Okay, I need a way to do this... basically I need it so one light turns on for like 3 seconds, then the other one does.
Thanks,
Douglas

 

What lights (LEDs, or 12 volt bulbs)? How much current needs to be switched?

Tim

 

Well, 12 volt would be optimal, but I could also have it power a relay if lower volatage makes it simpler. It won't be powering headlights or anything, so I doubt they need a lot of current.
Thanks,
Douglas

[This message has been edited by AmorgetRS (edited April 04, 2001).]

 

I posted a circuit recently in the GENERAL ENGINE forum under the heading How many amps do dual elect fans draw each?? describing a circuit to provide a 4 second delay. Since I do not have a scanner, I cannot show you a schematic, which would be the easiest way to describe this circuit.

From what you posted, I presume that you want a light to turn on for three seconds then turn off and remain off at which time another light which will come on and remain on until removal of power.

The circuit I described in the post I listed above will work fine. For it to work as you described, make the following changes.

1. Use this circuit but connect a double throw 12v relay to the output (pin 3).
2. Connect the common contact of this relay to the switched +12 that applies power to the IC (555).
3. Use an 27K and 100uF resistor/capacitor instead of the 18K/220uF combination.

The normally closed contact will be the output that will provide the initial three seconds of +12. The normally open contact will provide +12 for the time after this three second period.

Removal of +12 to the circuit will remove power to the relay and IC, thus removing power to the relays’ contacts and resetting the IC so it’ll be ready for another timing cycle. If the current through this relay is more than the switch can handle, use another relay to switch power to the common contact of this relay. This will remove +12 when the switch is off.

Using the formula I gave in the above mentioned post, a 27K resistor with a 100uF capacitor will give you about a three second timing period.

Circuit description:

Upon application of power to the 555, the output (pin 3) will remain low for the time period determined by the resistor/capacitor (RC) combination (27K and 100uF), or three seconds in this case. Since the relay coil is powered by this pin, the relay will remain off when first switched on.

Since the common contact of this relay is connected to the same switched +12 that powers this IC, you will have +12 here for the initial three seconds when switched on because its wired through the Normally Closed (NC) contacts. After three seconds, the output (pin 3) will go high, turning on the relay and closing the Normally Open (NO) contacts, turning on whatever is connected to the NO contacts. The output (pin 3) will remain high (and the relay closed) as long as power remains to the 555 IC. Removal of power will turn off both lights and reset the IC.

Remember that this IC can only source/sink about 300mA. If your relay draws more than ~200mA, you may want to use a transistor (or smaller relay) or use a smaller relay to drive a larger relay.

 

Stuart, thank you for the reply. That is not really what I want, I need them to keep going back and forth.
The part that pisses me off is that I had a way to do it some Junior Electronics kinda thingy that called it a stop light, where one would be on for a few seconds, then switch, then switch again, one being red, the other green.
Thanks agian,
Douglas

 

You want two lights to alternately illuminate at a 0.17Hz rate (3 seconds “on” then 3 seconds "off" for a 6 second cycle).

There are many ways to accomplish this. You’d probably get a different circuit for each person you talk to, with no one way the “correct” way. The circuit I propose uses a 555 timer (my favorite) and has only four components (minus the four LED components which would then total eight).

To drive two L.E.D.’s in an alternating fashion, simply wire the 555 as an astable timer. The 555 can source/sink 200mA which is plenty for an LED which typically draws ~20mA.

Resistors are 1/4 watt and the capacitor should be 16 volts minimum. No values are critical.

For the timing circuit, you’ll need:

1. one each 555 IC
2. one each 2K resistor
3. one each 33K resistor
4. one each 100uF 16 volt (min) capacitor

I do not have a scanner so I cannot provide a schematic. I’ll describe how to wire the circuit from the text below. It is simple enough that you should be able to follow it easily. You could use a perforated board or even solder directly to the legs of the IC if you’re careful. Wire the 555 IC as follows:

1. Connect pin 1 to ground.
2. Connect pins 2 and 6 together.
3. Connect pins 4 and 8 to positive 12.
4. Connect a 2K resistor to pins 4/8 and pin 7
5. Connect a 33K resistor to pin 7 and pins 2/6
6. Connect the 100uF capacitor to pins 2/6 and ground. The positive lead to pins 2/6, the negative lead to ground.

To alternately light two L.E.D.’s, you'll need the following:

1. two each 510 ohm resistors
2. two each L.E.D.'s

Connect as follows:

1. Connect a 510 ohm resistor to +12 (e.g. pins 4/8).
2. Connect the other side of this resistor to the anode (positive) side of the LED.
3. Connect the other side of this LED (cathode, or negative) to the output of the timer (pin 3)
4. Connect the anode (positive) side of a second LED to the output of the timer (pin 3).
5. Connect the other side of this LED (cathode or negative) to another 510 ohm resistor.
6. Connect the other side of this resistor to ground (e.g. pin 1).

If you want to directly drive two incandescent bulbs (each drawing <200mA), simply substitute each resistor/LED combination for a bulb.

The IC will operate from a supply voltage of 5-16 volts.

If you want to drive something more than 200mA, connect one relay to the output. The 555 can drive a relay directly if the coil is <200mA (many relays are). Connect the relay coil to ground and the output of the 555 (pin 3). Also connect a diode across the relay coil (which will act as a voltage snubber). The cathode, or banded (negative) side to the pin 3 side, the other side to ground. It doesn’t have to be physically at the relay coil – it can be at the IC (cathode to pin 3 and anode to ground). Using a double throw relay, connect the common contact to positive and the NC (normally closed) and NO (normally open) contacts to whatever you want to alternately power (lights, LED’s, motors, etc.). The timer will oscillate the relay (with the values given above ~3 seconds on, ~3 seconds off).

The 555 is inexpensive (25-40˘ through mail order – Radio Shack at several times that cost).

You can get the data sheet for the 555 at http://www.national.com/ds/LM/LM555.pdf which also has the formula for determining frequency and duty cycle (~49% duty cycle is obtained using the values listed above).

EDIT (this paragraph added): In this .pdf file, you’ll see a schematic of the timing circuit on page 7, figure 4. It is identical to the circuit above. Ra is 2K, Rb is 33K and C is the 100uF capacitor (positive lead to pin 2/6). You can omit the 0.01uF capacitor, leaving pin 5 unconnected (that capacitor is normally used for noise supression). The two LED’s (or light bulbs) are shown as RL, both connected to pin 3 with dashed lines. Just connect as described above.

If you want to vary the timing cycle, simply put a 100K potentiometer in place of the 33K resistor.

[This message has been edited by Stuart Moss (edited April 05, 2001).]

 

You rock! Thanks a ton!

 

Hey, I got all the parts, but I got a slight problem, I got the 100k Ohm Linear-Taper Pontentiometer and it has 3 little connections on it.... which ones do I use??
Thanks,
Douglas

 

Okay, man, I bought the parts and gave it a shot, and I really need to take a electronics class because when I tried to follow it all and everything I got so lost it wasn't even funny...
I am curious, would anyone be willing to make this for me? Of course I'd pay for parts and whatever and I am in no hurry, but I think this is over my head....

 

To use the potentiometer in place of the fixed resistor, use the middle connector (called the "wiper" because it does just that when you move the shaft!) and one of the other terminals. The middle-connector is the point where you obtain the variable resistance. The outer-connectors are the fixed resistance points.

The center-connecter or “wiper” is physically connected to the shaft of the potentiometer. Rotating the shaft will move this wiper toward one of the end-connectors, decreasing resistance the closer it gets to one of the end-contacts, and increasing resistance the farther away it gets from the other.

The resistance across the outer two connectors will be the resistance (not variable) of the potentiometer.

Depending upon which end-terminal you use, you can increase resistance with a CW (clockwise) or CCW (counter clockwise) rotation.

You can leave the other end-connector unconnected, or connected to the center terminal (usually a schematic will show these two points connected together). Electrially the result will be identical for this circuit.

If you have all the parts, I could assemble it if you sent it to me and included return postage (stamps or a money order for same).

Please coordinate via e-mail.