LM359M National Semiconductor, LM359M Datasheet - Page 13

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LM359M

Manufacturer Part Number
LM359M
Description
IC AMP DUAL PROG CUR MODE 14SOIC
Manufacturer
National Semiconductor
Datasheet

Specifications of LM359M

Amplifier Type
General Purpose
Number Of Circuits
2
Slew Rate
60 V/µs
Gain Bandwidth Product
400MHz
Current - Input Bias
8µA
Current - Supply
18.5mA
Current - Output / Channel
40mA
Voltage - Supply, Single/dual (±)
5 V ~ 22 V, ±2.5 V ~ 11 V
Operating Temperature
0°C ~ 70°C
Mounting Type
Surface Mount
Package / Case
14-SOIC (3.9mm Width), 14-SOL
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant
Output Type
-
-3db Bandwidth
-
Voltage - Input Offset
-
Other names
*LM359M

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Application Hints
DESIGN EXAMPLE
e
BW = 20 MHz, A
V
1. Basic circuit configuration:
2. Select I
3. Since the closed loop bandwidth will be determined by
IN
+
= 12V.
= 50 mV (MAX), f
that the closed loop bandwidth will be determined by R
and C
amplifier open loop gain of at least 20 dB at the desired
closed loop bandwidth of the circuit. For this example,
an I
20 MHz which will be sufficient. Using single resistor
programming for I
to obtain a 20 MHz bandwidth, both R
kept small. It can be assumed that C
of 1 pF to 5 pF for carefully constructed circuit boards to
insure stability and allow a flat frequency response. This
will limit the value of R
Also, for a closed loop gain of +10, R
R
s
+ r
SET
e
f
. To do this, the set current should program an
SET
of 0.5 mA will provide 26 dB of open loop gain at
where r
to provide adequate amplifier bandwidth so
V
= 20 dB, driving source impedance = 75Ω,
e
is the mirror diode resistance.
SET
IN
:
= 10 MHz (MAX), desired circuit
f
to be within the range of:
(Continued)
f
can be in the range
f
f
must be 10 times
and C
f
should be
00778820
f
13
4. So as not to appreciably load the 75Ω input termination
5. For A
6. The optimum output DC level for symmetrical AC swing
7. The DC feedback current must be:
8. R
9. Since R
10. As a final design step, C
resistance the value of (R
is:
DC biasing predictability will be insured because 640 µA
is greater than the minimum of I
For gain accuracy the total AC and DC mirror current
should be less than 2 mA. For this example the maxi-
mum AC mirror current will be:
therefore the total mirror current range will be 574 µA to
706 µA which will insure gain accuracy.
mirror current to be:
R
resistor in series with a 30Ω resistor which are standard
5% tolerance resistor values.
lower passband frequency corner of 8 Hz for this ex-
ample.
A larger value may be used and a 0.01 µF ceramic
capacitor in parallel with C
gain accuracy.
b
s
can now be found:
must be 750Ω–40Ω or 710Ω which can be a 680Ω
v
= 10; R
s
+ r
e
f
will be 750Ω and r
is set to 7.5 kΩ.
i
s
must be selected to pass the
i
+ r
will maintain high frequency
e
) is set to 750Ω.
SET
e
/5 or 100 µA.
is fixed by the DC
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