This particular timing circuit can be used to time one-shot events from a few seconds to a few hours. And in standby mode (ie, with RLY1 and LED1 off), its power consumption is very low. The heart of this circuit is a low-cost CMOS 4011 quad NAND gate, with IC1a & IC1b configured as a standard Set/Reset flip-flop. Briefly pressing switch S1 to sta
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rt the timing sequence pulls pin 1 of IC1a low and, as a result, pin 3 switches high. Two things happen while pin 3 is high: capacitor Cx begins charging via potentiometer Rx; and (2) pin 11 of IC1d will be low, which means that transistors Q3 and Q1 are both on. As a result, both LED 1 and relay RLY1 are also on. RLY1 and LED 1 remain on until Cx has been charged up to about 70% of Vcc (ie, the supply rail). At this point, pins 8 & 9 of IC1c are pulled high and so its pin 10 output goes low and resets the flip-flop by applying a low to pin 6 of IC1b. This causes pin 3 of IC1a to go low and so LED1 and RLY1 switch off and the timing period ends. At the same time, pin 4 of the flip-flop goes high and this turns on transistor Q2 while ever the flip-flop is held reset. This ensures that Cx is discharged, so that the circuit is ready the next time S1 is pressed. Diode D1 and its associated 10 µF capacitor reset the flip-flop when power is first applied, so that LED1 and RLY1 remain off until S1 is pressed. D4 is included to protect Q1 against the back-EMF that`s generated when the relay switches off. Choosing appropriate values for Cx & Rx for a given time delay is straightforward. The formula is T = 1. 24 x Rx x Cx, where T is the delay time in seconds. As an example, let`s assume that we require a time delay of 10s using a value of 100 µF for Cx. Now we just need to calculate the value of Rx as follows: In this case, an 82kO...
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