortion in the output stage, all the transistors
are biased ON slightly. This results in class AB operation
and reduces the crossover (notch) distortion of the class B
stage to a low level, (see performance curve, THD vs AB
BRIDGE AMPLIFIER
A switch can be added to convert a stereo amplifer to a
single bridge amplifer. The diagram below shows where the
switch and one resistor are added. When operating in the
bridge mode the output load is connected between the two
bias). The potentiometer, R , from pins 6–7 is adjusted to
B
give about 25 mA of current in the output stage. This current
is usually monitored at the supply or by measuring the volt-
Ý
Ý
2 is disconnected.
outputs, the input is V
1, and V
age across R .
E
IN
IN
Typical Applications (Continued)
Bridge Circuit Diagram
TL/H/7146–7
Output Transistors Selection Guide
Table A.
Driver Transistor
Output Transistor
Power
Output
PNP
NPN
PNP
NPN
@
20W 8X
MJE711
MJE171
D43C8
MJE721
MJE181
D42C8
TIP42A
2N6490
TIP41A
2N6487
@
30W 4X
@
40W 8X
MJE712
MJE172
D43C11
MJE722
MJE182
D42C11
2N5882
2N5880
@
60W 4X
7
Application Hints (Continued)
A 20W, 8X; 30W, 4X AMPLIFIER
Givens:
Solving for C :
f
1
t
e
C
f
7.8 mF; use 10 mF
Power Output
20W into 8X
30W into 4X
2qR f
f L
1
The recommended value for C is 5 pF for gains of 20 or
C
Input Sensitivity
Input Impedance
1V Max
100k
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
g
20 Hz–20 kHz 0.25 dB
Bandwidth
Equations (1) and (2) give:
e
e
e
e
20W/8X
30W/4X
V
V
17.9V
15.5V
I
I
2.24A
3.87A
OP
OP
OP
beta at I
, 3.87A. The TIP42, TIP41 complementary pair
Opeak
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
OP
Therefore the supply required is:
@
23V 2.24A, reducing to . . .
g
g
@
21V 3.87A
g
With 15% regulation and high line we get 29V from equa-
tion (3).
(higher f ) than the output transistors to insure that the R-C
T
on the output will prevent instability.
Sensitivity and equation (4) set minimum gain:
To find the heat sink required for each output transistor we
use equations (7), (9), and (10):
c
20
8
0
t
e
A
12.65
V
e
e
P
0.4 (30)
12W
(7)
1
D
b
150 C 55 C
§
§
s
e
e
i
7.9 C/W for T
55 C (9)
§
We will use a gain of 20 with resulting sensitivity of 632 mV.
§
JA
AMAX
e
7.9 2.1 1.0 4.8 C/W
12
Letting R equal 100k gives the required input impedance.
IN
For low DC offsets at the output we let R
for R gives:
s
b
b
i
(10)
§
SA
e
100k. Solving
f
2
If both transistors are mounted on one heat sink the thermal
resistance should be halved to 2.4 C/W.
f
1
§
The maximum average power dissipation in each driver is
found using equation (8):
e
e
R
100k
5.26k; use 5.1k
f
2
100k
e
R
f
1
b
20
1
12
e
e
400 mW
P
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:
DRIVER(MAX)
30
Using equation (9):
b
155 55
s
e
i
237 C/W
§
JA
20
0.4
e
e
4 Hz
f
L
5
e
c
e
5 100 kHz
f
20k
h
8
Application Hints (Continued)
Since the free air thermal resistance of the MJE711,
MJE721 is 100 C/W, no heat sink is required. Using this
The data points from the curve are:
Ê
§
information and equation (6) we can find the minimum value
e
e
e
23V, I
L
e
3A, IL 7A
V
M
60V, V
B
Using the dual slope protection formulas:
0.65
of R required to prevent thermal runaway.
E
e
e
100 (30) (0.002)
R
0.22X
E
t
e
R
0.19X
(6)
3
E
a
30
1
e
R
2
1k
We must now use the SOA data on the TIP42, TIP41 tran-
sistors to set up the protection circuit. Below is the SOA
curve with the 4X and 8X load lines. Also shown are the
desired protection lines. Note the value of V is equal to the
B
supply voltage, so we use the formulas in the table.
b
60 0.65
0.65
e
&
91k
R
1k
1
#
J
(/2(1.8) (23)
23
e
b
&
R
3
1k
1
24k
b
7(0.22) 0.65
#
J
Note that an R of 0.22X satisfies equation (6). The final
D.C. SOA of TIP42, TIP41
Transistors
E
schematic of this amplifier is below. If the output is shorted
the current will be 1.8A and V
AC, the average power is:
is 23V. Since the input is
CE
e
&
short P
21W
D
This power is greater than was used in the heat sink calcula-
tions, so the transistors will overheat for long-duration
shorts unless a larger heat sink is used.
TL/H/7146–8
Typical Applications (Continued)
20W-8X, 30W-4X Amplifier with 1 Second Turn-ON Delay
TL/H/7146–9
j
*Additional protection for LM391N; Schottky diodes and R
100X.
9
Application Hints (Continued)
A 40W/8X, 60W/4X AMPLIFIER
Given:
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 that is good for
E
the protection circuitry, we can then use equation (5) to find
the heat sink required for the drivers.
Power Output
40W/8X
60W/4X
Input Sensitivity
Input Impedance
1V Max
100k
The SOA characteristics of the 2N5882, 2N5880 transistors
are shown in the following curve along with a desired pro-
tection line.
g
20 Hz–20 kHz 0.25 dB
Bandwidth
Equations (1) and (2) give:
SOA 2N5882, 2N5880
e
e
e
e
40W/8X
60W/4X
V
V
25.3V
21.9V
I
I
3.16A
5.48A
OPeak
OPeak
OPeak
OPeak
Therefore the supply required is:
@
30.3V 3.16A, reducing to . . .
g
g
@
26.9V 5.48A
g
With 15% regulation and high line we get 38.3V using
equation (3).
The minimum gain from equation (4) is:
t
A
V
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.
e
e
e
e
C 5 pF
C
R
R
5.1k
R
100k
f
f
IN
1
2
TL/H/7146–10
e
10 mF
100k
C
f
The desired data points are:
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 ). The MJE712,
MJE722 driver pair are 80V transistors with a minimum beta
Ê
e
e
e
I
L
e
11A
V
80V
V
B
47V
3A
IL
M
Since the break voltage is not equal to the supply, we will
use two resistors to replace R and move V .
Opeak
3
B
Circuit Used
of 50 at 250 mA. This output combination guarantees I
with 5 mA from the LM391.
Opeak
Output transistor heat sink requirements are found using
equations (7), (9), and (10):
e
e
P
0.4 (60)
24W
(7)
D
b
200 55
s
e
e
55 C
i
6.0 C/W for T
§
(9)
§
JA
AMAX
24
s
b
b
e
6.0 1.1 1.0 3.9 C/W
i
(10)
§
For both output transistors on one heat sink the thermal
SA
resistance should be 1.9 C/W.
§
Now using equation (8) we find the power dissipation in the
driver:
TL/H/7146–11
24
Thevenin Equivalent
e
e
1.2W
P
(8)
(9)
DRIVER
20
b
150 55
s
e
i
79 C/W
§
JA
1.2
A
R3 RB3
e
Where: R
TH
ll
RA3
b
V
e
V
TH
A
R3
B
R3
Ð
(
a
TL/H/7146–12
10
Application Hints (Continued)
The formulas for R , R , and R do not change:
The easiest way to solve these equations is to iterate with
B
62k, then R3
E
1
2
standard values. If we guess RA3
e
e
47.12k;
0.65
3A
e
e
R
E
0.22X
use 47k. The Thevenin impedance comes out 26.7k, which
is close enough to 25.55k.
b
80 0.65
e
e
1
e
120k
Now we will use equation (5) to determine the heat sinking
requirements of the drivers to insure thermal stability:
R
1k
R
1k
2
0.65
The formula for R now gives R when the Va in the for-
3
TH
a
0.22 (20
1)
s
&
i
57 C/W
§
(5)
mula becomes V .
B
JA
40 (0.002)
V
B
This value is lower than we got with equation (9), so we will
use it in equation (10):
e
e
b
1
R
R
TH
2
Ê
b
w
I
L R
E
Ð
(
s
b
b
e
1 50 C/W
i
57
6
(10)
§
SA
47
b
e
25.55k
1k
1
This is the required heat sink for each driver. For low TIM
we add the 1 MX resistor from pin 3 to the output and a
910k resistor from pin 4 to ground. The complete schematic
is shown below.
b
11 (0.22) 0.65
Ð
0.76 R3
(
25.55k
V
get V .
is the additional voltage added to the supply voltage to
TH
B
a
e b
b
(V
B
e b
b
e b
17V
V
V
)
(47 30)
If the output is shorted, the transistor voltage is about 28V
and the current is 5A. Therefore the average power is:
TH
Now we must find RA3 and RB3 using the Thevenin formulas.
Putting V , Vb, and R into the appropriate formulas re-
e
e
70W
short PD
(/2(28) 5
TH
duces to:
TH
This is much larger than the power used to calculate the
heat sinks and the output transistors will overheat if the out-
put is shorted too long.
A
A
RB3
and
R3 RB3
e
e
ll
Typical Applications (Continued)
40W-8X, 60W-4X Amplifier
*High Frequency G
**Input Ground
***Speaker Ground
TL/H/7146–13
Note: All Grounds Should be Tied Together
Only at Power Supply Ground.
j
²
Additional protection for LM391N; Schottky diodes and R
100X.
11
Physical Dimensions inches (millimeters)
Molded Dual-In-Line Package (N)
Order Number LM391N-100
NS Package Number N16A
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