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Timed circuit for operating an automotive relay

L

lon

Jan 1, 1970
0
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
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
lon said:
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

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.
 
L

lon

Jan 1, 1970
0
I believe the circuit must supply approx 0.13A @ 12V to drive the relay.
(Bosch 12v 30 amp relay)
 
B

Bob Monsen

Jan 1, 1970
0
lon said:
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.

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.
 
T

Tom Biasi

Jan 1, 1970
0
lon said:
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
I'm guessing a headlight delay? There may be a setup already for your
vehicle.
 
T

Terry Pinnell

Jan 1, 1970
0
lon said:
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

This does what you describe:
http://www.terrypin.dial.pipex.com/Images/DelayedOffAuto.gif

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.
 
L

lon

Jan 1, 1970
0
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

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
 
E

ehsjr

Jan 1, 1970
0
lon said:
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

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

D1
+------|<------+
| |
+12 signal in ---------+--relay coil--+--Gnd
| |
+------||------+
C1
C2 R1
+---||--/\/\/---+
| |
+12---+---relaypoint--+--Bosch relay coil---Gnd
|
Existing wiring------+


Relay - RLY-407 from http://www.allelectronics.com/
D1 - 1N4001 diode
C1 - 1000 uF to 2200Uf (~5 sec to ~10 sec) 50v
C2 - .47 uF
R1 - 47 ohms 1/2 watt

Ed
 
E

ehsjr

Jan 1, 1970
0
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
 
J

Jamie

Jan 1, 1970
0
lon said:
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
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 ..
 
E

ehsjr

Jan 1, 1970
0
ehsjr said:

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
 
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