Down convert Frequency to a readable value

[stuhagen]

I have a digital wheel speed sensor that reads the speed of a turbine wheel on a turbo. It is set at factory to a peak of 4,666 events per second 4.6KHz (from 0-4666)
I need it to peak at 500 events per second. 500 Hz to be able to read it on the afermarket computer. The software in the computer can only read up to 500hz.
Reduce by 9.333 ~ 10? So what would I need to take this output and get it divided by 10 so it can read effectively through the computer software that runs the car. It is a PWM signal I am sure. I think it is a 12v-13.2v peak to peak. Although it might be a 5v peak to peak. I do not think this will be an issue as amplitude. It is the time frame that needs reducing to be readable.

Stu

 

[MikeMl]

Have you ever worked with digital logic, like CMOS? See the 4017 decade counter.

 

[stuhagen]

Have you ever worked with digital logic, like CMOS? See the 4017 decade counter.

Like Post #20 in this thread? https://www.electro-tech-online.com/threads/need-help-speedometer-signal-converter.128633/

Are these CMOS counters all stand alone ICs? I mean do they need passive devices RCs. etc to be calculated? Like what it says here: "4017" (a decade counter and 1-of-10 decoder in a single IC).

Maybe I will go pull a data sheet off and see if they have standard circuits for a 10x reducing circuit.

 

[Mikebits]

Have a read here. **broken link removed**

 

[dougy83]

Like Post #20 in this thread? https://www.electro-tech-online.com/threads/need-help-speedometer-signal-converter.128633/

Something like that. That one is a divide by 8 however.

Are these CMOS counters all stand alone ICs? I mean do they need passive devices RCs. etc to be calculated? Like what it says here: "4017" (a decade counter and 1-of-10 decoder in a single IC).

They are a stand-alone IC... but they need protection from spikes on the supply line and any input or outputs, hence the resistors and capacitors in the post you mentioned. The 4017 can divide by any value from 2 to 10.

 

[stuhagen]

I am having a hard time finding a specific design for this. I see this one but not sure what the IC1 is. Plus there are no R/C components. I do know that PIN 9 is the X10 reducer pinout. (Rest would be unused)
If someone can point me to a circuit design for a X10 reducer, I might be able to calculate the R/C components.

 

[MikeMl]

I am having a hard time finding a specific design for this. I see this one but not sure what the IC1 is. Plus there are no R/C components. I do know that PIN 9 is the X10 reducer pinout. (Rest would be unused)
If someone can point me to a circuit design for a X10 reducer, I might be able to calculate the R/C components.

This should do what you need. I designed it so you do not have to know what the amplitude of the input pulse train is. I also made it such that it should be relatively bullet-proof vis-a-vis the transients in the car's electrical system. The input signal is clamped such that a spike on the input shouldn't do anything, either. The output is set up for 0 to 5V; but it could be changed for 0 to 12V....

The simulation shows an input signal of 12V being clamped at 5V at the input of the counter. Write back if there is something you don't understand...

 

[stuhagen]

Awesome! If I knew more on how to use LT Splice simulation, I'd give it try. I will bread-board it and check it out on the scope. I know the power supply will be off Ign. somewhere. The output amplitude of the optical sensor is 5v. The PC reads it as a 5v input peaking at 500hz. So we will see how it down converts.

Stu

 

[MikeMl]

... So we will see how it down converts...

Minor nit, down converting usually refers to the heterodyne process, where Fo = Fin - Flo, where Flo is a local oscillator. Eg, 100.3MHz - 89.6Mhz = 10.7Mhz.
This is a frequency divider, where Fo = Fin/10.

 

[stuhagen]

Well, i will be honest. I graduated from EE school in 1978, and never really worked in the field. Ended up elsewhere! So these terms/design are familiar, but not ingrained!

 

[stuhagen]

This should do what you need. I designed it so you do not have to know what the amplitude of the input pulse train is. I also made it such that it should be relatively bullet-proof vis-a-vis the transients in the car's electrical system. The input signal is clamped such that a spike on the input shouldn't do anything, either. The output is set up for 0 to 5V; but it could be changed for 0 to 12V....

The simulation shows an input signal of 12V being clamped at 5V at the input of the counter. Write back if there is something you don't understand...

**broken link removed**

Do pins 13, 15 need tobe "open" (no connections). Also, it is Pin O0 that is the x10 pinout? (I thought it was pin 9) I am sure it is based on your simulator.

 

[MikeMl]

Actually, if the modulus of the counter is set up to divide by 10 (i.e. the counter is not short-cycled by hooking up some feedback to pins 13 and 15, any of the outputs will at 1/10 the clock frequency, so you can use any of them you like... Pins 13 and 15 must be tied to ground (Vss).

Graduated with BSEE in 1966...

 

[stuhagen]

So I had this simple version using solely a 555, but I cannot get it to simulate. Do you think a simple design like this might work as well? Only cause I have all the parts for this. Otherwise I will hit Digikey up for some of the items I do not have.

 

[MikeMl]

So I had this simple version using solely a 555, but I cannot get it to simulate. Do you think a simple design like this might work as well? Only cause I have all the parts for this. Otherwise I will hit Digikey up for some of the items I do not have.

That circuit can act as a 1/10 frequency divider over a very small range (~5%) of the frequency it is designed for. You need a divider that operates over a 10:1 frequency range (engine rpm from 600 to 6000).

 

[stuhagen]

I am aware of RC timing functions and know somewhat on how to adjust. But my question is, do you have a simple formula for this circuit that I can use to make it more flexible? I am going to upload some gerber files and have some proto boards made after I bread board it. So If there are some simple component changes I can make to convert to a 2x, 3x, 4x, etc. type divider, I may incorporate it in my boards. But I could be wrong that this type of I/C doesnt use RC timing. Cause I do not see what they could be other than the C1, or C2, and possibly R1.

Thanks

Stu

 

[dougy83]

No, the 4017 doesn't use RC timing. It's a Johnson Counter and will divide by any number between 2 and 10. As drawn in the circuit above, it's a divide by 10. To make it divide by a lesser factor, connect the RESET pin (pin 15) to the appropriate output; e.g. connecting O5 to R will create a divide by 5 counter. Using O3 creates a divide by 3, O7 a divide by 7 and so on.

 

[stuhagen]

No, the 4017 doesn't use RC timing. It's a Johnson Counter and will divide by any number between 2 and 10. As drawn in the circuit above, it's a divide by 10. To make it divide by a lesser factor, connect the RESET pin (pin 15) to the appropriate output; e.g. connecting O5 to R will create a divide by 5 counter. Using O3 creates a divide by 3, O7 a divide by 7 and so on.

So it "appears" as if Pin 15 is "floating"? Am I correct? It is set up now with no connection to Pin 15. Is Pin O0 the 10x pin then? I read the schematic as both 13 and 15 as unconnected to anything.

 

[MikeMl]

Here you go, divide by 2, 3, ... ,9, or 10. (Divide by 8 is shown in the simulation)

 

[dougy83]

So it "appears" as if Pin 15 is "floating"? Am I correct? It is set up now with no connection to Pin 15. Is Pin O0 the 10x pin then? I read the schematic as both 13 and 15 as unconnected to anything.

No, no and no respectively.

I'm referring to the schematic MikeMI posted. You'll notice that pin 15 (reset) is connected to ground. That allows /10. Connect pin 15 to the appropriate output to get a different divisor. O2...O9 allow for divide by 2...9 respectively. Only connect pin 15 to a single output or to ground. Don't connect it to O0 or O1, as they won't give you a useful output.

 

[stuhagen]

No, no and no respectively.

I'm referring to the schematic MikeMI posted. You'll notice that pin 15 (reset) is connected to ground. That allows /10. Connect pin 15 to the appropriate output to get a different divisor. O2...O9 allow for divide by 2...9 respectively. Only connect pin 15 to a single output or to ground. Don't connect it to O0 or O1, as they won't give you a useful output.

OOHH, sorry, I did not "grasp" that those were "ground" tags. ha..ha.. I thought there were arrows or something. So the below would be the various ways to connect for other dividers.
The Darlington output off 3 would always remain as is. I had to substitute a lesser value Zener cause Digikey did not have the other in stock. Difference was the one I chose was 1.2v vrs 1.5v
(1N5229)


Stu