LMV331M5X datasheet(11/32 Pages) TI

11 / 32 page

LMV331-N, LMV339-N, LMV393-N

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SNOS018G – AUGUST 1999 – REVISED FEBRUARY 2013

Figure 16. Squarewave Oscillator

To analyze the circuit, assume that the output is initially high. For this to be true, the voltage at the inverting input

Vc has to be less than the voltage at the non-inverting input Va. For Vc to be low, the capacitor C1 has to be

discharged and will charge up through the negative feedback resistor R4. When it has charged up to value equal

to the voltage at the positive input Va1, the comparator output will switch.

Va1 will be given by:

(9)

If:

R1 = R2 = R3

(10)

Then:

Va1 = 2VCC/3

(11)

When the output switches to ground, the value of Va is reduced by the hysteresis network to a value given by:

Va2 = VCC/3

(12)

Capacitor C1 must now discharge through R4 towards ground. The output will return to its high state when the

voltage across the capacitor has discharged to a value equal to Va2.

For the circuit shown, the period for one cycle of oscillation will be twice the time it takes for a single RC circuit to

charge up to one half of its final value. The time to charge the capacitor can be calculated from

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LMV331M5X 数据表(PDF) 11 Page - Texas Instruments