LM359J National Semiconductor, LM359J Datasheet - Page 12

• Manufacturer Part Number: LM359J
• Description: Dual/ High Speed/ Programmable/ Current Mode Norton Amplifiers
• Manufacturer: National Semiconductor

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Application Hints
DESIGN EXAMPLE:
e
BW = 20 MHz, A
75 , V
1. Basic circuit configuration:
2. Select I
3. Since the closed loop bandwidth will be determined by
to obtain a 20 MHz bandwidth, both R
small. It can be assumed that C
to 5 pF for carefully constructed circuit boards to insure sta-
bility and allow a flat frequency response. This will limit the
value of R
IN
= 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 pro-
gramming for I
+
SET
= 12V.
f
f
. To do this, the set current should program an
to be within the range of:
SET
of 0.5 mA will provide 26 dB of open loop gain at
to provide adequate amplifier bandwidth so
V
SET
= 20 dB, driving source impedance =
:
IN
= 10 MHz (MAX), desired circuit
(Continued)
f
can be in the range of 1 pF
f
and C
f
should be kept
DS007788-20
f
12
Also, for a closed loop gain of +10, R
+ r
4. So as not to appreciably load the 75
5. For A
6. The optimum output DC level for symmetrical AC swing
7. The DC feedback current must be:
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 maximum AC mirror
current will be:
therefore the total mirror current range will be 574 µA to 706
µA which will insure gain accuracy.
8. R
9. Since R
R
tor in series with a 30 resistor which are standard 5% toler-
ance resistor values.
10. As a final design step, C
A larger value may be used and a 0.01 µF ceramic capacitor
in parallel with C
s
e
must be 750 –40
resistance the value of (R
is:
ror current to be:
lower passband frequency corner of 8 Hz for this ex-
ample.
where r
b
can now be found:
v
= 10; R
e
s
+ r
is the mirror diode resistance.
e
i
will maintain high frequency gain accuracy.
will be 750
f
is set to 7.5 k .
or 710
i
s
SET
must be selected to pass the
and r
+ r
which can be a 680
/5 or 100 µA.
e
) is set to 750 .
e
is fixed by the DC mir-
f
must be 10 times R
input termination
resis-
s