7/25/2023 0 Comments Power on time delay relay circuit![]() So - with a 4700uF capacitor - the 1M resistor should give me just under 90 minutes. With the 470uF capacitor and the 1M resistor I got just under 9 minutes. There's no theoretical limit on the size of C5. If you want to shorten this discharge time - reduce the value of R11. To this must be added the initial 14 seconds - while Q1 discharges C5. So a 1M resistor and a 100uF capacitor will give about 120 seconds - or two minutes. I got roughly 12 seconds for every 100k/100uF combination. All you need is one reliable practical observation. However - because you're always using the same capacitor to activate the same input pin - the length of each step in your sequence should be fairly predictable. Manufacturing tolerances mean that your results are likely to be different from mine. Until Q1 switches off - the timing resistors cannot begin to charge C5. These times each include an initial delay of about 14 seconds - while Q1 discharges C5. And with the resistor values shown in the diagram - the events lasted 38 seconds, 67 seconds, 49 seconds and 9 minutes - respectively. I used a 470uF capacitor in the prototype. Then the relevant timing resistor takes over - and charges C5 up again. At the beginning of each event - Q1 discharges C5. The individual output times are controlled by the value of C5 - and the values of R5, R6, R7 & R8 respectively. You can also fix the total number of times your sequence will repeat - as well as the point in the sequence where the repetition will stop. ![]() You can produce More Complex Sequences - with overlapping and repeating events. You're not limited to a series of - "one after the other" - events. And the switch can be used to energize a relay - sound a buzzer etc. Each resistor will supply more than enough current to operate a transistor switch. The current available from each output pin is controlled by R1, R2, R3 & R4. The four outputs are taken from pins 3, 2, 4 & 7 - in that order. I've used a 12-volt supply in the diagram - but the circuit will work at anything from 5 to 15-volts. If you leave out D13 - the sequence will run only once. D13 causes the sequence to repeat continuously. And the length of each event is controlled independently. The number of events in the sequence may be increased to nine or ten. The relay stayed activated for more than 10 seconds! For more or less time, change the increase or decrease the capacitance of the series capacitor.This circuit uses a Cmos 4017 decade counter to create a sequence of four separate events. I managed to capture the voltage across the solenoid of the relay with the oscilloscope. As soon as the multimeter stopped beeping, I released the push button. As soon as I pressed and held the push button, the multimeter started beeping. I had my multimeter (in conductivity test mode) attached to the NO contacts of the relay. Surprisingly the relay remains activated for a long time. I used a push button to simulate the power switch. ![]() This is the circuit: Here is the assembled test circuit. I also placed a 10k resistor in parallel with the capacitor to bleed the charge when there is no power. ![]() I used a 10mF capacitance equivalent capacitor (three 3300uF capacitors in parallel). I made a test circuit with parts I had laying around and seems to work quite well. If the relay is small, a capacitor in series could work. It would likely turn right back on after it shutting itself off. Edit: The more I think about it, I don't think this will work as I envisioned. ![]() Refer to the 555 datasheet for additional details. Be careful about which 555 you use since they each have differences in the maximum output characteristics. Because you will be driving two relays, you will need to be mindful about not exceeding the 555's maximum output current. The one shown in the schematic can be very small since it is just powering itself and the components to the left of it. The relay shown it the schematic will likely have to be a different relay from the relay that you are going to power for 10 seconds. The bigger, the better since it only needs to 'bootstrap' the timer long enough to turn the relay on. You might need to experiment with the capacitor size to see what works best. In that case just put the capacitor in parallel with it. You can leave the 'start' switch if you need to activate the relay manually. The activation is controlled by the 'start' switch which can be replaced with a capacitor. In your case the capacitor will need to be much smaller and the pot can be removed if you don't need the timing to be adjustable. The timing is controlled by the resistors and capacitor on the left-hand side. In your case you would need to change the timing as well as how it is activated. Here's a circuit from Colin Mitchell's Talking Electronics website that can likely be changed to fit your needs. ![]()
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