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I am studying this time delay circuit and am curious to know the purpose of the 4.7kΩ resistor. As I see it, it has no purpose. I've built and tested this circuit with and without it and see no difference. enter image description hereAnyone have any insight on this?

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4 Answers 4

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One of the things it might be doing is reverse biasing the diode when SW1 is open, pulling it below cap voltage. This would reduce the likelyhood of stray fields forward biasing the diode and charging the (peak detector) cap, as long as 4.7k was strong enough to drain off the field charge and maintain reverse bias.

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The resistor makes the circuit behave predictably even if there is some leakage or shunt reactance across the switch.

For example, through capacitive coupling through long adjacent wires to the switch, or leakage due to the switch or wires getting wet.

Only a few hundred pF will actually create a noticeable (with care) effect at mains frequency, so it's not crazy.

You didn't test it with long wires or with leakage, so the issue didn't show up.

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The resistor prevents stray electric fields from charging up the 47 uF capacitor. Most of the time, it's not necessary, as the 150k resistor also performs this function, so it's really just insurance.

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The only purpose I can think of is to lightly load the transformer secondary during the timing operation. Without it the transformer is only loaded on positive half-cycles. Adding a 4k7 resistor across 24 V will only draw 5 mA or so but it may be enough to make the circuit less susceptible to mains noise.

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