Hi everyone. I've learned a lot from this site and have received help before with regards to BJTs, which I now understand.

I've been trying a uC in my car, despite all of the horror stories, etc. but that has somehow failed and I'm trying to replicate the circuit's functions using a simple 556 timer instead. Here's what I want to happen:

Timer is connected to permanent 12v supply. When a 12v trigger occurs from another source, I need a 12VDC relay to switch an external load for at least 5 seconds, and then switch back to normal operation (COM to NC pins on relay).

Attached is a schematic of what I have come up with so far. There is a TVS diode to protect the circuit from transient voltages: breakdown V is 18.9V MAX, so should protect the timer from excessive supply voltages.

I've fabricated multiple PCBs and wasted quite a bit of money so far on circuits that don't work, so I'd be very grateful if anyone could point out any errors I may have before I actually go ahead with this one.

Thanks for any help in advance.

 

What is the purpose of D4 going to the reset pin?
If you are not using the reset function (which you don't need here) you just tie it to V+.

 

Edited schematic:


The trigger input is floating. The circled connection dot seems to be an artifact.

Why are you using a dual timer if you're only going to use one?

Schematics are easier to read if you try to keep flow from left to right and top to bottom and avoid unnecessary wire jogs.

 

Edited schematic:
View attachment 135426

The trigger input is floating. The circled connection dot seems to be an artifact.

Why are you using a dual timer if you're only going to use one?

Schematics are easier to read if you try to keep flow from left to right and top to bottom and avoid unnecessary wire jogs.

Thanks for your input. The circled connection is simply tying the reset pin high as per the relevant datasheet.

The trigger input is also tied high at the +12v junction between the diode, 10k resistor and 10nF cap. This is meant to be grounded, i.e. taken low when the trigger activates the path to gnd via the BJT. The fact that you say it's floating has made me doubt myself, however. Please could you clarify?

I am actually using both sides of the timer, but only showed one on the schematic for ease of visual understanding- not that it helped with everything jumbled up as you say- sorry about that! The other side of the timer will be exactly the same as the one shown on my schematic.

Many thanks for your help.

 

What is the purpose of D4 going to the reset pin?
If you are not using the reset function (which you don't need here) you just tie it to V+.

Sorry for the confusion. D4's cathode is connected to the 12v rail, as is the reset pin.

 

The circled connection is simply tying the reset pin high as per the relevant datasheet.

I missed that dot, so the reset connection is okay.
I was finally able to follow your convoluted schematic and it seems to be okay.

In the future, if you expect anyone else to view your schematics, please draw them in a more conventional fashion with a minimum of jogs in the lines, and signal flow from left to right and power from top to bottom.
For example capacitors going to ground should generally be oriented vertically, not horizontally and resistors going to V+ should also be vertical.

 

The trigger input is also tied high at the +12v junction between the diode, 10k resistor and 10nF cap. This is meant to be grounded, i.e. taken low when the trigger activates the path to gnd via the BJT. The fact that you say it's floating has made me doubt myself, however. Please could you clarify?

Couldn't tell if there was a connection dot on that node because there's text over the crossing and there's no portion of the dot below the horizontal wire:


With 10M and 10uF, you should be getting a pulse of 11 seconds.

 

Thanks guys, and sorry for the confusion again, I'll know better next time

Couldn't tell if there was a connection dot on that node because there's text over the crossing and there's no portion of the dot below the horizontal wire:
View attachment 135450

With 10M and 10uF, you should be getting a pulse of 11 seconds.

It's actually 1Mohm and 10uF I have on there (at least that was my intention), as per page 10 on datasheet: http://www.ti.com/lit/ds/snas548d/snas548d.pdf
This is meant to delay a little under 10 seconds.

 

Thanks guys, and sorry for the confusion again, I'll know better next time

This is how I would have drawn it:

One wire crossing, no unnecessary wire bends, all components have designators, text doesn't obscure anything.

It's actually 1Mohm and 10uF I have on there (at least that was my intention), as per page 10 on datasheet: http://www.ti.com/lit/ds/snas548d/snas548d.pdf
This is meant to delay a little under 10 seconds.

On the previous page, they stated that the formula was 1.1RC.

 

This is how I would have drawn it:
View attachment 135459
One wire crossing, no unnecessary wire bends, all components have designators, text doesn't obscure anything.
On the previous page, they stated that the formula was 1.1RC.

Ah I see... So 1.1*1*10=11. Gotcha
So, for 5 secs approx a 1Mohm resistor with about 4.5 to 4.6uF should do it? Thanks so much guys, it makes a big difference knowing I've done something right this time- there's so much knowledge on this site and I have seen how many members give a lot of their time to help others.

And thanks for taking the time to draw a corrected schematic for me dl324, that's another thing I've learned to do properly!

 

So, for 5 secs approx a 1Mohm resistor with about 4.5 to 4.6uF should do it?

The closest standard value is going to be 4.7uF, but electrolytic capacitors are generally 10% tolerance. For finer control, you'll want to use the resistor.

thanks for taking the time to draw a corrected schematic for me dl324, that's another thing I've learned to do properly!

Your welcome. That's just one of many ways to draw it. You'll develop your own style. We just wanted to prevent some bad practices from becoming ingrained.

D4 is still unnecessary.

 

How much current does the relay coil draw at 12 V?

ak

 

How much current does the relay coil draw at 12 V?

ak

Coil resistance is 360 ohm +/- 10%. At a nominal 12V this should be around 33mA according to my limited knowledge.

 

D4 is still unnecessary.

Thanks for that, I got the original schematic from https://electronics.stackexchange.com/questions/222174/latching-with-a-timed-relay which uses a diode to clamp any voltage spikes when a push-switch is used on the trigger. I can see now that this isn't necessary for my circuit. That's another component I won't need to buy, thanks again!

 

The closest standard value is going to be 4.7uF, but electrolytic capacitors are generally 10% tolerance. For finer control, you'll want to use the resistor.

Oh, one more thing: I'm using ceramic capacitors as I wasn't aware of any real functional differences... I hope that's not going to pose any problems!

 

Coil resistance is 360 ohm +/- 10%. At a nominal 12V this should be around 33mA according to my limited knowledge.

In that case you can eliminate R4 and Q2, and have the 555 output drive the relay directly, with the other end tied to GND. Keep D1.

ak

 

In that case you can eliminate R4 and Q2, and have the 555 output drive the relay directly, with the other end tied to GND. Keep D1.

ak

That's a great point, thanks a lot! Also, one more question (sorry!).

I would like this circuit to only work when ignition is off. So I'm thinking an SPDT with the power rail of this circuit going through the NC pin, and the coil operated by a live ignition feed. When the power cuts off to this circuit altogether (i.e. ignition on, SPDT cuts from NC to NO and the timer loses its power supply), does the output of the timer still stay high because of the cap? Or does this function not work because the VCC of the timer is at 0v?

 

"the cap" - which cap. Reference designators are your friend.

As an alternative to another relay, consider connecting ignition power to the 555 reset pin by re-purposing Q2 and R4 and adding a 10 K resistor from Reset to Vcc. When the ignition power is high, the reset pin is low, the 555 output is forced low, deactivating relay L1 (should be K1).

ak