FIGURE 6. PRODUCING VERY LONG TIME-OUTS BY PULSE MODULATING THE OSCILLATOR

Oscillator

Components

R2

R1

CD4093

C1 = 0.1µF

-V

R3

C2

-

+

4

3

TRIG

OSC

5

6

2

1

8

7

OFF

ON

CAP

SYNC

T1

T2

Waveform A

LS7339

T1

≈ R1 C1

T2

≈ (R1 + R2)C1

DESCRIPTION:

EXAMPLE:

D1

T1

Period = T1 + T2

Pulse Duty Cycle = T1/Period

-V

Waveform B (Pulse Modulator)

Inverse of Pulse Duty Cycle = Period/T1 = 1 +T2/T1 = 2 + R2/R1

Waveform A

Waveform B

+V

D2

The Oscillator resistor is normally returned to the negative terminal of the DC Supply. If this resistor is returned to

a negative-going pulse instead, the Oscillator R-C becomes effectively multiplied by the inverse of the Pulse Duty

Cycle, because the Oscillator R-C can charge only while the pulse is present. (Pulse Modulation)

In Figure 6, the Oscillator R is R3, the Oscillator C is C2 and the Pulse Modulator is Waveform B through

isolation diode D2. The Multiplier, M = 2 + R2/R1 (the inverse of the Pulse Duty Cycle).

R3 = 5.1M

With M = 22, R2/R1 = 20. If R2 = 3M

Ω , R1 = 150kΩ .

7339-060201-5