Time Delay Relay

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[/i]Topic Posted + Image Uploaded By Scott35 For ECN Member "John Crighton"[/i]

This discussion may be found in the General Discussion Forum, under the topic:

12 volt motor delay

Text copied / pasted from relavent thread:
Quote:

LarryC, I agree with your estimate, except that the capacitor will probably want to be closer to 100,000 uF. For the benefit of

the curious, I offer a few more details.

A typical automotive relay, Omron G8W, has an 88-ohm coil and an 8-volt must-operate spec, so with the 20-ohm series resistor

it will operate reliably down to 9.8 battery volts.

We have to use the parallel value of the coil resistance and the series resistor to obtain "R" (88 || 20) = 16.3 ohms. The

actual RC time constant is then 16.3 ohms x 100000 uF = 1.63 seconds. But, the operating delay time will depend on the actual

relay pull-in voltage, the battery voltage, and component value tolerances.

With a 13.7-volt battery we get a steady-state coil voltage ("Vcoil") of (13.7 x 88/(20+88)) = 11.2 volts. Assuming nominal R

and C values and a typical pull-in value ("Voperate") of 6 volts, the delay time ("T") is calculated as:

T = -RC x ln ((Vcoil-Voperate) / Vcoil)

T = -1.63 x ln ((11.2-6) / 11.2)

= 1.25 seconds

(The ln () function is "natural logarithm." It's on your scientific pocket calculator.)

Here's the bad news:

1) The operating time will vary with battery voltage. At 11 volts, the same circuit will take 1.8 seconds.

2) The capacitor must be rated at least 16 volts. A 100,000uF 16V capacitor is big (about 1.5" diameter and 2.5" long) and

expensive (about $10-12 new).

3) As LarryC mentioned: Assuming the relay is released by simply opening the circuit, the R-C product becomes (88 ohms * 100000

uF) = 8.8 seconds. Assuming the relay drops out at about 2 volts, the release time is:

T = -RC x ln (Vrelease / Vcoil)

T = -8.8 x ln (2 / 11.2)

= a whopping 15.2 seconds!

You can see why the solid-state alternatives, such as the one suggested by TwinCitySparky, are so desirable.

JoeTestingEngr, the key to using the 555 in an application like this is to forget that it's a timer and just use it as a really

good inverting Schmitt trigger with a nice, high-current output. Tie the trigger and threshold pins together as an input, and

leave the discharge pin unconnected. The output goes high when the input is below 1/3 VCC and low when the input is above 2/3

VCC.

Oh, and I used GCADD for years before finally biting the bullet and buying AutoCAD LT Didn't they have a "Miss Manners"-like

character answering FAQs in their newsletter?

.....text continues.....

Joe, what SCR? Our 88-ohm example relay draws only about 150 mA, so the coil can be connected between the output pin and +V.

Once the timing cap voltage rises above 2/3 Vcc, the output goes low, the relay pulls in, and it stays that way until power is

removed.

A diode from the timing cap to +V ensures that it's discharged at the beginning of the cycle. Depending on the application

details, there may need to be a dummy load R from +V to ground to give it something to discharge into.

The .01 uF bypass cap on pin 5 has a 33 microsecond time constant, so it has no effect at all on the relay output.

All of which is purely academic, because I would never connect a 555 directly to vehicle power, which exhibits transients that

would knock your socks off. Like others here, I'd prefer a more robust solution.

Honestly, in this case I think I'd avoid silicon altogether and modify LarryC's approach: Use the R-C network on a pilot relay

to drive the power relay.

The pilot relay would be a reed type with an adequate contact rating to energize the power relay and a 5-volt coil. The series

resistor would then be just slightly higher than the coil resistance, maximizing the "R" (which, in turn, minimizes the

expensive "C"), and guaranteeing operation at low battery voltage. It also reduces the voltage across C, further reducing its

cost and size. I'd add a diode across the series-R to discharge the cap quickly when power is removed. Since the cap discharges

through the low-resistance power-relay coil while the pilot relay is still closed, it happens quickly.

Example: The COTO 9081-05-00 reed relay ($1.66) has a 500 ohm coil. Choose series-R = 620 ohms, 1/4W and C = 2200 uF, 10V (0.4"

diameter x 0.8" long, $0.39).

The pull-in delay is 1 second, and the dropout (of the pilot relay) is about 0.7 seconds.

Including a 30-amp automotive power relay such as the Potter&Brumfield 1432782-1 ($1.82), and a couple of 1N4001 diodes, the

total parts cost is less than $4.50.

Cheap enough yet?

Edit: The P&B relay has 30A contacts, not 40A.


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Relative Schematic Image Below
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Scott35
new 07.05.2006 @ 21:51:00
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