looking for a delay off circuit

which

supply. I

have the

relay

battery

with the

timer

i'm

Thanks

One really simple, really cheap way to do this is to get the mechanical timer out of a scrapped microwave oven. This is the kind you have to physically turn the knob to so many minutes, and it usually goes ding at the end of the cycle. These should be easily obtainable from a thrift store. But this has the disadvantage of having to be set manually.

I would also consider connecting power to the fans just by themselves. To keep the external power out of the PS, you can put a diode in series with the power lead from both the power supply and the external power supply, probably a wall wart. This is what's called a diode OR.

Here's one simple time delay. You can make the delay much longer by using another transistor as an emitter follower to the first one. With two transistors I've had no problem getting 2 or more minutes delay.

Hello, i've tried to get this circuit to work without much luck. I"m using the one on the right. I was hoping to use the 12v from the power supply which is hooked up to the relay to power this circuit but i'm having some wierd issues. For some reason the circuit never turns off unless i disconnect the connection to the base. i don't know if my delay is for some reason ridiculously long and something else is different (I'm using the same values as shown). two questions that i had was about the first diode right after the "ignition switch" does it matter what type of diode? also the diode around the relay, what purpose does it serve and again does the value mean much. Thanks for any hlep. I"m looking to get into electronics and actually looking for a good place to start, anyone recomend a book or something else to get my feet wet. I understand the basics of resistors capacitors voltage and current sources but active components give me issues. Thanks

Hello, i've tried to get this circuit to work without much luck. I"m using the one on the right. I was hoping to use the 12v from the power supply which is hooked up to the relay to power this circuit but i'm having some wierd issues. For some reason the circuit never turns off unless i disconnect the connection to the base. i don't know if my delay is for some reason ridiculously long and something else is different (I'm using the same values as shown). two questions that i had was about the first diode right after the "ignition switch" does it matter what type of diode? also the diode around the relay, what purpose does it serve and again does the value mean much. Thanks for any hlep. I"m looking to get into electronics and actually looking for a good place to start, anyone recomend a book or something else to get my feet wet. I understand the basics of resistors capacitors voltage and current sources but active components give me issues. Thanks

it sounds like you maybe using too high of a value cap at the base.

you can calculate the Time constant using the T = RC which means (R) Resistor feeding or draining the cap, (C) = Cap value in farads!

(T) would be the time it would take to reach a 63.2 % charge etc.

so lets assume for now your relay creates 50 ma at 12 volts DC on the coil. this would be 240 Ohm coil. now since your using a transistor (bipolar type btw), you need to use the Bata/hfe spec on the transistor to increase the simulated effect of the Resistor value since the transistor is going to aid in generating most of the current to the coil for you. lets assume your transistor is a 150 HFE/beta. HFE*COIL-R = 150*240=36,000 ohms;

now this is not the exact math to use but should be close enough to get you in the ball park. now lets assume your using a 100 uf cap.. T = 36000*0.000100 = 3.6 seconds to get to a 63% charge etc..

the diode on the input to the base is to isolate the circuit from the electronics in the auto to prevent it from influencing your cap at the base from other devices on your the ignition line. this the Cathode must be pointing to the cap and base and the anode from the ignition line. the diode going acrossed the relay coil is to protect your transistor since the release of voltage from the coil can generate a reverse polarity of potential voltage. if you have the time constant working, you may not even need it because a slow decay of the charge in the cold will keep the HV potential down/

now this only a rough note. you will need to experiment a bit.

If we are talking about the same circuit, the relay will stay energized (turned on) as long as 12 volts is connected to the base, so your circuit may be working properly. Connect your circuit to +12, with nothing connected to the left side (in the diagram) of the 1N4002 diode. Then, with a jumper wire, temporarily connect +12 volts to the left side of that diode. The relay should energize when you connect the jumper. Once it does, remove the jumper. The relay should de-energize in roughly 15 seconds. At this point, you don't care exactly how long it takes, just make sure it eventually does de-energize. You can work on setting the delay to 15 seconds later.

Just to make sure we're talking about the same thing:

It is not critical. Use a diode from the 1N400x family (1N4001 through 1N4007) for both diodes.

also the diode around the relay, what

The diode around the relay eliminates a "spike" that results when power to a relay is disconnected. It is good practice (and often necessary) to put a diode across the relay coil in DC circuits. In this particular circuit, it is not needed if the circuit is working properly. But don't omit it. When you first start out with electronics, do exactly what you are doing: ask about things! But follow the plans until you fully understand why changing them is a good idea.

By the way, the diode is marked with a band at one end. The banded end of the diode corresponds to end of the diode with the vertical line in the schematic symbol: ---| any hlep. I"m looking to get into electronics and actually looking for

What circuit??

--- Place your multimeter in "diode test" mode and connect it across a known good diode. With the diode conducting, the lead connected to the anode will be positive and the other negative.

With the multimeter still in diode test mode, connect the positive lead to the base of the transistor and the negative lead, first to the emitter and then to the collector. The multimeter should indicate a conducting diode in both cases. Now, connect the negative lead to the base and connect the positive lead, first to the emitter and then to the collector. The meter should indicate an open circuit in both cases. Finally, connect one of the leads to the emitter and the other to the collector and then reverse the connections. The meter should indicate an open circuit in both cases.

If the transistor passes all the tests it is most likely good. The possibility exists, however, that it has been damaged and its gain is not what it should be. In order to test that you can do something like this: (View in a non-proportional font like Courier)

C---[MILLIAMMETER]--+ +---[10K]---B | +| E |+ [1.5V] | [9V] | | | +-------------+-------------------+

Assuming a Vbe of about 0.7V for the transistor, there'll be about 80µA of base current in the transistor. If the transistor is good, the current indicated by the milliammeter (the collector current), divided by the base current, should lie within the range of beta specified for the device at that collector current.

-- John Fields Professional Circuit Designer

_____ ^ _______ /quotient of the\

thanks for all of the input, i've managed to get it working. Not to mention picking up a bit more of how things interact with one another. Again thanks

Perhaps one of the turbo timer circuits would be suitable - the gizmos that keep the car engine running for a short time after the ignition is switched off

David

izzi4 wrote:

that

switched off

... pre-ignition?

These guys have all kinds of timers with differnt time settings. Perhaps it's worth checking into

"quietguy" wrote in message news: snipped-for-privacy@REMOVE-TO-REPLYconfidential-counselling.com...

Here are some timers from National Controls

"quietguy" wrote in message news: snipped-for-privacy@REMOVE-TO-REPLYconfidential-counselling.com...

I don't think that can happen with a diesel, but....

David

D> "quietguy" wrote in

I don't think a diesel engine can run without it (or something similar) happening.

Turbo timers run the engine at idle for a few minutes to allow the turbocharger to cool, this allows the rotor to cool from orange-hot in the relatively cool idle exhaust gasses rather than cooling by conducttion up the shaft and cooking the bearings.

Bye. Jasen