Hi there. I need to design I circuit that when activated, by an input going from low to high (from zero to anything between 12V to 30V), will output iether high or low for approxomately 2 seconds.

This output will be powering a relay.

I need it to be as low a component count as possible and must be able to work upto 30V.

I originaly thaught about a nand gate monostable however the working voltages of CMOS put me off this, and, as mentioned it would be triggered from a posotive so a 555 monostable would be no good for this iether.

Thanks for any help, sorry for being vague.

Phil Walker.

 

a simple voltage divider can reduce the peak voltage of your 'input' pulse to anything you need realistically. a CE configured transistor can change the polarity of the pulse so that you can use the 555

 

Take this diagram for example:

What would happen if the switch was held closed, this would cause the a voltage difference across the resistor which would cause the resister to get hot over time, would this be the case? Thanks.

 

Take this diagram for example:

What would happen if the switch was held closed, this would cause the a voltage difference across the resistor which would cause the resister to get hot over time, would this be the case? Thanks.

As long as the resistor was big enough (~10-20k) it would not get even slightly warm.

 

At the highest value:

15v = I x 10kΩ
I = 1.5mA
P = 15V x 1.5mA = 22.5mW
​

A 1/8W resistor will handle this fine.

 

For some reason this circuit appears to be instantly burning the transister out. Any Ideas why? I cant see why it would.

Thanks, Phil.

 

Am I over complicating things, All I need to produce is something that when the ignition is switched on will have a 2 second high or low. Can I use a capacitor and its discharge properties across a resistor for example.

Thanks, Phil.

 

Is it required that the voltage drop nearly instantly or is it okay for it to slowly drop? If it's okay for it to drop slowly, then this circuit will work:

Click this link (you'll need Java.)

Adjust the timing resistor and capacitor to fine tune the time period. The default values time for about 3 seconds. At higher loads; the timing period falls; this is unavoidable, but can be compensated by increasing the resistor values or capacitor values. Both timing and discharge resistors should be the same. The discharge resistor ensures that when the ignition is turned off the capacitor discharges so it is ready to go next time. This design is suitable for up to 1 amp loads.

Since you're only powering a relay 1A will be more than enough. If you decide to use higher loads remember to heatsink the TIP122. The TIP122 contains all the circuitry to make a darlington pair, so the actual component count is very little.

Although I say only 10-15V in the schematic it should work up to 80V with appropriately rated components. 30V would not be a problem for this current design.

 

The main problem I see is a relay that will operate over a range of 12vdc to 30vdc, with a minimum of components, and switch something...or am I missing something? Maybe you could post a block diagram of your concept.

It might be easier to help solve your problem, than help solve your solution.

Ken

 

The main problem I see is a relay that will operate over a range of 12vdc to 30vdc, with a minimum of components, and switch something...or am I missing something? Maybe you could post a block diagram of your concept.

It might be easier to help solve your problem, than help solve your solution.

Ken

If 30V is only a temporary condition it would probably be okay.

 

If 30V is only a temporary condition it would probably be okay.

Don't understand what you mean!...temporary?...µSec?...minutes?

Again, how about a big picture view of what you are trying to do.

Ken

 

Don't understand what you mean!...temporary?...µSec?...minutes?

Again, how about a big picture view of what you are trying to do.

Ken

Temporary transient, probably less than 1 second? I don't know though - wait for the OP to reply.

 

Hi, sorry for the late reply, I basicly need to activate a relay (by taking an output to ground, not positive) for 2 secends when the ignition is switched on. The relay MUST trigger at the exact moment the ignition is switched on and needs to go off after 2 seconds.

The 'device' must be able to work on cars and lorry, so between 11V and upto 27V (Given extra for cranking and alternator voltage).

The components must be as descrete as possible.

That link to a capacitor discharge circuit would work, however it dascherges too slowly, and would cause the relay to bounce.

Thanks, Phil.

 

Hi, sorry for the late reply, I basicly need to activate a relay (by taking an output to ground, not positive) for 2 secends when the ignition is switched on. The relay MUST trigger at the exact moment the ignition is switched on and needs to go off after 2 seconds.

The 'device' must be able to work on cars and lorry, so between 11V and upto 27V (Given extra for cranking and alternator voltage).

The components must be as descrete as possible.

That link to a capacitor discharge circuit would work, however it dascherges too slowly, and would cause the relay to bounce.

Thanks, Phil.

It's worse than you think. Anything automotive must be rated to handle high voltage load dumps of 60V for hundreds of milliseconds.

Since the relay can't tolerate a slowly dropping voltage, then this solution, although a little more complex, would be an option. It times for approximately 3 seconds and has a very sharp turn-off characteristic. It is based around a voltage comparator, the LM311. (You will need a pull-up resistor to the supply voltage to use it properly, and you will need to ground the emitter pin of the LM311.)

Link to circuit

You'll also need to figure out how to handle the load dumps; 60V will probably fry the comparator. I suggest a zener and resistor along with a TVS.

 

As Tom says, an automotive electrical environment is very problematic for electronic circuits. Coupled with your requirement for operation over a wide range of voltage systems, makes a very simple, reliable solution with a minimum of discrete components almost impossible.

Attached is as simple as I can get.

Ken

 

Hi all, a little update. Ive managed to make/design a circuit to meet my requirements, Below is a diagram showing how ive achieved this. My only problem is that the output also connects to an output from another unit, which in turn both connect to one side of a relays coil, the other side of the coil connecting to +12V.

The output from my design sends an output of 0V for roughly 3 seconds then rises to an ouput of roughly 12V, this energises the relay for roughly 3 seconds when first triggered.

The additional unit that is also connected to this side of the coil usualy output 0V also, which isnt a problem, however it may, occasionaly output 12V, which would meen that there would be 0V on my ouput, and 12V on the other units output, ending in a problem.

To attempt to over come this I used a resistor, so that what my device does is act as a momentary pull down resistor.

A problem rises, however, when I add a resistor it appears that the Voltage drop is actualy over the resistor now and not the coil of the relay. The value of the resistor I used is 10KOhm and the resistance of the relay coil is only 12Ohms or so.

How would I overcome this problem, whilst still maintaining this pull down resistor?

Thanks, Phil.

 

The 10k resistor just makes your circuit useless, as you have found.

Another problem will be that time delay will vary with the different system voltages...12v to 28v. That's why I have the voltage regulator in my circuit. It might be possible to get by with a Zener diode and a resistor for timing regulation.

So, you're saying that the existing relay needs to be pulled to ground by "either" your delay circuit "or" by an existing circuit in the vehicles. To do that you need to "diode-OR" the existing relay connection. Remove the vehicle's existing grounding connection from the relay. Add two diodes with both their anodes to the relay's grounding connection. One diode's cathode goes to the existing grounding circuit, and the other diode's cathode to your output transistor's collector.

Is this intended to be a commercial product?

Ken

 

Others have already stated some of the problems associated with automotive stuff. I'm going to throw one more on the heap; temperature. I did a bunch of car alarms and car starters years ago. At the time I wasn't aware of the problems presented in simply starting the car, then throw in different voltage systems, the fact that the system can go negative (yup), jump starting and having to work when it's -40 degrees F to +140F...

Good luck...

Mike T.

 

I appriciate the problems faced with automotive electronics, and am taking them on board, thank you.

Ive attached a bigger picture of everything that is connecting to the proposed circuit.

Why exactly does the resistor cause the circuit to fail, is it due to the fact its resistance is waaaaay higher than the coil resistor?

The resistor is removed from the schematic below, but if it sits on the output of the left hand subsystem, refer to the above picture a few posts back for clarity.

Thanks again, Phil Walker.

Edit:

Somehow I missed your comment about Diode-Or'ing the outputs, I tried this and it works perfectly

Thank you very much, If anyone is interested I will post a Complete Schematic.

Thanks Again.

 

If the arrows are the diodes, they are drawn in the wrong direction.

Ken