LM359 datasheet(16/34 Pages) TI

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LM359

SNOSBT4C – MAY 1999 – REVISED MARCH 2013

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DESIGN EXAMPLE

eIN = 50 mV (MAX), fIN = 10 MHz (MAX), desired circuit BW = 20 MHz, AV = 20 dB, driving source impedance =

Ω, V+ = 12V.

1. Basic circuit configuration:

2. Select ISET to provide adequate amplifier bandwidth so that the closed loop bandwidth will be determined by

Rf and Cf. To do this, the set current should program an amplifier open loop gain of at least 20 dB at the

desired closed loop bandwidth of the circuit. For this example, an ISET of 0.5 mA will provide 26 dB of open

loop

gain

MHz

which

will

sufficient.

Using

single

resistor

programming

for

ISET:

3. Since the closed loop bandwidth will be determined by

to obtain a 20 MHz bandwidth,

both Rf and Cf should be kept small. It can be assumed that Cf can be in the range 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 Rf to be within the range of:

Also, for a closed loop gain of +10, Rf

must be 10 times Rs + re where re is the mirror diode resistance.

4. So as not to appreciably load the 75

Ω input termination resistance the value of (Rs + re) is set to 750Ω.

5. For Av = 10; Rf is set to 7.5 kΩ.

6. The optimum output DC level for symmetrical AC swing is:

7. The DC feedback current must be:

DC biasing predictability will be insured

because 640

μA is greater than the minimum of ISET/5 or 100 μA.

8. 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.

9. Rb can now be found:

10. Since Rs + re will be 750Ω and re is fixed by the DC mirror current to be:

Rs must

be 750

Ω–40Ω or 710Ω which can be a 680Ω resistor in series with a 30Ω resistor which are standard 5%

tolerance resistor values.

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