a simple delay for a limit switch - Page 1
Forum section: electronics / projects

author: semi.competent.builder
author_rank: Newbie
timestamp: February 25, 2012, 10:34:21 pm
content: I want to pull a 18V signal to ground at least 10ms after another circuit (via limit switch) goes high (or low).So far I have come up with using a RC delay on a limit switch in conjunction with a relay. When the limit switch is on, it charges a capacitor through a resistor. The capacitor voltage will appear across the coils providing a delay. Once there is enough voltage is across the coils/capacitor the relay contacts will short my other circuit to ground.I have measured the resistance across the coils and its 400 ohms, this will discharge the capacitor when limit switch is not activated. However, the RC resistor has to be around 400 ohms to provide a high enough voltage to turn the relay on. This means something like a 1mF capacitor is required for a time constant of 400ms. So eventually its a resistive divider and there is half the voltage across it.Is this solution any good and what voltages do relays typically turn on at?

author: firewalker
author_rank: Super Contributor
timestamp: February 25, 2012, 11:01:24 pm
content: You really should use a transistor to arm the relay.E.g.Alexander.

author: TerminalJack505
author_rank: Super Contributor
timestamp: February 25, 2012, 11:42:21 pm
content: Is the relay necessary or are you using it just to get the delay.If the relay isn't necessary then here's a transistor-based solution.Edited to add:The pulse in the simulation is to simulate the limit switch closing. Which, I'm sure you realize.

author: steve_w
author_rank: Regular Contributor
timestamp: February 26, 2012, 12:18:26 am
content: you could put the relay in series with the coil the capacitor will charge through the coil momentarily energising the relay. there are options from here in, you could have a self hold via a set of contacts to keep the relay on. When the capacitor is charged the relay will de energise not enough current to hold the relay on.The question I ask is why 10 mS?It takes most relays 10 to 25 mS to close the contacts so another 10 mS whats that all about?depending on your circumstances you can probably do this without any active components, just a relay and capacitors and resistors, if you need accurate times and mass production you will probably have go the an active device.

author: TerminalJack505
author_rank: Super Contributor
timestamp: February 26, 2012, 04:21:22 am
content: I played around with the circuit I posted earlier to get a faster rise time when the switch goes from closed to open.Here's what I came up with...

author: semi.competent.builder
author_rank: Newbie
timestamp: February 26, 2012, 08:54:28 am
content: Thanks for the help.Its for a motor controller (someone elses), once the motor moves back the original position I want it to repeat (by pulling the strobe signal low). I have tried the attached circuit, the output signal works fine but when I connect it the strobe signal is stuck at about 14V. I am sure that the internal circuitry of the motor controller is charging the cap and will not go low.Looking back now I see that I should swap the resistor and capacitor, so the strobe is tied to the resistor to ground?With firewalkers solution: wouldn't the PNP transistor blow up since the cap is initially 0V, it would exceed the Veb of the normal PNPs?TerminalJack505, will try this later, what simulation software are you using?

author: TerminalJack505
author_rank: Super Contributor
timestamp: February 26, 2012, 02:49:47 pm
content: QuoteThe simulation software is Qucs. It can be a bit of a pain sometimes but it's okay.