FIGURE 6. PRODUCING VERY LONG TIME-OUTS BY PULSE MODULATING THE OSCILLATOR
R2
D1
R1
Waveform A
Oscillator
Components
-V
4
3
2
1
TRIG V DD OFF ON
+V
LS7339
D2
R3
C2
-+
V SS OSC CAP SYNC
5
6
7
8
CD4093
C1 = 0.1µF
-V
T1
Waveform A
T2
Waveform B
Waveform B (Pulse Modulator)
T1
T1 ≈ R1 C1
T2 ≈ (R1 + R2)C1
Period = T1 + T2
Pulse Duty Cycle = T1/Period
Inverse of Pulse Duty Cycle = Period/T1 = 1 +T2/T1 = 2 + R2/R1
DESCRIPTION:
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).
EXAMPLE: R3 = 5.1MΩ , C2 = 1.0µF, Pulse Modulator = Waveform B and desired TD1 = 8 hours = 28,800 sec.
Since TD1 = 255 x M x R3 x C2, M calculates to be 22 for the selected Oscillator Components.
With M = 22, R2/R1 = 20. If R2 = 3MΩ , R1 = 150kΩ .
7339-060201-5