Timed circuit for operating an automotive relay

volts

the

output

is

It depends on how much current you need the circuit to supply. If it's small, then simply use a large capacitor. I've had a red LED flasher operate from a charged up half farad supercapacitor for hours.

Reply to
Watson A.Name - "Watt Sun, th
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You need to tell us what it's for. Otherwise, we can't even begin to tell you if it's possible, or easy. The problem is that you are going to power something with that 12V, and how much power it uses influences how to do this.

If there is 12V available elsewhere that can just be switched in, like in a car, that makes it easier. Actually storing the energy, and feeding it back is possible with a battery, or with a capacitor, depending again on what you are powering.

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Bob Monsen
Reply to
Bob Monsen

Hi there, wondering if anyone could help me out with a circuit.

I would like to make a circuit which basically continues to output 12 volts for 5-10 secs after an input signal (12 volts) is removed. If, during the

5-10 seconds the input voltage is restored there would be no loss of output voltage.

Hoping someone could lend me a hand with this. I have no idea if this is terribly difficult or not.

Thanks

lon

Reply to
lon

I'm guessing a headlight delay? There may be a setup already for your vehicle.

Reply to
Tom Biasi

I believe the circuit must supply approx 0.13A @ 12V to drive the relay. (Bosch 12v 30 amp relay)

Reply to
lon

This does what you describe:

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A breadboarded version worked OK. Let me know if the schematic and notes need further explanation, or if you want to see the output waveforms.

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Terry Pinnell
Hobbyist, West Sussex, UK
Reply to
Terry Pinnell

volts

output

This circuit is to activate an automotive relay. The current needed to drive the relay is approx 0.13A @ 12V. So this would be the requirements for signal out of the circuit.

Signal in is actually 2 12 volt lines but i figured i could combine them with diodes and a filter into one input line.

This input line is usually 12 volts, but it may go low for brief periods of time. During these transitions I want the circuit to output a continuous

12v signal. If however the input line goes low for an extended period (lets say 5 or 10s), I would like the circuit to produce 0 volts.

Sorry for any confusion and thank you to the replys.

lon

Reply to
lon

You can do it with a diode, relay 2 capacitors & a resistor. Here's a diagram - the parts list is below it.

D1 +------|

Reply to
ehsjr

After my previous post, it occurred to me that you might want an adjustable time period, so here's a new circuit. The existing Bosch relay circuit is assumed to be the top line of the diagram - everything below is what you add.

+12---o/ o--------------------+---BoschRelayCoil-----Gnd | +-------------+ | | ----- | / \\ D1 | --- | | | | | / e \\ R1 | \\ Q1 |--/\\/\\/\\---+---------->/ R3 c / | \\ | --- C1 / | --- | | | R2 | +12-------------+ +---/\\/\\/---+ | Gnd

C1 - 470 uF 63v D1 - 1N4001 Q1 - 2N2222 R1 - 4.7 K 1/4w R2 - 10 K 1/4w R3 - 50K pot

The length of time that the Bosch relay stays energized after the switch is turned off depends upon the specific relay drop out voltage and the specific transistor gain. R3 allows for varying the time from a minimum of about

2 seconds to a maximum of about 10 seconds. If you increase the value of C1, minimum on time will increase.

The circuit was tested with a relay that draws about 100 mA. Because your relay draws more, it is likley you will need to increase the value of C1.

Ed

Reply to
ehsjr

you could simply drive a heavy current type transistor using current source mode (the emitter feeding the 12 volts to the relay coil) drive the base from the 12 REF through a silicone diode for isolation into a cap that will be used to hold a charge, the - side of the cap will go to ground, from the + side of the cap which is also connected to the Cathode side of the isolation diode goes to the base of the transistor. the collector of the transistor comes from your main 12 supply. the emitter will drive one side of the relay coil, the other side of the relay coil will go to ground. basically the transistor is used to allow the use of a smaller cap which will hold a charge .. the relay it self will produce its own latching effects. its cheap and dirty but it works.

P.S. you should also use a silicone diode across the coil to suppress the fly back effects in the relay coil to prevent high voltage damage.. simply put the cathode of the diode (line side) on the same connection that is being connected to the emitter of the transistor, the anode side of the diode connected to the other leg of the relay coil. ... not using a transistor will require you to use a much larger cap to hold the relay in but with experimentation you may find that 1 diode and a cap could be sufficient..!, in this case the cap would be connected across the relay coil and thus will suppress the Flyback effects when the ref voltage is removed ..

Reply to
Jamie

Ooops - missing diode D2. Corrected in E-mail to op, figured I'd better correct it here, too:

+12---o/ o---------------------+---BoschRelayCoil-----Gnd | +-------------+ | | ----- --- / \\ D1 \\ / D2 --- ----- | | | | / e \\ R1 | \\ Q1 |--/\\/\\/\\---+---------->/ R3 c / | \\ | --- C1 / | --- | | | R2 | +12-------------+ +---/\\/\\/---+ | Gnd
Reply to
ehsjr

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