Very much a beginner here and trying to learn as much as I can. Hoping that the folks on this forum can help point out the error of my ways

I am (re)building a circuit to take the signals coming from a motorcycle's speed sensor and coil, dividing these by two, and feeding them to a cruise control module. I built this circuit before with no protection using the MC14024B and it lived for a couple years. This time I would like to add some protection in the hopes that it will live longer. The tach portion of my current board has become unstable, so I suspect that I have damaged the input to the clock pulse (CP) of the MC14024B ripple counter, or its output pin Q0.

Also thinking of changing the power supply circuit for conditioning the power to the board. I previously used a 7812 with the appropriate caps before and after and I think it did OK. The datasheet for the 7812 calls for 2V+ over its rated 12V regulation. That is very close to the voltage seen on this motorcycle. So this time instead of the 7812 I was thinking about using a different approach. Not sure if this is really necessary or if I should just stick with a 7812.

On the schematic below I have the two different options for conditioning/protecting the CP of U1 & U2. Option #1 uses a design I took from Digikey's website in a paper titled "Protecting Inputs in Digital Electronics". Option #2 uses a much simpler approach. Not sure if option #1 is overkill or if option #2 is too weak to provide the needed protection.

Any suggestions are greatly appreciated.

 

If you are using CMOS IC logic, a really cheap trick to gain that required 2 volt voltage-overhead is to run the logic at a lower voltage.
Five volts is very common but there are also 8 volt linear regulator ICs available from actual electronics distributors, not amazon.
That is the very simplest scheme that I am aware of,

 

How about using optoisolators? The ~12 volt signals drive the optoisolator LEDs through current limiting resistors to switch a 5 volt signal high or low.

 

If you are using CMOS IC logic, a really cheap trick to gain that required 2 volt voltage-overhead is to run the logic at a lower voltage.
Five volts is very common but there are also 8 volt linear regulator ICs available from actual electronics distributors, not amazon.
That is the very simplest scheme that I am aware of,

So are you suggesting that I use something like a 7810 to regulate down to 10v and run the ripple counters at 10v? The cruise control module I am feeding from this circuit is expecting a 12v square wave. But it might be happy with a 10v square wave. Would have to test that because I don’t have the specs on the cruise control module. The counter says it is happy with anything between 3v-18 so it won’t mind running at 10v. Thanks for the suggestion MisterBill2.

 

How about using optoisolators?

Well, I thought about that, but wouldn’t I still have to protect its input? Wouldn’t I just be kicking the can further down, or rather up the road

 

If you size the current limiting resistors for something like 10mA @ 12v, most optos will survive up to 20+ volts.

Going the other direction, the output for the tach can also be an opto isolator, with the input designed for 5 volts and the output for 12v.

Do check the specs of the opto isolator for the input current range needed.

 

After the CMOS logic then a simple one transistor stage can boost the output back to 12 volts, and also protect the logic. A darlington device, like the MPSA13 transistor works very well with a CMOS logic output to drive 12 or 15 volt devices. But the normal arrangement invertes the signal. That is OK for a square wave frequency, though.

 

Or a nice device in the form of a suitable Mosfet !

 

I appreciate the suggestions. Couple more questions. Is the reason for placing a transistor after the CMOS because the transistor has a better chance at surviving than the CMOS? And, would placing a diode on the output provide an additional level of protection for the output transistors?

 

I appreciate the suggestions. Couple more questions. Is the reason for placing a transistor after the CMOS because the transistor has a better chance at surviving than the CMOS? And, would placing a diode on the output provide an additional level of protection for the output transistors?

Many power Mosfets have the diode protection built in!

 

Many power Mosfets have the diode protection built in!

Some CMOS devices operating at lower voltages may not have an adequate voltage swing, or possibly not an adequate current output for fast switching of some Mosfet transistors.