-1

0

1

2

3

4

5

6

0

10

20

30

40

50

TIME (µs)

C001

VOUT

Va

Vc

CC

2

3

A2

1

2

3

V

(R

R )

V

R

(R

R )

R

R

CC

2

A1

2

1

3

V

R

V

R

R

R

˜

R

LMV761, LMV762, LMV762Q-Q1

SNOS998I – FEBRUARY 2002 – REVISED OCTOBER 2015

www.ti.com

Typical Application (continued)

Consider the output of Figure 23 is high to analyze the circuit. That implies that the inverted input (VC) is lower

than the noninverting input (VA). This causes the C1 to be charged through R4, and the voltage VC increases until

it is equal to the noninverting input. The value of VA at this point is calculated by Equation 4:

(4)

If R1 = R2 = R3, then VA1 = 2 VCC / 3

At this point the comparator switches pulling down the output to the negative rail. The value of VA at this point is

calculated by Equation 5:

(5)

If R1 = R2 = R3, then VA2 = VCC / 3.

The capacitor C1 now discharges through R4, and the voltage VC decreases until it is equal to VA2, at which point

the comparator switches again, bringing it back to the initial stage. The time period is equal to twice the time it

takes to discharge C1 from 2 VCC / 3 to VCC / 3, which is given by R4C1 × ln2. Hence, the formula for the

frequency is calculated by Equation 6:

F = 1 / (2 × R4 × C1 × ln2)

(6)

8.2.3 Application Curve

Figure Figure 25 shows the simulated results of an oscillator using the following values:

•

R1 = R2 = R3 = R4 = 100 kΩ

•

C1 = 100 pF, CL = 20 pF

•

V+ = 5 V, V– = GND

•

CSTRAY (not shown) from Va to GND = 10 pF

Figure 25. Square-Wave Oscillator Output Waveform

14

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